diff --git "a/c6h_checkpoint_2000.ndjson" "b/c6h_checkpoint_2000.ndjson" deleted file mode 100644--- "a/c6h_checkpoint_2000.ndjson" +++ /dev/null @@ -1,4548 +0,0 @@ -{ - "Problem": "Let $ n \\geq 2$ be a positive integer and $ \\lambda$ a positive real number. Initially there are $ n$ fleas on a horizontal line, not all at the same point. We define a move as choosing two fleas at some points $ A$ and $ B$, with $ A$ to the left of $ B$, and letting the flea from $ A$ jump over the flea from $ B$ to the point $ C$ so that $ \\frac {BC}{AB} \\equal{} \\lambda$. \r\n\r\nDetermine all values of $ \\lambda$ such that, for any point $ M$ on the line and for any initial position of the $ n$ fleas, there exists a sequence of moves that will take them all to the position right of $ M$.", - "Solution_1": "For $\\lambda\\ge1/(n-1)$ just repeatedly jump the leftmost flea over the rightmost flea.\n\nNow assume that $\\lambda<1/(n-1)$. Let the positions after the $k^{th}$ jump be $x_1 M $ then $ M(f) \\ge f_1 > M $. Thus, if the fleas were able to jump to infinity, then $ M \\rightarrow \\infty $. However, $ M $ is a monovariant! (It never increases). Indeed, if $ f_1 $ doesn't change then $ M $ decreases strictly, and if $ f_1 $ changes, $ M $ is easily computed to stay the same.\n\nThus, we're done.\n\nIf $ \\lambda \\ge \\frac{1}{n-1} $, then by making the last one jump over the first one, we compute that the average distance between consecutive fleas never decreases, and $ f_n$ always increases, thus the fleas jump to infinity.", - "Solution_3": "Hello! I was wondering if the following solution is correct? Here we replace lambda with c:\n\nAfter each step k, denote x(k)(1), x(k)(2), ..., x(k)(n) in increasing orders to be the positions of the fleas. Hence x(n)(k) is the maximum position. Note that the optimal way to increase the maximum position is by jumping the leftmost fly over the rightmost fly. \nNow, denote P(k) to be the maximum value of the positions after k steps (essentially the same as x(n)(k)). Note that P(k+1) - P(k) = c(x(n)(k) - x(n)(0)). After some algebra we get that c[(x(n)(k) - x(n)(1))] = [x(0)(n) - x(0)(1)][(n-1)(c)]^k, where x(0)(i) denotes the initial positions. Finally, let d = c(n-1). By collapsing the sum, we get that P(k) = (d^(k-1)+d^(k-2)+...+1)[x(0)(n) - x(0)(1)] = [(d^k - 1)/(d - 1)] * [x(0)(n) - x(0)(1)]. Note that [x(0)(n) - x(0)(1)] is constant, and we want that P(k) is unbounded as k approaches infinity. For all c < 1/(n-1), we see that P(k) is bounded above as k approaches infinity by 1/(1-d) * [x(0)(n) - x(0)(1)], and for c >= 1/(n-1) P(k) is unbounded, so we can move the fleas! Hence the desired answer is all c (lambda) >= 1/(n-1).\n\nThanks! :-D ", - "Solution_4": "Could anyone help?", - "Solution_5": "\\bump? Thanks", - "Solution_6": "Write $n = N+1$; we claim the answer is \\[\n\\boxed{\\left\\{\\lambda\\colon \\lambda \\geq \\frac 1N\\right\\}}.\n\\]Place the points on the number line; let $p$ denote the coordinate of point $P$. For a configuration $x_1\\leq x_2\\leq \\cdots \\leq x_{N+1}$ of points $X_1,X_2,\\ldots, X_{N+1}$, define its cap by $x_{N+1}$ and its score by $X_1X_{N+1} + X_2X_{N+1} + \\cdots + X_NX_{N+1}$. Notice now that we can get all the points past $M$ if and only if we can get the cap to exceed $m$; only if clearly holds, and for if just notice that if $X_{N+1}$ is past $M$, we just reflect everything over $X_{N+1}$ and win. So, we just want to find $\\lambda$ so that the cap can get arbitrarily large.\n\nLemma 1: $\\lambda \\geq \\frac 1N$ all work.\n\nProof: First, note that we can get a configuration in which no two fleas are at the same point. To do this, first take the leftmost point and reflect over the rightmost point. Then, we have a point strictly to the right of all other points. Suppose the points are $P_1,P_2,\\ldots, P_{N+1}$, respectively. Then, at each point, reflect the leftmost point over the rightmost one $N$ times, until we have $N+1$ points all at distinct places.\n\nSuppose our current configuration is $X_1,X_2,\\ldots, X_{N+1}$, and let $d_1 = X_1X_2$, $d_2 = X_2X_3, \\ldots$, and write $d = \\min(d_i)>0$. Reflect $X_1$ over $X_{N+1}$. Then, the configuration becomes $X_2,X_3,\\ldots, X_{N+1}, X_1'$, and the pairwise distances become $d_2,d_3,\\ldots, d_N, \\lambda(d_1+d_2+\\cdots+d_N)$, which are again all $\\geq d$. So, at each step, we know that the cap has coordinate increasing by at least $d$ every time, so we can get the cap to be arbitrarily large. $\\Box$\n\nLemma 2: $\\lambda < \\frac 1N$ all fail.\n\nProof: Define the value $V$ of a configuration to be \\[\nV = C + \\frac{\\lambda S}{1-N\\lambda},\n\\]where $C$ is the cap and $S$ is the score, respectively. We claim that the value is nonincreasing. Indeed, suppose we have a configuration $X_1,X_2,\\ldots, X_{N+1}$, where $x_1\\leq x_2\\leq \\cdots \\leq x_{N+1}$. Suppose we reflect $X_i$ over $X_j$ with $i 1$, this gives an upper bound. $\\blacksquare$\n\n[b][color=red]Claim:[/color][/b] When $\\lambda \\ge \\frac{1}{n-1}$, it suffices to always jump the leftmost flea over the rightmost flea.\n\n[i]Proof.[/i] If we let $x_i$ denote the distance travelled by $B_1$ in the $i$th step, then $x_i = a_i$ for $1 \\le i \\le n-1$ and $x_i = \\lambda(x_{i-1} + x_{i-2} + \\dots + x_{i-(n-1)})$.\n\nIn particular, if $\\lambda \\ge \\frac{1}{n-1}$ then each $x_i$ is at least the average of the previous $n-1$ terms. So if the $a_i$ are not all zero, then $\\{x_{n}, \\dots, x_{2n-2}\\}$ are all positive and thereafter $x_i \\ge \\min \\left\\{ x_n, \\dots, x_{2n-2} \\right\\} > 0$ for every $i \\ge 2n-1$. So the partial sums of $x_i$ are unbounded, as desired. $\\blacksquare$", - "Solution_8": "I claim the answer is $\\lambda \\ge \\frac{1}{n-1}$. To see this works, consider the strategy of hopping the leftmost flea over the rightmost. After $n-1$ moves, all fleas will now be at distinct positions, and define $m$ to be the minimum distance between two consecutive fleas. Now, the minimum distance is always at least $m$, since the distance made by hopping is the average of all the other distances. So, every hop, the configuration's leftmost point moves forward at least $m$, and all fleas will eventually pass $M$.\n\nIf $\\lambda <\\frac{1}{n-1}$, define a point $P$, which is left of all the fleas, and call the distance from the $n$th rightmost flea to $P$ $d_n$. Then, call the leakiness of the configuration to be $d_1-\\lambda(d_2+\\ldots+d_n)$. If $\\lambda<\\frac{1}{n-1}$, this number is clearly always positive. Now, note that if a flea jumps but does not go past the 1st flea, then the leakiness must decrease. If the $i$th flea jumps over the $j$th, and actually becomes the first flea, then the leakiness changes by $-(1+\\lambda)d_1+(d_j-d_i)\\lambda+d_j+\\lambda d_i=(1+\\lambda)(d_j-d_1)$. As $d_j\\le d_1$, this value is nonpositive, so the leakiness doesn't increase.\n\nThe minimum leakiness required to get across $M$ is $(1-(n-1)\\lambda)PM$, so if we move $M$ sufficiently far to the right, this quantity becomes larger than the leakiness of the initial configuration, as desired.", - "Solution_9": "I would like to point out that the problem solver should notice that the rightmost guy is special. We want the rightmost guy to the as far to the right as possible and all others as far left as possible. Then just from this we can make some monovariant.", - "Solution_10": "The answer is $\\lambda\\ge\\frac1{n-1}$.\n\n\n[color=blue][b]Proof of lower bound.[/b][/color] We will use this obvious strategy --- the leftmost flea always jumps over the rightmost flea. Consider the nondecreasing sequence $(x_i)_{i\\ge1}$ with $0n$, let $x_k=x_{k-1}(\\lambda+1)-x_{k-n}\\lambda$. After $k$ moves, the fleas have coordinates $x_{k+1}$, $\\ldots$, $x_{k+n}$. I will prove this sequence is unbounded.\n\nSumming over $x_k-x_{k-1}=\\lambda(x_{k-1}-x_{k-n})$, we have \\[x_k-x_n=\\lambda\\left(x_{k-1}+\\cdots+x_{k-(n-1)}\\right)-\\lambda(x_1+\\cdots+x_{n-1}).\\] In other words, $x_k-\\lambda\\left(x_{k-1}+\\cdots+x_{k-(n-1)}\\right)$ is a fixed constant $C>0$. Then since $x_{k-(n-1)}\\le x_{k-i}$ for $0\\le i\\le n-2$, we have \\begin{align*} \\frac{x_k+\\cdots+x_{k-(n-2)}}{n-1}&\\ge\\frac{x_k+\\cdots+x_{k-(n-1)}}n\\\\ &=\\frac{C+(\\lambda+1)\\left(x_{k-1}+\\cdots+x_{k-(n-1)}\\right)}n\\\\ &\\ge\\frac Cn+\\frac{x_{k-1}+\\cdots+x_{k-(n-1)}}{n-1}, \\end{align*} so $x_{k-1}+\\cdots+x_{k-(n-1)}$ is unbounded, and thus so is $(x_i)_{i\\ge1}$.\n\n\n[color=blue][b]Proof of upper bound.[/b][/color] At some point in time, let the fleas have coordinates $0M$, then \\begin{align*} I&=x_n-\\lambda(x_1+\\cdots+x_{n-1})\\\\ &>x_n-\\lambda(x_n+\\cdots+x_n)\\\\ &=x_n\\big[1-(n-1)\\lambda\\big]\\\\ &>M\\big[1-(n-1)\\lambda\\big], \\end{align*} which is larger than $I$ for sufficiently large $M$.", - "Solution_11": "We want to find all $\\lambda$ for which it is impossible to move all fleas infinitely rightward. I claim our answer is all $\\lambda \\geq \\tfrac{1}{n-1}$. \n\nWhen $\\lambda \\geq \\tfrac{1}{n-1}$, we can simply let the leftmost flea jump over the rightmost one on each move. After $n-1$ moves, we will have $n$ fleas in positions $p_1 < p_2 < \\ldots < p_n$. If we let $\\delta = \\min{(p_{i+1} - p_i)}$ for $i \\in [1, n-1]$, then it is easy to see that after each move, the distance between the leftmost and rightmost fleas is nondecreasing and thus so is $\\min{(p_{i+1} - p_i)}$, and that thus the new leftmost flea's position is at least $\\delta$ to the right of the old one. Hence the leftmost flea can go arbitrarily far to the right by moving $\\delta$ each time, and thus so can the rest of the fleas. \n\nWhen $\\lambda < \\tfrac{1}{n-1}$, I claim that the position of the rightmost flea is bounded. Indeed, suppose that after some $\\ell$ moves, the $n$ fleas are at positions\\[q_1 \\leq q_2 \\leq \\ldots \\leq q_n\\]with largest $m_{\\ell} = q_n$ and sum $s_{\\ell}$. If we were to perform a move on fleas on $q_i < q_j$, which must exist since all fleas can never be on the same point, then:[list]\n[*]the maximum changes from $m_k$ to $m_{k+1}$ by at most $\\lambda(x_j - x_i)$, which happens when $j = n$. \n[*]the sum changes from $s_k$ to $s_{k+1}$ by exactly $(x_j + [x_j + \\lambda(x_j - x_i)]) - (x_i + x_j) = (1 + \\lambda)(x_j - x_i)$\n[/list]\nTherefore, after any number of moves $K$, we see that\\[m_K \\leq m_0 + (\\lambda y_1 + \\ldots + \\lambda y_K) = m_0 + \\lambda(y_1 + \\ldots y_K)\\]and\\[s_K = s_0 + ([1 + \\lambda]y_1 + \\ldots + [1 + \\lambda]y_K) = s_0 + (1 + \\lambda)(y_1 + \\ldots + y_K)\\]where $y_i$ is the positive difference in the positions of the fleas moved in the $i$th move. Thus, we see that the ratio $\\tfrac{\\Delta m}{\\Delta s}$ is at most $\\tfrac{\\lambda}{1 + \\lambda}$ after any number of moves, and in fact we can solve\\[\\frac{m_K - m_0}{s_K - s_0} \\leq \\frac{\\lambda}{1 + \\lambda} \\implies m_K \\leq m_0 + \\frac{\\lambda}{1 + \\lambda}\\left(s_K - s_0\\right) \\leq m_0 + \\frac{\\lambda}{1 + \\lambda}(nm_K - s_0)\\]and obtain that $m_K \\leq \\tfrac{m_0 - \\mathcal{C}s_0}{1 - \\mathcal{C}n}$ which is a positive constant since $\\lambda < \\tfrac{1}{n-1} \\implies \\mathcal{C} = \\tfrac{\\lambda}{1 + \\lambda} < \\tfrac{1}{n}$. Thus, the fleas' positions are upper bounded by some positive constant, so if we make the starting points far left enough it is impossible to take them all to the right of a given position $M$. $\\blacksquare$", - "Solution_12": "[b]Setup[/b]\nWe claim that the answer is $\\lambda \\ge \\tfrac{1}{n-1}.$ In the proof, we shall be making use of two variables that change with the flea's position. We define $R$, the position of the right-most flea. Suppose we have fleas $A_1, A_2, \\dots, A_n$, and define $d(A_i,A_j)$ to be the positive difference between the positions of flea $A_i$ and flea $A_j.$ Define our second variable, $S$, to be $d(A_1,A_n)+d(A_2,A_n)+\\dots + d(A_{n-1},A_n).$ To make our proof easier, we note that if $R\\ge M$ at any point in our iteration of moves, we are done since any move that makes fleas jump over this right-most flea will bring that flea past $M$.\n\n[b]Bound[/b]\nTo prove our bound first, let $\\lambda<\\tfrac{1}{n-1}$ and we consider a single move, which WLOG starts from a configuration such that $A_1,A_2,\\dots, A_n$ are placed in that order from left to right. Then, define $d_i$ to be the initial value of $d(A_i,A_n).$ It is easy to see that $S=d_1+d_2+\\dots + d_{n-1}$. Now, suppose $A_i$ jumps over $A_j$, and we split this into several cases.\n\n[u]Case 1: $A_i$ ends up to the left of $A_n$[/u]\nIn this case, $S$ decreases and $R$ doesn't increase. We do not elaborate further.\n\n[u]Case 2: $A_i$ ends up to the right of $A_n$[/u]\nIn this case, $S$ and $R$ are both maximized when $A_2$ jumps over $A_n$. Indeed, we have $d(A_n,A_i)\\le \\lambda d_i.$ Thus, $R$ increases by at most $\\lambda d_i$ Note that $d(A_j,A_i)\\le d(A_j,A_n)+d(A_n,A_i)\\le d_j+\\lambda d_i$ so \\[S\\le d_1+\\dots + \\lambda (n-1)d_i+d_{i+1}+\\dots +d_{n-1}\\]\nThus, $S$ decreases by at least $(1-\\lambda (n-1)) d_i.$\n\nCombining both cases, for every $\\lambda>0$ we want to increase $R$ by, we must proportionally decrease at least $(1-\\lambda (n-1))>0$ from $S$. Since $S$ must be non-negative at all times, $R$ is bounded, so let $M$ be anything that $R$ cannot reach, and our bound is proved.\n\n[b]Construction[/b]\nOur construction for $\\lambda \\ge \\tfrac{1}{n-1}$ is as follows: take the left-most flea and jump it across the right-most flea. We again let $A_1,A_2,\\dots, A_n$ be placed in that order from left to right and define $d_i$ to be the initial value of $d(A_i,A_n).$ Initially, $S=d_1+d_2+\\dots + d_{n-1}.$ After the move, $d(A_i,A_1)$ becomes $d(A_i,A_n)+d(A_n,A_i)=d_i+\\lambda d_1$, so $S$ becomes \\[\\lambda(n-1)d_1+d_2+\\dots + d_{n-1}\\ge d_1+d_2+\\dots + d_{n-1}.\\] In particular, this means that the distance between the first and last flea is always at least $\\tfrac{S}{n-1}$ so $R$ increases by at least $\\tfrac{S}{(n-1)^2}$ each time. Therefore, $R$ is not bounded. We are done.", - "Solution_13": "The answer is all $\\lambda \\geq \\frac{1}{n-1}.$ To prove that this works, we let the leftmost flea jump over the rightmost on each move. If at some moment pairwise distances between consecutive fleas from left to right are $a_1,a_2,\\dots ,a_{n-1},$ then after the move they become $a_2,a_3,\\dots ,a_{n-1}, \\lambda \\sum a_i\\geq \\min a_i$. Therefore the average value of these distances doesn't decrease, so the distance between leftmost and rightmost fleas never decreases too and fleas can jump sufficently far.\n\nNow assume $\\lambda <\\frac{1}{n-1}.$ Numerate fleas by $1,2,\\dots ,n$ and let at the moment $x_i$ denotes the coordinate of $i-$th flea.\n\n[b]Claim.[/b] Number $F=2\\max x_i -\\sum_{j=1}^n x_j$ is a nonincreasing monovariant under moves.\n[i]Proof.[/i] Let $\\varphi$ be a move $x_i\\mapsto x_j+\\lambda (x_j-x_i)$ for $x_iM$ we obtain $$F>(1-\\lambda (n-1))x_n>(1-\\lambda (n-1))M\\implies M<\\frac{\\max F}{1-\\lambda (n-1)}.$$", - "Solution_14": "very cool\n[hide = solution]\nWe show that the answer is all real numbers $\\lambda \\ge \\frac{1}{n-1}$. Also in what follows, we'll assume the line to be the number line, with origin to the left of all fleas, so that they have positive coordinates (this is not necessary, but it has been done for convenience).\n\nWe'll show sufficiency for now. First, while there are fleas on the same position, we make them jump over the right most flea. Eventually all fleas are on distinct positions. Now let $\\delta$ be the minimum gap between consecutive fleas. If we repeatedly make the left most flea jump over the right most flea, then the barycenter of the fleas moves to the right by at least $\\delta$ every turn. Eventually the barycenter crosses the point $M$, and so some flea is beyond $M$. Then we make all the other fleas jump over this one, and all fleas are to the right of $M$, as required.\n\nNow it suffices to show that if $\\lambda < \\frac{1}{n-1}$ then there is some point $M$ which no flea can go beyond. The idea is to look at the position of the barycentre and the position of the right most flea after (say) $m$ turns. Let these positions be $a_m, b_m$ respectively. Suppose the operation was performed on fleas with positions $x < y$ in the $m$th turn. Then some algebra reveals that: \\[a_m-a_{m-1} = \\frac{(1+\\lambda)}{n} \\cdot (y-x)\\] and that $\\[b_m-b_{m-1} \\leq \\lambda(y-x)\\]$. It follows plainly that $\\frac{b_m-b_{m-1}}{a_m-a_{m-1}} \\leq \\frac{n\\lambda}{1+\\lambda}$. Let $K$ denote the previous constant. Observe that $K < 1$ and that $b_m-b_{m-1} \\leq K(a_m-a_{m-1})$. Adding up across all previous $m$, we obtain $b_m-b_0 \\leq K(a_m-a_0) < K(b_m-a_0)$ implying that $b_m < \\frac{b_0-Ka_0}{1-K}$, and so the point with coordinates $\\frac{b_0-Ka_0}{1-K}$ works for $M$.\n[/hide]", - "Solution_15": "[hide=Solution]We claim that it is possible iff $\\lambda \\ge \\frac{1}{n-1}$.\\\\\nLet $a_i$ be the distance between the $i$th frog from the left and the $i+1$th frog from the left.\nConsider the quantity $\\sum ia_i$. Note that if the $k$th frog from the left jumps in front of the rightmost frog to a distance $x$ in front of the rightmost frog, then the new distances are $b_1,\\cdots b_{n-1} = a_1,\\cdots,a_{k-2},a_{k-1}+a_k,a_{k+1},\\cdots,a_n,x$, where $a_0 = 0$. Note that $x \\le (n-1)\\lambda (a_k+\\cdots+a_{n-1})$ with equality if the $k$th frog from the left actually jumps over the rightmost frog.\\\\\n$[\\sum ia_i] - [\\sum ib_i] = a_k+\\cdots +a_{n-1}-(n-1)x \\ge a_k + \\cdots + a_{n-1} - (n-1)\\lambda (a_k + \\cdots + a_{n-1}) = (1-(n-1)\\lambda) (a_k+\\cdots+a_{n-1}))$.\\\\\nNote that if on the other hand the $k$th frog from the left jumps to in between the $j$th frog from the left and the $j+1$th frog from the right (and the distance between the $j$th frog from the left and the position the frog jumped to is $x$) then the new distances are\n$b_1,\\cdots,b_{n-1} = a_1,\\cdots,a_{k-2},a_{k-1}+a_k,a_{k+1},\\cdots,a_{j-1},x,a_j-x,a_{j+1},\\cdots,a_{n-1}$. Therefore,\\\\\n$[\\sum ia_i] - [\\sum ib_i] = a_k+a_{k+1}+\\cdots+a_{j-1}+ja_j-(j-1)x-j(a_j-x) = a_k+\\cdots+a_{j-1}+x$.\\\\\nWe now prove that if $\\lambda \\ge \\frac{1}{n-1}$ then it is possible.\\\\\nWe employ the strategy of having the leftmost frog jump over the rightmost frog. Note that in this case $[\\sum ia_i] - [\\sum ib_i] = (1-(n-1)\\lambda) (a_1+\\cdots +a_n))$. However, note that this is non-positive for $\\lambda \\ge \\frac{1}{n-1}$. Therefore, since $\\sum_{a_i} > \\frac{\\sum ia_i}{n}$, and $\\sum ia_i$ is increasing, $\\sum a_i$ is bounded below by a constant $C$, so after \n$n-1$ moves when the original rightmost frog is the leftmost, the new rightmost frog is at least a distance of $2C$ from the original leftmost frog, and in general in $k(n-1)$ moves the rightmost frog is at least $(k+1)C$ away from the rightmost frog. This can grow as large as required, as required.\\\\\nWe now prove that if $\\lambda < \\frac{1}{n-1}$ then it is not possible no matter the configuration.\nNote that if $\\lambda < \\frac{1}{n-1}$ then $\\sum ia_i$ is decreasing. Also note that if a frog doesn't jump in front of the rightmost frog, then $\\sum ia_i$ decreases by the distance that it travelled, and so since $\\sum ia_i < (n-1)(\\sum a_i)$, the frogs can jump a distance of at most\n$(n-1)(\\sum a_i)$ in this manner. Note on the other hand that if a the $k$th frog from the left jumps to a distance $x$ in front of the rightmost frog, then it has traveled a distance of $a_k+\\cdots+a_{n-1}+x$, and $\\sum ia_i$ decreases by $a_k+\\cdots+a_{n-1}-(n-1)x$, but \n$\\frac{a_k+\\cdots+a_{n-1}-(n-1)x}{a_k+\\cdots+a_{n-1}+x} = 1-\\frac{nx}{a_k+\\cdots+a_{n-1}+x} = 1-\\frac{n}{\\frac{a_k+\\cdots+a_{n-1}}{x}+1}$, which is minimised when $\\frac{n}{\\frac{a_k+\\cdots+a_{n-1}}{x}+1}$ is maximal, but this is maximised when $\\frac{a_k+\\cdots+a_{n-1}}{x}$ is minimal, but this is minimised when $x$ is maximal. The maximum value of $x$ is $\\lambda(a_k+\\cdots+a_n)$, so the minimum value of the fraction is\n$\\frac{(1-(n-1)\\lambda)(a_k+\\cdots+a_n)}{(1+\\lambda)(a_k+\\cdots+a_n)} = \\frac{1-(n-1)\\lambda}{1+\\lambda}$. Therefore, $\\frac{[\\sum ia_i]-[\\sum ib_i]}{\\text{distance travelled}} \\ge \\frac{1-(n-1)\\lambda}{1+\\lambda} \\implies \\sum ib_i \\le [\\sum ia_i] - \\frac{1-(n-1)\\lambda}{1+\\lambda}\\text{distance travelled}$. Therefore, the total distance that can be travelled by jumping over the rightmost frog is at most $\\frac{1+\\lambda}{1-(n-1)\\lambda}[\\sum ia_i]$. Therefore, the total distance the frogs can jump is bounded above by a constant in terms of the $a_i$ and $\\lambda$, and so the furthest distance a frog can jump is also bounded above by this constant, so we are done.[/hide]\n[hide = Comments] When I found the monovariant $\\sum ia_i$, I thought it would be pretty easy to finish from there, but it turned out that a lot more work had to be done to prove that $\\lambda < \\frac{1}{n-1}$ fails. Edit: after some algebra (turning the $a_i$ into the $f_i$) my invariant is actually almost the same as Juan Ortiz's invariant.[/hide]", - "Solution_16": "Could not find this invariant until hinted largely ;-;\n\nNote that this is equivalent to having one flea travel arbitrarily far.\n\n[b][color=pink]Claim:[/color][/b] We can achieve $\\lambda \\ge \\frac{1}{n - 1}$.\n[i]Proof.[/i] Let the fleas have initial starting positions $a_1, a_2, \\dots, a_n$. Let $C = \\underset{i}{argmin} (a_{i+1} - a_i)$ be the minimum difference between two consecutive fleas.\nThen $a_n - a_1 \\ge (n-1)C$ so jumping $a_1$ over $a_n$ to position $a_{n+1}$ makes it so that $a_{n+1} - a_n \\ge C$, so $C$ remains the minimum distance between any two fleas. Repeatedly jumping the last flea over the first flea gives the result. $\\blacksquare$\n\n[b][color=pink]Claim:[/color][/b] $\\lambda < \\frac{1}{n-1}$ is impossible.\n[i]Proof.[/i] Consider the value of $a_n - \\lambda (a_1 + \\dots + a_{n-1})$ where flies are re-labeled after each turn based on order.\nWe claim that this value is a monovariant that doesn't increase.\nUnless the farthest fly changes, the value is guarenteed to decrease. Else, suppose that $a_i$ jumps over $a_j$ to become the furthest fly. As $j$ increases, the value afterwards decreases. Thus, assume $j = n$ and that $a_i$ jumps to $(1 + \\lambda)a_n - \\lambda a_i$.\nHowever, then equality holds as \\[ (1 + \\lambda)a_n - \\lambda a_i - \\lambda (a_1 + \\dots + a_n - a_i) = a_n - \\lambda (a_1 + \\dots + a_{n-1}). \\]\nThen, it follows that for some $C$ and given initial config \\[ C \\ge a_n - \\lambda (a_1 + \\dots + a_{n-1}) \\ge a_n (1 - \\lambda (n-1)) \\] which implies $a_n \\le \\frac{C}{1 - \\lambda(n-1)}$ which bounds growth. $\\blacksquare$\n" -} -{ - "Problem": "Characterize the subgroups of $GL_n(R)$ such that every matrix in this subgroup has all its eigenvalues of modulus 1.", - "Solution_1": "Anyone, any idea or reference? I amp very interested on this thing, but I cannot find anything relevant on the literature. Moubi,do you know some references?", - "Solution_2": "If I find a reference I' ll tell you.\r\n\r\nFrom wich oral of ENS is this related ?", - "Solution_3": "Thanks, Moubi. It started from a problem from oral ENS asking for the continuous morphisms from $GL_n(C)$ into $R$. That one is not very difficult, but this one...", - "Solution_4": "Ok, till now I think I've got the form of subgroups of $GL_n(C)$ which consist only of matrices of eigenvalues of modulus 1. After 5 days of continuous work, without noticing ( idiot!) that all results were just in front of me. Again Leva's great ideas were the essential part. \r\n So, let's prove (I will call it theorem, I have no idea whether it is classical or not, but it surely deserves this title):\r\n Theorem\r\n A subgroup of $GL_n(C)$ all of whose matrices have eigenvalues on the unit circle is conjugated to a subgroup of the matrices having triangular form by blocks and on the main diagonal having as blocks some unitary matrices of fixed dimensions $ m_1,...,m_p$ with $ m_1+...+m_p=n$.\r\n Proof:\r\n We will prove this by induction on the dimension of the matrices. So, let's prove the inductive step. Let $G$ be such a group of $nxn$ matrices. Consider the algebra generated by the elements of $G$. It is a subalgebra of $ L(C^n)$ thus by Burnside's theorem it is either all $ L(C^n)$ or it has an invariant space. If it is the last case, then let $V'$ a suplementary of $V$ and look at the matrices of $G$ in the base of $C^n$ obtained by superposition of two basis in $V,V'$. Now, apply the inductive hypothesis to the parts of the matrices of $G$ corresponding to $V,V'$. \r\n Thus it remains the case when the algebra generated by the elements of $G$ is the whole $M_n(C)$. Take $A_1,...,A_k$ (with $k=n^2$) a basis of $ M_n(C)$ with elements from $G$. Observe that the application sending the matrix $B\\in M_n(C)$ into the $k$-tuple $ (Tr(A_1B),...,Tr(A_kB))$ is bijective. Indeed, it is linear and injective, since if $Tr(A_j B)=0$ for all $j$ then $ Tr(XB)=0$ for all $X$ and thus $B=0$. This means that the application is also surjective (the two spaces having the same dimension). Applying surjectivity for the $k-$ tuples $ (1,0,0,...,0), (0,1,0,...,0),...,(0,0,...,1)$ we obtain the existence of some matrices $ B_1,...,B_k$ such that $ Tr(B_i A_j)=1$ only for $i=j$ and $0$ otherwise. Next, observe that these relations also imply that $ B_1,...,B_k$ are linearly independent. Indeed, if $ x_1 B_1+...+x_k B_k=0$ then multiply this with $A_j$ and take the trace. It will follow that all $x_k$ are 0. Thus, $B_1,...,B_k$ is a base in $M_n(C)$. \r\n Consider now a matrix $X$ and decompose it in the form $X=x_1 B_1+...+x_k B_k$. Take the trace in the relation $ XA_j= x_1 B_1 A_j+...+x_k B_k A_j$. You'll find $ x_j=Tr(XA_j)$. Now, observe that if $X\\in G$ then we have $| Tr(A_jX)|\\leq n$ since all eigenvalues of $A_jX$ are of modulus 1. Thus, by taking a norm $||||$ on $M_n(C)$ we observe that $ ||X||\\leq n(||B_1||+...+||B_k||)$, which means that $G$ is bounded in $M_n(C)$ and so it's closure is a compact subgroup in $GL_n(C)$. But it is well-known that such a subgroup is conjugated to a subgroup of the unitary matrices ( I think Alekk posted it for $M_n(R)$ with orthogonal matrices, but it's practically the same proof). Thus, $G$ is conjugated to a subgroup of the unitary matries $U_n(C)$.\r\n Could anyone tell me whether this works or not? Thanks. :) :)\r\n PS: what is the latex command for making a new paragraph?" -} -{ - "Problem": "Several positive integers are written on a blackboard. One can erase any two distinct integers and write their greatest \r\ncommon divisor and least common multiple instead. Prove that eventually the numbers \r\nwill stop changing. \r\n\r\n\r\n\r\n\r\nP.S. it puzzles me and i cant understand the solution given.", - "Solution_1": "If I'm not mistaken....\r\n\r\n[hide]Lemma: If we have a list of positive integers $ a_1, \\ldots, a_n$ written on a blackboard, and at each stage, for any $ a_i$ and $ a_j$, we replace $ a_i$ with $ \\min(a_i, a_j)$ and $ a_j$ with $ \\max(a_i, a_j)$, then the numbers on the board will eventually remain the same. \n\nProof: Suppose we define our sequence of pairs $ (i, j)$ with $ i < j$ to operate on to be $ s_1, s_2, \\ldots$, and call this operation operation S. We say that we apply operation S to a term in this sequence $ s_n$, we are applying operation $ S$ on $ a_i$ and $ a_j$, where $ s_n \\equal{} (a_i, a_j)$.\n\nObserve that if we perform operation S on a pair with $ i < j$ and $ a_i < a_j$, the numbers on the board will stay fixed. Observe that if at some point in the sequence, applying operation S to $ s_{n\\plus{}1}, s_{n\\plus{}2}, \\ldots$ does nothing, we will be done (since our board will remain the same after these operations.)\n\nDefine the number of inversions of this sequence to be the number of pairs $ a_i$, $ a_j$ such that $ i < j$ and $ a_i > a_j$. We claim that if an application of operation S to some term $ s_n$ swaps two terms of a sequence, the number of inversions decreases. Indeed, if $ i < j$ and $ a_i > a_j$, then $ a_i$ and $ a_j$ will have their positions swapped. I could go through this all more rigorously, but I'm lazy, and really, it's easy to verify that the number of inversions does decrease. \n\nIf there were indeed infinitely many terms in $ \\{s_n\\}$ that changed the board, then we would have an infintiely decreasing sequence of integers, a contradiction. Thus, our board must eventually remain constant, as desired. \n\nLet our integers be $ a_1, a_2, \\ldots, a_n$. We now prime factorize each of the integers. Suppose that for some prime $ p$ dividing some integer on the blackboard, we take another blackboard and write down in order $ v_p(a_1), v_p(a_2), \\ldots, v_p(a_n)$ (where $ v_p(n)$ denotes the greatest exponent of $ p$ dividing $ n$). Note that when we take $ a_i$ and $ a_j$, and replace them with their greatest common factor and least common multiple, respectively, we are replacing $ v_p(a_i)$ and $ v_p(a_j)$ with $ \\min(v_p(a_i), v_p(a_j))$ and $ \\max(v_p(a_i), v_p(a_j))$, respectively. Applying our lemma, we see that at some point, the exponents of $ p$ in our sequence will stay fixed when this operation is performed on them. Since we can use this argument on every prime factor, it follows that all the numbers on the blackboard must eventually stay fixed as well. [/hide]", - "Solution_2": "[quote=\"Zhero\"]Lemma: If we have a list of positive integers $ a_1, \\ldots, a_n$ written on a blackboard, and at each stage, for any $ a_i$ and $ a_j$, we replace $ a_i$ with $ \\min(a_i, a_j)$ and $ a_j$ with $ \\max(a_i, a_j)$, then the numbers on the board will eventually remain the same. [/quote] This result is not actually the one you used. (Your lemma as stated is just silly: replacing $ \\{a_i, a_j\\}$ with their min and max doesn't ever change the set of numbers ;).) Just the inversions monovariant argument is fine :)", - "Solution_3": "Good point. :blush:\r\n\r\nI think I meant to say the sequence of numbers on the board will stay the same, i.e., taking order into account.", - "Solution_4": "Well the gcd of two distinct numbers is less than both unless one is a multiple of the other. But if one is a multiple of the other the two numbers will just stay the same, so we can assume that is not the case. So we keep picking two numbers, neither of which is a multiple of the other, and the smaller of the two keeps decreasing. As long as there exists two numbers, neither of which is a multiple of the other, we will keep decreasing some of the numbers until they can't decrease anymore (1 is the smallest natural), and then move on to the rest, and this process will countinue until for any two numbers, one is a multiple of the other, and then they all stay the same.\r\nIs this rigorous?", - "Solution_5": "Perhaps some of those numbers could be made more explicit?\r\n\r\nAnother invariant argument, because $ \\text{gcd}(a,b) \\le a,b \\le \\text{lcm}(a,b)$ and $ \\text{gcd}(a,b)\\text{lcm}(a,b) \\equal{} ab$, we (because, for example $ \\text{lcm}(a,b) \\minus{} \\text{max}(a,b) \\equal{} \\frac{\\text{max}(a,b)}{\\text{gcd}(a,b)}[\\text{min}(a,b) \\minus{} \\text{gcd}(a,b)]$, etc) have $ \\text{gcd}(a,b) \\plus{} \\text{lcm}(a,b) \\ge a \\plus{} b$, equality iff equality held in the earlier condition. So $ S \\equal{} \\sum a_i$, which is a positive integer, will strictly increase, but it is bounded above [say, by $ S \\le n\\text{lcm}(a_1, \\ldots, a_n)$].", - "Solution_6": "[quote=\"Zhero\"]Proof: Suppose we define our sequence of pairs $ (i, j)$ with $ i < j$ to operate on to be $ s_1, s_2, \\ldots$, and call this operation operation S. We say that we apply operation S to a term in this sequence $ s_n$, we are applying operation $ S$ on $ a_i$ and $ a_j$, where $ s_n \\equal{} (a_i, a_j)$.\n\nObserve that if we perform operation S on a pair with $ i < j$ and $ a_i < a_j$, the numbers on the board will stay fixed. Observe that if at some point in the sequence, applying operation S to $ s_{n \\plus{} 1}, s_{n \\plus{} 2}, \\ldots$ does nothing, we will be done (since our board will remain the same after these operations.)\n\nDefine the number of inversions of this sequence to be the number of pairs $ a_i$, $ a_j$ such that $ i < j$ and $ a_i > a_j$. We claim that if an application of operation S to some term $ s_n$ swaps two terms of a sequence, the number of inversions decreases. Indeed, if $ i < j$ and $ a_i > a_j$, then $ a_i$ and $ a_j$ will have their positions swapped. I could go through this all more rigorously, but I'm lazy, and really, it's easy to verify that the number of inversions does decrease. \n\nIf there were indeed infinitely many terms in $ \\{s_n\\}$ that changed the board, then we would have an infintiely decreasing sequence of integers, a contradiction. Thus, our board must eventually remain constant, as desired. \n\nLet our integers be $ a_1, a_2, \\ldots, a_n$. We now prime factorize each of the integers. Suppose that for some prime $ p$ dividing some integer on the blackboard, we take another blackboard and write down in order $ v_p(a_1), v_p(a_2), \\ldots, v_p(a_n)$ (where $ v_p(n)$ denotes the greatest exponent of $ p$ dividing $ n$). Note that when we take $ a_i$ and $ a_j$, and replace them with their greatest common factor and least common multiple, respectively, we are replacing $ v_p(a_i)$ and $ v_p(a_j)$ with $ \\min(v_p(a_i), v_p(a_j))$ and $ \\max(v_p(a_i), v_p(a_j))$, respectively. Applying our lemma, we see that at some point, the exponents of $ p$ in our sequence will stay fixed when this operation is performed on them. Since we can use this argument on every prime factor, it follows that all the numbers on the blackboard must eventually stay fixed as well. [/hide][/quote]\r\n\r\ni really dont understand your proof but i did understand what the book had. What is an inversion exactly?i dint understand para 3,5. someone please elaborate :(", - "Solution_7": "suppose $(a,b)$ is transformed to $(gcd(a,b),lcm(a,b))=(c,d)$. $d \\ge max(a,b) $ it can be easily shown that $c+d \\ge a+b$ .So we can say that the sum is non decreasing as well as bounded.As the sum is bounded at some point d will stop changing and as ab=cd so c will stop changing. Hence the numbers will stop changing." -} -{ - "Problem": "8. How many 3-digit numbers between 100 and 300 are there for which one of the digits equals the sum of the other two? \r\n\r\nI don't know how to solve this besides going through each number? Can anyone help?", - "Solution_1": "Going through every number isn't that bad. Since the hundreds digit would have to be 1 or two, the other two digits would have a difference of 1 or 2, or the other two digits would add up to the hundreds digit.\r\n\r\n101-2 (110, 101)\r\n112-2 (112, 121)\r\n123-2\r\n134-2\r\n145-2\r\n...\r\n189-2\r\n9*2=18\r\n\r\nthen\r\n\r\n202-2\r\n213-2\r\n...\r\n279-2\r\n8*2=16\r\n\r\n18+16=34\r\n\r\nthen there is 211\r\n\r\nso 34+1=35\r\n\r\nIf I messed up somewhere, just ignore me...", - "Solution_2": "According to the answers, you got it right. Thanks. :D", - "Solution_3": "[hide]\nThe hundreds digit is either 1 or 2.\n\nFor 1, \nThe other two digits have a difference of 1\n0,1\n1,2\n2,3\n...\n8,9\nThat is 9 pairs. You can order in two ways\n\nFor 2,\nThe other two digits have a difference of 2\n0,2\n1,3\n2,4\n...\n7,9\n\nThat is 8 pairs. You can order in two ways.\n\nSo total it is $9\\cdot2+8\\cdot2=18+16=34$\n\nThere's also 211\n35 total[/hide]" -} -{ - "Problem": "New poll", - "Solution_1": "Yeah two people is what we discussed, and it's better anyway. Btw, I accidentally hit 3 in the poll, so consider that.", - "Solution_2": "is this ever going to happen?", - "Solution_3": "here is a better poll:\r\n\r\nHOW MANY TIMES HAS THIS QUESTION BEEN POLLED??!?!?!??!!?", - "Solution_4": "I actually think that two is better, but then we might have a 16 team competition.", - "Solution_5": "only one smart person is needed. if you have too many people u have a wide mem expansion" -} -{ - "Problem": "I need some serious help to make a collection of ray diagrams through different type of lenses!\r\nAssume do > C, I have to give a set of below answers for concave mirror, convex mirror, converging lens and diverging lens.\r\n\r\ndo=7.5cm\r\nho=1cm\r\ndi=?\r\nhi=?\r\nM=?\r\n\r\nHow can I get those information above and how do I draw those pictures?", - "Solution_1": "Use the mirror and lens equations to get type of image and image distance etc.", - "Solution_2": "More specifically, the useful equations are:\r\n\r\n$ \\frac 1{di} \\plus{} \\frac 1{do} \\equal{} \\frac 1f$ Where [i]f[/i] is the focal distance.\r\n\r\nAlso: \r\n$ \\frac {\\minus{}di}{do}\\equal{} \\frac {hi}{ho}$" -} -{ - "Problem": "On a n \u00d7 n table we place natural numbers so that any two neighboring numbers \r\ndiffer by at most 1. Prove that there is a number that appears on the table at least \r\nn times.", - "Solution_1": "on number have 9 neighboring numbers or onl 4?!\r\n\r\nWe can make a beautifull solution of that using John Conway's Game of Life, don't we?\r\n\r\nIf so, what is the rules of the game?", - "Solution_2": "@yagaron: I think 4 at most but I am not sure. You can solve it for 4 :)", - "Solution_3": "if is $4$ the answer is very simple, i think\r\n\r\nLet's a Example for $n=5$\r\n\r\n${\\left( \\begin{array}{lllll}1 & \\square & \\square & \\square & 1 \\\\ \\square & \\square & \\square & \\square & \\square \\\\ \\square & \\square & 1 & \\square & \\square \\\\ \\square & \\square & \\square & \\square & \\square \\\\ 1 & \\square & \\square & \\square & 1 \\end{array}\\right)}$\r\n\r\nThe anothers spaces can be replaced by all numbers diferent from those picked, since there is no limitation for the numbers to place\r\nAnd it's easy to see that $n>3$\r\n\r\nI'm making the repeted number $1$ just to simplify, but could be another\r\n\r\nWe can generalize it to.\r\n\r\n${\\left( \\begin{array}{lllllllll}1 & \\square & \\square & \\square & \\square & \\square & \\square & \\square & 1 \\\\ \\square & \\ddots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & .\\cdot{}^{\\cdot}& \\square \\\\ \\square & \\ldots & 1 & \\square & \\square & \\square & 1 & \\ldots & \\square \\\\ \\square & \\ldots & \\square & \\square & \\square & \\square & \\square & \\ldots & \\square \\\\ \\square & \\ldots & \\square & \\square & 1 & \\square & \\square & \\ldots & \\square \\\\ \\square & \\ldots & \\square & \\square & \\square & \\square & \\square & \\ldots & \\square \\\\ \\square & \\ldots & 1 & \\square & \\square & \\square & 1 & \\ldots & \\square \\\\ \\square & .\\cdot{}^{\\cdot}& \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\ddots & \\square \\\\ 1 & \\square & \\square & \\square & \\square & \\square & \\square & \\square & 1 \\end{array}\\right)}$\r\n\r\nIt's just valid for even $n$.\r\nBut with a similar construction we can make it for odd $n$'s too.\r\nIf we can't complete the numbers in those $2$ diagonals we can put is in another place distant from the both diagonals\r\n\r\nPS: the number of neighboring could be 4 or 8. not 9\r\n\r\nsorry" -} -{ - "Problem": "Prove that if $ a,b,c > 0$ then\r\n\\[ \\frac {(4a \\plus{} b \\minus{} c)^2}{2a^2 \\plus{} (b \\plus{} c)^2} \\plus{} \\frac {(4b \\plus{} c \\minus{} a)^2}{2b^2 \\plus{} (c \\plus{} a)^2} \\plus{} \\frac {(4c \\plus{} a \\minus{} b)^2}{2c^2 \\plus{} (a \\plus{} b)^2}\\geq8.\r\n\\]", - "Solution_1": "Nobody is interested in this? I have tried the same method used for USA 2003, but I failed. Homogenous method does not work for this because we can not turn each of the three expression on the left side into an expression of a single variable. Say, $a+b+c=3$, ....\r\n\r\n\r\nMaybe brute force can help but I am too lazy to do this. Or if anyone can end this topics by giving a counterexample.", - "Solution_2": "[quote=\"pvthuan\"]Prove that if $ a,b,c > 0$ then\n\\[ \\frac {(4a \\plus{} b \\minus{} c)^2}{2a^2 \\plus{} (b \\plus{} c)^2} \\plus{} \\frac {(4b \\plus{} c \\minus{} a)^2}{2b^2 \\plus{} (c \\plus{} a)^2} \\plus{} \\frac {(4c \\plus{} a \\minus{} b)^2}{2c^2 \\plus{} (a \\plus{} b)^2}\\geq8.\n\\]\n[/quote]\r\nBy Cauchy Schwarz Inequality, we get\r\n\\[ \\sum \\frac {(4a \\plus{} b \\minus{} c)^2}{2a^2 \\plus{} (b \\plus{} c)^2} \\ge \\sum \\frac {(4a \\plus{} b \\minus{} c)^2}{2(a^2 \\plus{} b^2 \\plus{} c^2)}\r\n\\]\r\nIt suffices to prove that\r\n\\[ \\sum (4a \\plus{} b \\minus{} c)^2 \\ge 16\\sum a^2\r\n\\]\r\nwhich is obviously true because\r\n\\[ \\sum (4a \\plus{} b \\minus{} c)^2 \\equal{} \\sum (16a^2 \\plus{} 8 a(b \\minus{} c) \\plus{} (b \\minus{} c)^2) \\equal{} 16\\sum a^2 \\plus{} \\sum (b \\minus{} c)^2 \\ge 16\\sum a^2\r\n\\]\r\nWe have done.\r\n:)" -} -{ - "Problem": "This is 14.15, on p.81 of greenberg and harper's algebraic topology book.\r\n\r\nLet $ X \\equal{} X_1 \\cup X_2$, and $ A \\equal{} X_1 \\cap X_2$. Using the exact sequences of triples, show that if the inclusion $ (X_1, A) \\rightarrow (X,X_2)$ induces an isomorphism in homology then the same holds for the inclusion $ (X_2,A) \\rightarrow (X,X_1)$.\r\n\r\nI don't really know how I can use the exact sequences of triples. Can anyone enlighten me on this?", - "Solution_1": "I'm not sure if I can bump this yet, anyway.. can anyone help?", - "Solution_2": "[quote=\"Soarer\"]This is 14.15, on p.81 of greenberg and harper's algebraic topology book.\n\nLet $ X = X_1 \\cup X_2$, and $ A = X_1 \\cap X_2$. Using the exact sequences of triples, show that if the inclusion $ (X_1, A) \\rightarrow (X,X_2)$ induces an isomorphism in homology then the same holds for the inclusion $ (X_2,A) \\rightarrow (X,X_1)$.[/quote]\n\nI have got something, but I don't really believe it's true - check it carefully, please.\n\nWe do not need $ X = X_1 \\cup X_2$, and $ A = X_1 \\cap X_2$. It is enough to have $ X_1 \\cup X_2\\subset X$, and $ A \\subset X_1 \\cap X_2$.\n\nHowever, I hope that by saying \"isomorphism in homology\", you mean \"isomorphism in homologies of every degree\"; else, I don't have a proof (but I haven't searched for counterexamples either).\n\nConsider the sequence $ H_n\\left(X_1,A\\right)\\to H_n\\left(X,A\\right)\\to H_n\\left(X,X_2\\right)$, where both arrows are induced by inclusions. The composition of these transformations is induced by the inclusion $ \\left(X_1,A\\right)\\to\\left(X,X_2\\right)$, and therefore (by a condition of the problem) it is an isomorphism, so that it is injective. Hence, its \"first arrow\" $ H_n\\left(X_1,A\\right)\\to H_n\\left(X,A\\right)$ must also be injective. Hence, the kernel of the arrow $ H_n\\left(X_1,A\\right)\\to H_n\\left(X,A\\right)$ is zero.\n\nThe long exact sequence for the triple $ \\left(X,X_1,A\\right)$ looks like this:\n\n$ ...\\to H_{n + 1}\\left(X,X_1\\right)\\to H_n\\left(X_1,A\\right)\\to H_n\\left(X,A\\right)\\to H_n\\left(X,X_1\\right)$\n$ \\to H_{n - 1}\\left(X_1,A\\right)\\to ...$.\n\nWe know that the kernel of the arrow $ H_n\\left(X_1,A\\right)\\to H_n\\left(X,A\\right)$ is zero here - but this kernel is the image of the preceding arrow $ H_{n + 1}\\left(X,X_1\\right)\\to H_n\\left(X_1,A\\right)$. Hence, the image of the arrow $ H_{n + 1}\\left(X,X_1\\right)\\to H_n\\left(X_1,A\\right)$ must be zero, so that this arrow is simply the zero homomorphism. Repeating this argument with $ n - 1$ instead of $ n$, we see that $ H_n\\left(X,X_1\\right)\\to H_{n - 1}\\left(X_1,A\\right)$ is the zero homomorphism. Thus, our sequence becomes\n\n$ ...\\to H_{n + 1}\\left(X,X_1\\right)\\substack{0 \\\\\n\\to} H_n\\left(X_1,A\\right)\\to H_n\\left(X,A\\right)\\to H_n\\left(X,X_1\\right)$\n$ \\substack{0 \\\\\n\\to} H_{n - 1}\\left(X_1,A\\right)\\to ...$.\n\nThis readily yields the exact sequence\n\n$ 0\\to H_n\\left(X_1,A\\right)\\to H_n\\left(X,A\\right)\\to H_n\\left(X,X_1\\right)\\to 0$.\n\nOn the other hand, the long exact sequence for the triple $ \\left(X,X_2,A\\right)$ yields\n\n$ H_n\\left(X_2,A\\right)\\to H_n\\left(X,A\\right)\\to H_n\\left(X,X_2\\right)$.\n\nAn easy diagram chase yields the following fact: If $ A$, $ B$, $ X$, $ Y$, $ U$ are (say) abelian groups such that $ 0\\to A\\to U\\to B\\to 0$ and $ X\\to U\\to Y$ are exact sequences and such that the composition of the arrows $ A\\to U$ and $ U\\to Y$ is an isomorphism, then the compositions of the arrows $ X\\to U$ and $ U\\to B$ is an isomorphism as well.\n\nThis applies to our problem, because the composition of the arrows $ H_n\\left(X_1,A\\right)\\to H_n\\left(X,A\\right)$ and $ H_n\\left(X,A\\right)\\to H_n\\left(X,X_2\\right)$ is an isomorphism (namely, the one induced by the inclusion $ \\left(X_1,A\\right)\\to\\left(X,X_2\\right)$), and thus we conclude that the composition of the arrows $ H_n\\left(X_2,A\\right)\\to H_n\\left(X,A\\right)$ and $ H_n\\left(X,A\\right)\\to H_n\\left(X,X_1\\right)$ is an isomorphism as well, i. e. the inclusion $ \\left(X_2,A\\right)\\to\\left(X,X_1\\right)$ induces an isomorphism. We are done.\n\nDid this all make sense? (I'm not asking whether this could have been done on 2 lines as well, because most likely the answer is yes.)\n\n darij", - "Solution_3": "Yeah it works, thanks. :)" -} -{ - "Problem": "Question : 10 x 3 rectangle (float)\r\nA clown walks around the edge of the float...\r\n\r\nFloat goes at 1m/s\r\nClown moves at 2m/s (clockwise) around the float...\r\n\r\nWhat is the general equation?\r\n\r\nMy Work Done: \r\n\r\nI worked out the angles and the distances around the float but I need a\r\ngeneral equation.", - "Solution_1": "its a piece wise function so split it up into parts. it should be easy after that" -} -{ - "Problem": "I don't understand why real analysis and algebra-questions are posted in advanced fields? I don't look in advanced fields because I don't know what a lot of the questions are about, so I miss a lot of questions about things I know about. \r\n\r\nI don't think that topology, measure, integration and stuff like that belong in advanced fields.", - "Solution_1": "Where else should topology go\u00bf\r\nAlgebra and integration are just misplaced there...", - "Solution_2": "Maybe I don't look at advanced fields as often as I should, but when I do see obvious misplacements, I move the topics. Topology and functional analysis I usually leave there. \r\n\r\nFeel free to point out to me things you think ought to be moved. (I might not always agree.)", - "Solution_3": "A lot depends on the perceived difficulty of the question. I view \"Advanced fields\" as the place for problems that either do not fit in any other category, or are perceived as much more difficult or requiring more advanced knowledge than normal. As Kent said, feel free to PM any of the College Playground moderators if you think that something should be moved; just be specific in your requests and don't expect us to always agree with your point of view :)." -} -{ - "Problem": "I don't know if this has been asked before but anyways.\r\nSuppose I don't want to read a particular topic ever again. How do I remove that topic from my list of post???", - "Solution_1": "What list of posts? :huh:", - "Solution_2": "The list of 'view your posts'...", - "Solution_3": "... Just delete it by pressing the right button (the \"x\" in the right corner under the post).", - "Solution_4": "You cannot delete any topics. Only a moderator or admin can delete topics. You can make your own personal list of topics by adding them to your favourite topics and then clicking on the favorites link.", - "Solution_5": "I'm not talking about deleting topics. I'm just asking how to remove the topic from the 'View Your posts' list????", - "Solution_6": "Maybe you didn't read my above reply. You cannot delete those topics and they will always appear in that list. If you want a personal list of your own with the topics you choose click on the Favorites link. You can add and delete there any topic by easily clicking on \"Add this topic to Favorites\" link.", - "Solution_7": "[quote=\"Rushil\"]I'm just asking how to remove the topic from the 'View Your posts' list????[/quote]\r\n\r\nThe reason it's called \"View Your Posts\" is because it is a complete and exclusive list of your posts. This means that a topic appears there if and only if you make a post in that topic, and that furthermore there are no other ways of adding or removing items from that list.\r\n\r\nIf there has not been a reply to a post you've made, then you can delete it by clicking on the X button located in the lower right-hand corner of the post. Otherwise, notify a moderator of that forum and if one is not available, try an administrator.\r\n\r\nFor future reference, \"delete\" as far as posts and topics are considered means to erase them. What you wish to do is to modify a list of shortcuts. Providing as much information as possible (including page names, URLs, etc.) helps us answer your question the first time around.", - "Solution_8": "Thanks... I'll keep that in mind!!!1" -} -{ - "Problem": "Let $f:[0,1]\\to[0,1]$ be a continuous function s.t. $f(0)=0$ with the property that for each $x\\in[0,1]$ there is a positive integer $n=n(x)$ s.t. $f^{(n)}(x)=x$, where $f^{(n)}$ means $f$ composed with itself $n$ times. Show that $f(x)=x,\\ \\forall x\\in[0,1]$.", - "Solution_1": "*$f$ is surjective\r\n*$f$ is injective: if $f(x)=f(y)$ then \r\n$f^{(n_xn_y)}(x)=f^{(n_x)}(y)=f^{(n_x)}(x)=x$ and $f^{(n_xn_y)}(x)=f^{(n_y)}(x)=f^{(n_y)}(y)=y$\r\n*$f(0)=0$, $f$ is continuous,$f:[0,1] \\to [0,1]$ . So $f$ is strictly increasing.\r\nSuppose that $f(x)>x$(resp. $f(x)>x$) then $f^{(n)}(x)>x$(resp. $f^{(n)}(x)0 ? :huh:", - "Solution_3": "[quote=\"@petko\"]Very nice :D Thank you!\nBut isn't it true only for t>0 ? :huh:[/quote]\r\nWhen t<0, the answer is zero. When t=0, the integral does not converge, however it is reasonable to set its value $ \\pi$.\r\nYou can see the conclusion easily from the following (rough) derivation\r\n$ \\int_{ \\minus{} \\infty}^{ \\plus{} \\infty}\\frac {e^{ix t} }{1 \\plus{} i x}dx \\equal{} \\int_{ \\minus{} \\infty}^{ \\plus{} \\infty}\\int_{ \\minus{} \\infty}^t e^{i x u \\plus{} u \\minus{} t}du dx \\equal{} \\int_{ \\minus{} \\infty}^t 2\\pi\\delta(u)e^{u \\minus{} t} du$", - "Solution_4": "[b]Buffalo[/b] could you explain in more detail the very last part of your equation, please?\r\nI mean:\r\n\r\n$ \\int\\limits_{ \\minus{} \\infty}^{ \\plus{} \\infty}\\int\\limits_{ \\minus{} \\infty}^t e^{i x u \\plus{} u \\minus{} t}du dx \\equal{} \\int\\limits_{ \\minus{} \\infty}^t 2\\pi\\delta(u)e^{u \\minus{} t} du \\stackrel{\\text{is it true that...?}}{\\Rightarrow} \\int\\limits_{ \\minus{} \\infty}^{ \\plus{} \\infty} e^{i x u} dx \\equal{} 2 \\pi \\delta(u)$\r\n\r\n:?: :huh:", - "Solution_5": "I think that if t=0 the integral is $ 2\\pi$." -} -{ - "Problem": "There are n triangles of positive area that have one vertex A(0,0) and the other two vertices whose coordinates are drawn independently with replacement from (0,1,2,3,4). Compute n", - "Solution_1": "[hide=\"solution\"]We need to find the number of possible coordinates for the other two vertices of the triangle, and the coordinates are chosen from the first 5 whole numbers. There are $ 5*4\\equal{}20$ possible coordinates, and we need to choose two of them. $ \\binom{20}{2}\\equal{}\\boxed{190}$[/hide]", - "Solution_2": "Consider the pair $ (1, 1)$ and $ (2, 2)$, then modify your solution :)", - "Solution_3": "What does this mean??\r\n\r\n[quote=\"firecricket91\"]of positive area [/quote]\r\n\r\n\r\n\r\n :)", - "Solution_4": "It means \"nondegenerate.\"", - "Solution_5": "If lets say B is at (3,2) and C is at (2,3).\r\n\r\nIs that traingle different than a triangle that has C at (3,2) and B and (2,3)?", - "Solution_6": "[hide]\nFirst these points are not coincident, so in this $ 5\\times 5$ grid, with $ (0,0)$ alreday chosen, we have $ \\binom{24}{2}\\equal{}276$ triangles.\nHowever, we also need to eliminate the cases when the three points are collinear. Within this grid, among all the rays started from origin, we have three rays with four lattice points and two rays with two lattice points. Thus, we need to subtract $ 3\\binom{4}{2}\\plus{}2\\binom{2}{2}\\equal{}20$ triangles, so $ n\\equal{}256$.\n\n[/hide]", - "Solution_7": "When it means n trianlges does it mean the vertices can be interchanged?\r\n\r\nBecause it gave A as (0,0) doenst that kinda imply the verticies matter", - "Solution_8": "A triangle doesn't care what you name its vertices." -} -{ - "Problem": "What is Cramer's Rule? How is it used in solving systems? What do the determinants in Cramer's method for solving systems represent?\r\n\r\nP.S.: I don't know where to post this topic. Maybe the moderators will move it to another section.", - "Solution_1": "i think this belongs to Advanced Algebra section.\r\n\r\nYou can find it in almost every linear algebra book.\r\nYou can also have a look at\r\nhttp://mathworld.wolfram.com/CramersRule.html", - "Solution_2": "I know this is in the wrong section, but I didn't want to begin a new thread. Maybe this can be moved.\r\n\r\nCramer's Rule states the following:\r\n\r\n$x_i=\\frac{\\det A_i(\\mathbf{x})}{\\det A}$\r\n\r\nHere's an example:\r\n\r\n$x_1+2x_2=5$\r\n$3x_1+7x_2=17$\r\n\r\nI will compute $x_1$.\r\n\r\n$A=\\begin{bmatrix}1 & 2 \\\\ 3 & 7\\end{bmatrix}$\r\n\r\nThe meaning of $A_i$ is that you replace the \"i\"th column of A with the vector x, which in this case is <5,17>.\r\n\r\n$A_1=\\begin{bmatrix}5 & 2 \\\\ 17 & 7\\end{bmatrix}$\r\n\r\nSo $\\det A=7-6=1$ and $\\det A_1(\\mathbf{x})=35-34=1$.\r\n\r\n$x_1=\\frac{\\det A_1(\\mathbf{x})}{\\det A}=1$\r\n\r\nMaybe you can try to find $x_2$ with this method. It is actually very simple and quick. I can provide a proof of the rule as well if you need it.\r\n\r\nAlex.", - "Solution_3": "Cramer's Rule is: If we have a set of equations\r\n\\[ \\\\a_1x+b_1y+c_1z=k_1\\\\a_2x+b_2y+c_2z=k_2\\\\a_3x+b_3y+c_3z=k_3 \\]\r\n$x=\\frac{\\left|\\begin{matrix}k_1&b_1&c_1\\\\k_2&b_2&c_2\\\\k_3&b_3&c_3\\end{matrix}\\right|}{\\Delta}$ $y=\\frac{\\left|\\begin{matrix}a_1&k_1&c_1\\\\a_2&k_2&c_2\\\\a_3&k_3&c_3\\end{matrix}\\right|}{\\Delta}$ $z=\\frac{\\left|\\begin{matrix}a_1&b_1&k_1\\\\a_2&b_2&k_2\\\\a_3&b_3&k_3\\end{matrix}\\right|}{\\Delta}$ $\\Delta=\\left|\\begin{matrix}a_1&b_1&c_1\\\\a_2&b_2&c_2\\\\a_3&b_3&c_3\\end{matrix}\\right|$\r\n\r\nOf course, this also works for higher order determinants.\r\n\r\nP.S. I'm 14 years old, so I don't know if this is around what you wanted to know. :)\r\n\r\nMasoud Zargar", - "Solution_4": "What if our system has coeficients not in $R$(or $C$), but in a field $K$? If $K$ is commutative, it is clear, Cramer is still available. But what if it is not commutative?", - "Solution_5": "Zero'th problem: fields are commutative for me ;)\r\n\r\nFirst problem for noncommutative case:\r\nA linear equation is not necessary of type $a_1 x_1 + a_2 x_2 +...=d$, but can contain more than one linear term of type $a_1 x_1 b_1$ with $x_1$.\r\n\r\nNext problem: the definition of determinant looses it's symmetry of changing rows or columns, thus even interchanging the equations or variables could give us another determinant.\r\n\r\nThird problem: even the division in the rule is not defined on which side to be taken.\r\n\r\nConclusion: by some random constants, e.g. from the quaternions, there should be a counterexample.\r\n\r\nPS: moderators, please move this thread..." -} -{ - "Problem": "prove that with a,b,c>0 we have\r\n(2* (a^2+b^2+c^2))^(3/2)>=3*sqrt(3)*(a+b)(b+c)(c+a); :D", - "Solution_1": "Oh,It's a easy ineq in Mathlinks.\r\nAssume that $a+b+c=1$\r\n$(a+b)(b+c)(c+a)=(1-a)(1-b)(1-c) \\le (\\frac{3-a-b-c}{3})^3=\\frac{8}{27}$\r\n$a^2+b^2+c^2 \\ge \\frac{1}{3}$\r\nAnd we are done!" -} -{ - "Problem": "Let $x$ and $y$ be positive integers such that$3x^2 + x = 4y^2 + y$. Prove that\r\n$x - y$is a square.", - "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=32858\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=38983\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=38350\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=30699\r\n\r\n darij", - "Solution_2": "1. I'm not there to search for problems that have been posted $101$ times.\r\n2.: e.g.: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=square+x-y&t=38983" -} -{ - "Problem": "Can we order online? Or do we have to print that order form and send it to the address?", - "Solution_1": "Print the order form and mail in along with payment, or\r\ncall 800-527-3690, between the hours of 9:00 am - 4:00 pm (Central time) \r\nMonday through Friday, and order using a credit card. Orders are usually filled in no more than 2 business days and mailed via U.S. postal mail.\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions", - "Solution_2": "Yup. I ordered a copy of that CD. The only ways you can order it is by mail or fax.", - "Solution_3": "It contains all the AMC 8 from 2001-2006, and the AMC 10 A&B, AMC 12 A&B, AIME I & II, USAMO and the IMO Team tests from 2001-2007.\r\n\r\nDoes it have the solutions or do we have to buy them separte", - "Solution_4": "As well as the contests themselves, the complete and full solution booklets for all contests, exactly as printed, are on the CD. This includes the solutions for the USAMO problems too.\r\n\r\n-- Steve Dunbar\r\nMAA Director, American Mathematics Competitions" -} -{ - "Problem": "Have no idea how to solve this kind of equation $ 2^x\\equal{}x^3$. :?:", - "Solution_1": "The equation has two roots. That's obvious from the graph of these functions. The question of finding them is tough: it can be found approximately using Taylor's expansion of LHS, or another approximation-iteration process, but I do not believe that exact value of it can be found. Mathematica 6's answer is not closed:\r\n\r\nx == -(3 ProductLog[-Log[2]/3])/Log[2] || \r\n x == -(3 ProductLog[-1, -Log[2]/3])/Log[2]\r\n\r\nOr, numerically \r\n\r\nx == 1.37347 || x == 9.93954\r\n\r\nThat's why I believe that the equation doesn't have a closed form of solution, and that the problem lies in the domain of numerical methods.", - "Solution_2": "There are probably an infinite number of complex roots since the equation has the infinite polynomial representation\r\n\r\n1+xln2+(xln2)^2/2!+(xln2)^3/3!+... - x^3 = 0", - "Solution_3": "Yes, you're correct. There are three infinite families of solutions. But for the sake of education :lol: I closed the domain on reals in my analysis...", - "Solution_4": "Thanks everybody. It has two solutions, so the answer says. Actually I thought a little bit and understood that I have to use the graphs of the functions. \r\n\r\nI got another problem now...\r\n\r\nCan't solve this one $ (0,2)^{x\\plus{}1}\\equal{}\\sqrt{35\\plus{}5x}.$\r\n\r\nI know that the answer is -2, but don't know how to prove it mathematicaly.", - "Solution_5": "Try tu use [url=http://en.wikipedia.org/wiki/Lambert_W_function]Lambert's $ W$ function[/url] in your solution.", - "Solution_6": "[quote=\"@petko\"]Try tu use [url=http://en.wikipedia.org/wiki/Lambert_W_function]Lambert's $ W$ function[/url] in your solution.[/quote]\r\nI don't know this function but I think you are trying to complicate a very simple problem :wink:\r\n\r\n\r\nTry to consider the function $ f(x)\\equal{}(0,2)^{x \\plus{} 1} \\minus{}\\sqrt {35 \\plus{} 5x}$.\r\nIt is continuous and strictly decreasing in it's domain $ [\\minus{}7,\\infty)$.\r\nWe have $ f(\\minus{}7)>0$, $ f(0)<0$.\r\nHence (Darboux property) there is exactly one $ t$ s.t. $ f(t)\\equal{}0$.\r\n\r\nAnd now you may try to guess this number. It's not hard to guess $ t\\equal{}\\minus{}2$.\r\nThe proof above shows that $ \\minus{}2$ is the only one such number, so it is the only solution to Elena's equation", - "Solution_7": "[b]knoppix[/b] well... I suggested using W function as rather general solution :whistling: . You are right - some equations are simple as this one, but... some aren't ;)", - "Solution_8": "Thanks guys! I don't think I have to use this W equation, because this test is just for school. It shouldn't be that complicated.\r\nAnyway thanks a lot!" -} -{ - "Problem": "Se considera ecuatia: a1^2/(x-b1)+a2^2/(x-b2)+.....+an^2/(x-bn)=c ,unde ak, bk sunt din R,k=1..n, n>=11 si c din R* .Aratati ca ecuatia are toate radacinile rationale.", - "Solution_1": "Ecuatia este: $ \\frac {{a_1}^2}{x \\minus{} b_1} \\plus{} \\frac {{a_2}^2}{x \\minus{} b_2} \\plus{} \\dots \\plus{} \\frac {{a_n}^2}{x \\minus{} b_n} \\equal{} c$, unde $ a_k;b_k\\in\\mathbb{R}$ pentru $ k\\in\\{1;2;\\dots ;n\\}$ si $ c\\in\\mathbb{R}^*$, iar $ n\\ge 11$.\r\nAm impresia ca totusi enuntul corect ar fi: [b]Aratati ca ecuatia are toate radacinile reale![/b]" -} -{ - "Problem": "Does there exist a natural number with the sum of digits of its $ kth$ power being\r\nequal to $ k$, if a) $ k \\equal{} 2004$; b) $ k \\equal{} 2006?$", - "Solution_1": "a) $ n^2004$ is congruent to $ 6 mod 9$\r\n\r\nn obviously has to be relatively prime to 9, since otherwise $ n^2004$ is congruent to 0 mod 9, but then since phi(9) is 6, $ n^6$ is congruent to 1 mod 9, so $ n^2004$ is congruent to 1 mod 9, so no solution.\r\n\r\nb) $ n^2006$ is congruent to 8 mod 9\r\n\r\nagain, n is relatively prime, blah, blah, blah, $ n^2$ is congruent to 8 mod 9. But the quad. res. mod 9 are 1,4,0,7, contradiction!" -} -{ - "Problem": "hey there can you change your signature?\r\n\r\nI guess nobody would like that sort of language on the site! :D :D\r\n\r\nso ... change it!!! \r\n\r\ncheers!", - "Solution_1": "i know my opinion on this topic isnt reqd but neway,wot does the message say,i was just curiou.i think \"merde\" in french was \"s**t\",if i remember correctly.but vialli's italian.but i think i see the connection\r\n.and i i'm thinking correctly,vialli SHOULD change the signature", - "Solution_2": "yea what was the signature anyway?", - "Solution_3": "People, stop reviving old threads please." -} -{ - "Problem": "Show that\r\n$\\displaystyle (a^2+b^2+c^2)(a^3+b^3+c^3)\\geq 4abc(\\frac{a^2}{b+c}+\\frac{b^2}{a+c}+\\frac{c^2}{a+b})^2$\r\n\r\n$\\forall a,\\ b,\\ c >0$ \r\n\r\n :)", - "Solution_1": "Sorry but I'm a beginner. So can I do this?\r\nIt's homogeneous, so we can assume $abc = 1$.\r\n\r\n$\\frac{a^2}{b+c} \\leq \\frac{a^2}{2\\sqrt{bc}} = \\frac{1}{2}a^{\\frac{5}{2}}$\r\n\r\nThus now the inequality becomes: $(a^{\\frac{5}{2}}+b^{\\frac{5}{2}}+c^{\\frac{5}{2}})^2 \\leq (a^2+b^2+c^2)(a^3+b^3+c^3)$ which is true by Cauchy.", - "Solution_2": "[quote=\"hendrata01\"]Sorry but I'm a beginner. So can I do this?\nIt's homogeneous, so we can assume $abc = 1$.\n\n$\\frac{a^2}{b+c} \\leq \\frac{a^2}{2\\sqrt{bc}} = \\frac{1}{2}a^{\\frac{5}{2}}$\n\nThus now the inequality becomes: $(a^{\\frac{5}{2}}+b^{\\frac{5}{2}}+c^{\\frac{5}{2}})^2 \\leq (a^2+b^2+c^2)(a^3+b^3+c^3)$ which is true by Cauchy.[/quote]\r\n\r\n\r\nExcellent!! :clap: \r\n\r\n\r\nMy solution involved twice the Cauchy-Schwarz ineq.... :)" -} -{ - "Problem": "Show that if N = 2p - 1 with p a prime, then N divides 2N-1 - 1.", - "Solution_1": "there exists a k for which (2^p)-k divides p\r\n\r\nso name (2^p) - k = A then you get (2^A)-1 / (2^p)-1\r\n\r\nand i guess this is well-known...", - "Solution_2": "and i just got msg from arne:\r\n\r\nindeed, FLT says k can just be replaced with 2 here :P" -} -{ - "Problem": "Can anyone help me set up this problem. \r\n\r\nA pond initially contains 1,000,000 gal of water and an unknown amount of undesirable chemical. Water containing 0.01 gram of this chemical per gallon flows into the pond at a rate of 300 gal/hr. The mixture flows out at the same rate, so the amount of water in the pond remains constant. Assume that the chemical is uniformly distributed throughout the pond. \r\n\r\na). Write a differential equation for the amount of chemical in the pond at any time. I am confused with the units, including gal and grams. Am I supposed to work with grams or concentration? \r\n\r\nAny help would be greatly appreciated. THx", - "Solution_1": "My suggestion: make your main function, $y(t),$ be the mass of the the contaminant in the whole pond, measured in grams.\r\n\r\nThe concentration in the pond at any time is then $10^{-6}y(t),$ with the units of measurement there being grams per gallon. (Yes, that's a truly weird unit of concentration - but I've actually seen it. Just go along with it.)\r\n\r\nNow, what happens during a small interval of time $\\Delta t?$ (Use hours as the unit of measurement for $\\Delta t.$)\r\n\r\nThe amount of the chemical flowing into the pond is $300\\cdot 0.01\\cdot \\Delta t.$ The units of measurement there are\r\n\r\n$\\frac{\\text{gal}}{\\text{hr}}\\cdot\\frac{\\text{gram}}{\\text{gal} }\\cdot \\text{hr}=\\text{gram}$\r\n\r\nNow, what is the amount flowing out of the pond? The volume of water flowing out is $300\\cdot\\Delta t.$ The concetnration varies with time, but taking $\\Delta t$ to be a very short interval, we approximate it as $10^{-6}\\cdot y(t).$ Then the amount flowing out is approximately $10^{-6}\\cdot y(t)\\cdot 300\\cdot\\Delta t.$ Once again the units work out to be grams. So:\r\n\r\n$y'(t)\\cdot\\Delta t\\approx 300\\cdot 0.01\\cdot \\Delta t-10^{-6}\\cdot y(t)\\cdot 300\\cdot\\Delta t$\r\n\r\n$y'(t)=3-(3\\times 10^{-4})y(t).$\r\n\r\nI'll leave you to solve that equation yourself. It is both linear and separable and hence can be solved in more than one way.\r\n\r\nTwo questions:\r\n\r\n1. What is the steady-state condition of the pond? In particular, what is $\\lim_{t\\to\\infty}y(t)$ (or the concentration you could derive from that)? Hint: the \"common sense\" value is the correct value, and could be seen in this autonomous differential equation without bothering to find the solution.\r\n\r\n2. What is the time scale? Here's an interesting alternative: imagine setting up a movable impermeable membrane between the inlet and the outlet so that the input solution never mixes with the original contents of the pond at all. It would take $\\frac{10^6}{300}$ hours to completely replace the original contents of the pond. So that - about 333 hours - is some kind of fundamental time scale here. Now for my question: for the mixing process, what is the half-life? (Yes, that is a meaningful question, if you look at it correctly.)" -} -{ - "Problem": "I have attached a file with the questions in .doc format. Sorry for the inconvienience", - "Solution_1": "Here they are in the forum. I think there's a problem with the first question.\r\n\r\n[b]Problem 1:[/b] \r\n\r\nShow that $ \\lim_{x \\to \\infty} \\frac{(1\\plus{}x)^{1/x} \\minus{}e\\plus{}\\frac{ex}{2}}{x^2} \\equal{} \\frac{11}{24} e$\r\n\r\n[b]Problem 2:[/b]\r\n\r\nCompute $ \\lim_{n \\to \\infty} (a^n\\plus{}b^n)^{1/n}$, given that $ a>b>0$.", - "Solution_2": "[url=http://www.mathlinks.ro/viewtopic.php?p=1476334#1476334]See[/url]", - "Solution_3": "No no the question is correct. So people , solution for the 1st one?" -} -{ - "Problem": "Show that for all $ x,y,z\\ge 0$ we have\r\n\\[ \\frac {x}{x^2 \\plus{} y \\plus{} 1} \\plus{} \\frac {y}{y^2 \\plus{} z \\plus{} 1} \\plus{} \\frac {z}{z^2 \\plus{} x \\plus{} 1}\\le 1\r\n\\]\r\nand the equality holds iff x=y=z=1.\r\n\r\nThis problem seems strange because it is non-homogeneous and there is no relationship between variables. However, it is easy. Can you prove it?\r\n\r\nIf you can, please try with a harder one: Find the maximum, for all $ x,y,z\\ge 0$, of the functions\r\n\\[ f(x,y,z) \\equal{} \\frac {x}{x^2 \\plus{} y^2 \\plus{} 1} \\plus{} \\frac {y}{y^2 \\plus{} z^2 \\plus{} 1} \\plus{} \\frac {z}{z^2 \\plus{} x^2 \\plus{} 1}\r\n\\]\r\nand\r\n\\[ g(x,y,z) \\equal{} \\frac {x}{x^2 \\plus{} \\sqrt {y} \\plus{} 1} \\plus{} \\frac {y}{y^2 \\plus{} \\sqrt {z} \\plus{} 1} \\plus{} \\frac {z}{z^2 \\plus{} \\sqrt {x} \\plus{} 1}.\r\n\\]\r\n\r\n(Hint: the maximum is attained when x=y=z).", - "Solution_1": "[quote=\"Hatucdao\"]Prove that for all $ x,y,z\\ge 0$ we have[/quote]\r\nStep1 we prove that\r\n$ A \\equal{} x/(x \\plus{} 2z) \\plus{} y/(2x \\plus{} y) \\plus{} z/(2y \\plus{} z) > \\equal{} 1$\r\nbecause $ (x \\plus{} y \\plus{} z)^{2} \\leq A(x(x \\plus{} 2z) \\plus{} y(2x \\plus{} y) \\plus{} z(2y \\plus{} z))$\r\nStep 2\r\nwe see that $ x^{2} \\plus{} y \\plus{} 1 \\geq 2x \\plus{} y;...$\r\nStep 3 Do by yourself", - "Solution_2": "Great desmrhai! As you see, the problem is in fact homogeneous one :D Could you please try two others?", - "Solution_3": "[quote=\"Hatucdao\"]Show that for all $ x,y,z\\ge 0$ we have\n\\[ \\frac {x}{x^2 \\plus{} y \\plus{} 1} \\plus{} \\frac {y}{y^2 \\plus{} z \\plus{} 1} \\plus{} \\frac {z}{z^2 \\plus{} x \\plus{} 1}\\le 1\n\\]\nand the equality holds iff x=y=z=1.\n\nThis problem seems strange because it is non-homogeneous and there is no relationship between variables. However, it is easy. Can you prove it?\n\nIf you can, please try with a harder one: Find the maximum, for all $ x,y,z\\ge 0$, of the functions\n\\[ f(x,y,z) \\equal{} \\frac {x}{x^2 \\plus{} y^2 \\plus{} 1} \\plus{} \\frac {y}{y^2 \\plus{} z^2 \\plus{} 1} \\plus{} \\frac {z}{z^2 \\plus{} x^2 \\plus{} 1}\n\\]\nand\n\\[ g(x,y,z) \\equal{} \\frac {x}{x^2 \\plus{} \\sqrt {y} \\plus{} 1} \\plus{} \\frac {y}{y^2 \\plus{} \\sqrt {z} \\plus{} 1} \\plus{} \\frac {z}{z^2 \\plus{} \\sqrt {x} \\plus{} 1}.\n\\]\n(Hint: the maximum is attained when x=y=z).[/quote]\r\n2) \\[ \\sum \\frac{x}{x^2\\plus{}y^2\\plus{}1} \\le \\frac{1}{2} \\sum \\frac{x}{\\sqrt{x^2\\plus{}y^2}} \\le \\frac{3}{2\\sqrt{2}}\\]\r\n:)\r\nI will think of 3), and post it if I can... :)", - "Solution_4": "Hi Can, you are fast and correct ! :) Similar to the first one, the main point in the second inequality is also of changing into the homogeneous cases. Such change is natural and always does not meet any risk because the original one (non-homogeneous) is in fact equivalent to the later one (homogeneous). Anyway, could you please explain some words about the estimate?\r\n\\[ \\sum \\frac {x}{\\sqrt {x^2 \\plus{} y^2}} \\le \\frac {3}{\\sqrt {2}}\r\n\\]\r\nI am looking forward to seeing your treatment for 3).", - "Solution_5": "[quote=\"Hatucdao\"]Hi Can, you are fast and correct ! :) Similar to the first one, the main point in the second inequality is also of changing into the homogeneous cases. Such change is natural and always does not meet any risk because the original one (non-homogeneous) is in fact equivalent to the later one (homogeneous). Anyway, could you please explain some words about the estimate?\n\\[ \\sum \\frac {x}{\\sqrt {x^2 \\plus{} y^2}} \\le \\frac {3}{\\sqrt {2}}\n\\]\nI am looking forward to seeing your treatment for 3).[/quote]\r\nI think that you prove $ \\sum \\frac {1}{\\sqrt {1 \\plus{} a^{2}}}$ for $ abc \\equal{} 1$\r\nOr $ \\sum \\frac {1}{\\sqrt {1 \\plus{} a}}$ for $ abc \\equal{} 1$", - "Solution_6": "[quote=\"desmrhai\"] I think that you prove $ \\sum \\frac {1}{\\sqrt {1 \\plus{} a^{2}}}$ for $ abc \\equal{} 1$\nOr $ \\sum \\frac {1}{\\sqrt {1 \\plus{} a}}$ for $ abc \\equal{} 1$[/quote]\r\nYes, of course they are equivalent, but they are not easier :D. Could you give a simple solution for them?", - "Solution_7": "[quote=\"Hatucdao\"]Hi Can, you are fast and correct ! :) Similar to the first one, the main point in the second inequality is also of changing into the homogeneous cases. Such change is natural and always does not meet any risk because the original one (non-homogeneous) is in fact equivalent to the later one (homogeneous). Anyway, could you please explain some words about the estimate?\n\\[ \\sum \\frac {x}{\\sqrt {x^2 \\plus{} y^2}} \\le \\frac {3}{\\sqrt {2}}\n\\]\nI am looking forward to seeing your treatment for 3).[/quote]\r\n\r\nThis is problem 113 in the book \"Old and New Inequalities\". I'll rewrite the proof later :)", - "Solution_8": "[quote=\"Hatucdao\"]Show that for all $ x,y,z\\ge 0$ we have\n\\[ \\frac {x}{x^2 \\plus{} y \\plus{} 1} \\plus{} \\frac {y}{y^2 \\plus{} z \\plus{} 1} \\plus{} \\frac {z}{z^2 \\plus{} x \\plus{} 1}\\le 1\n\\]\nand the equality holds iff x=y=z=1.\n\n.[/quote]\r\nWe have a question : Find the [b]minimum[/b] value of expression $ P \\equal{} \\frac {x}{x^2 \\plus{} y \\plus{} 1} \\plus{} \\frac {y}{y^2 \\plus{} z \\plus{} 1} \\plus{} \\frac {z}{z^2 \\plus{} x \\plus{} 1}$ if we add the condition $ x \\plus{} y \\plus{} z \\equal{} 3$ ?\r\nI think it's an interesting one. :)", - "Solution_9": "[quote=\"Hatucdao\"]Hi Can, you are fast and correct ! :) Similar to the first one, the main point in the second inequality is also of changing into the homogeneous cases. Such change is natural and always does not meet any risk because the original one (non-homogeneous) is in fact equivalent to the later one (homogeneous). Anyway, could you please explain some words about the estimate?\n\\[ \\sum \\frac {x}{\\sqrt {x^2 \\plus{} y^2}} \\le \\frac {3}{\\sqrt {2}}\n\\]\n[/quote]\r\nI proved it by using Cauchy Schwarz :)\r\n\\[ \\left( \\sum \\frac{x}{\\sqrt{x^2\\plus{}y^2}}\\right)^2 \\le \\left( \\sum (x^2\\plus{}z^2)\\right) \\left( \\sum \\frac{x^2}{(x^2\\plus{}y^2)(x^2\\plus{}z^2)}\\right) \\equal{}\\frac{4(x^2\\plus{}y^2\\plus{}z^2)(x^2y^2\\plus{}y^2z^2\\plus{}z^2x^2)}{(x^2\\plus{}y^2)(y^2\\plus{}z^2)(z^2\\plus{}x^2)} \\le \\frac{9}{2}\\]\r\nwhere the last inequality is valid by AM-GM Inequality. :)", - "Solution_10": "[quote=\"dduclam\"][quote=\"Hatucdao\"]Show that for all $ x,y,z\\ge 0$ we have\n\\[ \\frac {x}{x^2 \\plus{} y \\plus{} 1} \\plus{} \\frac {y}{y^2 \\plus{} z \\plus{} 1} \\plus{} \\frac {z}{z^2 \\plus{} x \\plus{} 1}\\le 1\n\\]\nand the equality holds iff x=y=z=1.\n\n.[/quote]\nWe have a question : Find the [b]minimum[/b] value of expression $ P \\equal{} \\frac {x}{x^2 \\plus{} y \\plus{} 1} \\plus{} \\frac {y}{y^2 \\plus{} z \\plus{} 1} \\plus{} \\frac {z}{z^2 \\plus{} x \\plus{} 1}$ if we add the condition $ x \\plus{} y \\plus{} z \\equal{} 3$ ?\nI think it's an interesting one. :)[/quote]\r\nIf we want to find minimum, it is very easy. Notice that\r\n\\[ x^2\\plus{}y\\plus{}1 \\le x^2\\plus{}(3\\minus{}x)\\plus{}1 \\le 3x\\plus{}(3\\minus{}x)\\plus{}1\\equal{}2x\\plus{}4 \\le 10\\]\r\nHence\r\n\\[ P \\ge \\frac{x}{10} \\plus{}\\frac{y}{10}\\plus{}\\frac{z}{10}\\equal{}\\frac{3}{10}\\]\r\nWe have equality for $ x\\equal{}3,y\\equal{}z\\equal{}0$. Therefore $ \\min P\\equal{}\\frac{3}{10}.$ ;) :)", - "Solution_11": "[quote=\"can_hang2007\"] I proved it by using Cauchy Schwarz :)\n\\[ \\left( \\sum \\frac {x}{\\sqrt {x^2 \\plus{} y^2}}\\right)^2 \\le \\left( \\sum (x^2 \\plus{} z^2)\\right) \\left( \\sum \\frac {x^2}{(x^2 \\plus{} y^2)(x^2 \\plus{} z^2)}\\right) \\equal{} \\frac {4(x^2 \\plus{} y^2 \\plus{} z^2)(x^2y^2 \\plus{} y^2z^2 \\plus{} z^2x^2)}{(x^2 \\plus{} y^2)(y^2 \\plus{} z^2)(z^2 \\plus{} x^2)} \\le \\frac {9}{2}\n\\]\nwhere the last inequality is valid by AM-GM Inequality. :)[/quote]\r\nGreat work! This is exactly what I have waited for! I don't like the answers such as \"the solution is in the book XYZ\" :D I was surprised if you can find it in a short time (I have needed a long time :) ). \r\n\r\nThe problem of finding minimum when x+y+z=3 is homogeneous one, and it is not in this series. Now, try the third one, please!", - "Solution_12": "[quote=\"Hatucdao\"][quote=\"desmrhai\"] I think that you prove $ \\sum \\frac {1}{\\sqrt {1 \\plus{} a^{2}}}$ for $ abc \\equal{} 1$\nOr $ \\sum \\frac {1}{\\sqrt {1 \\plus{} a}}$ for $ abc \\equal{} 1$[/quote]\nYes, of course they are equivalent, but they are not easier :D. Could you give a simple solution for them?[/quote]\r\n$ (\\sqrt {\\frac {2}{1 \\plus{} a^{2}}} \\plus{} \\sqrt {\\frac {2}{1 \\plus{} b^{2}}})^{2} \\leq 2(\\frac {2}{1 \\plus{} a^{2}} \\plus{} \\frac {2}{1 \\plus{} b^{2}}) \\leq \\frac {8c}{c \\plus{} 1}$\r\nbut\r\n$ 2\\sqrt {\\frac {2c}{c \\plus{} 1}} \\plus{} \\sqrt \\frac {2}{c^{2} \\plus{} 1} \\leq 3$\r\nUnderstand? And you? your soluton? Don't say that u don't know", - "Solution_13": "[quote=\"Hatucdao\"]\n$ g(x,y,z) \\equal{} \\frac {x}{x^2 \\plus{} \\sqrt {y} \\plus{} 1} \\plus{} \\frac {y}{y^2 \\plus{} \\sqrt {z} \\plus{} 1} \\plus{} \\frac {z}{z^2 \\plus{} \\sqrt {x} \\plus{} 1}.$\n[/quote]\r\nUsing Am-Gm: $ \\sum\\ \\frac {x^2}{x^4 \\plus{} y \\plus{} 1} \\leq\\ \\sum\\ \\frac {x^2}{x^3 \\plus{} x \\plus{} y} \\leq\\ \\sum\\ \\frac {1}{2} \\sum\\ \\sqrt []{\\frac {x}{x \\plus{} y}} \\leq\\ \\frac {3}{2\\sqrt []{2}}$\r\nBut equality doesn't occurs :wink:", - "Solution_14": "[quote=\"nguoivn\"] Using Am-Gm: $ \\sum\\ \\frac {x^2}{x^4 \\plus{} y \\plus{} 1} \\leq\\ \\sum\\ \\frac {x^2}{x^3 \\plus{} x \\plus{} y} \\leq\\ \\sum\\ \\frac {1}{2} \\sum\\ \\sqrt []{\\frac {x}{x \\plus{} y}} \\leq\\ \\frac {3}{2\\sqrt []{2}}$\nBut equality doesn't occurs :wink:[/quote]\nIt seems you are trying to change the problem into a homogeneous one. This argument works for the first two inequalities but sometime it is impossible. For example, let me propose more a problem\n\n[b]Problem 4)[/b] Prove that for all $ a,b,c\\ge 0$ then\n\\[ \\frac {a}{a^2 \\plus{} \\frac {3}{2}b \\plus{} 1} \\plus{} \\frac {b}{b^2 \\plus{} \\frac {3}{2}c \\plus{} 1} \\plus{} \\frac {c}{c^2 \\plus{} \\frac {3}{2}a \\plus{} 1}\\le \\frac {6}{7}.\n\\]\nThe equality holds when $ a \\equal{} b \\equal{} c \\equal{} 1$.\n\nThis problem seems similar to the first problem in this topic, but it is in fact much more interesting. One cannot using estimate $ a^2 \\plus{} 1\\ge 2a$ again, because now the supremum of\n\\[ \\frac {a}{2a \\plus{} \\frac {3}{2}b} \\plus{} \\frac {b}{2b \\plus{} \\frac {3}{2}c} \\plus{} \\frac {c}{2c \\plus{} \\frac {3}{2}a}\n\\]\nis 1, which is larger than $ \\frac {6}{7}$. Now please try 4) before return to 3). \n\n[quote=\"desmrhai\"] Understand? And you? your soluton? Don't say that u don't know[/quote]\r\nTo desmrhai: Please be more polite before talking about mathematics.", - "Solution_15": "$ \\frac {x}{x^2 + y + 1} + \\frac {y}{y^2 + z + 1} + \\frac {z}{z^2 + x + 1}\\le 1$\r\n\r\nsince $ x^2 + 1 \\geq 2x$ and with similarly, cyclic permutations for $ y$ and $ z$. It is enough to prove:\r\n\r\n$ \\frac {x}{2x + y} + \\frac {y}{2y + z} + \\frac {z}{2z + x }\\le 1$\r\n\r\n$ \\frac {2x}{2x + y} + \\frac {2y}{2y + z} + \\frac {2z}{2z + x }\\le 2$\r\n\r\n$ \\frac {2x + y - y}{2x + y} + \\frac {2y + z - z}{2y + z} + \\frac {2z + x - x}{2z + x }\\le 2$\r\n\r\n$ 1 - \\frac {y}{2x + y} + 1 - \\frac {z}{2y + z} + 1 - \\frac {x}{2z + x }\\le 2$\r\n\r\n$ 3 - \\frac {y}{2x + y} - \\frac {z}{2y + z} - \\frac {x}{2z + x }\\le 2$\r\n\r\nequivalently:\r\n\r\n$ \\frac {y}{2x + y} + \\frac {z}{2y + z} + \\frac {x}{2z + x } \\geq 1$\r\n\r\nUsing cauchy-schwarz:\r\n\r\n$ (2xy + y^2 + 2yz + z^2 + 2zx + x^2)(\\frac {y}{2x + y} + \\frac {z}{2y + z} + \\frac {x}{2z + x }) \\geq (x + y + z)^2$\r\n\r\n$ \\frac {y}{2x + y} + \\frac {z}{2y + z} + \\frac {x}{2z + x } \\geq \\frac {(x + y + z)^2}{x^2 + y^2 + z^2 + 2xy + 2yz + 2zx} = \\frac {(x + y + z)^2}{(x + y + z)^2} = 1 $", - "Solution_16": "[quote=\"Hatucdao\"][quote=\"can_hang2007\"] I proved it by using Cauchy Schwarz :)\n\\[ \\left( \\sum \\frac {x}{\\sqrt {x^2 + y^2}}\\right)^2 \\le \\left( \\sum (x^2 + z^2)\\right) \\left( \\sum \\frac {x^2}{(x^2 + y^2)(x^2 + z^2)}\\right) = \\frac {4(x^2 + y^2 + z^2)(x^2y^2 + y^2z^2 + z^2x^2)}{(x^2 + y^2)(y^2 + z^2)(z^2 + x^2)} \\le \\frac {9}{2}\n\\]\nwhere the last inequality is valid by AM-GM Inequality. :)[/quote]\nGreat work! This is exactly what I have waited for! I don't like the answers such as \"the solution is in the book XYZ\" :D I was surprised if you can find it in a short time (I have needed a long time :) ). \n\nThe problem of finding minimum when x+y+z=3 is homogeneous one, and it is not in this series. Now, try the third one, please![/quote]\r\n$ f(x)=\\sqrt{x}$ is concave. Hence\r\n$ \\sum \\sqrt{\\frac{a}{a+b}}=\\sum \\frac{a+c}{2(a+b+c)}\\cdot \\sqrt{\\frac{4a(a+b+c)^2}{(a+b)(a+c)^2}}\\leq \\sqrt{\\sum \\frac{2a(a+b+c)}{(a+c)(b+c)}}$. Hence it suffices to prove $ \\sum \\frac{2a(a+b+c)}{(a+c)(b+c)}\\leq \\frac{9}{4}$ which is equivalent to $ (a+b)(b+c)(c+a)\\geq \\frac{8}{9}(a+b+c)(ab+bc+ca)$ which is well known.", - "Solution_17": "Hatucdao, i think your third inequality can be solved using calculator only.\r\n\\[ g(x,y,z) \\equal{} \\frac {x}{x^2 \\plus{} \\sqrt {y} \\plus{} 1} \\plus{} \\frac {y}{y^2 \\plus{} \\sqrt {z} \\plus{} 1} \\plus{} \\frac {z}{z^2 \\plus{} \\sqrt {x} \\plus{} 1}\r\n\\]\r\nAs you said in your hint maximum is attained when $ x \\equal{} y \\equal{} z$ (unfortunately i don't know how to prove that, this inequality attains maximum at $ x \\equal{} y \\equal{} z$). Therefore, we should find maximum of\r\n\\[ g(x) \\equal{} \\frac {3x}{x^2 \\plus{} \\sqrt {x} \\plus{} 1}\r\n\\]\r\nwhich is attained at $ x \\equal{} 1.3154751$ and the maximum value of the given function is [b]$ 1.0314267$[/b]\r\n\r\nCan someone give elementary proof to the this question? Using Cauchy-Schwarz maybe?\r\n\r\nThanks! :lol:", - "Solution_18": "[quote=\"Hatucdao\"][b]Problem 4)[/b] Prove that for all $ a,b,c\\ge 0$ then\n\\[ \\frac {a}{a^2 \\plus{} \\frac {3}{2}b \\plus{} 1} \\plus{} \\frac {b}{b^2 \\plus{} \\frac {3}{2}c \\plus{} 1} \\plus{} \\frac {c}{c^2 \\plus{} \\frac {3}{2}a \\plus{} 1}\\le \\frac {6}{7}.\\]\nThe equality holds when $ a \\equal{} b \\equal{} c \\equal{} 1$.[/quote]\r\nSee here: http://can-hang2007.blogspot.com/2010/01/inequality-32-p-t-nam.html" -} -{ - "Problem": "Solve equation: \r\n$ 8(\\log_2 x)^3\\plus{}12(\\log_2 x)^2\\plus{}46 \\log_2 x \\minus{}48\\sqrt{2} x\\plus{}69\\equal{}0$", - "Solution_1": "Just observing gives one of the solutions is $ x \\equal{} \\sqrt{2}$." -} -{ - "Problem": "Find all functions $ f: \\mathbb R\\rightarrow \\mathbb R$ satisfying $ f(f(x)\\plus{}y)\\equal{}2x\\plus{}f(f(y)\\minus{}x)$ for every $ x,y\\in \\mathbb R$", - "Solution_1": "We first prove the $ f(x)$ is injective. Suppose there is $ f(b)\\equal{}f(c)\\equal{}d$ with $ b\\neq c$.\r\n$ P(b,y): f(d\\plus{}y)\\equal{}2b\\plus{}f(f(y)\\minus{}b)$\r\n$ P(c,y): f(d\\plus{}y)\\equal{}2c\\plus{}f(f(y)\\minus{}c)$.\r\nSo we have for any $ y$,\r\n$ f(f(y)\\minus{}b)\\minus{}f(f(y)\\minus{}c)\\equal{}2c\\minus{}2b$ ---------------(1)\r\n$ P(x,b): f(f(x)\\plus{}b)\\equal{}2x\\plus{}f(d\\minus{}x)$\r\n$ P(x,c): f(f(x)\\plus{}c)\\equal{}2x\\plus{}f(d\\minus{}x)$\r\nSo we have for any $ x$.\r\n$ f(f(x)\\plus{}b)\\equal{}f(f(x)\\plus{}c)$ ----------------------(2)\r\nIf we let $ u\\equal{}f(x)\\plus{}y$ and $ v\\equal{}f(y)\\minus{}x$, we have $ f(u)\\minus{}f(v)\\equal{}2x$. This implies we can always find $ x_0,y_0$ such that $ f(x_0)\\minus{}f(y_0)\\equal{}\\minus{}b\\minus{}c$. Plug this in (1), we have\r\n$ f(f(y_0)\\minus{}b)\\minus{}f(f(y_0)\\minus{}c)\\equal{}f(f(x_0)\\plus{}c)\\minus{}f(f(x_0)\\plus{}b)\\equal{}2c\\minus{}2b\\neq0$, a contradiction with (2) when we plug in $ x_0$.\r\nAssume $ f(0)\\equal{}a$, \r\n$ P(0,y): f(a\\plus{}y)\\equal{}f(f(y))$.\r\nSince $ f(x)$ is injective, we have $ f(y)\\equal{}y\\plus{}a$." -} -{ - "Problem": "I was read an algebra book and I found the definition of function field $F/K$ of one variable. One of the conditions says that $K$ is algebraically closed in $F$. I have learnt that the algebraic closure of given field is unique so does it mean that if $G$ is an algebraic closure of $F/K$, then $G\\subset F$ or what?", - "Solution_1": "There is no algebraic extension of $K$ contained in $F$, e.g. $K(X)$, where $K$ is not necessarily algebraically closed.", - "Solution_2": "\"$K$ is algebraically closed in $F$\" means just that every element of $F\\setminus K$ is transcendental over $K$. (The set of all elements of $F$ which are algebraic over $K$ is called the algebraic closure of $K$ in $F$, and $K$ being algebraically closed in $F$ means that it equals its algebraic closure in $F$.) owk" -} -{ - "Problem": "Consider an equilateral triangle. Divide each side of it into $4$ equal segments. If you count all the subdivision points and the three vertices, you find that you have $12$ points. Draw a small circle around each of these $12$ points so that no two circles overlap. The numbers $1, 2, 3, 4, ..., 12$ are arranged, one per circle so the sum of the numbers on each side of the triangle are the same. What is the least sum possible?", - "Solution_1": "[hide]So each side has 5 circles\n\nthen one side would have a + b + c + d + e\nnext would have e + f + g + h + i\nlast would have i + j + k + l + a\n\nmake the overlaps as small as possible\n\na = 1, e = 2, i = 3\n\nFirst row has 3\nSecond row has 5\nthird row has 4\n\nNow numbers 4 thru 12 are left.\n\nThis is a sum of 72.\nDivided evenly, you'd have 24 per row. But some of the rows are uneven.\nSo add 25 for first row\n23 for second row\n24 for last row.\n\nThis gives a sum of 28 for each side of triangle.\n[/hide]", - "Solution_2": "[hide]The smallest numbers have to be on the outside, so 1, 2, and 3 are the corners. Then I put 12 and 7 on one side, 11 and 8 on another, and 10 and 9 on the last. So far the sides' sums are 24, 23, and 22. So I put 6 on the side with a sum of 22, 5 on the side with a sum of 23, and 4 on the side with a sum of 24. So the sum of each side is[color=green] 28[/color].[/hide]" -} -{ - "Problem": "Find all finite set S of points in the plane with the following property: for any three distinct points A,B and C in S, there is a fourth point D in S such that A,B,C and D are the vertices of a parallelogram(in some order).\r\n\r\n(I've read a solution for this. It's long and I failed to understand any thing, can anyone post a solution with clear explanation? thanks)", - "Solution_1": "Have a look here \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?search_id=1793448355&t=148838\r\n\r\nPierre.", - "Solution_2": "I've seen that and fail to understand, anybody can explain to me?\r\nThanks" -} -{ - "Problem": "Let number $x^4+x^{2}.y^{2}+y^4$ can be divisible with number 11 where x,y positive numbers.Show that this number also divisible with 14641.", - "Solution_1": "Look at the quadratic residues $\\bmod 11$ and note that $x \\equiv y \\equiv 0 \\bmod 11$." -} -{ - "Problem": "If I'm given a set of functions that are solutions to some differential equation, what process can I use to find the simplest differential equation that fits those functions? Constant coefficients are simpler than non-constant, lower order is simpler than higher order.", - "Solution_1": "I really need some help on this so don't even worry about the \"simplest\" part. Just explain to me a process where I can backwards engineer a differential when I'm given a set of particular solutions.\r\n\r\nFor example, if I'm given the set of functions $ \\{x^2\\}$, one possible differential equation is $ y'''\\equal{}0$, but the simplest differential equation is $ y'\\minus{}\\frac{2}{x}y\\equal{}0$. Both of these equations are obvious fits because of such a simple set, but what about sets like $ \\{x, e^x, sin(x)\\}$ or even $ \\{\\ln(x)\\}$?", - "Solution_2": "[quote=\"epkid08\"]the simplest differential equation is $ y' \\minus{} \\frac {2}{x}y \\equal{} 0$[/quote] Not $ y \\minus{} x^2 \\equal{} 0$?\r\n\r\nI don't think there's very much to be said in general. What's the motivation for this question?\r\nFor the example $ \\{x, e^x, \\sin x\\}$ (note \\sin, just like \\ln, \\lim, etc.), here are two possible ways to proceed:\r\n\r\n1) $ e^x \\equal{} e^{1 \\cdot x}$; $ x \\equal{} x \\cdot e^{0 \\cdot x}$, and $ \\sin x$ comes as the imaginary part of $ e^{i x}$. So, we need a polynomial with roots $ 0, 0, 1, i$. This strongly suggests $ t^2 \\cdot (t \\minus{} 1) \\cdot (t^2 \\plus{} 1) \\equal{} t^5 \\minus{} t^4 \\plus{} t^3 \\minus{} t^2$. So the equation we want is $ y^{(5)} \\minus{} y^{(4)} \\plus{} y^{(3)} \\minus{} y'' \\equal{} 0$.\r\n\r\n2) We have $ a y \\plus{} b y' \\plus{} cy'' \\plus{} \\ldots \\equal{} 0$ for $ y \\equal{} x$, $ y \\equal{} e^x$ and $ y \\equal{} \\sin x$. The first suggests $ ax \\plus{} b \\equal{} 0$, the second suggests $ a \\plus{} b \\plus{} \\ldots \\equal{} 0$ and the third suggests $ a \\minus{} c \\plus{} \\ldots \\equal{} 0$ and $ b \\minus{} d \\plus{} \\ldots \\equal{} 0$. Now just make some simple choices. How about $ a \\equal{} 1$, $ b \\equal{} \\minus{} x$? Okay, then we can choose $ d \\equal{} \\minus{} x$ and we have to solve $ a \\plus{} b \\plus{} c \\plus{} d \\plus{} e \\equal{} 0$ and $ a \\minus{} c \\plus{} e \\equal{} 0$ for $ c$ and $ e$, and this leads to an equation with only four derivatives but nonconstant coefficients.\r\n\r\n\r\nNote that the first method is restricted to very particular sets of functions: we can only work with exponentials, polynomials, sine and cosine, and products or sums of these things.", - "Solution_3": "Thanks so much! Your first method will help a bunch for most of my problems, actually all but one.\r\n\r\n\r\n\r\n[quote=\"JBL\"]\n2) We have $ a y \\plus{} b y' \\plus{} cy'' \\plus{} \\ldots \\equal{} 0$ for $ y \\equal{} x$, $ y \\equal{} e^x$ and $ y \\equal{} \\sin x$. The first suggests $ ax \\plus{} b \\equal{} 0$, the second suggests $ a \\plus{} b \\plus{} \\ldots \\equal{} 0$ and the third suggests $ a \\minus{} c \\plus{} \\ldots \\equal{} 0$ and $ b \\minus{} d \\plus{} \\ldots \\equal{} 0$. Now just make some simple choices. How about $ a \\equal{} 1$, $ b \\equal{} \\minus{} x$? Okay, then we can choose $ d \\equal{} \\minus{} x$ and we have to solve $ a \\plus{} b \\plus{} c \\plus{} d \\plus{} e \\equal{} 0$ and $ a \\minus{} c \\plus{} e \\equal{} 0$ for $ c$ and $ e$, and this leads to an equation with only four derivatives but nonconstant coefficients.[/quote]\n\nFor my last problem, I need to use the set $ \\{\\ln(x)\\}$, so $ a y \\plus{} b y' \\plus{} cy'' \\plus{} \\ldots \\equal{} 0$, for $ y \\equal{} \\ln(x)$, but what equation of $ \\{a, b,...,n\\}$ does this \"suggest?\"\n\n\n\n\n[quote=\"JBL\"]the simplest differential equation is $ y' \\minus{} \\frac {2}{x}y \\equal{} 0$ Not $ y \\minus{} x^2 \\equal{} 0$?\n[/quote] \r\n\r\nUnfortunately, I don't think my professor would be all too happy about me putting that type of equation down for every one of my answers. :D", - "Solution_4": "Well, again, for sets of a single function, I personally like $ y \\minus{} f(x) \\equal{} 0$. But, of course, $ y' \\minus{} f'(x) \\equal{} 0$ works fine, too, as do dozens of possible variations on this theme. In the particular case of $ f(x) \\equal{} \\ln x$, you can get a simple-looking second-order equation $ a y \\plus{} b y' \\plus{} cy'' \\equal{} 0$ where $ a, b, c$ are all polynomial.", - "Solution_5": "In particular you can get $ xy' \\equal{} 1$.", - "Solution_6": "[quote=\"t0rajir0u\"]In particular you can get $ xy' \\equal{} 1$.[/quote]\r\n\r\nI'm not sure, but I think this is also against the rules. I don't have the paper with me, but I think it might have to be a homogeneous equation and can't be separable.", - "Solution_7": "So you can get $ (xy')' \\equal{} xy'' \\plus{} y' \\equal{} 0$. (One of your earlier examples is also separable, so I don't really see the point of that restriction. This is a very artificial problem.)", - "Solution_8": "[quote=\"t0rajir0u\"]So you can get $ (xy')' \\equal{} xy'' \\plus{} y' \\equal{} 0$. (One of your earlier examples is also separable, so I don't really see the point of that restriction. This is a very artificial problem.)[/quote]\r\n\r\nAlright, I'll give it a shot, thanks for the help." -} -{ - "Problem": "An electronic oscillator produces 4.5 volts effective AC at a frequency of 1850 kiloherts, feeding it into a 12 picofarad capacitor. How much current flows", - "Solution_1": "$ I=\\frac{dq}{dt}$\r\n\r\n$ I_{ef}=q_{ef}\\omega=cU_{ef}\\omega=2\\pi CUf$", - "Solution_2": "very nice Namiqfizik. Now all you have to do is plug in these values" -} -{ - "Problem": "In the following $ R_j ,j\\in \\{1,2,...,26\\}, $ denotes perfect one ohm resistors. Consider that between two points $ A , B $ we have following network\r\n\\[A\\bullet-R_1-R_2-R_3-\r\n\\begin{array}{|lcr|}\r\n --------&R_4&------ \\\\\r\n --------&R_5&------ \\\\\r\n --------&R_6&------ \\\\\r\n --------&R_7&------ \\\\\r\n\r\n --------&R_8&------ \\\\\r\n --------&R_9&------ \\\\\r\n --------&R_{10}&------ \\\\\r\n --R_{11}--R_{12}-&\\cdots& ---R_{26}- \\\\\r\n\\end{array}-\\bullet B\r\n\\]\r\n Let [b]r[/b] be the resistance, measured in ohms, between $ A $ and $ B $ .[b] Find r with five decimals. [/b]\r\n\r\nKey words: Continued fractions\r\n\r\nSome references:\r\n[1] Amer.Mat.Monthly (1974), Problem E 2459, proposed by A.A. Mullin.\r\n[2] Matematicki Vesnik , Problem 393,13(28)1976, solution in 15(1978) pag.298.", - "Solution_1": "plz post the soln.........\r\n :oops:", - "Solution_2": "This looks like a problem requiring as much math as every 10-year old should know. If they were not requiring 5 decimal places, the answer could be calculated as 3.1 ohms without a calculator.\r\n$ R_1$, $ R_2$, and $ R_3$ are in series and add up to an equivalent resistance of $ 3$ ohms.\r\nIn the set of resistance between the two vertical lines, there is the series group of $ R_{11}$ through $ R_{26}$ that adds up to $ 16$ ohms, and is in parallel with the other seven resistors. The total resistance between the two vertical lines, R, is about 1/7 ohm and can be calculated with:\r\n$ 1/R \\equal{} 1/1 \\plus{} 1/1 \\plus{} 1/1 \\plus{} 1/1 \\plus{} 1/1 \\plus{} 1/1 \\plus{} 1/1 \\plus{} 1/16 \\equal{} 7 \\plus{} 1/16 \\equal{} 7.0625$\r\n$ R \\equal{} 1/7.0625 \\equal{} 0.14159$ ohm\r\nAdding that $ R$ to the $ 3$ ohms for the $ R_1$ to $ R_3$ set you have\r\n$ r \\equal{} 3 \\plus{} 0.14159 \\equal{} 3.14159$ ohm", - "Solution_3": "[quote=\"KMST\"]This looks like a problem requiring as much math as every 10-year old should know. [/quote]\r\nThen you've missed the point entirely. The circuit encodes a convergent of the continued fraction of $ \\pi$ ;) It's the structure of the answer, not the answer itself, that is of interest." -} -{ - "Problem": "Does anyone know how to draw a border around a chunk of text? I have a figure that consists of some sample text, and I would like to have a border around it to distinguish it from the body of the document. I'm looking for something that is like [code]\\boxedverbatim[/code] only without the verbatim part.\n\nHere's what my figure looks like:\n[code]\n\\begin{figure}[htp]\n\\small\nThe bordered text should be here.\n\\caption{\\small\nSome caption.}\n\\label{fig:myfig}\n\\end{figure}\n[/code]\r\n\r\nThanks!", - "Solution_1": "$\\boxed{some random text}$\r\n\r\n\\boxed{TEXTHERE}", - "Solution_2": "That doesn't seem to work. First, it doesn't allow math-mode inside of the brackets. Second, if even if there are no math commands, the spaces inside disappear (I get text like \"hellothisistext\" instead of \"hello this is text\") and the text stays on one line and goes off the edge of the page.", - "Solution_3": "\\boxed is for text that is [i]already[/i] in math mode such as $\\boxed{\\sin x}$. If you want a box round text mode then use \\fbox as in $\\fbox{this is framed text}$. You can also change to math mode inside the box.\r\n\r\nBut that is for just one line of text. If you want more than one line of text then use \\parbox inside \\fbox as in $\\fbox{\\parbox{2.5cm}{More than \\\\ one line of text}}$", - "Solution_4": "Is there a way to make the text auto-wrap when using the fbox and parbox?", - "Solution_5": "It does auto-wrap so I didn't need the newline \\\\ in the example I gave :)\r\n$\\fbox{\\parbox{2.5cm}{More than one line of text}}$\r\nor\r\n$\\fbox{\\parbox{1.2cm}{More than one line of text}}$\r\nthe difference just being the width of the parbox you set, which you will need to experiment with.\r\nYou can have double boxes\r\n$\\fbox{\\fbox{\\parbox{1.5cm}{More than one line of text}}}$\r\nand you can change the thickness of the box lines \r\n$\\setlength{\\fboxrule}{0.5mm}\\fbox{\\parbox{2.5cm}{More than one line of text}}$\r\nand the amount of space between the frame and the text\r\n$\\setlength{\\fboxsep}{3mm}\\fbox{\\parbox{2.5cm}{More than one line of text}}$", - "Solution_6": "Thank you very much!" -} -{ - "Problem": "How many tetrahedrons can be formed with the vertices of a cube?", - "Solution_1": "uhh...not moems!!!Probably HSB!", - "Solution_2": "You may be right, i thought about that too, but actually this is not a space geometry problem but rather a combinatoric one... :maybe:", - "Solution_3": "Uh, isn't it just $ \\binom{8}{4}\\minus{}6$?\r\nSince any 4 points will make a tetrahedron, as long as they're not on the same plane, and there are only 6 sets that are on the same plane...", - "Solution_4": "Not quite...In your solution you also have to eliminate the 2 \"diagonal\" planes which would imply an answer of $ 62$ tetrahedrons, which was my answer too but it seems that there are $ 58$..at least this was the answer(without demonstration) I found :maybe:", - "Solution_5": "nope, there's six of those \"diagonal\" planes. Each edge determines one, but each are double-counted, so there's 6. :)", - "Solution_6": "Meh, so it's just $ \\binom{8}{4}\\minus{}12$ then..." -} -{ - "Problem": "Prove that the set $ \\{1,2,3,\\dots\\}$ iis a disjoint union of an infinite number of Fibonacci-like sequnces (i.e. sequences $ (a,a\\plus{}b,2a\\plus{}b,3a\\plus{}2b,5a\\plus{}3b,\\dots)$, each term is a sum of two precedding terms, with some positive integers $ a,b$). I remember that there was a short clear construction, but can not remember it.", - "Solution_1": "Really beatyfullproblem!\r\nLet's start with a well-known lemma as follows:\r\nLemma:Each positive integer number could be represented as a sum of distinct Fibonaci numbers and no neighborings are used simultaneously.\r\n Furthermore,this representation is unique.\r\nProof:\r\n For the existence,just induction!\r\n For the uniqueness,assuming that:$ F_n < x < F_{n \\plus{} 1}$.We will show that $ F_n$ must appear in the disired rep.\r\n Indeed,else then $ n\\leq F_{n \\minus{} 1} \\plus{} F_{n \\minus{} 3} \\plus{} F_{n \\minus{} 5} \\plus{} ... \\plus{} F_{2(1)}$\r\nbut : $ F_n \\equal{} F_{n \\minus{} 1} \\plus{} F_{n \\minus{} 3} \\plus{} F_{n \\minus{} 5} \\plus{} ... \\plus{} F_{2(1)}$.QED\r\n Now return back our problem,let's describe each positive integer as a vector in $ \\mathbb{Z}_2^{\\infty}$.For instance,$ 17 \\equal{} 1 \\plus{} 3 \\plus{} 13 \\equal{} (1,0,0,1,0,0,1,0,0,0,\\cdots)$\r\n ANd we just arrange number whose rep. has the same type:For exemple\r\n$ 27 \\equal{} 1 \\plus{} 5 \\plus{} 21 \\equal{} (0,1,0,0,1,0,0,1,0,0,0\\cdots )$ and $ 17 \\equal{} 1 \\plus{} 3 \\plus{} 13 \\equal{} (1,0,0,1,0,0,1,0,0,0,\\cdots )$ has same type and 17 is the first number of the sequence.Each number is constructed by adding one $ 0$ in the left-most of the previous number!\r\n Note that each contructed sequence forms a Fibonaci-type sequence.\r\nAnd we are done!" -} -{ - "Problem": "What are all ordered pairs of integers (x,y) which satisfy\r\n$x^{3}+y^{3}-3x^{2}+6y^{2}+3x+12y+6=0?$", - "Solution_1": "$x^{3}+y^{3}-3x^{2}+6y^{2}+3x+12y+6=0$\r\n$\\iff x^{3}+y^{3}-3x^{2}+6y^{2}+3x+12y+7=1$\r\n$\\iff (x+y+1)(x^{2}+y^{2}-xy-4x+5y+7)=1$\r\n\r\n\r\n\r\nCase 1\r\n\r\n$\\left \\{ \\begin{array}{l}x+y+1 = 1 \\\\ x^{2}+y^{2}-xy-4x+5y+7 = 1 \\\\ \\end{array}\\right.$\r\n$\\iff \\left \\{ \\begin{array}{l}y =-x \\\\ x^{2}+y^{2}-xy-4x+5y+7 = 1 \\\\ \\end{array}\\right.$\r\n$\\iff \\left \\{ \\begin{array}{l}y =-x \\\\ x^{2}+(-x)^{2}-x(-x)-4x+5(-x)+7 = 1 \\\\ \\end{array}\\right.$\r\n$\\iff \\left \\{ \\begin{array}{l}y =-x \\\\ 3x^{2}-9x+6 = 0 \\\\ \\end{array}\\right.$\r\n$\\iff \\left \\{ \\begin{array}{l}y =-x \\\\ 3(x-1)(x-2) = 0 \\\\ \\end{array}\\right.$\r\n$\\iff (x, y) = (1,-1), (2,-2)$\r\n\r\n\r\nCase 2\r\n\r\n$\\left \\{ \\begin{array}{l}x+y+1 =-1 \\\\ x^{2}+y^{2}-xy-4x+5y+7 =-1 \\\\ \\end{array}\\right.$\r\n$\\iff \\left \\{ \\begin{array}{l}y =-x-2 \\\\ x^{2}+y^{2}-xy-4x+5y+7 =-1 \\\\ \\end{array}\\right.$\r\n$\\iff \\left \\{ \\begin{array}{l}y =-x-2 \\\\ x^{2}+(-x-2)^{2}-x(-x-2)-4x+5(-x-2)+7 =-1 \\\\ \\end{array}\\right.$\r\n$\\iff \\left \\{ \\begin{array}{l}y =-x-2 \\\\ 3x^{2}-3x+1 =-1 \\\\ \\end{array}\\right.$\r\n$\\iff (x, y) = (\\frac{3-i\\sqrt{15}}{6}, \\frac{-15+i\\sqrt{15}}{6}), (\\frac{3+i\\sqrt{15}}{6}, \\frac{-15-i\\sqrt{15}}{6})$\r\n\r\n\r\n\r\nThus $(x, y)=(1,-1), (2,-2)$", - "Solution_2": "[hide=\"my solution\"]\nthat would equal\n$(x-1)^{3}+(y+2)^{3}-1=0$\n$(x-1)^{3}+(y+2)^{3}=1$\nit's a circle! *wishes I knew Asymptote*\ncenter= (1,2), radius 1\ntherefore, the integer pairs are\n(0,2), (2,2), (1,3), (1,1)\n[/hide]\r\nnice avatar, Kouichi", - "Solution_3": "mgao, I think you made a small error. The center of the circle is really (1,-2)\r\nThis makes the points (0,-2) (2,-2) (1,-1) and (1,-3) solutions, although (0,-2) and (1,-3) turn out to be extraneous." -} -{ - "Problem": "hi,\r\nlets assume that ABC a triangle .$AB = c ,BC=a,AC=b$ and \r\n\r\n$a+b+c=1$ \r\n\r\nprove that : $\\frac{1}{2}\\geq a^2+b^2+c^2+4abc$", - "Solution_1": "[quote=\"anonyme\"]hi,\nlets assume that ABC a triangle .$AB = c ,BC=a,AC=b$ and \n\n$a+b+c=1$ \n\nprove that : $\\frac{1}{2}\\geq a^2+b^2+c^2+4abc$[/quote]\r\n\r\n$a=x+y, b=y+z, c=z+x$ and $x+y+z=\\frac{1}{2}$\r\n\r\nRHS becomes: \r\n$(x+y)^2+(y+z)^2+(z+x)^2+4(x+y)(y+z)(z+x)$\r\n\r\nIt is equal to: \r\n$(x+y)^2+(y+z)^2+(z+x)^2+4(\\frac{1}{2}-z)(\\frac{1}{2}-x)(\\frac{1}{2}-y)=(x+y+z)^2+\\frac{1}{2}+x^2+y^2+z^2+2xy+2yz+2zx-(x+y+z)-4xyz=2(x+y+z)^2-(x+y+z)+\\frac{1}{2}-4xyz=\\frac{1}{2}+\\frac{1}{2}-\\frac{1}{2}-4xyz=\\frac{1}{2}-4xyz$\r\n\r\nInequality becomes:\r\n$xyz\\geq0$ which is true", - "Solution_2": "that is good ;) .", - "Solution_3": "see: \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=126074&highlight=a%5E2%2Bb%5E2%2Bc%5E2%2B4abc#126074\r\n\r\nanonyme, this inegality was in the last morrocan olympiads are you solved it? :roll:", - "Solution_4": "this problem is very easy, this inegality was in the last morrocan [color=red]junior [/color] olympiads : $a 0$\r\nsay $ab+bc+ca =\\frac{(a+b+c)^2 -(a^2+b^2+c^2)}{2}$", - "Solution_5": "could you tchebytchev show us Morrocan Olympiad \r\n\r\nProblems ,pleaze ?" -} -{ - "Problem": "slove for x\r\n where [x] is the largest integers which is smaller than x", - "Solution_1": "Clearly, x is an integer (it is a difference of integers). We have x + x/6 = x/2 + 2x/3. Then,\r\n\r\n[logx] = {x/6} - {x/2} - {2x/3}.\r\n\r\nBut this is less than one. So [logx] = 0 and logx < 1. Now we have only ten possible values.", - "Solution_2": "[quote=\"feliz\"]Clearly, x is an integer (it is a difference of integers). We have x + x/6 = x/2 + 2x/3. Then,\n\n[logx] = {x/6} - {x/2} - {2x/3}.\n\nBut this is less than one. So [logx] = 0 and logx < 1. Now we have only ten possible values.[/quote]\r\nThanks you very much!" -} -{ - "Problem": "In triangle $ABC$ AM is the median, BL is the bisector of angle B and CH is perpendicular to AB. Prove that if AM,BL and CH intersect in one point then $c^{2}(c-a)=(b^{2}-a^{2})(a+c)$ where AB=c,BC=a and CA=b", - "Solution_1": "let the intersect point be N. BM=MC=a/2. MB/BA=MN/NA=(a/2)/c now, by menelaos theroem: (CM/CB).(BH/HA).(AN/NM)=1. so, (BH/HA)=a/c. \r\nby ratio properties, BH/a=HA/c=(BH+HA)/(a+c).therefore, BH=ac/(a+c),HA=(c^2)/(a+c). since CHB and CHA are right triangles, CB^2-BH^2=AC^2-HA^2. this is equvalent to our relation." -} -{ - "Problem": "The expression typed below is also located on page 150 of aops vol 2.\r\nV is a vector with elements v1 and v2, and W is a vector with elements w1 and w2. i is the commonly known as i vector (1 0) and j is the commonly known as j vector (0 1). A is a transformation matrix\r\n\r\n(v1Ai + v2Aj) x (w1Ai + w2Aj). \r\n\r\nSomehow this becomes,\r\n(v1w2 - w1v2)(Ai x Aj) + v1w1(Ai x Ai) + v2w2(Aj x Aj).\r\n\r\nI dont see how the former expression can become to second expression, but rather, when I computed out the former expression, I got\r\n\r\nA(v1w2 - v2w1)k, where k is the vector (0 0 1). This expression is mentioned a paragraph down on the page as being the more simplified version. For the purpose of knowledge and understanding, can someone please show me how the first expression I typed is equal to the second one?", - "Solution_1": "[quote=\"jelyman\"]\n(v1Ai + v2Aj) x (w1Ai + w2Aj). \n\nSomehow this becomes,\n(v1w2 - w1v2)(Ai x Aj) + v1w1(Ai x Ai) + v2w2(Aj x Aj).\n\nA(v1w2 - v2w1)k, where k is the vector (0 0 1). [/quote]\r\n\r\n(v1Ai + v2Aj) x (w1Ai + w2Aj)=v1Ai x w1Ai + v1Ai x w2Aj + v2Aj x w1Ai + v2Aj x w2Aj. For the third term, we remember that b x a= -a x b, so now we collect like terms to get the second equation. Now, we can evaluate the second equation to get what you got later on by using the determinant rule. just remember that for any vector a, a x a = 0, and that i x j = k." -} -{ - "Problem": "Let $ x,y,z$ be positive integers. If $ 7$ divides $ (x\\plus{}6y)(2x\\plus{}5y)(3x\\plus{}4y)$ than prove that $ 343$ also divides it.", - "Solution_1": "This is completely straightwforward: if $ 7$ divides any factor, it divides all of them.\r\nThe generalisation is obvious:\r\nLet $ p$ be prime and let $ S\\equal{} \\prod_{k\\equal{}1} ^{\\frac{p\\minus{}1}2} (kx \\plus{} (p\\minus{}k)y)$ for some $ x,y$. Then if $ p|S$, then $ p^{\\frac{p\\minus{}1}2}|S$.", - "Solution_2": "[b]Lemma:[/b] If $ p$ is prime, then $ p | ab \\implies p | a \\text{ or } p | b$.\r\n\r\n[b]Corollary:[/b] $ p | a^n \\implies p | a \\implies p^n | a^n$.\r\n\r\nNote that $ a$ only has to be defined $ \\bmod p$. Here $ a \\equiv x \\minus{} y \\bmod 7$. The given expresson is $ 6(x \\minus{} y)^3 \\bmod 7$.", - "Solution_3": "Notice that $ x\\plus{}6y\\equiv 2x\\plus{}5y \\equiv 3x\\plus{}4y \\pmod7$ (since $ x\\plus{}6y\\equiv x\\minus{}y \\pmod7$). But $ 7$ must divide at least one of $ (x\\plus{}6y)$, $ (2x\\plus{}5y)$, $ (3x\\plus{}4y)$, so it divides all of them. Therefore $ 7\\times7\\times7\\equal{}343$ divides $ (x\\plus{}6y)(2x\\plus{}5y)(3\\plus{}4y)$. Q.E.D.", - "Solution_4": "[quote=vedran6]Let $ x,y,z$ be positive integers. If $ 7$ divides $ (x\\plus{}6y)(2x\\plus{}5y)(3x\\plus{}4y)$ than prove that $ 343$ also divides it.\n(I am searcing for some kind of generalisation :D )[/quote]\n\n$Lemma$:$(a+b)|(ax + by)$ $\\implies$ $(x-y) \\equiv 0$ $(mod$ $a+b)$ where $a,b>0$.\n$Proof$:$ax + by \\equiv 0$ $(mod$ $a + b)$ $\\implies$ $ax + (a + b - a)y \\equiv 0$ $(mod$ $a + b)$ or $ax - ay \\equiv 0 $ $(mod$ $a+b)$ and since $a$ doesn't divide $(a+b)$ we get $(x-y) \\equiv 0$ $(mod$ $a+b)$.\nWe apply this $Lemma$ and we get that $x \\equiv y$$(mod$ $7)$ $\\implies$ $7$ divides $(x+6y)$,$(2x+5y)$,$(3x+4y)$ $\\implies$ $7^3 = 343$ divides $(x+6y)(2x+5y)(3x+4y)$ and we are done.", - "Solution_5": "$x+6y\\equiv 2x+5y\\equiv 3x+4y\\equiv (x\u2013y) ,2(x\u2013y), 3(x\u2013y) (mod7)$ all bears $x\u2013y$\nBut$ 7| P(x,y,z)$\n So it must divide at least one of them if it divides one of the given no it will divide all since same$ mod7$ ,claim\n$7|abc\\implies 7|a,7|b,7|c$ if same mods\n", - "Solution_6": "x+6y mod 7 is 6(y-x) mod 7, similarly the others are 5(y-x) and 4(y-x) mod 7. Then at least one of these must be 0 mod 7, or y-x must be 0 mod 7, so all of them are multiples of 7, product is 0 mod 343." -} -{ - "Problem": "How do you solve problems like this?\r\nA circle with diameter 12 units has a chord AB with length 8 units. What is the distance between AB and a parallel diameter?", - "Solution_1": "Look at the diagram and then solve it using the pythagoream theorem" -} -{ - "Problem": "Let $ n$ and $ N$ be natural number. Prove that for any $ \\alpha$\u000b, $ 0\\le\\alpha\\le N$, and any real $ x$, it holds that \\[{ |\\sum_\f\f\f\f\f{k=0}^n}\\frac{\\sin((\\alpha+k)x)}{N+k}|\\le\\min\\{(n+1)|x|, \\frac{1}{N|\\sin\\frac{x}{2}|}\\}\\]", - "Solution_1": "give your solutions" -} -{ - "Problem": "What is the smallest composite number whose prime factorization contains a number other than 2, 3, 5, or 7?", - "Solution_1": "it should be next smallest prime number^2.\r\n\r\nso it's 11^2, which is 121.\r\n\r\nanswer ; 121", - "Solution_2": "Actually, it says [b]a[/b] number besides 2, 3, 5, or 7, meaning that only one factor has to not be included in these. Therefore, the answer is 2*11=22." -} -{ - "Problem": "In chess, a knight moves in an L-shaped manner -- two spaces in one direction and one space in a direction perpendicular to the first direction -- as shown. Beginning in and including the upper left corner (marked *), what is the most number of squares in a $4\\times 4$ chessboard that a knight can visit without visiting any square more than once?\r\n\r\n[img]http://i96.photobucket.com/albums/l167/adasarathy/chessboard.gif[/img]", - "Solution_1": "i might be wrong but I found 9", - "Solution_2": "I found 15...", - "Solution_3": "Yes 15 is possible.", - "Solution_4": "15 is the answer, can you prove that 16 isn't possible?", - "Solution_5": "Is there a way you could show that? :maybe:", - "Solution_6": "[quote=\"anirudh\"]Is there a way you could show that? :maybe:[/quote]\r\nyes, there is a way, that was meant to be a challenge.", - "Solution_7": "can someone show how you get 15?\r\n\r\nand can you move the one space, then turn first? \r\n\r\n-jorian", - "Solution_8": "[quote=\"jhredsox\"]can someone show how you get 15?\n\nand can you move the one space, then turn first? \n\n-jorian[/quote]\r\nRead my post...\r\nI know the proof, but I've seen it before so if no one gets it in a few days I'll show it.\r\nHere is a hint:\r\n[hide=\"hint\"]Consider parity[/hide]", - "Solution_9": "[quote=\"bpms\"]15 is the answer, can you prove that 16 isn't possible?[/quote]\r\n\r\nWell, if you reached 16 then you had to hit the square you started off with again, which is not allowed.\r\n\r\nSo if 15 is possible than 15 is the answer.", - "Solution_10": "[quote=\"mathgeniuse^ln(x)\"][quote=\"bpms\"]15 is the answer, can you prove that 16 isn't possible?[/quote]\n\nWell, if you reached 16 then you had to hit the square you started off with again, which is not allowed.\n\nSo if 15 is possible than 15 is the answer.[/quote]\r\nBut you include the corner you start at when you count out of 16...", - "Solution_11": "[hide]i got 15, i just tried to waste as little space as possible. I dont think there is any systemiatic way except guess and check.[/hide]", - "Solution_12": "[hide=\"proof of 16\"]down right right; down left left; down right right; right up up; up left left[/hide]", - "Solution_13": "[quote=\"nutz_for2.718281828\"][hide=\"proof of 16\"]down right right; down left left; down right right; right up up; up left left[/hide][/quote]\r\n:roll: you only count the ones you land on for each move.\r\nIf no one posts the proof in a couple days I'll do it.", - "Solution_14": "Oh, you do...\r\n\r\nI thought otherwise...that's why i didn't get 15" -} -{ - "Problem": "Hi everyone,\r\nI just bought a new Macbook Pro and I was wondering if there was a Mac OS X compatible LaTex or something similar for the Mac.\r\n\r\n\r\n\r\nThanks", - "Solution_1": "Yes, there is. [url=http://www.tug.org/mactex/]Here[/url] is a complete Mac distribution. I use TexShop for the editing, which is included in that package." -} -{ - "Problem": "I took a contest and the questions are supposed to be based on triangle similarity and proportionality theorems like the angle bisector theorem. And well, I couldn't get the answer to this question during the contest, but I brought it home and I [i]think[/i] I solved it. The thing is, I didn't use similarity or anything.. and it looks like an angle bisector problem... so I'm thinking there is probably an easier way to do it :\r\n\r\nIn triangle PQR, PQ has length 6, QR has length 15, and angle PQR is 60 degrees. S is a point on PR such that QS bisects angle PQR. Find the length of PS.\r\n\r\nTry it first, but here's how I did it (and I don't even know if this is right...\r\n[hide]First I found the triangle's area = (45 :sqrt: 3)/2. Then I dropped two altitudes, from P to QS and from S to QR, so we've got 2 30-60-90 triangles. since PQ = 6 then the altitutude to QS = 3. Let QS = x so the altitude from S to QR = x/2. Then I just added the areas (3*x)/2 + ((x/2)*15)/2 and set that equal to the total area, (45 :sqrt: 3)/2. Solving for x I got x = (30 :sqrt: 3)/7\n\n(ok, so I don't have to keep saying this, let point T be the point on QS where the altitude from P lands). From the 30-60-90 triangle, we know QT = 3 :sqrt: 3 so TS = (30 :sqrt: 3)/7 - 3 :sqrt: 3 = (9 :sqrt: 3)/7. Then I used the pythagorean theorem and got [b]PS = (6 :sqrt: 19)/7[/b] Any simpler ways?? This was a no calculator contest, too, so all those high radicals were fun :P [/hide]\r\n\r\nAnd I'm hungry, so I'll post the second question after someone replies to this one.", - "Solution_1": "[hide]\n\nlaw of cos: $(PR)^2=15^2+6^2-2*6*15*\\cos(60)$, so $x$=$3\\sqrt{19}$. Then use angle bisector theorem. $\\frac{6}{PT}$=$\\frac{15}{3\\sqrt{19}-PT}$, solve for $PT$.[/hide][/code]", - "Solution_2": "woah thats.. pretty cool :) I've never used the law of cosines before, but I've known it from those crazy animations on the left bar :P \r\n\r\nOk thanks, the second one is this:\r\n\r\nAB is a diameter of a circle O. DA and BC are tangents circle O. DB and AC intersect at point E, which is a point on circle O. If AD has length 10 and BC has length 8, what is the length of BE?\r\n\r\n\r\nNow this one I didn't get an answer to. I've had some scattered thoughts, but I doubt they'll mean much.\r\n[hide]Um, well, at first I thought it was a two-poles problem, so I found that the distance from point E to AB is 40/9. Then I had lots of similar triangles.. but I didn't get anywhere. And I found that the key point in this problem (or so I think) is that E is on the circle... but that still didn't mean much. \n\nBut now I just had a thought. hmm. oh, I might have figured it out. hold on :lol: \n\nEDIT: haha, yeah, I figured it out while I was typing this :P but A, maybe there's an easier way, and B, maybe I\"m wrong. Here's what I got for my answer. \n[hide](8 :sqrt: 5)/3[/hide][/hide]", - "Solution_3": "[quote=\"djshowdown2\"]woah thats.. pretty cool :) I've never used the law of cosines before, but I've known it from those crazy animations on the left bar :P \n\nOk thanks, the second one is this:\n\nAB is a diameter of a circle O. DA and BC are tangents circle O. DB and AC intersect at point E, which is a point on circle O. If AD has length 10 and BC has length 8, what is the length of BE?\n\n\nNow this one I didn't get an answer to. I've had some scattered thoughts, but I doubt they'll mean much.\n[hide]Um, well, at first I thought it was a two-poles problem, so I found that the distance from point E to AB is 40/9. Then I had lots of similar triangles.. but I didn't get anywhere. And I found that the key point in this problem (or so I think) is that E is on the circle... but that still didn't mean much. \n\nBut now I just had a thought. hmm. oh, I might have figured it out. hold on :lol: \n\nEDIT: haha, yeah, I figured it out while I was typing this :P but A, maybe there's an easier way, and B, maybe I\"m wrong. Here's what I got for my answer. \n[hide](8 :sqrt: 5)/3[/hide][/hide][/quote]\r\n\r\nI'm getting the ans. as 40/9..is that correct??", - "Solution_4": "[quote=\"djshowdown2\"]AB is a diameter of a circle O. DA and BC are tangents circle O. DB and AC intersect at point E, which is a point on circle O. If AD has length 10 and BC has length 8, what is the length of BE?\n[/quote]\r\n\r\nIt is \"power of a point to the circle\" problem. If you have a circle (O) and and a point P (inside or outside of the circle) and if you draw 2 secants intersecting the circle at points K, L (the first secant) and M, N (the second secant), then the product (called power of the point P to the circle)\r\n\r\n$PK.PL = PM.PN$\r\n\r\ndoes not depend on the secant, but only on the circle and the point. To see this, just realize that the quadrilateral KLMN is cyclic, $\\angle KLM = 180^o - \\angle NMK$ and consequently, the triangles $\\triangle PKM \\sim \\triangle PNL$ are similar and\r\n\r\n$\\frac{PK}{PM} = \\frac{PN}{PL}$\r\n\r\nOne of the secants can be a tangent and power of the point becomes square of the tangent length from the point. In your problem, power of the point D to the circle with diameter AB is\r\n\r\n$DA^2 = DE.DB = (DB - BE).DB$\r\n\r\nOn the other hand the lines AD, BC are parallel, the triangles $\\triangle DAE \\sim \\triangle BCE$ are similar and\r\n\r\n$\\frac{BE}{CB} = \\frac{DE}{DA} = \\frac{DB - BE}{DA}$\r\n\r\n$DB - BE = BE\\ \\frac{DA}{CB}$\r\n\r\n$DB = BE \\left ( \\frac{DA}{CB} + 1 \\right )$ \r\n\r\nSubstituting this to the \"power of the point D to the given circle\" equation,\r\n\r\n$DA^2 = BE^2\\ \\frac{DA}{CB} \\left ( \\frac{DA}{CB} + 1 \\right )$\r\n\r\n$BE = \\frac{DA.CB}{\\sqrt{DA^2 + DA.CB}} = \\frac{80}{\\sqrt{100 + 80}} = \\frac{40}{3 \\sqrt{5}} = \\frac {8 \\sqrt{5}}{3}$", - "Solution_5": "Note that there is a right angle at E because it's inscribed in a semicircle. Also, A and B are right angles because they are formed from a centerline and tangents. You have a bunch of right triangles that are similar, so make a bunch of proportions.\r\nI got: [hide=\"click here\"]$\\frac {8\\sqrt 5}{3}$ as well.[/hide]" -} -{ - "Problem": ":ninja: What is the 1st time after 3.30pm(or am)(analogue clock) in which the distance from the small hand to 6 (in clock) is the same as distance between the large hand to 6.\r\nWith solution. (taking into consideration that the small and the large hand are the same lenght :spider:", - "Solution_1": "is there something i'm not getting?\r\nat 3:30, the large handle IS on 6, so the distance is zero...aint?", - "Solution_2": "Wait, should both the hands be touching $6$? Because then wouldn't it just be $6:30$? Or did you mean that at what time after $3:30$ would the two hands be touching each other.", - "Solution_3": "at $6:30$, the both hands ARE NOT touching the $6$.\r\nbecause with every one minute movement of the big hand, the smaller hand has to move $\\frac{1}{60}$, thus at $6:30$, the smaller hand is between 6, and 7. :huuh:", - "Solution_4": "in understanding the question. i shall repeat the question again. What id the 1st time after 3.30 when the distance from the large hand to the 6 is the same as distance from small hand to 6. DID YOU UNDERSTAND? :D You see, when the large hand (minute hand passes 6) the distance it has just moved is====== to the distance of the small hand(hour hand) from 6 got it?\r\n ;) and.... iversonfan2005 at 6.30 both hands wont be at 6 got it?\r\n\r\n\r\n\r\n\r\n\r\n\r\nTHINK LOGICALLY! ;)", - "Solution_5": "Well...here's what I got...\r\n3:41 and 7/13 of a minute...\r\nYou know that at 3:36, the minute hand is 6 marks away from the 6, the small hand is 12 marks away, and you know that the large hand moves at 12 times the rate of the small hand moves...So then you know that the difference in the distances is 6 minute marks, so then the hour hand must move 6/13 of a mark, and the minute hand 72/13 of a mark, totaling 6 marks, with the hour hand moving 1/12th the distance that the minute hand moves. So...add...and you get what I said, 3:41:32.307692 :D", - "Solution_6": "Yup thats cool.... but I did I by an angle way :D \r\nBut the answers are the same. ;) \r\nAnyway how long did it take you?\r\nJust asking", - "Solution_7": "I got the same answer - $3$:$41$ and $\\frac {7}{13}$ of a minute. The next time the hands will be equidistant from the $6$ is either when the hands touch or when the hands form equal angles with a vertical line drawn through the center of the clock. Since the minute hand starts on the $6$ and the hour hand starts to the right of the $6$, and the minute hand moves faster than the hour hand, the hands will form equal angles before they touch. The hour hand starts $75\\deg$ away from the $6$ and moves $\\frac {1}{2}\\deg$ each minute, so the measure of the angle it forms can be described by $75- \\frac {m}{2}$. The minute hand starts at the $6$ and moves $6\\deg$ each minute, so the measure of the angle it forms can be described by $6m$. Equating the two expressions,\r\n\\begin{eqnarray*}\r\n75- \\frac {m}{2}&=&6m\\\\\r\n150-m&=&12m\\\\\r\n150&=&13m\\\\\r\n\\frac {150}{13}&=&m\r\n\\end{eqnarray*}", - "Solution_8": "yes i did it similarly but with a bit of algebra\r\nAt 3:30 the hour hand is 12.5 ticks away from 6 \r\nand the minute hand is 0 ticks away from 6 (meaning it is on 6)\r\n\r\nthe hour hand moves 1/12 ticks/minute \r\nand the minute hand moves 1 tick/minute\r\nso in x minutes the hour hand will be $12.5-\\frac{x}{12}$ away from 6 (x relatively small)\r\nand in x minutes the minute hand will be $0+x$ ticks away from 6\r\n\r\nset them equal and solve\r\n$12.5-\\frac{x}{12}=0+x$\r\n$x=11.53846...$\r\nSo it will be 11 minutes after 3:30 which is $3:41.53846...$\r\n\r\nedit:i started posting before i saw that matt276eagles posted", - "Solution_9": "Didn't really time myself, but probably somewhere in the range of 30-60 seconds, without a calculator.", - "Solution_10": "[quote=\"qweretyq\"]\n\nedit:i started posting before i saw that matt276eagles posted[/quote]\r\nDon't worry about it! More solutions are definitely welcome. :lol:", - "Solution_11": "Yup you are all correct :D" -} -{ - "Problem": "The compound $ \\text{MON\\minus{}0585}$ is a non-toxic, biodrgradable larvicide (highly selective against mosquito larvae). Synthesize this compound.\r\n\r\nhttp://ctd.mdibl.org/detail.go;jsessionid=F8289B9243A077E36E3782881A0F7655?type=chem&acc=C006114\r\n\r\nThis link shows what the compound looks like.", - "Solution_1": "[hide=\"Answer\"]Here's a possible synthesis from phenol:\n\n1) $ C_6H_5OH \\plus{} C_6H_5COCl/pyr \\longrightarrow C_6H_5\\minus{}O\\minus{}CO\\minus{}C_6H_5$\n\n2) $ C_6H_5\\minus{}O\\minus{}CO\\minus{}C_6H_5 \\plus{} AlCl_3/25^oC \\longrightarrow p\\minus{}C_6H_5\\minus{}CO\\minus{}C_6H_4\\minus{}OH$\n\n3) $ p\\minus{}C_6H_5\\minus{}CO\\minus{}C_6H_4\\minus{}OH \\plus{} NaHO/H_2O/Cl_2 \\longrightarrow\\,\\,\\,sodium\\,\\,4\\minus{}benzoyl\\minus{}2,6\\minus{}dichlorophenoxide$\n\n4) $ sodium\\,\\,4\\minus{}benzoyl\\minus{}2,6\\minus{}dichlorophenoxide \\plus{} CH_3I \\longrightarrow 4\\minus{}benzoyl\\minus{}2,6\\minus{}dichloroanisole$\n\n5) $ 4\\minus{}benzoyl\\minus{}2,6\\minus{}dichloroanisole \\plus{} HOCH_2CH_2OH/H^\\plus{} \\longrightarrow 4\\minus{}benzoyl\\minus{}2,6\\minus{}dichloroanisole ketal$\n\n6) $ 4\\minus{}benzoyl\\minus{}2,6\\minus{}dichloroanisole ketal \\plus{} 2Mg/THF,\\,\\,then 2CO_2,\\,\\, then H_3O^\\plus{} \\longrightarrow 2\\minus{}metoxi\\minus{}5\\minus{}benzoylisophthalic acid$\n\n7) $ 2\\minus{}metoxi\\minus{}5\\minus{}benzoylisophthalic acid \\plus{} excess\\,Me_3Al \\longrightarrow O\\minus{}methyl\\, MON\\minus{}0585$\n\n8) $ O\\minus{}methyl\\, MON\\minus{}0585 \\plus{} conc.HI \\longrightarrow MON\\minus{}0585$.[/hide]", - "Solution_2": "By the way, the question allows you to use benzene as well. The aromatic rings can come from benzene, phenol or both.", - "Solution_3": "I quite dislike the synthesis given in the book. My primary concern is that it doesn't appear, to me, to be phsically applicable (well who am I to say that) as it involved Friedel Craft Alkylation which is hard to control (and I think they are a bit too specific).\r\n\r\nI will post it if wanted.", - "Solution_4": "Yes, post the book's synthesis, but hidden, if you don't mind.", - "Solution_5": "[hide=\"Synthesis\"]\n\nStart with benzene:\n\n1. $ \\ce{CH(CH3)2Cl}$ and $ \\ce{AlCl3}$ to make $ \\text{isopropyl benzene}$ (is that the name?)\n\n2. $ \\ce{(PhCO2)2}$ and $ \\ce{CCl4}$ and $ NBS$ to achieve benzylic bromination\n\n3. $ \\ce{FeBr3}$ and phenol to link the previou molecule to the para position of the phenol\n\n4. $ \\ce{C(CH3)3Cl}$ and $ \\ce{AlCl3}$ to introduce $ \\text{tert\\minus{}butyl}$ groups to both positions ortho to the hydroxyl group\n[/hide]\r\n\r\nWouldn't it be hard to be specific about where you want to introduce those alky branches?", - "Solution_6": "The first step would be tricky, since polyalkylation is an important side reaction.", - "Solution_7": "Exactly, hence my dislike of this method. I am still trying to fully understand your method :oops: . Anyway thanks for the alternative method.\r\n\r\nIn general can we control the degree of branching during Friedel-Craft Alkylation?", - "Solution_8": "I don't think so. Anyway, that's not needed, since we have the Friedel-Crafts Acylation reaction as an alternative." -} -{ - "Problem": "Let $ x_1,x_2,x_3$ are three roots of the equation: $ x^3-3x^2+1=0$\r\nIn the $ Oxy$ plane given a curve $ y=\\frac{x^2}{2}-3x-\\frac1x$. Take $ A=(x_1,y_1);B=(x_2,y_2);C=(x_3,y_3)$\r\nWithout finding $x_1,x_2,x_3$\r\nWrite the general equation of the circle $ (C)$ so that $ A,B,C \\in (C)$\r\nAnd please show me the general way to solve problem like this one???", - "Solution_1": "Hello, darkmaster!\r\n\r\nI can get you started on this one . . .\r\n\r\n[quote]Let $ x_1,x_2,x_3$ are three roots of the equation: $ x^3-3x^2+1=0$\nIn the $ Oxy$ plane given a curve $ y=\\frac{x^2}{2}-3x-\\frac1x$.\nTake $ A=(x_1,y_1);B=(x_2,y_2);C=(x_3,y_3)$\n\nWithout finding $x_1,x_2,x_3$, write the general equation of the circle $ (C)$ so that $ A,B,C \\in (C)$\nAnd please show me the general way to solve problem like this one???[/quote]\r\nFrom the study of Polynomial Equations, we know the following:\r\n$.\\quad[1]\\;\\;\\;x_1 + x_2 + x_3 = 3,\\;\\;\\;x_1x_2 + x_2x_3 + x_1x_3 = 0,\\;\\;\\;x_1x_2x_3 = -1$\r\n\r\nLet the center of the circle be $O(h,k).$\r\nThen $OA = OB = OC = r$ . . . And we have:\r\n $(h-x_1)^2 + (k-y_1)^2 \\;= \\;(h-x_2)^2 + (k-y_2)^2 \\;= \\;(h-x_3)^2 + (k-y_3)^2 \\;= \\;r^2$\r\n\r\nYou should be able to solve for $h$ and $k$ in terms of $x_1,\\;x_2,\\;x_3$\r\nThen use [1] to simplify the answers even further.\r\n\r\nI'll wait in the car . . .", - "Solution_2": "Hi Soroban!\r\nThis way is so complicated! I agree that both h,k and $ r^2$ can be written as symmetric terms of $ x_1,x_2,x_3$. But After solving it, I found a huge difficulty in placing (1) into those term. ( Because the expression is so long and complicated).\r\nThis is just a test in class, so we must pay attention to time.\r\n\r\nI'll wait in a bus :lol: :lol:" -} -{ - "Problem": "I know this was posted here before, but I don't want to revive an old, old thread.\r\n\r\nWho wants an account? I could invite like 100 people.\r\n\r\nFeatures:\r\n\r\n- Over 2,500 megabytes (two gigabytes) of free storage \r\n- Built-in Google search that instantly finds any message you want \r\n- Automatic arrangement of messages and related replies into \r\n\"conversations\" \r\n- Text ads and related pages that are relevant to the content of your \r\nmessages \r\n\r\nIt's awesome.", - "Solution_1": "i want one!", - "Solution_2": "i'll have one... :|", - "Solution_3": "Cool where is it?", - "Solution_4": "Leave your name and email and someone will invite you. You will receive an link in your email box to sign up.", - "Solution_5": "I also have about 100 invites, if anyone is interested.", - "Solution_6": "My e-mail is alpha_gaurav_beta@yahoo.co.uk", - "Solution_7": "My e-mail is sparklingstar_suntana@hotmail.com", - "Solution_8": "I also have about 100 invites. As if there arn't enough floating around.", - "Solution_9": "It's really good, but htese days its a little slow, and its not just me, but also other ppl i know", - "Solution_10": "okay....\r\nsheng_jiang2003@yahoo.com", - "Solution_11": ":huh: sometimes it seems to me that Gmail is super super slow... sometimes I can log in on one account but not on another account.", - "Solution_12": "you cant do both accounts at the same time, i dont think", - "Solution_13": "Gmail is awesome for me. I don't ever remember it being slow.\r\n\r\nAlthough sometimes it says the server's down and to \"cross my fingers and try again later\". :P" -} -{ - "Problem": "hello everybody, I hope you are ok, please help me with this:\r\n\r\nLet $ U,V\\subseteq{\\mathbb{R}^m}$ be open subsets, $ \\varphi: U\\rightarrow{V}$ a diffeomorphism of class $ C^2$, $ f\\in{C^2(V)}$, $ a\\in{U}$ and $ b\\equal{}\\varphi(a)$. If $ b$ is a critic point of $ f$, show that the Hessians matrix of $ f$ at the point $ b$ and $ f\\circ{\\varphi}$ at the point $ a$, have the same rank.\r\n\r\nThanks a lot", - "Solution_1": "Let's put $ f(b)\\equal{}0$ and $ a\\equal{}b\\equal{}0$ to simplify notation; this is a matter of subtracting constants.\r\n\r\nWrite $ f(y)\\equal{}\\sum_{i,j} H_{i,j}y_iy_j\\plus{}o(|y|^2)$. Now plug in $ y\\equal{}\\varphi(x)$: \r\n$ f(\\varphi(x))\\equal{}\\sum_{i,j} H_{i,j}\\varphi_i(x)\\varphi_j(x)\\plus{}o(|x|^2)$. Here $ \\varphi_i(x)\\equal{}\\sum_{k}c_{ik}x_k\\plus{}o(|x|)$, where $ (c_{ik})$ is the derivative matrix of $ \\varphi$ at $ 0$. Hence $ f(\\varphi(x))\\equal{}\\sum_{i,j,k,l} H_{i,j}c_{ik}c_{jl}x_kx_l\\plus{}o(|x|^2)$. The Hessian matrix of $ f\\circ\\varphi$ is the Hessian of $ f$ multiplied by $ D\\varphi$ on both sides (transposed on the left). Multiplication by an invertible matrix does not change the rank." -} -{ - "Problem": "I want to show that if $ a_{1}, a_{2}, a_{3}\\cdots$ is a sequence of real numbers that converges to $ a$, then\r\n\r\n$ \\lim_{n\\to\\infty}(\\frac{{\\sum_{i = 1}^{n}}{a_{i}}}{n}) = a$.\r\n\r\nThis is what I have so far:\r\n\r\nSince the original sequence converges to $ a$, for all $ \\epsilon > 0$ there exists a natural number $ N$ such that $ |a_{n}-a| <\\epsilon$ for all $ n > N$.\r\n\r\nLet $ \\frac{{\\sum_{i = 1}^{n}}{a_{i}}}{n}= S_{n}$. Then, using algebra and the triangle inequality, we get:\r\n${{{ {|S_{n}-a| =|\\frac{1}{n}(a_{1}+\\cdots+a_{n})-a| =|\\frac{1}{n}\\sum_{i = 1}^{n}}{a_{i}-a}|\\leq\\frac{1}{n}\\sum_{i = 1}^{n}}{|a_{i}-a|}\\leq\\frac{1}{n}(|{a_{1}-a}|+\\dots+|a_{n}-a|) <\\frac{\\epsilon}{n}+\\frac{1}{n}\\sum_{i = 1}^{n-1}}|{a_{i}-a}|}$\r\n\r\nSo: ${ {|S_{n}-a| <\\frac{\\epsilon}{n}+\\frac{1}{n}\\sum_{i = 1}^{n-1}}|{a_{i}-a}|}$\r\nI don't know why it cut off to the second line.\r\n\r\nAnyway, this is what I've done so far, and I feel like I went about it the wrong way. I feel like I started out correctly, but then missed an important step, since the last quantity doesn't really help me.\r\n\r\nAny ideas, tips, hints, anything? Thanks", - "Solution_1": "Expressions that are too long get split up by the site's $ \\text{\\LaTeX}$ parser- that's better than running off the screen as they do offline. You can spot this in the preview, and break it up yourself instead.\r\n\r\nThe key here: $ n$ and $ N$ are different, and you're not going to get anything by stopping at $ N+1$.\r\n$ |S_{n}-a|\\le\\frac{1}n\\sum_{i=1}^{n}|a_{i}-a|=\\frac{1}n\\sum_{i=1}^{N}|a_{i}-a|+\\frac{1}n\\sum_{i=N+1}^{n}|a_{i}-a|$\r\n$ \\le\\frac{1}n\\sum_{i=1}^{N}|a_{i}-a|+\\frac{n-N}{n}\\epsilon$\r\n(I split that into two lines because it was a bit too long)\r\n\r\nWhat can you do with that estimate?", - "Solution_2": "The estimate tends to $ \\epsilon$, but I'm not sure how to use that.\r\n\r\nWe want $ |S_{n}\\minus{}a|<\\epsilon$ for all $ n$ greater than a natural number $ N$.\r\n\r\nI haven't had a chance to really sit down and work with it yet, but the only thing I can come up with is to use induction on $ n$, starting with $ n\\equal{}N\\plus{}1$ and show that it's less than $ c\\epsilon$ for all $ n$. Seems overkill though.", - "Solution_3": "If it goes to $ \\epsilon$, that means there's some $ N'>N$ such that $ |S_{n}\\minus{}a|<2\\epsilon$ for all $ n>N'$. There was never any reason to expect you'd be able to use the same $ (\\epsilon,N)$ pair for the sequence of averages as for the original sequence.", - "Solution_4": "Use Stolz lemma for\r\n$ x_{n}\\equal{}\\sum_{i\\equal{}1}^{n}a_{i}$\r\n$ y_{n}\\equal{}n$\r\n$ \\lim_{n\\to\\infty}\\frac{\\sum_{i\\equal{}1}^{n}}{n}\\equal{}\\lim_{n\\to\\infty}\\frac{x_{n\\plus{}1}\\minus{}x_{n}}{n\\plus{}1\\minus{}n}\\equal{}a$", - "Solution_5": "jmerry already provided the \"solution\", so far we have $ \\forall\\epsilon >0\\,\\exists N$ such that $ \\forall n\\geq N:\\,|S_{n}\\minus{}a|\\leq\\frac{1}{n}\\sum_{i\\equal{}1}^{N}|a_{i}\\minus{}a|\\plus{}\\frac{n\\minus{}N}{N}\\epsilon$\r\n\r\ne.g. now choose $ N^{*}\\geq N$ with $ \\sum_{i\\equal{}1}^{N}|a_{i}\\minus{}a| 0$, we have $ y \\minus{} x \\minus{} 42 \\equal{} 2$ and $ y \\plus{} x \\plus{} 42 \\equal{} 122$; the latter equation implies that $ x \\plus{} y \\equal{} \\boxed{080}$. \n\nIndeed, by solving, we find $ (x,y) \\equal{} (18,62)$ is the unique solution. \n[/hide]", - "Solution_11": "[quote=\"alkjash\"]Yea, I used difference of squares too, only I substituted $ z \\equal{} x \\plus{} 42$ for ease.[/quote]\r\n\r\nMe too, but I kept getting $ (y\\minus{}z)(y\\plus{}z)\\equal{}234$ and wondering why $ y\\minus{}z$ and $ y\\plus{}z$ had different parities. Wasted way too much time on that problem...", - "Solution_12": "[quote=\"mathnerd314\"][quote=\"alkjash\"]Yea, I used difference of squares too, only I substituted $ z \\equal{} x \\plus{} 42$ for ease.[/quote]\n\nMe too, but I kept getting $ (y \\minus{} z)(y \\plus{} z) \\equal{} 234$ and wondering why $ y \\minus{} z$ and $ y \\plus{} z$ had different parities. Wasted way too much time on that problem...[/quote]\r\nI subtracted wrong the first time too...i got =$ 254$ and then i had to try to find what's wrong because it was impossible to get it with only 1 power of 2 if both factors had to be even since diff of squares was even.", - "Solution_13": "I took the AIME unofficially and I couldn't get this problem... I spent a long time on it, playing with perfect squares modulo 3, 4, and even 5... lol I got up this morning (the morning after) and I suddenly realized distracted's solution and I solved the problem sucessfully. So I got it overnight. :)", - "Solution_14": "my solution (not very \"elegant\")\r\n\r\n[hide]moving $ y$ to the left, we get $ x^2 + 84x + 2008 - y^2 = 0$\nUsing the quadratic formula, where $ c = 2008 - y^2$, we get:\n$ x = \\frac { - b\u00b1\\sqrt {b^2 - 4ac}}{2a}$ please ignore the \u00c5, this is the best I could get with latex\n$ x = \\frac { - 84\u00b1\\sqrt {7056 - 4(2008 - y^2)}}{2a}$\nwhich simplifies to\n$ x = \\frac { - 84\u00b1\\sqrt { - 976 + 4y^2}}{2}$\n\nIf we let $ z^2$ equal the number inside the square root. Note that since $ \\sqrt {z^2}$ must be even to make $ x$ an integer, $ z$ itself must be even. So we have the equation:\n$ z^2 = - 976 + 4y^2$\n$ z^2 - 4y^2 = - 976$\n$ z^2 - (2y)^2 = - 976$\n$ (z + 2y)(z - 2y) = - 976$\n$ 976$ factors into $ (2^4)(61)$.\nthere are many ways to split -976 into a pair of integer factors, but only one pair of these will result into $ x$ and $ y$ into positive integers. This pair is $ - 4$ and $ 244$.\nIf $ z + 2y$ is $ 244$and $ z - 2y$ is $ - 4$, when adding the two equations,\n$ 2z = 240$ ==> $ z = 120$\nso $ \\sqrt {z^2}$ = $ \\sqrt {120^2}$ = $ 120$.\n\nGoing back to the quadratic formula, replacing$ \\sqrt {b^2 - 4ac}$ with $ 120$,\n$ x = \\frac { - 84\u00b1120}{2}$\nClearly the \u00b1 must be +, so\n$ x = \\frac { - 84 + 120}{2}$ so $ x = 18$.\n\n$ 18^2 + 84(18) + 2008 = y^2$\n$ 3844 = y^2$ ==> $ y = 62$ ==> $ 18 + 62 = [80]$\n[/hide]\r\nA lot of work, but since I don't know trig and calculus I couldn't do several of the problems, so I had a lot of time :D", - "Solution_15": "The plus-or-minus symbol is just [b]\\pm[/b]; for example:\r\n\\[ \\frac{\\minus{}b \\pm \\sqrt{b^2\\minus{}4ac}}{2a}\\]", - "Solution_16": "[hide=\"New post\"]\nWe complete the square to obtain\n$ (x \\plus{} 42)^2 \\plus{} 244 \\equal{} y^2$.\n\nTherefore $ y^2 \\minus{} (x \\plus{} 42)^2 \\equal{} 244$. We use the fact that the difference between two consecutive squares is the sum of their square roots to obtain values for $ y$ and $ x$. A quick check reveals that $ (60 \\plus{} 61) \\plus{} (61 \\plus{} 62) \\equal{} 244$, so $ y \\equal{} 62$ and $ x \\plus{} 42 \\equal{} 60$, or $ x \\equal{} 18$. We are given that the solution is unique, so we have\n\n$ x \\plus{} y \\equal{} \\fbox{80}$[/hide]", - "Solution_17": "[hide=\"Quick bounding\"]\n$(x+42)^2 < x^2+84x+2008 < (x+45)^2$ is pretty obvious.\n\nSo $x^2+84x+2008=(x+43)^2$ or $(x+44)^2$.\n\nThe former clearly doesn't work (just consider modulo 2), so we have $x^2+84x+2008=(x+44)^2=x^2+88x+1936 \\implies 4x=72 \\implies x=18$ and then $x+y=2x+44=\\boxed{080}$[/hide]", - "Solution_18": "We can start by completing the square, $(x+42)^2 + 244 = y^2$.\n\nWe can rewrite the equation like this $y^2 - (x+42)^2 = 244$ which equals $(y-x-42)(y+x+42) = 244$.\n\nNow we can say that $y - x - 42 = 2$ and $y + x + 42 = 122$. \n\nSo we get that $y = 62$ and $x = 18$ so our answer is $62 + 18 = 080$", - "Solution_19": "redacted", - "Solution_20": "I just wrote up a quick solution cause I was practicing typing solutions in latex.", - "Solution_21": "redacted", - "Solution_22": "[quote=aop2014]I understand, but it is generally not helpful to fill up the forums with tons of ancient problem threads, try adding solutions to the wiki instead or just writing them for yourself.[/quote]\n\nThis is actually appropriate behavior for such forums. Users are encouraged to use old problem threads to write solutions to such problems.", - "Solution_23": "redacted", - "Solution_24": "[quote=aop2014]Ok,\n1) I've been told off before for reviving ancient threads in any way and generally avoid it\n2) It really does clutter it up sometimes, not as much anymore\n3) I'll keep this in mind, thanks for the advice.[/quote]\n\nno its actually what you're meant to do....you did nothing wrong....naman12 is correct(as always)", - "Solution_25": "redacted", - "Solution_26": "[hide=Solution]\nWe complete the square to get $x^2+84x+2008=(x+42)^2+244$.\n\nWe substitute to get $y^2-(x+42)^2=244$\n\nWe use difference of squares to get $(y+x+42)(y-x-42)=244$.\n\nWe need $2$ numbers of the same parity that multiply to $244$.\n\nWe are strategically lazy so we will pick the most obvious pair $122$ and $2$.\n\nWe are strategically lazy so we will not check other pairs for more solutions.\n\nWe know that $x+y+42=122$ so $x+y=\\boxed{080}$.\n[/hide]", - "Solution_27": "[quote=chickendude]There exist unique positive integers $ x$ and $ y$ that satisfy the equation $ x^2 \\plus{} 84x \\plus{} 2008 \\equal{} y^2$. Find $ x \\plus{} y$.[/quote]\n\n[hide=Solution]We complete the square. We get that it would be $x^2+84+1764+244=y^2.$ We then get that $(x+42)^2=y^2-244.$ Therefore, we get that two squares are $244$ apart from each other, so we can let $x+42=z$. We get that $z^2-y^2=244$ where we see that $z>42$. We get that the prime factorization of $244$ is $61 \\times 2^2$. Therefore, we see that $(z+y)(z-y)=61 \\times 2^2.$ We get that we have multiple possibilities. We see that $z+y$ cannot equal $61$, since then $z<42$. We try $z+y=122$. We see that $z-y=2$. Therefore, $z=62$ and $y=60$ This works. We see that $z+y$ cannot equal 244 for parity. Therefore, we get that $z=62$ or $x+42=62$, so $x=20$. We get that the answer is $20+60=\\boxed{080}$.", - "Solution_28": "oops I am not completing the problems in the packet in the correct order :/ well they start off easy so am getting teensy bit bored.\n\n[hide=Solution]Rewrite as $(x+42)^2+244=y^2$, or $y^2-(x+42)^2=244 \\implies (y+x+42)(y-x-42)=244$. The sum of $y+x+42$ and $y-x-42$ is $2y$, and the product is even, so each of $y+x+42$ and $y-x-42$ is even. This means $y+x+42=122, y-x-42=2$, so $2y=124 \\implies y = 62$ and $x=18$. Adding gives $\\boxed{80}$.[/hide]", - "Solution_29": "[quote=OlympusHero]oops I am not completing the problems in the packet in the correct order :/ well they start off easy so am getting teensy bit bored.\n\n[hide=Solution]Rewrite as $(x+42)^2+244=y^2$, or $y^2-(x+42)^2=244 \\implies (y+x+42)(y-x-42)=244$. The sum of $y+x+42$ and $y-x-42$ is $2y$, and the product is even, so each of $y+x+42$ and $y-x-42$ is even. This means $y+x+42=122, y-x-42=2$, so $2y=124 \\implies y = 62$ and $x=18$. Adding gives $\\boxed{80}$.[/hide][/quote]\n\nDude just do all the past aime 6-15s lol so you don't get bored", - "Solution_30": "thankfully, I can get these done quickly, so it'll only take like a day or less to get there plus I don't really want to miss any problems I can learn from, who knows maybe one will be confusing. but yeah the later ones will be more helpful.", - "Solution_31": "@above,thank you so much for being more generous!!", - "Solution_32": "Anyways link the packet plz", - "Solution_33": "[hide=Solution]So $\\sqrt{x^2+84x+2008}$ must be a positive integer. If $y\\le{x+42}$, then $y^2\\le{x^2+84x+1764}x^2+84x+2008$ Therefore, $y=x+43$ or $x+44$. Note that $x$ and $y$ are of the same parity so $y=x+44$. \n\nThe equation now is $x^2+84x+2008=x^2+88x+1936 \\implies 4x=72 \\implies x=18$. So $x+y=2x+44=\\boxed{080}$. [/hide]", - "Solution_34": "[hide=Difference of Squares]$x^2+84x+1764+244=y^2$\n$y^2-(x+42)^2=244$\n$(y-x-42)(y+x+42)=244$\nWe have to factorize 244 into two divisors of same parity (here, even) and the only way to do that is $244=2*122$. \nThis gives $y=\\frac{124}{2}=62, x=18$\nSo $x+y=62+18=80$. \nSo the answer is $080$.[/hide]\nHope it helps! :) ", - "Solution_35": "My solution is also just similar to one of the quadratic solution i have factored my equation into x^2+84x+2008=k^2\nAs (k+2y)(k-2y)=-976 \nI thought that there would be many factors so I checked for gcd of the two products and got it as 4y but if you see that if k is odd then the gcd of the two products is one and if it's even then theor gcd is 2\nSo I have made two cases \nCase 1 when gcd is even \n(K+2y)(k-2y)=-976 can only be factored in two ways \nAs 61. -2^4\n976. -1 \nBut if you see both ways do not work for y \n\n\nSecond case when gcd is two \nThen it has three cases \n8.61. -2\n4.61. -4\n2.61. -8 \nOnly value that's it's satisfying is the middle one getting y as 62 then putting it back in eq 1 and getting x=18\n", - "Solution_36": "u have one hint: square off", - "Solution_37": "2017 AIME || P6. Basically, the same thing, but with different numbers.", - "Solution_38": "[b]Solution:[/b] This just screams complete the square. So, doing that, we get $(x+42)^2 + 244 = y^2$. (Let $z=x+42$ for ease of solution writing.) Basically, you add $244$ to one square to get another square. After trial and error, it is easy to see that $y$ is $z+2$ and so then clearly $2z + 1 + 2z + 3 = 4z + 4 = 244 \\Rightarrow z=60 \\Rightarrow x=18, y=62$. So our answer is $\\boxed{080}$." -} -{ - "Problem": "The digits 1, 2, 3, 4, 5 are each used once to compose a five digit number ABCDE. The three digit number ABC id divisible by 4, BCD is divisible by 5, and CDE is divisible by 3. Find the five-digit number ABCDE.", - "Solution_1": "[hide]43251[/hide]\r\n\r\ni can explain later if you want, but this is an ASAP thign... so i decided to HURRY!", - "Solution_2": "[hide]BC is equal to either 12, 24, 32, or 52. D=5. C+E+5 is divisible by 3.\n$12453$ works too.[/hide]", - "Solution_3": "chess64, not only is that another solution, but it is the unly solution...the original solution of 43251 by uglyman is incorrect because 251 is not a multiple of 3...", - "Solution_4": "yes 12453 is the only solution", - "Solution_5": "12453 is the only solution", - "Solution_6": "[quote=\"math92\"]The digits 1, 2, 3, 4, 5 are each used once to compose a five digit number ABCDE. The three digit number ABC id divisible by 4, BCD is divisible by 5, and CDE is divisible by 3. Find the five-digit number ABCDE.[/quote]\r\n[hide]\nThe tens digit is 5, since BCD is divisible by 5, the hundreds digit is 4, since ABC is divisible by 4.\n5+4=9, So the last digit is 3. The answer is $\\framebox{12453}$, since the last 2 digits have to be divisible by 4 in ABC and 14 is not divisible by 4.[/hide]", - "Solution_7": "[hide]12453. hee hee, that's fun to type :D [/hide]", - "Solution_8": "12453\r\n :winner_first:", - "Solution_9": "[quote=\"math92\"]The digits 1, 2, 3, 4, 5 are each used once to compose a five digit number ABCDE. The three digit number ABC id divisible by 4, BCD is divisible by 5, and CDE is divisible by 3. Find the five-digit number ABCDE.[/quote]As Chess pointed out, since 4|ABC, at this point A can be any number, any multiple of 100 is divisible by 4. BC is a two digit number divisible by 4. Candidates are {12, 24, 32, 52}. Since 5|BCD, we know D = 5, because 0 is not on the list of digits. Since 3|CDE, $(C+5+E) \\equiv 0\\mod{3} \\Rightarrow (C+E) \\equiv 1 \\mod{3}$. If $C = 2$ then $E \\equiv 2 \\mod{3}$, however, the only two numbers in the list $\\equiv 2\\mod{3}$ have already been used. Therefore BC = 24. $C = 4 \\Rightarrow E \\equiv 0 \\mod{3}$. The only number $\\equiv 0 \\mod{3}$ on the list is 3; therefore E =3. That leaves A = 1 as the only possibility.\r\n\r\n$\\boxed{ABCDE = 12453}$", - "Solution_10": "12453 is the only possibility :idea:", - "Solution_11": "12345! :10:" -} -{ - "Problem": "Hi \r\n\r\nI am new so I wanted to know who all have gone to 12th this time and will be giving JEE in 2009.", - "Solution_1": "I am. Arijit(arijit91) and Sanjith(thealchemist) are as well.", - "Solution_2": "me tooo(gone 2 12th)-- but i m not an aspirant 4 jee", - "Solution_3": "@rituraj are you not writing JEE?\r\n\r\ni'm writing JEE 2009 :)", - "Solution_4": "So howz it going ppl. I guess you ppl must be through with every topic unlike me I have hardly done anything except Mechanics,wave,thermo excluding heat ,Co-geo,some algebra,trig eq. :D :blush: and ofc nothing in chem except redox/mole ,gas and bonding. :)\r\n\r\nWhat about u :maybe:", - "Solution_5": ":o :o you have finished so much!!", - "Solution_6": "ya iam mukunth from PS. Sr , IIT-2009 aspirant ~", - "Solution_7": "[quote=\"anandghegde\"]@rituraj are you not writing JEE?\n[/quote]\r\nno\r\n& i mayyy not b writing JEE 2009 as well :oops: \r\n\r\nofc i have not even started yet", - "Solution_8": "do u have any plans of writing? or are you not interested in engg? :) \r\n\r\nhey how do i quote someone? :maybe: \r\n\r\nAnd how do i hide some text if i want to? :maybe:", - "Solution_9": "every post has a \"quote\" option at end and u can use backspace in the post to hide...... :blush: \r\n\r\ni m a medical student :( :(", - "Solution_10": "why such sad faces?? :)", - "Solution_11": "u kno all that mugging up and trbls like that :P \r\nbut doesn't matter-in the end i love it :D :D", - "Solution_12": "I am writing jee-09 too", - "Solution_13": "JEE 2010 FOR ME ALONG WITH NNNNNNNNNNNNNNN", - "Solution_14": "OMG.......\r\n\r\nthey are going to give JEE 2009 :o \r\ni thought them to have given JEE 2008" -} -{ - "Problem": "\u03a1\u03b5 \u03c0\u03b1\u03b9\u03b4\u03b9\u03ac \u03bc\u03b5 \u03c4\u03bf stfu \u03c4\u03b9 \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9? \u0388\u03c7\u03c9 \u03b1\u03c0\u03bf \u03c4\u03b7\u03bd \u03a0\u03ad\u03bc\u03c0\u03c4\u03b7 \u03c0\u03bf\u03c5 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03ce \u03bd\u03b1 \u03ba\u03b1\u03c4\u03b5\u03b2\u03ac\u03c3\u03c9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03b2\u03b9\u03b2\u03bb\u03af\u03b1 \u03b1\u03bb\u03bb\u03ac \u03b4\u03b5\u03bd \u03ba\u03b1\u03c4\u03b5\u03b2\u03b1\u03af\u03bd\u03b5\u03b9 \u03c4\u03af\u03c0\u03bf\u03c4\u03b1 :| \u039c\u03ae\u03c0\u03c9\u03c2 \u03be\u03ad\u03c1\u03b5\u03c4\u03b5 \u03c4\u03b9 \u03c3\u03c5\u03bc\u03b2\u03b1\u03af\u03bd\u03b5\u03b9 ?", - "Solution_1": "\u039c\u03b1\u03c2 \u03c0\u03ae\u03c1\u03b1\u03bd \u03c7\u03b1\u03bc\u03c0\u03ac\u03c1\u03b9 \u03ba\u03b1\u03b9 \u03b6\u03ae\u03c4\u03b7\u03c3\u03b1\u03bd \u03bd\u03b1 \u03ba\u03bb\u03b5\u03af\u03c3\u03b5\u03b9 \u03b7 \u03b9\u03c3\u03c4\u03bf\u03c3\u03b5\u03bb\u03af\u03b4\u03b1 :( \r\n\r\n\u0398\u03b1 \u03ba\u03bf\u03b9\u03c4\u03ac\u03be\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03c4\u03b7 \u03bc\u03b5\u03c4\u03b1\u03c6\u03ad\u03c1\u03bf\u03c5\u03bc\u03b5 \u03ba\u03ac\u03c0\u03bf\u03c5 \u03b1\u03bb\u03bb\u03bf\u03cd....", - "Solution_2": "\u03c0\u03bf\u03b9\u03bf\u03b9 (\u03b1\u03bd\u03c4\u03b5 \u03bc\u03b7\u03bd \u03b2\u03bf\u03bc\u03bf\u03bb\u03bf\u03c7\u03ae\u03c3\u03c9 \u03c0\u03ac\u03bb\u03b9) \u03c0\u03ac\u03bb\u03b9??", - "Solution_3": "\u0394\u03b5 \u03be\u03ad\u03c1\u03bf\u03c5\u03bc\u03b5 \u0391\u03bc\u03b2\u03c1\u03cc\u03c3\u03b9\u03b5. \u03a0\u03c1\u03bf\u03c6\u03b1\u03bd\u03ce\u03c2 \u03ad\u03b3\u03b9\u03bd\u03b5 \u03b1\u03bd\u03ce\u03bd\u03c5\u03bc\u03b1...", - "Solution_4": "[b][color=red]\u03a0\u03a1\u039f\u03a3\u039f\u03a7\u0397 !!! [/color][/b]\r\n\r\n \u038c\u03c4\u03b1\u03bd \u03bc\u03b5 \u03c4\u03bf \u03ba\u03b1\u03bb\u03bf \u03b2\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03ad\u03bf \u03c7\u03ce\u03c1\u03bf \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b9\u03c3\u03c4\u03bf\u03c3\u03b5\u03bb\u03af\u03b4\u03b1 , \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c4\u03b7 \u03b4\u03af\u03bd\u03b5\u03c4\u03b5\u03b9 [b]\u03c0\u03bf\u03c4\u03ad [/b]link \u03c3\u03b5 \u03ac\u03bb\u03bb\u03bf\u03c5\u03c2 \u03b5\u03ba\u03c4\u03cc\u03c2 \u0395\u03bb\u03bb\u03ac\u03b4\u03b1\u03c2 , \u03bf\u03cd\u03c4\u03b5 \u03c3\u03b5 PM , \u03b4\u03b9\u03cc\u03c4\u03b9 \u03c4\u03bf mathlinks \u03c4\u03bf \u03c0\u03b1\u03c1\u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03bf\u03cd\u03bd \u03bf\u03b9 \u03b5\u03ba\u03b4\u03cc\u03c4\u03b5\u03c2. \u0395\u03af\u03bd\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c4\u03b9\u03bc\u03cc\u03c4\u03b5\u03c1\u03bf \u03c3\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03c6\u03af\u03bb\u03bf \u03c3\u03b1\u03c2 \u03bd\u03b1 \u03c4\u03bf\u03c5 \u03c3\u03c4\u03b5\u03af\u03bb\u03bb\u03b5\u03c4\u03b5 \u03bc\u03b5 e mail \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03b2\u03b9\u03b2\u03bb\u03af\u03bf , \u03b1\u03bb\u03bb\u03ac \u03c0\u03bf\u03c4\u03ad \u03bc\u03ad\u03c3\u03c9 \u03c4\u03bf\u03c5 mathlinks.\r\n\r\n[u]\u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2 [/u]", - "Solution_5": "\u0393\u03b5\u03b9\u03b1 \u03c7\u03b1\u03c1\u03ac \u03c3\u03b5 \u03cc\u03bb\u03bf\u03c5\u03c2! \u039c\u03c0\u03c1\u03ac\u03b2\u03bf \u03b3\u03b9\u03b1 \u03c4\u03bf site \u03c0\u03bf\u03c5 \u03c0\u03ac\u03c4\u03b5 \u03bd\u03b1 \u03be\u03b1\u03bd\u03b1-\u03b1\u03bd\u03b5\u03b2\u03ac\u03c3\u03b5\u03c4\u03b5, \u03c0\u03b1\u03c1\u03ac \u03c4\u03b9\u03c2 \u03c0\u03b9\u03ad\u03c3\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c5 \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd. \u039c\u03b9\u03b1 \u03c0\u03c1\u03cc\u03c4\u03b1\u03c3\u03b7 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03ad\u03c7\u03b5\u03c4\u03b5 \u03be\u03b1\u03bd\u03ac \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03b1\u03bd\u03b5\u03b2\u03ac\u03b6\u03b5\u03c4\u03b5 \u03c4\u03b1 \u03b2\u03b9\u03b2\u03bb\u03af\u03b1 \u03c3\u03b5 rapidshare. \u0392\u03c1\u03b5\u03af\u03c4\u03b5 \u03ae \u03bc\u03bf\u03b9\u03c1\u03b1\u03c3\u03c4\u03b5\u03af\u03c4\u03b5 \u03ad\u03bd\u03b1 premium account (\u03b3\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03b4\u03b9\u03b1\u03b3\u03c1\u03ac\u03c6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c4\u03b1 \u03b2\u03b9\u03b2\u03bb\u03af\u03b1 \u03bc\u03b5\u03c4\u03ac \u03b1\u03c0\u03cc \u03ad\u03bd\u03b1 \u03bc\u03ae\u03bd\u03b1) \u03ba\u03b1\u03b9 \u03b5\u03af\u03c3\u03c4\u03b5 \u03ba\u03cd\u03c1\u03b9\u03bf\u03b9. \u03a4\u03b7\u03bd \u03b9\u03c3\u03c4\u03bf\u03c3\u03b5\u03bb\u03af\u03b4\u03b1 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c4\u03b5 \u03bd\u03b1 \u03c4\u03b7\u03bd \u03ad\u03c7\u03b5\u03c4\u03b5 \u03c3\u03b5 \u03ad\u03bd\u03b1 blog. \u03a4\u03bf \u03ba\u03ac\u03bd\u03b5\u03b9 \u03c0\u03bf\u03bb\u03cd\u03c2 \u03ba\u03cc\u03c3\u03bc\u03bf\u03c2 \u03ba\u03b1\u03b9 \u03bb\u03b5\u03b9\u03c4\u03bf\u03c5\u03c1\u03b3\u03b5\u03af \u03c0\u03bf\u03bb\u03cd \u03ba\u03b1\u03bb\u03ac \u03c9\u03c2 \u03c4\u03ce\u03c1\u03b1.\r\n\r\n\u0393\u03b9\u03b1 \u03c4\u03bf Stfu \u03ad\u03bc\u03b1\u03b8\u03b1 \u03c3\u03ae\u03bc\u03b5\u03c1\u03b1 \u03ba\u03b1\u03b9 \u03b5\u03af\u03bc\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03c0\u03b5\u03c1\u03af\u03b5\u03c1\u03b3\u03bf\u03c2 \u03bd\u03b1 \u03b4\u03c9 \u03c4\u03b9 \u03c5\u03bb\u03b9\u03ba\u03cc \u03ad\u03c7\u03b5\u03c4\u03b5 \u03c3\u03c4\u03b1 \u0395\u03bb\u03bb\u03b7\u03bd\u03b9\u03ba\u03ac. \u0395\u03bb\u03c0\u03af\u03b6\u03c9 \u03bd\u03b1 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03c5\u03bb\u03b9\u03ba\u03cc \u03bc\u03b5 \u03b5\u03bb\u03b5\u03cd\u03b8\u03b5\u03c1\u03b1 \u03b4\u03b9\u03ba\u03b1\u03b9\u03ce\u03bc\u03b1\u03c4\u03b1 \u03b3\u03b9\u03b1 \u03bc\u03c0\u03bf\u03c1\u03ce \u03bd\u03b1 \u03c4\u03bf \u03b2\u03ac\u03bb\u03c9 \u03c3\u03b5 \u03bc\u03b9\u03b1 \u03c3\u03b5\u03bb\u03b9\u03b4\u03bf\u03cd\u03bb\u03b1 \u03c0\u03bf\u03c5 \u03ad\u03c7\u03c9 \u03c6\u03c4\u03b9\u03ac\u03be\u03b5\u03b9 (math.stuff.gr).\r\n\r\n\u039a\u03b1\u03bb\u03ae \u03c3\u03c5\u03bd\u03ad\u03c7\u03b5\u03b9\u03b1 \u03ba\u03b1\u03b9 \u03ba\u03b1\u03bb\u03ae \u03b4\u03cd\u03bd\u03b1\u03bc\u03b7!", - "Solution_6": "\u039a\u03b1\u03bb\u03b7\u03c3\u03c0\u03ad\u03c1\u03b1,\r\n\r\n\u0394\u03c5\u03c3\u03c4\u03c5\u03c7\u03ce\u03c2 \u03b4\u03b5\u03bd \u03ad\u03c7\u03c9 \u03c0\u03bf\u03bb\u03cd \u03c7\u03c1\u03cc\u03bd\u03bf \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b1\u03c3\u03c7\u03bf\u03bb\u03b7\u03b8\u03ce \u03bb\u03cc\u03b3\u03c9 \u03c4\u03bf\u03c5 \u03bc\u03b5\u03c4\u03b1\u03c0\u03c4\u03c5\u03c7\u03b9\u03b1\u03ba\u03bf\u03cd \u03ba\u03b1\u03b9 \u03af\u03c3\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03bf \u03a3\u03c4\u03ad\u03bb\u03b9\u03bf\u03c2 \u03bd\u03b1 \u03ad\u03c7\u03b5\u03b9 \u03c4\u03c1\u03ad\u03be\u03b9\u03bc\u03bf. \u0395\u03af\u03bd\u03b1\u03b9 \u03b4\u03cd\u03c3\u03ba\u03bf\u03bb\u03bf \u03ba\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03c7\u03c1\u03bf\u03bd\u03bf\u03b2\u03cc\u03c1\u03bf \u03bd\u03b1 \u03b1\u03bd\u03b5\u03b2\u03ac\u03b6\u03b5\u03b9\u03c2 \u03ad\u03bd\u03b1 - \u03ad\u03bd\u03b1 \u03c4\u03b1 \u03b2\u03b9\u03b2\u03bb\u03af\u03b1 \u03c0\u03bf\u03c5 \u03b5\u03af\u03c7\u03b1\u03bc\u03b5 \u03c3\u03c4\u03bf stfu , \u03c3\u03c4\u03b7 \u03c3\u03b5\u03bb\u03af\u03b4\u03b1 \u03c0\u03bf\u03c5 \u03c0\u03c1\u03bf\u03c4\u03b5\u03af\u03bd\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03bf \u03c1\u03af\u03c3\u03ba\u03bf \u03bd\u03b1 \u03c3\u03b2\u03ae\u03c3\u03bf\u03c5\u03bd \u03bc\u03b5 \u03c4\u03b7\u03bd \u03c0\u03ac\u03c1\u03bf\u03b4\u03bf \u03b5\u03bd\u03cc\u03c2 \u03bc\u03ae\u03bd\u03b1. \u03a0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b2\u03c1\u03bf\u03cd\u03bc\u03b5 \u03ba\u03ac\u03c4\u03b9 \u03c0\u03b9\u03bf \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03cc. \r\n\r\n\u03a4\u03b1 \u03b2\u03b9\u03b2\u03bb\u03af\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03c0\u03bb\u03b5\u03b9\u03bf\u03c8\u03b7\u03c6\u03af\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03be\u03b5\u03bd\u03cc\u03b3\u03bb\u03c9\u03c3\u03c3\u03b1 (\u03b1\u03b3\u03b3\u03bb\u03b9\u03ba\u03ac) \u03ba\u03b1\u03b9 \u03c3\u03b5 \u03ba\u03b1\u03bc\u03af\u03b1 \u03c0\u03b5\u03c1\u03af\u03c0\u03c4\u03c9\u03c3\u03b7 \u03bd\u03cc\u03bc\u03b9\u03bc\u03b1! \u0391\u03c6\u03bf\u03c1\u03bf\u03cd\u03bd \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\u03cd\u03c2 \u03ba\u03b1\u03b9 \u03bf\u03bb\u03c5\u03bc\u03c0\u03b9\u03ac\u03b4\u03b5\u03c2 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ce\u03bd \u03b1\u03c0\u03cc \u03b4\u03b9\u03ac\u03c6\u03bf\u03c1\u03bf\u03c5\u03c2 \u03b5\u03ba\u03b4\u03bf\u03c4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03bf\u03af\u03ba\u03bf\u03c5\u03c2. \u0391\u03c5\u03c4\u03cc\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03bf \u03bb\u03cc\u03b3\u03bf\u03c2 \u03c0\u03bf\u03c5 \u03b4\u03b5 \u03c4\u03b7 \u03b4\u03b9\u03b1\u03c6\u03b7\u03bc\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03b9 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03bf\u03cd\u03bc\u03b5 \u03ba\u03b1\u03c4\u03ac \u03c4\u03bf \u03b4\u03c5\u03bd\u03b1\u03c4\u03cc\u03bd, \u03bd\u03b1 \u03c4\u03b7\u03bd \u03ba\u03c1\u03b1\u03c4\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03bc\u03c5\u03c3\u03c4\u03b9\u03ba\u03ae. (\u03a6\u03c5\u03c3\u03b9\u03ba\u03ac \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03ba\u03b1\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03c0\u03bf\u03c5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03bf\u03bc\u03b9\u03bc\u03cc\u03c4\u03b1\u03c4\u03b1...)\r\n\r\n\u0395\u03bb\u03c0\u03af\u03b6\u03c9 \u03bd\u03b1 \u03bc\u03c0\u03bf\u03c1\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03b2\u03c1\u03bf\u03cd\u03bc\u03b5 \u03c3\u03cd\u03bd\u03c4\u03bf\u03bc\u03b1 \u03bc\u03af\u03b1 \u03ba\u03b1\u03bb\u03ae \u03bb\u03cd\u03c3\u03b7 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c3\u03c5\u03bd\u03b5\u03c7\u03af\u03c3\u03b5\u03c4\u03b5 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03bb\u03b1\u03bc\u03b2\u03ac\u03bd\u03b5\u03c4\u03b5 \u03c4\u03b1 \u03b2\u03b9\u03b2\u03bb\u03af\u03b1 \u03c4\u03b1 \u03bf\u03c0\u03bf\u03af\u03b1 \u03b5\u03af\u03c7\u03b1\u03bc\u03b5 \u03ba\u03b1\u03b8\u03ce\u03c2 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \u03ba\u03b1\u03b9 \u03b1\u03c5\u03c4\u03ac \u03c0\u03bf\u03c5 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03b2\u03ac\u03bb\u03bf\u03c5\u03bc\u03b5 \u03b1\u03ba\u03cc\u03bc\u03b7. (\u039a\u03b1\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03bf\u03cd\u03c4\u03c3\u03b9\u03ba\u03b1...)\r\n\r\n\u0391\u03bb\u03ad\u03be\u03b1\u03bd\u03b4\u03c1\u03bf\u03c2", - "Solution_7": "\u0394\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03af\u03c3\u03c9\u03c2 , \u03bd\u03b1 \u03c4\u03b1 \u03b2\u03ac\u03bb\u03bf\u03c5\u03bc\u03b5 \u03c3\u03b5 \u03ad\u03bd\u03b1 \u03bb\u03bf\u03b3\u03b1\u03c1\u03b9\u03b1\u03c3\u03bc\u03cc e-mail \u03c9\u03c2 \u03c3\u03c5\u03bd\u03bd\u03c5\u03bc\u03ad\u03bd\u03b1 \u03b1\u03c1\u03c7\u03b5\u03af\u03b1 , \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03c4\u03b1 \u03ba\u03b1\u03c4\u03b1\u03b2\u03ac\u03b6\u03b5\u03b9 \u03bc\u03b5\u03c4\u03ac \u03bf \u03ba\u03b1\u03b8\u03ad\u03bd\u03b1\u03c2 \u03bc\u03b1\u03c2 , \u03ce\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03af\u03bd\u03b4\u03c5\u03bd\u03bf\u03c2 \u03bd\u03b1 \u03c3\u03b2\u03b7\u03c3\u03c4\u03bf\u03cd\u03bd\u03b5?\r\n\u0391\u03bb\u03bb\u03ac \u03b8\u03b1 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c3\u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03b2\u03c1\u03b5\u03b8\u03b5\u03af \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1 \u03ba\u03b1\u03ba\u03cc\u03b2\u03bf\u03c5\u03bb\u03bf \u03ac\u03c4\u03bf\u03bc\u03bf \u03bd\u03b1 \u03c4\u03b1 \u03c3\u03b2\u03ae\u03c3\u03b5\u03b9 \u03bc\u03cc\u03bd\u03bf.", - "Solution_8": "\u0394\u03c5\u03c3\u03c4\u03c5\u03c7\u03ce\u03c2 \u03ba\u03b1\u03b9 \u03b3\u03c9 \u03b1\u03c5\u03c4\u03ae \u03c4\u03b7\u03bd \u03c0\u03b5\u03c1\u03af\u03bf\u03b4\u03bf \u03b4\u03b5\u03bd \u03ad\u03c7\u03c9 \u03c3\u03c7\u03b5\u03b4\u03cc\u03bd \u03ba\u03b1\u03b8\u03cc\u03bb\u03bf\u03c5 \u03c7\u03c1\u03cc\u03bd\u03bf \u03bd\u03b1 \u03b1\u03c3\u03c7\u03bf\u03bb\u03b7\u03b8\u03ce \u03bc\u03b5 \u03c4\u03b7\u03bd \u03b9\u03c3\u03c4\u03bf\u03c3\u03b5\u03bb\u03af\u03b4\u03b1. \u03a0\u03ac\u03c1\u03b1\u03c5\u03c4\u03b1, \u03b1\u03bd \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03ad\u03c7\u03b5\u03b9 \u03bd\u03b1 \u03c0\u03c1\u03bf\u03c4\u03b5\u03af\u03bd\u03b5\u03b9 \u03ad\u03bd\u03b1 site \u03cc\u03c0\u03bf\u03c5 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03bc\u03b5 \u03bd\u03b1 \u03b1\u03bd\u03b5\u03b2\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b1 \u03b1\u03c1\u03c7\u03b5\u03af\u03b1 ([u]\u03c7\u03c9\u03c1\u03af\u03c2[/u] \u03bd\u03b1 \u03c3\u03b2\u03ae\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9) , \u03bc\u03b5 \u03ba\u03b1\u03bb\u03cc uploading , \u03b1\u03c2 \u03bc\u03bf\u03c5 \u03c4\u03b7\u03bd \u03c0\u03c1\u03bf\u03c4\u03b5\u03af\u03bd\u03b5\u03b9 \u03c3\u03b5 private message. To rapidshare \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03cc\u03c3\u03bf \u03ba\u03b1\u03c4\u03ac\u03bb\u03bb\u03b7\u03bb\u03bf \u03b3\u03b9\u03b1 \u03bc\u03b1\u03c2, \u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03bb\u03cc\u03b3\u03bf\u03c5\u03c2 \u03c0\u03bf\u03c5 \u03b1\u03bd\u03ad\u03c6\u03b5\u03c1\u03b5 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03ba\u03b1\u03b9 \u03bf \u0391\u03bb\u03ad\u03be\u03b1\u03bd\u03b4\u03c1\u03bf\u03c2.\r\n\r\n\u039f\u03c0\u03bf\u03b9\u03b1\u03b4\u03ae\u03c0\u03bf\u03c4\u03b5 \u03c0\u03c1\u03cc\u03c4\u03b1\u03c3\u03ae \u03c3\u03b1\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03c5\u03c0\u03c1\u03cc\u03c3\u03b4\u03b5\u03ba\u03c4\u03b7. \u0397 \u03b9\u03c3\u03c4\u03bf\u03c3\u03b5\u03bb\u03af\u03b4\u03b1 \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03c0\u03ac\u03bb\u03b9 up \u03c4\u03bf \u03c3\u03c5\u03bd\u03c4\u03bf\u03bc\u03cc\u03c4\u03b5\u03c1\u03bf\u03bd \u03b4\u03c5\u03bd\u03b1\u03c4\u03cc\u03bd \u03ba\u03b1\u03b9 \u03b8\u03b1 \u03c3\u03c5\u03bd\u03b5\u03c7\u03af\u03c3\u03b5\u03b9 \u03bd\u03b1 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9, \u03ad\u03c3\u03c4\u03c9 \u03ba\u03b1\u03b9 \u03ba\u03ac\u03c4\u03c9 \u03c5\u03c0\u03cc \u03b1\u03c5\u03c4\u03ad\u03c2 \u03c4\u03b9\u03c2 \u03b4\u03cd\u03c3\u03ba\u03bf\u03bb\u03b5\u03c2 \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b5\u03c2 \u03c0\u03bf\u03c5 \u03b1\u03bd\u03b1\u03ba\u03cd\u03c0\u03c4\u03bf\u03c5\u03bd \u03ba\u03b1\u03c4\u03ac \u03b4\u03b9\u03b1\u03c3\u03c4\u03ae\u03bc\u03b1\u03c4\u03b1. \u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03bf\u03cd\u03bc\u03b5 !! :)", - "Solution_9": "\u03c0\u03b1\u03b9\u03b4\u03b9\u03ac \u03ad\u03c7\u03c9 \u03bc\u03b9\u03b1 \u03b9\u03b4\u03ad\u03b1\r\n\u03b1\u03c3 \u03bc\u03bf\u03c5 \u03c3\u03c4\u03b5\u03af\u03b5\u03bb\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf\u03c2 \u03b5\u03bc\u03b1\u03b9\u03bb \u03bd\u03b1 \u03c4\u03b7\u03bd \u03c0\u03ce \u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03b4\u03ce \u03c5\u03c0\u03ac\u03c1\u03c7\u03bf\u03c5\u03bd \u03c1\u03bf\u03c5\u03c6\u03b9\u03ac\u03bd\u03bf\u03b9 \u03c0\u03bf\u03c5 \u03bc\u03b1\u03c2 \u03c6\u03b8\u03bf\u03bd\u03bf\u03cd\u03bd \r\nadios \u03c1\u03bf\u03c5\u03c6\u03b9\u03ac\u03bd\u03bf\u03b9", - "Solution_10": "\u03c4\u03bf stfu \u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9? \u03ad\u03c7\u03c9 \u03ba\u03ac\u03c4\u03b9 \u03c3\u03c4\u03bf \u03bc\u03c5\u03b1\u03bb\u03cc \u03bc\u03bf\u03c5 \u03b1\u03bb\u03bb\u03ac \u03c3\u03b5 \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03bd\u03b1 \u03ba\u03bf\u03bb\u03bb\u03ac\u03b5\u03b9:-p", - "Solution_11": "\u03a4\u03bf Stfu \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03ac\u03bb\u03bb\u03bf\u03bd \u03ae\u03c4\u03b1\u03bd \u03bc\u03af\u03b1 \u03b9\u03c3\u03c4\u03bf\u03c3\u03b5\u03bb\u03af\u03b4\u03b1 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03bf\u03c0\u03bf\u03af\u03b1 \u03ba\u03b1\u03c4\u03ad\u03b2\u03b1\u03b6\u03b5\u03c2 \u03c0\u03b1\u03c1\u03b1 \u03c0\u03bf\u03bb\u03bb\u03ac \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd\u03c4\u03b1 \u03b2\u03b9\u03b2\u03bb\u03af\u03b1 \u03bc\u03b5 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b1\u03bb\u03bb\u03ac \u03b4\u03c5\u03c3\u03c4\u03c5\u03c7\u03ce\u03c2 \u03bc\u03b1\u03c2 \u03c4\u03b7\u03bd \u03ad\u03ba\u03bb\u03b5\u03b9\u03c3\u03b1\u03bd . \u039f \u0391\u03bb\u03ad\u03be\u03b1\u03bd\u03b4\u03c1\u03bf\u03c2 (cretanman ) \u03b5\u03af\u03c7\u03b5 \u03c4\u03b7 \u03c3\u03b5\u03bb\u03af\u03b4\u03b1 \u03ba\u03b1\u03b9 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03c0\u03b5\u03b9 \u03c0\u03b5\u03c1\u03b9\u03c3\u03c3\u03bf\u03c4\u03b5\u03c1\u03b1 .", - "Solution_12": "\u03ba\u03b1\u03bb\u03ac \u03ba\u03b1\u03b9 stfu \u03c4\u03b7\u03bd \u03bf\u03bd\u03cc\u03bc\u03b1\u03c3\u03b1\u03bd? \u03ad\u03bb\u03b5\u03bf\u03c2!!", - "Solution_13": "[quote=\"thatin\"]\u03ba\u03b1\u03bb\u03ac \u03ba\u03b1\u03b9 stfu \u03c4\u03b7\u03bd \u03bf\u03bd\u03cc\u03bc\u03b1\u03c3\u03b1\u03bd? \u03ad\u03bb\u03b5\u03bf\u03c2!![/quote]\r\n\u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03c0\u03c9\u03c2 \u03c4\u03bf \u03cc\u03bd\u03bf\u03bc\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03ac \u03ba\u03b1\u03bb\u03cc \u03b3\u03b9\u03b1 \u03c4\u03bf\u03bd \u03c3\u03ba\u03bf\u03c0\u03cc \u03c4\u03b7\u03c2 \u03c3\u03b5\u03bb\u03af\u03b4\u03b1\u03c2 \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03bd\u03b1 \u03b2\u03bf\u03b7\u03b8\u03ac\u03b5\u03b9 \u03bc\u03b1\u03b8\u03b7\u03c4\u03ad\u03c2 \u03c0\u03bf\u03cd \u03b1\u03c3\u03c7\u03bf\u03bb\u03bf\u03cd\u03bd\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\u03cd\u03c2 \u03b1\u03c6\u03bf\u03cd\r\nstfu \u03c3\u03b7\u03bc\u03b1\u03af\u03bd\u03b5\u03b9 solving tips for you :wink:", - "Solution_14": "\u03b1\u03bb\u03bb\u03b9\u03ce\u03c2 \u03c4\u03bf \u03ad\u03c7\u03b5\u03b9 \u03b7 wikipedia:-p" -} -{ - "Problem": "Prove that if the trace of a square matrix $ A$ is $ 0$, then there exist two square matrices $ X$ and $ Y$ of the same size as $ A$ such that $ A\\equal{}XY\\minus{}YX$.", - "Solution_1": "[quote=\"Valentin Vornicu\"]Prove that if the trace of a square matrix $ A$ is $ 0$, then there exist two square matrices $ X$ and $ Y$ of the same size as $ A$ such that $ A \\equal{} XY \\minus{} YX$.[/quote]\r\n\r\nHere the proof from the book \r\nRoger A.Horn & Charles R.Johnson, \r\nTopics in Matrix Analysis, Cambridge University Press\r\n\r\nChapter 4, Matrix equations and the Kronecker product \r\ntheorem 4.5.2 p\r\nIf trA=0, then the matrix equation XY-YX=A can be solved.To do so \r\nwe may replace A by any matrix similar to A because the property \r\nof being a commutator is similarity invariant.\r\nA is unitarily to a matrix whose diagonal entries are all equal to 1/n.trA\r\nLet A=(a(i,j)) with a(1,1)=...=a(n,n)=0\r\nChoose X=diag(x_1,...,x_n), x_i are pairwise distinct, with X fixed and A\r\ngiven , XY-YX=A becomes a linear matrix equation in Y=(y(i,j))\r\nXY-YX=((x_i-x_j)y(i,j))\r\n\r\ny(i,j)= c(i,j)/(x_i-x_j) if i=/=j\r\ny(i,j) arbitrary if i=j\r\ngives a solution since x_i=/=x_j when i=/=j and a(i,i)=0\r\n\r\nM.Omarjee", - "Solution_2": "Nice solution.\r\nAlso much much shorter than the one I knew before.\r\nOf course that solution was written in an elementary style for students whose knowledge on linear algebra raises to Hamilton-Cayley and not a bit further. For those students similarity in matrices is SF, so that solution had to prove everything.\r\nThe bottom line is that I couldn't understand that solution because it was to explicit and I too young, but I understand yours.\r\n\r\nI think Valentin must now find a way to make this topic a [b]no sticky[/b], normal post.", - "Solution_3": "Cool solution Moubi! This problem was proposed to the 11-graders at that contest I've mentioned b4, Gheorghe Titeica. I spent a lot of time on it in the 11'th grade (not in the contest; I tried to solve it for training), but got nowhere. When I eventually decided to look at bthe solution in a book, I got scared (it spanned over several pages), closed the book and ran away! :D :D :D", - "Solution_4": "I know the feeling grobber. That was my first reaction to that proof too.", - "Solution_5": "[quote=\"Moubinool\"]\nA is unitarily to a matrix whose diagonal entries are all equal to 1/n.trA\n[/quote]\r\n\r\nI am stupid. How do you prove it? I may prove it only for real case." -} -{ - "Problem": "For fixed integer m, let Xn have Binomial (n, 1/2) distribution. Let Yn = (Xn mod m) take values in Im = {0,1,2,..,m-1}\r\nSHow that the total variation distance between the distribution of Yn and uniform distribution on Im converges to 0 as n -> infinity.\r\n\r\nFirst of all, what is Yn???? Say Xn has probability (n choose k) (1/2)^n....does it mean Yn has probability from one of : (1 choose K)(1/2) ...... (m choose k) (1/2)^m???\r\n\r\nAnd what does it mean by Yn take values in Im???", - "Solution_1": "What we actually have is this. For $ 0\\le k\\le m \\minus{} 1,$\r\n\r\n$ P(Y_n \\equal{} k) \\equal{} \\sum_{j \\equal{} 0}^{\\lfloor (n \\minus{} k)/m\\rfloor}P(X_n \\equal{} k \\plus{} jm) \\equal{} \\sum_{j \\equal{} 0}^{\\lfloor (n \\minus{} k)/m\\rfloor}\\binom{n}{k \\plus{} jm}\\frac1{2^n}.$\r\n\r\nMy gut instinct is to seek a transform method. I'll find the Fourier transform (on $ \\mathbb{Z}$) of the binomial distribution, then use Poisson summation ideas to find the discrete Fourier transform of the distribution of $ Y_n$ (which can be seen to be periodic).", - "Solution_2": "Expanding on that idea:\r\n\r\nLet $ f_n(x)\\equal{}\\sum_{k\\equal{}0}^nP(X_n\\equal{}k)e^{\\minus{}2\\pi ikx}.$\r\n\r\nBy the binomial theorem, we can compute that $ f_n(x)\\equal{}\\frac1{2^n}(1\\plus{}e^{\\minus{}2\\pi ix})^n.$\r\n\r\nNow let $ g_n(\\ell)\\equal{}\\sum_{k\\equal{}0}^{m\\minus{}1}P(Y_n\\equal{}k)e^{\\minus{}2\\pi ik\\ell/m}.$\r\n\r\nThe theorem I'll quote (the proof is just a manipulation of sums) is that\r\n\r\n$ g_n(\\ell)\\equal{}f_n\\left(\\frac{\\ell}{m}\\right).$ Hence, $ g_n(\\ell)\\equal{}\\frac1{2^n}(1\\plus{}e^{\\minus{}2\\pi i\\ell/m})^n.$\r\n\r\nBut $ |1\\plus{}e^{2\\pi i\\ell/m}|\\equal{}2\\left|\\cos\\left(\\frac{\\pi \\ell}{m}\\right)\\right|.$ Since this is strictly less than $ 2$ for all $ \\ell\\ne0,$ we have this:\r\n\r\n$ \\lim_{n\\to\\infty}g_n(\\ell)\\equal{}\\begin{cases}1,&\\ell\\equal{}0\\\\0&\\ell\\ne 0\\end{cases}.$\r\n\r\nThis is exactly the discrete Fourier transform we would get from the \"discrete uniform\" distribution that has probability function $ \\frac1m$ at each value in $ \\{0,1,\\dots,m\\minus{}1\\}.$\r\n\r\nAs for the nature of the convergence: all norms are equivalent in a finite dimensional vector space.\r\n\r\n---\r\n\r\nNow I claim that there was nothing particularly special at all about the binomial distribution with $ p\\equal{}\\frac12.$\r\n\r\nClaim: Suppose $ X_j$ is an i.i.d. sequence of integer-valued random variables satisfying [condition to be named later]. Fix $ m.$ Define $ Y_n\\equiv\\sum_{j\\equal{}1}^nX_j\\pmod{m}.$ Then $ Y_n$ converges (in any reasonable sense) to the \"uniform discrete\" distribution on $ \\{0,1\\dots,m\\minus{}1\\}$ as $ n\\to\\infty.$\r\n\r\nCan someone fill in a reasonable [condition to be named later] and prove this? The techniques of this post should be sufficient.", - "Solution_3": "Sorry for my stupidness,\r\nbut where does the summation come from?\r\nYn = Xn (mod m)\r\nSo how does the P(Yn=k) is a sum?\r\nDo you mind to explain this bit a little more?", - "Solution_4": "Let's suppose $ m\\equal{}3$ and $ n\\equal{}8.$ Then $ X_8$ takes on the values $ 0,1,2,3,4,5,6,7,8$ with probabilities $ \\frac1{256},\\frac8{256},\\frac{28}{256},\\frac{56}{256},\\frac{70}{256},\r\n\\frac{56}{256},\\frac{28}{256},\\frac8{256},\\frac1{256}.$\r\n\r\nNow, $ Y_8\\equiv X_8\\pmod{3}.$ $ Y_8$ takes on only the value $ 0,1,2.$\r\n\r\nWe have $ Y_8\\equal{}0$ whenever $ X_8\\equiv 0\\pmod{3},$ which is whenever $ X_8\\equal{}0,3,$ or $ 6.$ Hence:\r\n\r\n$ P(Y_8\\equal{}0)\\equal{}P(X_8\\equal{}0)\\plus{}P(X_8\\equal{}3)\\plus{}P(X_8\\equal{}6)\\equal{}\\frac{1\\plus{}56\\plus{}28}{256}\r\n\\equal{}\\frac{85}{256}.$\r\n\r\nWe have $ Y_8\\equal{}1$ whenever $ X_8\\equiv 1\\pmod{3},$ which is whenever $ X_8\\equal{}1,4,$ or $ 7.$ Hence:\r\n\r\n$ P(Y_8\\equal{}1)\\equal{}P(X_8\\equal{}1)\\plus{}P(X_8\\equal{}4)\\plus{}P(X_8\\equal{}7)\\equal{}\\frac{8\\plus{}70\\plus{}8}{256}\r\n\\equal{}\\frac{86}{256}.$\r\n\r\nWe have $ Y_8\\equal{}2$ whenever $ X_8\\equiv 2\\pmod{3},$ which is whenever $ X_8\\equal{}2,5,$ or $ 8.$ Hence:\r\n\r\n$ P(Y_8\\equal{}2)\\equal{}P(X_8\\equal{}2)\\plus{}P(X_8\\equal{}5)\\plus{}P(X_8\\equal{}8)\\equal{}\\frac{28\\plus{}56\\plus{}1}{256}\r\n\\equal{}\\frac{85}{256}.$" -} -{ - "Problem": "Rounded to the nearest integer, on average, five cars passed through the intersection of Q and 16th per minute on Tuesday. On that day, what is the largest number of cars that could have passed through the intersection?", - "Solution_1": "i don't quite understand the problem :blush:", - "Solution_2": "Ehhh it was phrased kinda badly. Let me rephrase it:\r\n\r\n\r\nOn average, $ x$ cars passed through the intersection of Q and 16th per minute on Tuesday. Given that $ x$ rounded to the nearest integer is 5, find the largest number of cars that could have passed through the intersection that day.", - "Solution_3": "[hide]Would it be 7,919, because there are 1,440 minutes in a day, and 5.5 cars is the lowest number that would round to 6 cars. $ 5.5\\times 1,440 = 7,920$. The highest number below that is 7,919.[/hide]", - "Solution_4": "[hide]\n\n$ \\text{There are}$, $ 60\\cdot24= 1440$ $ \\text{minutes in a day. The lowest number that can be rounded to}$ $ 5$ $ \\text{is}$ $ 5.4999999\\dots$\n$ \\text{Thus, gives the answer: }$\n\n$ 1440\\cdot5.49999999999999\\dots$\n$ =7919.9999\\dots856$\n\n$ \\text{Because, the average}, x ,\\text{was rounded down, we have to round this down too.}$ \n\n$ \\text{Therefore, we get}$ $ 7919$\n\n[u][b]Answer to the Traffic Problem:[/b][/u]$ \\boxed{7919\\text{ cars}}$\n\n$ \\text{So yeah, I got the same answer as vamathletes.}$\n[/hide]" -} -{ - "Problem": "Let $x=2^{12}3^{6}$ and $y=2^{8}3^{8}$. Find an integer $z$ such that $x^{x}y^{y}=z^{z}$.", - "Solution_1": "I enjoyed that contest...\r\n\r\n[hide]$2^{11}3^{7}$[/hide]\r\n\r\nThat one took a while to solve. I'll post the proof later; need to go somewhere now...\r\n\r\nEdit: I just realized that wasn't the problem you posted. I was thinking $2^{12}3^{6}$ and $2^{8}3^{8}$.", - "Solution_2": "The answer is $2{}^{1}{}^{1}*3^{7}$ (The answer to the actual question on the contest) I solved it using logarithms.", - "Solution_3": "I solved it by plugging in and GUESS AND CHECKING IT", - "Solution_4": "I solved it by raising the numbers to the powers and splitting up the exponents and grouping like bases and factoring the exponents... the other five problems on the contest were easy.", - "Solution_5": "Well the buses problem was easy if you followed intuition (which I did). But today I managed to prove that expected amount of time you would have to wait for each bus is one half of the bus's \"period.\"", - "Solution_6": "My favorite was number 5 which I edited and called \"monkey numbers\" and posted on this forum.\r\n\r\nSince it was the NYML, I assume you're from New York? If so, what school?", - "Solution_7": "[quote=\"miyomiyo\"]My favorite was number 5 which I edited and called \"monkey numbers\" and posted on this forum.\n\nSince it was the NYML, I assume you're from New York? If so, what school?[/quote]\r\nit was the CAML too, so I'm assuming that the math league thingy is for all US states", - "Solution_8": "That makes more sense. I was surprised how many kids were from new york. :wink:", - "Solution_9": "Further support for the claim: It was on the NEML as well.", - "Solution_10": "Oh yeah, you're right. My work has the right answer, I just put the wrong one. Here's how I solved $x=2^{12}3^{6}$ and $y=2^{8}3^{8}$\r\n\r\n[hide]\n$x=2^{12}3^{6}$\n$y=2^{8}3^{8}$\n$x^{x}y^{y}=z^{z}$\n\n$(2^{12}3^{6})^{x}(2^{8}3^{8})^{y}=z^{z}$\n$(2^{12x}3^{6x})(2^{8y}3^{8y})=z^{z}$\n$(2^{12x+8y})(3^{6x+8y})=z^{z}$\n$(12x+8y)\\log{2}+(6x+8y)\\log{3}=z\\log{z}$\n$(2^{2}3^{1}x+2^{3}y)\\log{2}+(2^{1}3^{1}x+2^{3}y)\\log{3}=z\\log{z}$\n$(2^{14}3^{7}+2^{11}3^{8})\\log{2}+(2^{13}3^{7}+2^{11}3^{8})\\log{3}=z\\log{z}$\n$(2^{3}2^{11}3^{7}+3^{1}2^{11}3^{7})\\log{2}+(2^{2}2^{11}3^{7}+3^{1}2^{11}3^{7})\\log{3}=z\\log{z}$\n$(2^{3}+3^{1})(2^{11}3^{7})\\log{2}+(2^{2}+3^{1})(2^{11}3^{7})\\log{3}=z\\log{z}$\n$(2^{3}+3^{1})\\log{2}+(2^{2}+3^{1})\\log{3}=\\frac{z\\log{z}}{(2^{11}3^{7})}$\n$8\\log{2}+3\\log{2}+4\\log{3}+3\\log{3}=\\frac{z\\log{z}}{(2^{11}3^{7})}$\n$11\\log{2}+7\\log{3}=\\frac{z\\log{z}}{(2^{11}3^{7})}$\n$(2^{11}3^{7})\\log{2^{11}3^{7}}=z\\log{z}$\n${(2^{11}3^{7})}^{(2^{11}3^{7})}=z^{z}$\n\n[/hide]\r\n\r\nNow I'm trying to find some general method for problems like this.", - "Solution_11": "[quote=\"drunner2007\"]Well the buses problem was easy if you followed intuition (which I did). But today I managed to prove that expected amount of time you would have to wait for each bus is one half of the bus's \"period.\"[/quote]\r\n\r\nWas it not just $\\int_{0}^{a}xp(x) dx$ where a is the max wait time and p(x) is the probability density function for that bus?", - "Solution_12": "[quote=\"Elemennop\"][quote=\"drunner2007\"]Well the buses problem was easy if you followed intuition (which I did). But today I managed to prove that expected amount of time you would have to wait for each bus is one half of the bus's \"period.\"[/quote]\n\nWas it not just $\\int_{0}^{a}xp(x) dx$ where a is the max wait time and p(x) is the probability density function for that bus?[/quote]\r\n\r\nRight. So what would $p(x)$ be for bus A.", - "Solution_13": "I got stuck on that problem :wallbash_red:\r\n\r\nI tried using logarithms but got nowhere. Maybe I was just too sleep-deprived... Applause to Potato Theory. :clap:\r\n\r\nMathleague was in Missouri, too, and based on these special cases I shall generalize and jump to the conclusion that it was given in all 50 states. :D", - "Solution_14": "[quote=\"mathnerd314\"]I got stuck on that problem :wallbash_red:\n\nI tried using logarithms but got nowhere. Maybe I was just too sleep-deprived... Applause to Potato Theory. :clap:\n\nMathleague was in Missouri, too, and based on these special cases I shall generalize and jump to the conclusion that it was given in all 50 states. :D[/quote]\r\nrofl I was humming AFI in my mind during the test, so I couldn't focus PLUS it was during lunch and I was SUPER hungry, so i didn't bother doing #6 and just left early to get lunch :rotfl:", - "Solution_15": "How did you're schools do. My school had 5 5's.", - "Solution_16": "[quote=\"drunner2007\"][quote=\"Elemennop\"][quote=\"drunner2007\"]Well the buses problem was easy if you followed intuition (which I did). But today I managed to prove that expected amount of time you would have to wait for each bus is one half of the bus's \"period.\"[/quote]\n\nWas it not just $\\int_{0}^{a}xp(x) dx$ where a is the max wait time and p(x) is the probability density function for that bus?[/quote]\n\nRight. So what would $p(x)$ be for bus A.[/quote]\r\n\r\nIf bus A is the 16 minute one, then $p(a)=\\frac{1}{16}$.", - "Solution_17": "[quote=\"Elemennop\"][quote=\"drunner2007\"]Well the buses problem was easy if you followed intuition (which I did). But today I managed to prove that expected amount of time you would have to wait for each bus is one half of the bus's \"period.\"[/quote]\n\nWas it not just $\\int_{0}^{a}xp(x) dx$ where a is the max wait time and p(x) is the probability density function for that bus?[/quote]\r\n\r\nUh-oh, calculus... :huh:\r\n\r\nHow did you prove that the expected amount of time... etc.? I just accepted that as fact. The answer was A, wasn't it... :?: (highly doubtful)", - "Solution_18": "We had the same problem in PA math league.\r\n\r\n[hide=\"Hint\"]Let $z=2^{a}3^{b}$.[/hide]", - "Solution_19": "[quote=\"mathnerd314\"]I got stuck on that problem :wallbash_red:\n\nI tried using logarithms but got nowhere. Maybe I was just too sleep-deprived... Applause to Potato Theory. :clap:\n\nMathleague was in Missouri, too, and based on these special cases I shall generalize and jump to the conclusion that it was given in all 50 states. :D[/quote]\r\n\r\nThere was mathleague in Wisconsin too, in fact I have the official website :D \r\n\r\n[url]http://mathleague.com/[/url]\r\n\r\nThis is a list of all the leagues that do mathleague\r\n\r\nhttp://mathleague.com/reports/report0506_1.htm\r\n\r\nI got a five, along with one other person from my school, and there were 7 fours. The only problem that I didn't get was this one :rotfl: .\r\n\r\nedit: accidentally pressed hide instead of URL", - "Solution_20": "I think we got two sixes in our school (one of them mine) and a couple of fives. It's good to see actual difficulty on the Math League tests, as there was a massive tie last year for the winning school, all of which had perfect scores.", - "Solution_21": "I got a 5, i was about 10 seconds away from finishing this problem when time ran out...\r\n\r\nI actually just combined all the exponents with your normal exponent rules, like adding for multiplication and multiplying for exponents, etc.\r\n\r\nBtw, there were two other 5's in my school, and a boatload of 4's...", - "Solution_22": "I got a five along with the majority of the school. There were probably 20 or so sixes.", - "Solution_23": "There were at least 3 6s. I don't know about the number of 5s...", - "Solution_24": "haha, I got a [hide]3[/hide]\r\nrofl I was extremely hungry and it was during lunch, so I left pretty early to get lunch PLUS I got an AFI song stuck in my head and I couldn't concentrate", - "Solution_25": "I got a [size=50]3[/size] to......\r\n\r\nARGH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!", - "Solution_26": "the bus one was incredibly easy o_O, i don't see how anybody could have gotten it wrong\r\n\r\nthis one, however, is tricky >.>\r\n\r\ni did it correctly but for some reason i wrote $2^{7}3^{11}$", - "Solution_27": "[quote=\"Treething\"]the bus one was incredibly easy o_O, i don't see how anybody could have gotten it wrong\n[/quote]\r\n\r\nI had the hardest time with the bus one\r\n\r\nmaybe cause I SUCK at probability", - "Solution_28": "The bus one...wow. Take it easy.\r\n\r\nFor the expected total time, add the time it takes to walk (constant) and the expected (average) time for the waiting at the stop. Just divide the interval by two, since you can wait a min of 0 minutes or a max of the interval length. So...\r\n\r\nExpected value= time it takes to walk + (1/2)interval length.\r\n\r\n14.5>14, so A is shorter...", - "Solution_29": "I'm amazed that there were three people in my school that got a score of zero. Some of the problems seemed like they belonged on a mathcounts chapter countdown round. x^1002+2x^2324=3, what is x???????", - "Solution_30": "[quote=\"jhcreinhardt\"]The bus one...wow. Take it easy.\n\nFor the expected total time, add the time it takes to walk (constant) and the expected (average) time for the waiting at the stop. Just divide the interval by two, since you can wait a min of 0 minutes or a max of the interval length. So...\n\nExpected value= time it takes to walk + (1/2)interval length.\n\n14.5>14, so A is shorter...[/quote]\r\n\r\nyour right about dividing the \"period\" of each bus by 2, but try proving that that gives the expected amount of time.", - "Solution_31": "[quote=\"drunner2007\"][quote=\"jhcreinhardt\"]The bus one...wow. Take it easy.\n\nFor the expected total time, add the time it takes to walk (constant) and the expected (average) time for the waiting at the stop. Just divide the interval by two, since you can wait a min of 0 minutes or a max of the interval length. So...\n\nExpected value= time it takes to walk + (1/2)interval length.\n\n14.5>14, so A is shorter...[/quote]\n\nyour right about dividing the \"period\" of each bus by 2, but try proving that that gives the expected amount of time.[/quote]\r\nWouldn't a geometric proof suffice? The probability distribution (for buses arriving every $n$ minutes) is the function $f(x)=1/n$ for the interval $[0,n]$ and $f(x)=0$ everywhere else. The average is the x-coordinate where there is an area of $0.5$ to the left of it, which is easy to find on such a simple distribution $x=n/2$.", - "Solution_32": "I brute forced the exponents and got it right, and solved the bus problem with integrals. But despite the fact that I can do calculus and exponential algebra that takes up two whole pages of scratch paper, I somehow forgot that 2, 2, and 1 satisfy the Triangle Inequality, put 7 for that problem, and got it wrong. :rotfl:", - "Solution_33": "for the bus one you could just be like $\\frac{1}{a}\\left((b)+(b+1)+(b+2)+\\cdots+(b+c)\\right)$ where $a$ and $b$ are numbers that i don't remember\r\n\r\nit's not exact but you get the answer (A=14.5 and B=15 so A)", - "Solution_34": "For the bus one:\r\n\r\nA - Shortest wait is 6 min. walk + 0 min. wait = [u]6 minutes[/u]\r\nA - Longest wait is 6 min. walk + 16 min. wait = [u]22 minuts[/u]\r\nA - Average wait is (22 + 6) / 2 = 28 / 2 = [b]14[/b]\r\n\r\nB - Shortest wait is 10 min. walk + 0 min. wait = [u]10 minutes[/u]\r\nB - Longest wait is 10 min. walk + 9 min. wait = [u]19 minutes[/u]\r\nB - Average wait is (10 + 19) / 2 = 29 / 2 = [b]14.5[/b]\r\n\r\n14 < 14.5\r\n\r\nA is shorter", - "Solution_35": "This was an ok mathleague I guess... i made a ton of mistakes though and only got 3.\r\n\r\nFor example, on the AB + A + B one, I factored it just like I think everyone else would have as (a +1)(b+1), but I forgot the minus 1 at the end and wrote 59 instead of 58! Arghenflugen!\r\n\r\nAnd on the bus question, I stupidly forgot to add the 6 and 10, as I had completely disregarded their relevance.\r\n\r\nAnd I too tried a log solution to the last problem, but my paper looked so bad at the end that I decided I couldn't find an answer anyway...", - "Solution_36": "I took that on the NEML in Massachusetts too. Our school (Boston Latin) got 3 6's (of which I was one) and a bunch of 5's. But we are a pretty big school. I solved that last problem by dividing it up into powers of 2 and 3, and then playing around with the exponents until I got something in the form $z^{z}$. \r\nI did the bus one by intuition, but I later proved rigorously that if it comes every x minutes, you wait an average of x/2 minutes. I used a method basically like integrating. \r\nBut you can see the result intuitively without too much work. Imagine the possible arrival times as a line segment, with the previous bus stop time on the left and the coming one on the right. The amount of time you will wait is the distance between your location on the line and the right endpoint. Look at the midpoint of your interval, directly between two bus stop times. If you land at the midpoint itself, you will wait x/2 minutes. But every point to the right of the midpoint has a mirror point, its reflection over the midpoint. If you average the wait times for these two points together, you get x/2. Then, to find the overall average, you average together this infinite number of x/2's, after which you of course get x/2. You can actually turn this into a rigorous proof by dividing the line into 2n equal segments and then letting n approach infinity.", - "Solution_37": "My school had 4 6's, myself being one of them, and one of the others is also an AoPS member.\r\n\r\nWe also had a lot of 5's.", - "Solution_38": "the exponent one....\r\n\r\n\r\nit is obvious that $z=2^{a}\\cdot 2^{b}$, then we get\r\n\r\n$2^{14}3^{7}+2^{11}3^{8}=a(2^{a}+2^{b})$\r\n$2^{13}3^{7}+2^{11}3^{8}=b(2^{a}+2^{b})$\r\nthen $\\frac{a}{b}=\\frac{11}{7}$\r\n$a=11$, $b=7$ looks nice (and obviously works)" -} -{ - "Problem": "Let $ F(x,y,z)$ be polynomial with integer coefficients, for which it is known, that there exists real $ x_0,y_0$ and $ z_0$ so that $ F(x_0,y_0,z_0)\\equal{}0$. Prove, that for some real [i]algebraic[/i] numbers $ x_1,y_1$ and $ z_1$ it is true that $ F(x_1,y_1,z_1)\\equal{}0$.", - "Solution_1": "Here's a possible approach: consider the Jacobian $ \\left(\\frac {\\partial F}{\\partial x}, \\frac {\\partial F}{\\partial y}, \\frac {\\partial F}{\\partial z}\\right)$ of $ F$ at the give root $ (x_0, y_0, z_0)$. If this is not the zero vector, assume wlog that $ \\frac {\\partial F}{\\partial z}(x_0, y_0, z_0) \\neq 0$. By the implicit function theorem there's a function $ g(x,y)$ such that $ F(x,y,g(x,y)) \\equal{} 0$ whenever $ (x,y)$ are in some small neighborhood of $ (x_0, y_0)$. In particular, you can certainly find a point $ (x_1,y_1)$ such that $ (x_1,y_1)$ are rational and $ F(x_1,y_1,z_1)$=0 where $ z_1 \\equal{} g(x_1,y_1)$. But now $ z_1$ is a root of $ p(Z) \\equal{} F(x_1,y_1,Z)$ which is rational polynomial, so $ z_1$ is algebraic and the point $ (x_1,y_1,z_1)$ is as required (with the added bonus that $ x_1$ and $ y_1$ are actually rational).\r\n\r\nIf the given solution $ P \\equal{} (x_0, y_0, z_0)$ is a singular point so that the Jacobian vanishes there, then $ P$ is also a root of the polynomials $ \\frac {\\partial F}{\\partial x}, \\frac {\\partial F}{\\partial y}, \\frac {\\partial F}{\\partial z}$, which cannot all vanish unless $ F \\equal{} 0$. So one of those is a genuine polynomial of which $ P$ is a root, and it has a lower total degree, so we can finish off the argument by induction on the degree.", - "Solution_2": "I think the problem is easier than you might think. Since there exists a real solution $ \\left(x_0,y_0,z_0\\right)$ of $ F\\left(x_0,y_0,z_0\\right) \\equal{} 0$, the polynomial $ F$ is not constant (unless it is identically $ 0$, in which case the problem is trivial). Consider the polynomial $ F\\left(X,Y,Z\\right)$ as an element of the polynomial ring $ \\mathbb{Q}\\left(X,Y,Z\\right)$. Its three (formal!) partial derivatives $ \\frac {d}{dX}F\\left(X,Y,Z\\right)$, $ \\frac {d}{dY}F\\left(X,Y,Z\\right)$ and $ \\frac {d}{dZ}F\\left(X,Y,Z\\right)$ cannot all be zero (otherwise, a simple comparison of coefficients would show that $ F$ is constant), so we can WLOG assume that $ \\frac {d}{dZ}F\\left(X,Y,Z\\right)\\neq 0$ in the polynomial ring $ \\mathbb{Q}\\left(X,Y,Z\\right)$. Then, there exist rationals $ \\alpha$ and $ \\beta$ such that $ \\frac {d}{dZ}F\\left(\\alpha,\\beta,Z\\right)\\neq 0$ in the (smaller) polynomial ring $ \\mathbb{Q}\\left(Z\\right)$ (because $ \\frac {d}{dZ}F\\left(\\alpha,\\beta,Z\\right) \\equal{} 0$ is a polynomial identity in $ \\alpha$ and $ \\beta$ over the integral domain $ \\mathbb{Q}\\left(Z\\right)$, so if it holds for all $ \\left(\\alpha,\\beta\\right)\\in\\mathbb{Q}\\times\\mathbb{Q}$, it must hold as a polynomial identity, because $ \\mathbb{Q}$ is infinite). This yields that the polynomial $ F\\left(\\alpha,\\beta,Z\\right)$ is nonconstant, hence has an algebraic root for $ Z$.\r\n\r\nThe result generalizes: If $ K\\subseteq L$ is a field extension, and $ I\\subseteq K\\left[X_1,X_2,...,X_n\\right]$ is an ideal such that the polynomials from $ I$ have a common root in $ \\overline{L}$, then these polynomials also have a common root in $ \\overline{K}$. This can be seen by one of the following two arguments:\r\n1) Elimination theory. This can be used to actually construct such a root, and its construction only requires finite extensions of field, so if the polynomials were defined over $ K$, they must have a common root in $ \\overline{K}$.\r\n2) Hilbert's Nullstellensatz. This doesn't yield the assertion as trivially as 1), but is far better known (unfortunately). Using the Nullstellensatz, the existence of a common root becomes equivalent to the ideal being the whole polynomial ring, and what remains is to show that an ideal $ I$ of $ K\\left[X_1,X_2,...,X_n\\right]$ is the whole $ K\\left[X_1,X_2,...,X_n\\right]$ if and only if the ideal $ I\\cdot L\\left[X_1,X_2,...,X_n\\right]$ of $ L\\left[X_1,X_2,...,X_n\\right]$ is the whole $ L\\left[X_1,X_2,...,X_n\\right]$. While $ \\Longrightarrow$ is trivial, what we need is $ \\Longleftarrow$. This, again, is trivial by taking a basis of $ L$ over $ K$ (or, if we wish to avoid Hamel bases, by a basis of a part of $ L$, namely the one required to ensure that $ I\\cdot L\\left[X_1,X_2,...,X_n\\right] \\equal{} L\\left[X_1,X_2,...,X_n\\right]$).\r\n\r\n darij", - "Solution_3": "I think it's doesn't have to concern any $ \\frac {\\partial F}{\\partial X}$\r\nBy induction on $ n$ we can prove for any field $ Q\\subseteq k\\subseteq R$ and any n-variable polynomial $ F\\in Q[X]$ the proposition holds.\r\nn=1 is trivial.\r\nsupposing it holds for n=h-1. If $ x_1 \\equal{} t$ is transcendent on $ k$, by induction the other $ x_i$ is algebra on $ k(t)$, set $ E \\equal{} k(t,x_2,\\cdots ,x_n) \\equal{} k(t,y)$ where y is algebra on k(t), $ irr(k(t),y) \\equal{} h_t(y)$. Mutiply all conjugates of $ F$ (as a k(t)[y] polynomial), we get a polynomial of $ t$, since $ t$ is transcendent this polynomial must be 0. Thus for any $ t\\in k$ and any root of h_t(y) there is a corresponding solution of $ F$, choose $ t'$ so close to $ t$ that $ h_t(y)$ still has a real root, but this time $ y$ is algebra and so all $ x_i$ is algebraic on $ k$.\r\n\r\nby the way, darij: how dose your generalization represent \u2026\u2026F has not only algebraic root but $ real$ algebraic root. Dose $ \\bar{K}$ means algebra closed field of $ K$ ?", - "Solution_4": "Ah, I see. I misread the \"real\" in the \"real algebraic\", and the title \"algebraic geometry\" didn't help me either, since algebraic geometry has not much to do with $ \\mathbb{R}$ for me. Of course, this makes my post completely irrelevant to this topic. My apologies.\r\n\r\n darij", - "Solution_5": "[quote=\"randomgraph\"]\n\nIf the given solution $ P \\equal{} (x_0, y_0, z_0)$ is a singular point so that the Jacobian vanishes there, then $ P$ is also a root of the polynomials $ \\frac {\\partial F}{\\partial x}, \\frac {\\partial F}{\\partial y}, \\frac {\\partial F}{\\partial z}$, which cannot all vanish unless $ F \\equal{} 0$. So one of those is a genuine polynomial of which $ P$ is a root, and it has a lower total degree, so we can finish off the argument by induction on the degree.[/quote]\r\n\r\nHello,What do you mean by genuine polynomial ?And How to use induction then?Thank you" -} -{ - "Problem": "In a class of $ n\\geq 4$ some students are friends. In this class any $ n \\minus{} 1$ students can be seated in a round table such that every student is sitting next to a friend of him in both sides, but $ n$ students can not be seated in that way. Prove that the minimum value of $ n$ is $ 10$.", - "Solution_1": "Why nobody is interested? I think this is a nice problem. It doesn't involve a special idea I know, but it is enjoying to deal with cases.", - "Solution_2": "Nice problem!\r\nSolution:\r\n\r\nCase 1:$ n\\le 6$\r\nIf n satisfies the conditon,\r\nBecause every $ n\\minus{}1$ students forms a circle, so each student has at least $ 2$ friends. In addition if a student has only $ 2$ friends, then consider the $ n\\minus{}1$ students that only includes $ 1$ friends of him/her, the $ n\\minus{}1$ students cannot form a circle, a contradiction. Therefore each student has at least $ 3$ friends.\r\nChoose a student arbitrarily and call him/her $ A$, the other $ n\\minus{}1$ students forms a circle. Since $ n\\minus{}1\\le 5$ and $ A$ has at least $ 3$ friends, $ A$ must have $ 2$ friends who are adjacent on the circle, and then we can add $ A$ to the circle i.e. we get a circle of the all $ n$ students, a contradiction. Case 1 is done.\r\n\r\nCase 2:$ n\\equal{}7$\r\nIf $ n\\equal{}7$ satisfies the conditon,\r\nThe same as in Case 1 we have that each student has at least $ 3$ friends. But $ 7$ is an odd number, so there exist a student who has at least $ 4$ friends. Call him $ B$. Consider the circle formed by other $ 6$ students,.Since $ B$ has at least $ 4$ friends, he must have $ 2$ friends who are adjacent on the circle, and then we can add $ B$ to the circle i.e. we get a circle of the all $ 7$ students, a contradiction. Case 2 is done.\r\n\r\nCase 3:$ n\\equal{}8$\r\nIf $ n\\equal{}8$ satisfies the conditon,\r\nThe same as in Case 1 we have that each student has at least $ 3$ friends. Choose a student arbitrarily and call him/her $ C$, then $ C$ has at least $ 3$ friends. Due to the condition the other $ 7$ students form a circle.Assume these students are $ 1,2,3,4,5,6,7$, respectively. And they are the same order on the circle. If $ C$ has at least $ 4$ friends, then there will be $ 2$ of them who are adjacent on the circle, and then we can add $ C$ to the circle i.e. we get a circle of the all $ 8$ students, a contradiction. So $ C$ has exactly $ 3$ friends. Since $ C$ is chosen arbitrarily, each student has exactly $ 3$ friends.\r\n\r\nw.l.o.g we can assume $ C$'s friends are $ 2,4,6$, then $ C,2,4,6$ have no more friends, and $ 1,3,5,7$ each has one more friend. Because $ 1$ and $ 7$ are already friends, so there are only two possible cases:\r\n\r\n(1) $ 1$ and $ 5$, $ 3$ and $ 7$ are friends respectively. But then we have a circle of all $ 8$ students: $ 6\\minus{}C\\minus{}2\\minus{}3\\minus{}4\\minus{}5\\minus{}1\\minus{}7\\minus{}6$, a contradiction.\r\n(2) $ 1$ and $ 3$, $ 5$ and $ 7$ are friends respectively. But then we have a circle of all $ 8$ students: $ 6\\minus{}C\\minus{}4\\minus{}3\\minus{}2\\minus{}1\\minus{}7\\minus{}5\\minus{}6$, a contradiction, too.\r\nCase 3 is done.\r\n\r\nCase 4:$ n\\equal{}9$\r\n\r\nIf $ n\\equal{}9$ satisfies the conditon,\r\nwe have that each student has at least $ 3$ friends and $ 9$ is an odd number, so there exists a student who has at least $ 4$ friends.Call him/her $ D$.\r\nDue to the condition the other $ 8$ students form a circle.Assume these students are $ 1,2,3,4,5,6,7,8$, respectively. And they are the same order on the circle.\r\nIf $ D$ has at least $ 5$ friends, similar to previous case we get that two of them are adjacent on the circle which leads to a contradiction.\r\nSo we assume $ D$ has exactly $ 4$ friends, and w.l.o.g we assume that they are $ 1,3,5,7$, respectively. Next color students $ 1,3,5,7,D$ black and color students $ 2,4,6,8$ white. Then all known pair of friends are one black student and one white student. Take $ 5$ black students and $ 3$ white students. Due to the condition they form another circle. Since there are only $ 3$ white students on the latter circle, we have that there exists two adjacent students on this circle who are both black. w.l.o.g there are only two possible relative position of them on the former circle: (1) $ 1$ and $ 3$ ; (2) $ 1$ and $ 5$.\r\n(1) $ 1$ and $ 3$ are friends.But then we have a circle of all $ 9$ students:$ 2\\minus{}D\\minus{}8\\minus{}7\\minus{}6\\minus{}5\\minus{}4\\minus{}3\\minus{}1\\minus{}2$, a contradiction.\r\n(2) $ 1$ and $ 5$ are friends.But then we have a circle of all $ 9$ students:$ 8\\minus{}D\\minus{}4\\minus{}3\\minus{}2\\minus{}1\\minus{}5\\minus{}6\\minus{}7\\minus{}8$, a contradiction, too.\r\nCase 4 is done.\r\n\r\nNow we have proved that $ n$ is at least $ 10$.Finally we construct an example to show that $ n\\equal{}10$ satisfies the condition.\r\nThe example is:\r\nTen students are $ 1,2,3,4,5,6,7,8,9,10$, respectively. $ 1\\minus{}2\\minus{}3\\minus{}4\\minus{}5\\minus{}6\\minus{}7\\minus{}8\\minus{}9\\minus{}1$ is a circle. All other pairs of friends are $ 10\\minus{}3$, $ 10\\minus{}6$, $ 10\\minus{}9$, $ 1\\minus{}5$, $ 2\\minus{}7$, $ 4\\minus{}8$.It is not hard to check that this example satisfies the condition.\r\n\r\nEND", - "Solution_3": "Cute and nice problem.\nI think I found an alternative that gives more intricate details about the degrees and skips the casework mostly.\nFirstly let $n \\le 9$.\nLet the number of edges in the graph be $e$.\nThen consider the vertex $v$ with the greatest degree and the cycle of the remaining $n-1$ vertices.\nWe see that the our vertex cannot be connected with two adjacent vertices in the $n-1$ cycle.\nThus we see that by the pigeonhole principle, $deg(v) \\le [\\frac{n+1}{2}]$\n\nTherefore, $e \\le [\\frac{n}{2}[\\frac{n+1}{2}]]$\nBy considering two chosen cycles that a vertex is a part of ,we see that $\\delta(G) \\ge 3$.\n\nWe now claim that $G$ contains a 4-cycle.\nFor this consider the vertex with smallest degree $u$.\nCall three of its adjacent vertices $A,B,C$.\nNow, as $e \\le [\\frac{n}{2}[\\frac{n+1}{2}]] < [\\frac{n^2}{4}]$,\n$G$ is triangle free.\nThus no two of $A,B,C$ are adjacent.\nNow, we prove that at least two $A,B,C$ have a common neighbour apart from $u$ which will give the claim.\nFor this assume the contrary.\nThen we would get that since their neighborhoods apart form $u$ are disjoint\n$n \\ge 2+2+2+4=10$ ,a contradiction.\nThus $G$ contains a $4-cycle$.\nThus, by a well known result\n$e > \\frac{n}{4} (1+ \\sqrt{4n-3})$\nHowever this just gives that $n$ upto $8$ dont' work.\nSo,we for $n=9$ we just proceed as above or can adapt an alternative since we got more structure.\nThis bound just implies $n=9$, then there are atleast $5$ vertices of degree $4$ which is just extra info it gives(may or may not use for the casework ).\nFInish by giving a construction,which isn't too difficult." -} -{ - "Problem": "1. A store charges $28 for a certain type of sweater. This price is 40 percent more than the amount it costs the store to buy one of these sweaters/ At an end of season sale, store employees can urchase any remaining sweaters at 30 percent off the store's cost. How much would it cost an employee to purchase a sweater of this type at this sale?\r\n\r\n2. In rectangle ABCD, point E is the midpoint of line BC. If the area of quadrilateral ABED is 2/3, what is the area of rectangle ABCD?", - "Solution_1": "[hide=\"1\"]Isn't it just 28*0.7=19.60? [/hide]\n\n[hide=\"2\"]Draw lines ED, AE, and EF, where F is the midpoint of AD. There are four congruent triangles created. ABED is three of them. The are fo the rectangle is 4/3*2/3=8/9. [/hide]", - "Solution_2": "I'm not sure I understand the wording on #1. When it says the store's cost, I assume it's referring to the 28 dollar price? If so, why does it bother to include the 40% increase in what the store pays for it? And if it's referring to the price that the store bought the sweater for, it isn't very clear...\r\n\r\n#2. (Note: $[ABC...]$ denotes the area of polygon $ABC...$.) $E$ is the midpoint of rectangle $ABCD$, and $E$ is the midpoint of $BC$. If we take the midpoint of $AD$ and label it $F$, then $[ABEF]=[FCDE]=\\frac{1}{2}\\cdot [ABCD]$. $ED$ divides rectangle $ECDF$ in half, since it is a diagonal, which means that $[EDF]=[ECD]=\\frac{1}{2}\\cdot [ECDF]$. Therefore, we have $\\frac{1}{2}\\cdot [ABCD]+\\frac{1}{2}\\cdot \\frac{1}{2}\\cdot [ABCD]=\\frac{3}{4}\\cdot [ABCD]=\\frac{2}{3}$. Therefore, $[ABCD]=\\frac{8}{9}$.\r\n\r\nI hope you understood that. If not, please tell me what needs clarification.", - "Solution_3": "I think the 40% was just a trick? They do that a lot on tests, but not usually on SAT's.", - "Solution_4": "but the answer is $14.00", - "Solution_5": "Q1: let x be the cost of a sweater for the store.\r\n\r\nthen (28 - x)/x =0.4\r\n\r\nat the end, you will find out x = 20\r\n\r\nthe answer: x * 70% = 14", - "Solution_6": "Cheerful Coffin, welcome to AoPS! And your solution is correct.", - "Solution_7": "[quote=\"woohooxd\"]but the answer is 14.00[/quote]\r\n\r\nOkay, so they mean that the employees can buy a sweater for 30% off the price that the store pays to buy the sweaters. You can find the price that the store pays, although it isn't necessary, but I'll explain it that way first. The store sells sweaters for 28 dollars, which is 40% more than they buy it for. Therefore, $1.4x=28$, where $x$ is the price the store pays. Solving that gives $x=20$, so the store pays 20 dollars. The employees get 30% off of that price, meaning that $y=.7\\cdot x \\Rightarrow y=14$, so the employees pay 14 dollars.\r\n\r\nThe other, quicker way of solving it is the following: $1.4x=28$, $.7x=y \\Rightarrow 1.4x=2y \\Rightarrow 2y=28 \\Rightarrow y=14$. :)" -} -{ - "Problem": "Let $ A,B \\in\\mathcal{M}_n(\\mathbb{C})$, such that $ AB\\equal{}0$, show that $ A$ and $ B$ are simultaneously triangularizable.\r\n\r\nP.S.: Not sure it is already on this forum.", - "Solution_1": "[hide=\"Hint\"]I posted a first step to solve this problem at the end of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=208827]Commutative transforms.[/url][/hide]\n[hide=\"Second hint\"]The next step is to induct over $ n.$[/hide]" -} -{ - "Problem": "I'm a bit clueless here on how to start this exercise. \r\n\r\nLet $X$ be a nonempty set with the discrete metric and show that $C(X,\\mathbb{R})$ is separable $\\iff X$ is finite. \r\n\r\nHere $C(X,\\mathbb{R})$ is the set of bounded continuous mappings from $X$ into $\\mathbb{R}$.", - "Solution_1": "Show that the only dense set in the discrete metric is the whole space.", - "Solution_2": "[quote=\"jmerry\"]Show that the only dense set in the discrete metric is the whole space.[/quote]\r\n\r\nThank you for the help, but I'm afraid I'm still unable to see a solution. \r\n\r\nIf $X$ is infinite, then by a previous exercise we can take $X=\\mathbb{N}^+$ and define $f(n)=0$ if $|f_n(n)|\\ge 1$ and $f(n)=|f_n(n)|+1$ if $|f_n(n)|<1$ where $(f_n)$ is a sequence in $C(X,\\mathbb{R})$, for then $\\|f-f_n \\|\\ge 1$ for all $n$. Hence $C(X,\\mathbb{R})$ isn't separable since it doesn't even have an everywhere dense subset. \r\n\r\nHowever, I must be missing something here for I cannot see this fails just because $X$ is finite.", - "Solution_3": "[quote=\"Jimmy_K\"]However, I must be missing something here for I cannot see this fails just because $X$ is finite.[/quote]Think about what it is you're doing. Let $\\{f_n\\}_{n \\in \\mathbf{N}}$ be a countable subset of $\\mathcal{C}(\\mathbf{N},\\mathbf{R})$. You want to construct a function $f$ which is 'far away' enough from every $f_n$. In order to guarantee this, you pick the value of $f(1)$ so that $|f - f_1| > 1$, the value of $f(2)$ so that $|f - f_2| > 1$, etc. In order for this process to not terminate, you need to have $f$ defined on some infinite set -- otherwise you'd run out of degrees of freedom for your function $f$.\r\n\r\nTo prove that $\\mathcal{C}(X,\\mathbf{R})$ is separable when $X$ is finite, just use that $\\mathbf{R}$ is separable.", - "Solution_4": "Thank you. I realized it later on and its pretty obvious now. It's very similar to the diagonal construction in the proof of the uncountability of $\\mathbb{R}$." -} -{ - "Problem": "Let $ ABCD$ be a rectangle with $ AB\\equal{}b, AD\\equal{}a$. A circle with center $ A$ and radius $ AC$ intersect line $ BD$ and $ E$ and $ F$. Bisector of angle $ ECF$ intersect chord $ EF$ at $ N$. Calculate $ \\dfrac{EN}{NF}$.", - "Solution_1": "first apply the angle bisector theorem and thentry inversion through :) $ C$ then the solution follows easily", - "Solution_2": "Please post your solution if you have already solved it.\r\n\r\nI have a really ugly solution using coordinates: :oops: \r\n\r\n[hide=\"My Solution\"]From the angle bisector theorem, $ \\frac {EN}{NF} = \\frac {CE}{CF}$. Consider the homothety that sends $ A$ to $ A'$ such that $ \\overrightarrow{CA'} = 2\\overrightarrow{CA}$. This sends rectangle $ ABCD$ to rectangle $ A'B'C'D'$ and we can restate our problem as follows: $ A'B'CD'$ is a rectangle and $ A,B,D$ are the midpoints of $ CA',CB'$ and $ CD'$ respectively. $ BD$ intersects the circumcircle of $ A'B'CD'$ at $ E$ and $ F$. Then we need to find $ \\frac {CE}{CF}$.\n\nConsider a coordinate system with $ A$ as the origin, and the axes along $ AD$ and $ AB$. Then $ D = (a,0),B = (0, - b), C = (a, - b)$. Hence the equation of $ BD$ (or $ EF$) is $ bx - ay = ab$ and the circumcircle of $ A'B'C'D'$ is $ x^2 + y^2 = a^2 + b^2$. From these we find that $ E = \\left(\\frac {a(b^2 + \\sqrt {D})}{a^2 + b^2},\\frac {b( - a^2 + \\sqrt {D})}{a^2 + b^2}\\right)$ and $ F = \\left(\\frac {a(b^2 - \\sqrt {D})}{a^2 + b^2},\\frac {b( - a^2 - \\sqrt {D})}{a^2 + b^2}\\right)$. Here $ D = a^4 + a^2b^2 + b^4$.\n\nThus (after simplification) $ CE = \\frac {\\sqrt {b^2(b^2 + \\sqrt D)^2 + a^2(a^2 - \\sqrt D)^2}}{a^2 + b^2}$ and ${ CF = \\frac {\\sqrt {a^2(a^2 + \\sqrt D)^2 + b^2(b^2 - \\sqrt D)^2}}{a^2 + b^2}}$. And finally $ \\frac {CE}{CF} = \\sqrt {\\frac {b^2(b^2 + \\sqrt D)^2 + a^2(a^2 - \\sqrt D)^2}{{a^2(a^2 + \\sqrt D)^2 + b^2(b^2 - \\sqrt D)^2}}}$.\n\nI haven't found any nice expressions. :blush: [/hide]", - "Solution_3": "I don't think that my ratio is quite well. But at least it is a ratio with $ a$ and $ b$.\r\n\r\n[color=red]Here goes my solution,[/color]\r\nConsider a dilatation with center $ C$ and ratio $ 2$ which maps the rectangle $ ABCD$ to $ A'B'CD'$.\r\nNow using angle bisector theorem we get, ratio,$ k \\equal{} \\frac {EN}{NF} \\equal{} \\frac {EC}{CF} \\equal{} \\frac {\\sin \\angle {EFC}}{\\sin \\angle{CEF}} \\equal{} \\frac {\\sin \\angle {EA'C}}{\\sin \\angle{CA'F}}$\r\n\r\n$ B'EFD'$ is an isosceles trapezoid (as it is a cyclic one). So, $ B'E \\equal{} D'F$ and $ \\angle B'A'E \\equal{} \\angle FA'D'$\r\n\r\nLet $ \\angle BB'A \\equal{} B$ and $ \\angle ACD \\equal{} C$\r\n\r\nBy easy angle chasing we get that, $ \\angle EA'C \\equal{} C \\minus{} B/2$ and $ \\angle CA'F \\equal{} \\pi/2 \\minus{} C \\minus{} B/2$\r\nso,\r\n$ \\frac {\\sin \\angle {EA'C}}{\\sin \\angle{CA'F}} \\equal{} \\frac {\\sin (C \\minus{} B/2)}{\\cos (C \\plus{} B/2)} \\equal{} \\frac {\\sin C.\\cos{B/2} \\minus{} \\cos C.\\sin{B/2}}{\\cos C.\\cos{B/2} \\minus{} \\sin C.\\sin{B/2}}$\r\n\r\nLet $ AC \\equal{} r \\equal{} \\sqrt {a^2 \\plus{} b^2}$\r\nSo,\r\n$ \\sin {B/2} \\equal{} \\sqrt {\\frac {r \\minus{} a}{2r}}$ and $ \\cos B/2 \\equal{} \\sqrt {\\frac {r \\plus{} a}{2r}}$ and $ \\sin C \\equal{} \\frac {a}{r}$ and $ \\cos C \\equal{} \\frac {b}{r}$\r\nso,\r\n$ \\frac {\\sin C.\\cos{B/2} \\minus{} \\cos C.\\sin{B/2}}{\\cos C.\\cos{B/2} \\minus{} \\sin C.\\sin{B/2}} \\equal{} \\frac {{\\frac {a}{r}}\\sqrt {\\frac {r \\plus{} a}{2r}} \\minus{} {\\frac {b}{r}}\\sqrt {\\frac {r \\minus{} a}{2r}}}{{\\frac {b}{r}}\\sqrt {\\frac {r \\plus{} a}{2r}} \\minus{} {\\frac {a}{r}}\\sqrt {\\frac {r \\minus{} a}{2r}}}$\r\nAfter cancelling common terms we get,\r\n$ k \\equal{} \\frac {a\\sqrt {r \\plus{} a} \\minus{} b\\sqrt {r \\minus{} a}}{b\\sqrt {r \\plus{} a} \\minus{} a\\sqrt {r \\minus{} a}}$\r\n\r\n$ Q.E.D$\r\n\r\np.s. If the solution looks unclear please have a look at the diagram.\r\n\r\n[url=http://img146.imageshack.us/my.php?image=figure2dm8.png][img]http://img146.imageshack.us/img146/6676/figure2dm8.th.png[/img][/url]", - "Solution_4": "I have another trignometric solution (wasnt as trivial as I thought first)\r\nlet the diagonals meet each other at $ K$\r\nnow $ \\frac {EN}{NF} \\equal{} \\frac {CE}{CF}$ and $ K$ bisects $ AC$ let angle subtended by $ CE$ be $ x$ and that by $ CF$ be $ y$ therefore $ \\frac{CE}{CF}\\equal{} \\frac{ \\sin{x}}{ \\sin{y}}$\r\nby simple angle chasing angle $ EAC \\equal{} 2y$ and angle $ ACE \\equal{} 90^{0} \\minus{} y$\r\nas $ EK$ is a median to triangle $ CEA$ we get $ \\cos{x \\plus{} y} \\equal{} 2 \\sin{x} \\sin{y}$\r\nwhich implies $ \\cos{x}\\cos{y} \\equal{} 3 \\sin{x} \\sin{y}$ which implies $ 8 \\sin^{2}{x} \\sin^{2}{y} \\plus{} sin^{2}x \\plus{} sin^{2}y \\equal{} 1$ __________________________$ eq^{n} 1$\r\nalso $ AC \\cos{x \\plus{} y} \\equal{} c \\cos{x \\plus{} y} \\equal{} \\frac {ab}{c}$\r\nthis implies $ \\frac {ab}{c^{2}} \\equal{} \\sin{x} \\sin{y}$\r\ndividing $ eq^{n} 1$ by the one found now and substituting $ \\frac {ab}{c^{2}} \\equal{} \\sin{x} \\sin{y}$ \r\nyou get a quadratic in the ratio required", - "Solution_5": "i can't see anything...... :( please correct your latex code", - "Solution_6": "fixed\r\n :oops:", - "Solution_7": "[quote=\"gauravpatil\"]I have another trignometric solution (wasnt as trivial as I thought first)\nlet the diagonals meet each other at $ K$\nnow $ \\frac {EN}{NF} \\equal{} \\frac {CE}{CF}$ and $ K$ bisects $ AC$ let angle subtended by $ CE$ be $ x$ and that by $ CF$ be $ y$ therefore $ \\frac {CE}{CF} \\equal{} \\frac {\\sin{x}}{\\sin{y}}$\nby simple angle chasing angle $ EAC \\equal{} 2y$ and angle $ ACE \\equal{} 90^{0} \\minus{} y$\nas $ EK$ is a median to triangle $ CEA$ we get $ \\cos{x \\plus{} y} \\equal{} 2 \\sin{x} \\sin{y}$\nwhich implies $ \\cos{x}\\cos{y} \\equal{} 3 \\sin{x} \\sin{y}$ which implies $ 8 \\sin^{2}{x} \\sin^{2}{y} \\plus{} sin^{2}x \\plus{} sin^{2}y \\equal{} 1$ __________________________$ eq^{n} 1$\nalso $ AC \\cos{x \\plus{} y} \\equal{} c \\cos{x \\plus{} y} \\equal{} \\frac {ab}{c}$\nthis implies $ \\frac {ab}{c^{2}} \\equal{} \\sin{x} \\sin{y}$\ndividing $ eq^{n} 1$ by the one found now and substituting $ \\frac {ab}{c^{2}} \\equal{} \\sin{x} \\sin{y}$ \nyou get a quadratic in the ratio required[/quote]\r\n\r\nCould you please write a bit clearly. most probably I am missing something or you have missed something. :maybe:", - "Solution_8": "I :) have modified it a bit\r\nlet the diagonals meet each other at $ K$ let $ c$ be third side\r\nnow $ \\frac {EN}{NF} \\equal{} \\frac {CE}{CF}$ and $ K$ bisects $ AC$ let angle subtended by $ CE$ be $ x$ and that by $ CF$ be $ y$ therefore $ \\frac {CE}{CF} \\equal{} \\frac {\\sin{x}}{\\sin{y}}$ by sine rule\r\nby simple angle chasing angle $ EAC \\equal{} 2y$ and angle $ ACE \\equal{} 90^{0} \\minus{} y$\r\nas $ EK$ is a median to triangle $ CEA$ we get $ \\cos{x \\plus{} y} \\equal{} 2 \\sin{x} \\sin{y}$ substituting $ \\cos{x \\plus{} y} \\equal{} \\cos{x} \\cos{y} \\minus{} \\sin{x} \\sin{y}$\r\nwhich implies $ \\cos{x}\\cos{y} \\equal{} 3 \\sin{x} \\sin{y}$ which implies $ 8 \\sin^{2}{x} \\sin^{2}{y} \\plus{} sin^{2}x \\plus{} sin^{2}y \\equal{} 1$ ______$ eq^{n} 1$\r\nnow we calculate distance between a and bd in 2 ways $ AE \\cos{x \\plus{} y} \\equal{} c \\cos{x \\plus{} y} \\equal{} b \\sin{ABD} \\equal{} \\frac {ab}{c}$\r\nthis implies $ \\frac {ab}{c^{2}} \\equal{} \\cos{x \\plus{} y} \\equal{} \\sin{x} \\sin{y}$\r\ndividing $ eq^{n} 1$ by the one found now and substituting $ \\frac {ab}{c^{2}} \\equal{} \\sin{x} \\sin{y}$ \r\nyou get a quadratic in the ratio required or rather an equation of this form $ ratio \\plus{} \\frac {1}{ratio} \\equal{} r$ where $ r$ is a known constant in $ a,b$ and $ c$\r\natleast i see nothing missing", - "Solution_9": "[size=117][color=darkblue][b]Very nice application of the [u]Generalized Pythagoras' Theorem[/u] (G.P.T.) ![/b][/color][/size]\r\n\r\n[quote=\"Ahiles\"] [color=darkred]Let $ ABCD$ be a rectangle with $ AD = a$ , $ AB = b$ . The circle $ w = C(A,r)$ , \n\nwhere $ r = \\sqrt {a^2 + b^2}$ , intersects $ BD$ in $ E$ and $ F$ so that $ B\\in (ED)$ . Calculate $ \\frac {CE}{CF}$ .[/color] [/quote]\r\n[color=darkblue][b][u]Proof.[/u][/b] Denote $ \\|\\begin{array}{c} BE = x \\\\\n \\\\\nDF = y\\end{array}$ . Observe that $ \\|\\begin{array}{c} \\cos\\widehat {ABD} = \\frac br \\\\\n \\\\\n\\cos\\widehat {ADB} = \\frac ar\\end{array}$ . Apply [b]G.P.T.[/b] to :\n\n$ \\|\\begin{array}{cccc} \\triangle\\ ABE\\ : & r^2 = x^2 + b^2 + 2bx\\cdot \\frac br & \\implies & rx^2 + 2b^2x - ra^2 = 0 \\\\\n \\\\\n\\triangle\\ ADF\\ : & r^2 = y^2 + a^2 + 2ay\\cdot\\frac ar & \\implies & ry^2 + 2a^2y - rb^2 = 0\\end{array}\\|$ $ \\implies$ $ \\|\\begin{array}{c} rx = \\sqrt {\\Delta} - b^2 \\\\\n \\\\\nry = \\sqrt {\\Delta} - a^2\\end{array}\\|\\ \\ (1)$ \n\nwhere $ \\boxed {\\ \\Delta = a^4 + a^2b^2 + b^4\\ }$ . Observe that $ \\|\\begin{array}{c} \\cos\\widehat {CBD} = \\frac ar \\\\\n \\\\\n\\cos\\widehat {CDB} = \\frac br\\end{array}$ . Apply [b]G.P.T.[/b] to :\n\n$ \\|\\begin{array}{cc} \\triangle\\ BCE\\ : & CE^2 = x^2 + a^2 + 2ax\\cdot\\frac ar \\\\\n \\\\\n\\triangle\\ DCF\\ : & CF^2 = y^2 + b^2 + 2by\\cdot\\frac br\\end{array}\\|$ $ \\implies$ $ (\\frac {CE}{CF})^2 = \\frac {rx^2 + 2a^2x + ra^2}{ry^2 + 2b^2y + rb^2}\\stackrel {(*)}{\\ = \\ }$ $ \\frac {(ra^2 - 2b^2x) + 2a^2x + ra^2}{(rb^2 - 2a^2y) + 2b^2y + rb^2} =$\n\n$ \\frac {ra^2 + (a^2 - b^2)x}{rb^2 - (a^2 - b^2)y} =$ $ \\frac {r^2a^2 + (a^2 - b^2)\\cdot rx}{r^2b^2 - (a^2 - b^2)\\cdot ry} =$ $ \\frac {r^2a^2 + (a^2 - b^2)(\\sqrt {\\Delta } - b^2)}{r^2b^2 - (a^2 - b^2)(\\sqrt {\\Delta} - a^2)} =$ $ \\frac {a^4 + b^4 + (a^2 - b^2)\\sqrt {\\Delta}}{a^4 + b^4 - (a^2 - b^2)\\sqrt {\\Delta}} =$\n\n$ \\frac {[(a^4 + b^4) + (a^2 - b^2)\\sqrt {\\Delta}]^2}{(a^4 + b^4)^2 - (a^2 - b^2)^2(a^4 + a^2b^2 + b^4)} =$ $ \\frac {[(a^4 + b^4) + (a^2 - b^2)\\sqrt {\\Delta}]^2}{(a^4 + b^4)^2 - (a^2 - b^2)(a^6 - b^6)} =$ $ \\frac {[(a^4 + b^4) + (a^2 - b^2)\\sqrt {\\Delta}]^2}{a^2b^2(a^2 + b^2)^2}$ .\n\nIn conclusion, $ \\boxed {\\ \\frac {CE}{CF} = \\frac {|a^4 + b^4 + (a^2 - b^2)\\sqrt {\\Delta}|}{ab(a^2 + b^2)}\\ }$ . Remark that $ x - y = \\frac {a^2 - b^2}{\\sqrt {a^2 + b^2}}$ .[/color]" -} -{ - "Problem": "I am reading my algebra-book, which says the following.\r\n\r\nif $ n$ is a $ a$-pseudoprime, but the strong $ a$-pseudoprime sequence for $ n$ looks like $ \\{...,b,1,...,1\\}$ with $ b\\equiv 1$ or $ \\minus{} 1 \\pmod{n}$ being FALSE, then $ n$ fails the strong $ a$-pseudoprime test, so $ n$ must be composite.\r\n\r\nIt then goes on to say the following cryptic statement:\r\n\r\n[color=red]The number $ n$ cannot be prime because $ 1$ has at least three square roots, $ 1$, $ \\minus{} 1$, and $ b\\pmod{n}$[/color]\r\n\r\nWhat is this link between n being prime and the roots it has?\r\n\r\n\r\nThanks in advance.", - "Solution_1": "[b]Legendre's Theorem:[/b] A polynomial of degree $ n$ with coefficients $ \\bmod p$ (for $ p$ prime) has at most $ n$ roots. \r\n\r\n[b]Special case:[/b] The polynomial $ x^2 \\equiv 1 \\bmod p$ has at most $ 2$ roots. When $ p > 2$, these roots are $ 1, \\minus{} 1$.\r\n\r\nIf another square root of $ 1$ exists $ \\bmod n$, then $ n$ cannot be prime.\r\n\r\n[hide=\"Direct proof of the cryptic statement\"] Suppose there exists $ b \\neq 1, \\minus{}1$ that is a square root of $ 1 \\bmod n$. Then $ n | b^2 \\minus{} 1$. Suppose further that $ n$ is prime. Then $ n | (b \\minus{} 1)(b \\plus{} 1)$, so either $ n | b\\minus{}1$ or $ n | b\\plus{}1$, but we assumed that $ b \\neq 1, \\minus{}1 \\bmod n$ - contradiction. [/hide]", - "Solution_2": "oh t0rajir0uu, you know every proof so it comes directly from the hand. That makes me heartened (is heartened a word? disheartened is..) :0)" -} -{ - "Problem": "Let $k, m, n$ be integers such that $1m\n\nThen let S={x,x+1...k}\n\nNotice that it has the desirable (but inconclusive) property that if I add in another element to this set, it no longer satisfies the condition. It can be proven easily for n=2, and probably not too hard for n=3 either. (n=1 it is not true).\n\nAny ideas? very curious about this.", - "Solution_2": "Someone reminded me of this problem and so I have given it another shot :)\n\nLet $x(n,m,k)$ be the number $x$ as given in the above post. Let $f(n,m,k)$ be the size of the set $S$ in the above post.\n\nNotice: $x(n,m,k) f(n,m,k)>f(n',m',k)$\n\nNow suppose $N>2$ we make the inductive hypothesis that $f(n,m,k)$ is indeed the answer to the problem for $n=2,3...N-1$ (the base case $n=2$ is easy and I will omit it).\n\nSuppose we have chosen $f(N,m,k)+1$ numbers. Then we have some number $y$ that is less than $x(N,m,k)$. Let $m'=m-y$. Now remove $y$ from our set. We have $f(N,m,k)$ numbers in our set. Now we want to show that $f(N,m,k)>f(N-1,m',k)$ and we will do this if we can show that $x(N,m,k)x(N,m,k)$ Hence $f(N-1,m',k)1$, while for $n=1$ it is $k-1$ (choose every number other than $m$)\n\nThis number is easy to compute hence I declare it a suitable solution to the problem :P" -} -{ - "Problem": "PLease help me to prove the following statments. I have searched the topic similar to this one but get nothing. Thanks in advance. \r\n\r\nLet H be a nonempty subset of a group G. The set N(H)={a $\\in G | aHa^{-1}=H$} is called normalizer of H in G.\r\n(i) Prove that N(H) is a subgroup of G.\r\n Suppose H is a subgroup of G.\r\n(ii) Prove that H is normal in G iff N(H)=G.\r\n(iii) Prove that H is normal in N(H).\r\n(iv) Prove that N(H) is the largest subgroup of G in which H is normal, i.e., if H is normal in a subgroup K of G , then K is a subset of N(H). \r\nIs the last one is straight forward from the (ii) and (iii)?", - "Solution_1": "All of this is not very hard.\r\nPlease type your definition of normal group, and of subgroup. You should solve these problems almost immediately then.", - "Solution_2": "Today i have thought a lot about it and here what i got:\r\n\r\n [hide=\" For (i):\"]\nI see that e $\\in N(H)$. But How to show that the inverse of the element belongs to N(H)? \nIf $a\\in N(H)$ then $\\forall h' \\in H \\text{ we have }aha^{-1}=h' \\implies a^{-1}h'a=h \\Rightarrow a^{-1}Ha=H$ and therefore $a^{-1}\\in N(H)$ too. Now let $a,b \\in N(H)$. Then $abHb^{-1}a^{-1}=aHa^{-1}=H$ that's why $ab \\in N(H)$ and hence N(H) is a group.\nIs that correct? :lol: [/hide]\n[hide=\"For (ii):\"]\n[color=blue]Theorem: Let H be a subgroup of a group G.Then H is a normal in G iff $aHa^{-1}$ is a subset of H for all $a \\in G$.[/color]\n\nLet H be a normal subgroup ,then $aHa^{-1}$ is a subset of H for all $a \\in G$. Since $|aHa^{-1}|=|H|$ we have $aHa^{-1}=H$. (this result i have got today!) and we get that N(H)=G. \nConversely, let N(H)=G. Then $aHa^{-1}=H$ for all $a \\in G$ and that's why $aHa^{-1}$ is a subset of H for all $a \\in G$ . Hence H is normal.\n :lol: [/hide]\n[hide=\"For (iii):\"]\nUsing the definition of normalizer we get that $aHa^{-1}$ is a subset of H for all $a \\in N(H)$ and by the theorem the result follows. :oops: [/hide]\n[hide=\"For (iv):\"] Let H be a normal subgroup in a subset K of G ,then $aHa^{-1}$ is a subset of H for all $a \\in K$. Since $|aHa^{-1}|=|H|$ we have $aHa^{-1}=H$ for all $a \\in K$. Therefore Kis a subset of N(H). :oops: \n[/hide]\r\nWhen i first saw this problem i did not realized that we can extend the theorem above given in my book. That was why i couldn't solve it. I am studying Abstract Algebra in summer without anybody's help eccept Mathlinkers, thank you all a lot!", - "Solution_3": "Hmm, \r\nsome corrections or comments\r\n\r\nyou are frequently using the size |H| of a group. What makes you sure they are finite.\r\n$(\\mathbb{Z},+)$ is an infinite subgroup of $(\\mathbb{Q},+)$\r\nI can assure you they have the same \"size\" but obviously the latter is strictly bigger!\r\n\r\nI would advise you to prove this little lemma first, it will HELP you :wink: \r\n\r\nThe following statements about a subgroup $H$ in $G$ are equivalent :\r\n\r\n1. $H$ is a normal subgroup of $G$\r\n2. $a^{-1}H a=H \\ \\forall a\\in G$ (you may find it interesting to know that $a^{-1}H a$ is often written as $H^{a}$)\r\n3. $a^{-1}H a\\subseteq H \\ \\forall a\\in G$\r\n4. $\\forall a\\in G: a H = H a$", - "Solution_4": "[quote=\"fredbel6\"]...you are frequently using the size |H| of a group. What makes you sure they are finite.....I can assure you they have the same \"size\" but obviously the latter is strictly bigger![/quote]\r\n\r\n\r\nIf i change the word the same size to equipolent, will the rest be correct?", - "Solution_5": "Hm, what exactly would you mean by equipollent?", - "Solution_6": "Is there one-one fuction from $a^{-1}Ha$ onto H?", - "Solution_7": "You mean a bijection\r\n\r\nYes, just map every $h\\in H$ onto $h^{a}=a^{-1}h a$\r\nBut that will NOT be a correct proof. you cannot say one is in the other and there is a bijection :thus the sets are equal.\r\nJust check my example with $\\mathbb{Q}$ I gave up here.", - "Solution_8": "Then how can i prove the $H^{a}=H$ iff H is normal?\r\nZ and Q are equipolent. Sorry I think i cant get your question sorry :(", - "Solution_9": "Two comments :\r\n\r\n1. what is your definition of normal subgroup?? That would really clear things up! :o \r\n\r\n2. Once we have that sorted out, you should try this again :\r\n[quote] The following statements about a subgroup $H$ in $G$ are equivalent :\n\n1. $H$ is a normal subgroup of $G$\n2. $a^{-1}H a=H \\ \\forall a\\in G$ (you may find it interesting to know that $a^{-1}H a$ is often written as $H^{a}$)\n3. $a^{-1}H a\\subseteq H \\ \\forall a\\in G$\n4. $\\forall a\\in G: a H = H a$[/quote]", - "Solution_10": "My defenition:\r\n $H$ is a normal subgroup of $G$ if $\\forall a\\in G: a H = H a$.\r\nThen there is a theorem:\r\n$a^{-1}H a\\subseteq H \\ \\forall a\\in G$ iff $H$ is a normal subgroup of $G$.\r\nAnd I haven't seen $a^{-1}H a=H \\ \\forall a\\in G$ ,that's what i am trying to prove proving incorrectly the |H|=|H^a|", - "Solution_11": "Assume that : \r\n\r\n$H^{x}\\subseteq H \\forall x \\in G$\r\n\r\nWell then we also have :\r\n\r\n$H^{a^{-1}}\\subseteq H\\forall a\\in G$\r\n\r\nnow if $H^{a^{-1}}\\subseteq H$\r\nthen $(H^{a^{-1}})^{a}\\subseteq (H)^{a}$\r\n\r\nor $H \\subseteq (H)^{a}$ as well", - "Solution_12": "Oh, thanx a lot. It was easy again . :blush: :ewpu:" -} -{ - "Problem": "Fie:\r\n\r\n$S^`=\\{ (x,y,z,t) \\ \\&\\mbox{\\mid}x+y-z-t=0\\}$ $\\ \\& \\mbox{si}$\r\n$S^=\\{ (x,y,z,t) \\ \\&\\mbox{\\mid}x-y-z+t=0\\}$\r\n\r\ndoua subspatii in $R^4$ ; determinati intersectia lor si demonstrati identitatea:\r\n\r\n$S^`+S^=R^4$", - "Solution_1": "nici o solutie?", - "Solution_2": "Fie $v_{1}= \\left( 1,0,0,1 \\right)^{T}, \\, v_{2}= \\left( 0,1,0,1 \\right)^{T}, \\, v_{3}= \\left( 0,0,1,-1 \\right)^{T}, \\, v_{4}= \\left( 1,0,0,-1 \\right)^{T}$.\r\n\r\nObserv\u0103m c\u0103 $v_{1},v_{2},v_{3}\\in S^\\prime, \\, v_{4}\\in S^{\\prime \\prime}$ \u015fi c\u0103 $\\det \\left( v_{1}\\, v_{2}\\, v_{3}\\, v_{4}\\right) \\neq 0$." -} -{ - "Problem": "Consider the triangle $ ABC$. The incircle of the triangle $ ABC$ touches $ (BC)$ at the point $ D$.Let $ [DK]$ be the diameter in the incircle of the triangle ABC.\r\nConsider also the point $ T$ on $ BC$, where the $ A$-excircle touches the side $ (BC)$.Prove that $ A,K$ and $ T$ are collinear.\r\n\r\nAdditional task: Prove that $ \\frac{AK}{AT}\\equal{}\\frac{h_a\\minus{}2r}{h_a}$, where $ h_a$ is the usual notation of lenght of altitude in the triangle ABC drawn from $ A$.", - "Solution_1": "let the line perpendicolar to DK on K meet AB on E and AC on F, then the triangles $ \\triangle AEF$, $ \\triangle ABC$ are homothetic with center A and ratio the ratio of the antitude, so $ \\frac{AK}{AT}\\equal{}\\frac{h_a\\minus{}2r}{h_a}$. But K is the tangency point of the a-excircle of $ \\triangle AEF$ so A,K,T are collinear by the homothety." -} -{ - "Problem": "Plot it out. Its logical, and it certainly looks true to me.", - "Solution_1": "The basic transformations (what happens when you add or subtract to x or y, or when you multiply x or y by a constant) are always the same, no matter what the graph is.", - "Solution_2": "By the way, I suggest you stop using step and start using the proper floor or just square brackets.. I think you said ages ago thats how you learnt it but if you used it in a competition or something I don't think its official enough to be understood properly..\r\n\r\nBut yes, those transformations are exactly the same, no matter what graph you have. Even if it some crazy graph like f(x) is the [x]th digit of pi. Or anything." -} -{ - "Problem": "AMC 12 2001. \r\n16. A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe?\r\n\r\nA. 8! B.(2^8)8! C. (8!)^2 D. 16! /(2^8) E. 16!", - "Solution_1": "[hide=\"solution\"]\nSuppose that there are n legs, then there are n socks and n shoes.\n\nImagine that the shoes and socks can be put on in any order: then the number of orders will be (2n)!.\n\nThere is an inherent symmetry to the orders in that for any given order where sock_i is put on before shoe_i, there is an identical ordering where the two have swapped places (shoe_i is put on instead of sock_i and sock_i is put on instead of shoe_i). Only one of those orderings is valid since socks must be put on before shoes, so 1/2 of the orderings will have sock_1 put on before shoe_1. Since there are n sock-shoe pairs, we need to divide by 2^n.\n\nThus, the answer would be (2n)! / 2^n.\n\nYour guess for (8!)^2 is incorrect because it presumes that ALL socks are put on before ANY shoes, when that is not necessarily the case; nothing in the problem specifies this additional restriction.\n[/hide]", - "Solution_2": "label each foot:$ a,b,c,d,e,f,g,h$\r\nthen problem is equivalent to arranging 2 of each of the letters. (e.g. $ aabbccddeeffgghh$)\r\n\r\nYou just consider the first time a letter appears to be the event of putting on the sock and the second time putting on shoe. (i think this is same general argument used above, but maybe a little shorter?)", - "Solution_3": "[hide]There are $ 16!$ ways to put on shoes and socks in any manner. BUt for each foot, we must multiply by $ \\frac{1}{2}$ since there is a 1/2 chance that socks are put on first. \nD[/hide]" -} -{ - "Problem": "Please refer to the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10857]Getting Started Series Challenge Problems Introduction[/url]for an explanation of my objectives for this series of messages. \r\n\r\nI hope this is fairly easy for most people visiting this forum, but let's make sure by trying to solve the problem. I can find much harder (and somewhat easier) problems if need be, so please let me know if this level is appropriate for you, by public post or by private message. This problem is adapted from Beginning Algebra 8th Edition by Margaret L. Lial and John Hornsby. Very similar, and even harder, problems can be found in the fourth grade volume of the Hua Loo-Keng School Mathematics Textbook series. \r\n\r\nProblem One: \r\n\r\nOn the first day of summer tokenadult flies southwest from Minneapolis on an airplane that took off at the same time as A+MATH's airplane, which flies northeast. The plane flying A+MATH to the northeast flies 50 miles per hour faster than tokenadult's airplane. In thirty minutes, the planes are 275 miles apart. What is the speed of each airplane? \r\n\r\n(For extra challenge, try to solve the whole problem without using any literal algebra, using only the block diagram problem-solving taught in Singapore, for example.) \r\n\r\nPlease show steps to your solution, and please put your solutions and steps in spoiler text. Have fun.", - "Solution_1": "[hide]This is the same as if tokenadult was staying put and a+math was flying at 2x+50 mph. 2x+50 = 275*2. x+25=275. x=250. Therefore a+maths speed is 300. Tokenadult's is 250. [/hide]", - "Solution_2": "I thought the simpifying (?) assumption at the start of Scrambled's solution was interesting." -} -{ - "Problem": "Find all positive integer solutions to $(a+1)^{x}-a^{y}=1$", - "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=130446" -} -{ - "Problem": "\u0391\u03bd \\[x,y,z>0\\] \u03bc\u03b5 \\[\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=\\frac{3}{2}\\] \u03bd\u03b1 \u03b4\u03b5\u03b9\u03be\u03b5\u03c4\u03b5 \u03bf\u03c4\u03b9 \\[\\frac{x}{x+yz}+\\frac{y}{y+zx}+\\frac{z}{z+xy}\\geq 1\\]", - "Solution_1": "H dosmeni sxesi ginetai $2(xy+yz+xz)=3xyz \\ \\ (1)$\r\n\r\nH pros apodeiksi anisotita grafetai $\\frac{x^{2}}{x^{2}+xyz}+\\frac{y^{2}}{y^{2}+xyz}+\\frac{z^{2}}{z^{2}+xyz}\\geq 1$\r\n\r\nOmws apo tin anisotita Andreescu exoume oti \r\n\r\n\\begin{eqnarray}\\frac{x^{2}}{x^{2}+xyz}+\\frac{y^{2}}{y^{2}+xyz}+\\frac{z^{2}}{z^{2}+xyz}&\\geq&\\frac{(x+y+z)^{2}}{x^{2}+y^{2}+z^{2}+3xyz}\\nonumber \\\\ &=&\\frac{x^{2}+y^{2}+z^{2}+2(xy+yz+xz)}{x^{2}+y^{2}+z^{2}+3xyz}\\nonumber \\\\ &\\stackrel{(1)}{=}& 1 \\nonumber \\end{eqnarray}\r\n\r\nAlexandros", - "Solution_2": "\u03a3\u03c9\u03c3\u03c4\u03b1!\u039c\u03b1\u03bb\u03bb\u03bf\u03bd \u03c0\u03b1\u03c1\u03b1\u03b7\u03c4\u03b1\u03bd \u03b5\u03c5\u03ba\u03bf\u03bb\u03b7... :oops:", - "Solution_3": "\u039c\u03bc\u03bc! \u0394\u03b5\u03bd \u03ae\u03c4\u03b1\u03bd \u03ba\u03b1\u03b8\u03cc\u03bb\u03bf\u03c5 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b7! \u0395\u03cd\u03ba\u03bf\u03bb\u03b7 \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b5 \u03cc\u03c0\u03bf\u03b9\u03bf\u03bd \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b5\u03b9 \u03ae \u03c4\u03b7 \u03b4\u03b5\u03b9 \u03bb\u03c5\u03bc\u03ad\u03bd\u03b7.\u03a6\u03c5\u03c3\u03b9\u03ba\u03ac \u03b4\u03b5\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5 \u0399\u039c\u039f , \u03b1\u03bb\u03bb\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03bf\u03bb\u03cd \u03c9\u03c1\u03b1\u03af\u03b1 \u03ba\u03b1\u03b9 \u03b4\u03b9\u03b4\u03b1\u03ba\u03c4\u03b9\u03ba\u03ae. \u039c\u03b1\u03ba\u03ac\u03c1\u03b9 \u03bd\u03b1 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03c3\u03b1\u03bd \u03bd\u03b1 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03bf\u03c5\u03bd \u03bf\u03b9 \u03bc\u03b9\u03c3\u03bf\u03af \u03b1\u03c0\u03cc \u03c4\u03bf\u03c5\u03c2 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03bf 1/1000 \u03c4\u03c9\u03bd \u03bc\u03b1\u03b8\u03b7\u03c4\u03ce\u03bd \u03c4\u03bf\u03c5 \u039b\u03c5\u03ba\u03b5\u03af\u03bf\u03c5!!!\r\n\r\n \u0391\u03c5\u03c4\u03ac \u03c4\u03b1 \u03b3\u03c1\u03ac\u03c6\u03c9 \u03b3\u03b9\u03b1\u03c4\u03af \u03c0\u03bf\u03c4\u03ad \u03b4\u03b5\u03bd \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c5\u03c0\u03bf\u03c4\u03b9\u03bc\u03ac\u03c4\u03b5 \u03c4\u03b9\u03c2 \u03b9\u03ba\u03b1\u03bd\u03cc\u03c4\u03b7\u03c4\u03ad\u03c2 \u03c3\u03b1\u03c2.\u03a3\u03c4\u03b1 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03ac \u03ba\u03b1\u03b9 \u03b7 \u03c0\u03b9\u03bf \u03b1\u03c0\u03bb\u03ae \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03ba\u03c1\u03cd\u03b2\u03b5\u03b9 \u03c4\u03b7 \u03b4\u03c5\u03c3\u03ba\u03bf\u03bb\u03af\u03b1 \u03c4\u03b7\u03c2.\u0391\u03bd \u03bb\u03c5\u03b8\u03b5\u03af , \u03b1\u03c5\u03c4\u03cc \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03ac\u03bd\u03c4\u03b1 \u03bc\u03b5\u03b3\u03ac\u03bb\u03b7 \u03b5\u03c0\u03b9\u03c4\u03c5\u03c7\u03af\u03b1 , \u03ba\u03b1\u03b9 \u03b1\u03c5\u03c4\u03cc\u03c2 \u03c0\u03bf\u03c5 \u03c4\u03b7 \u03bb\u03cd\u03bd\u03b5\u03b9 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03bb\u03b1\u03cd\u03c3\u03b5\u03b9 \u03c4\u03bf \u03b1\u03c0\u03bf\u03c4\u03ad\u03bb\u03b5\u03c3\u03bc\u03b1 \u03c4\u03b7\u03c2 \u03c0\u03c1\u03bf\u03c3\u03c0\u03ac\u03b8\u03b5\u03b9\u03ac\u03c2 \u03c4\u03bf\u03c5 , \u03b1\u03bd\u03b1\u03be\u03ac\u03c1\u03c4\u03b7\u03c4\u03b1 \u03b1\u03c0\u03cc \u03c4\u03bf \u03b2\u03b1\u03b8\u03bc\u03cc \u03b4\u03c5\u03c3\u03ba\u03bf\u03bb\u03af\u03b1\u03c2 \u03c4\u03b7\u03c2 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7\u03c2.\r\n \u03a3\u03c5\u03bc\u03c0\u03ad\u03c1\u03b1\u03c3\u03bc\u03b1 :\r\n [color=red] \u039b\u03a5\u03a3\u03a4\u0395 \u039a\u0391\u0399 \u0391\u03a0\u039f\u039b\u0391\u03a5\u03a3\u03a4\u0395 [/color]\r\n\r\n \u03ba\u03ac\u03b8\u03b5 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7. \u039a\u03ac\u03b8\u03b5 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03bb\u03cd\u03c3\u03b5\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03cd\u03c3\u03ba\u03bf\u03bb\u03b7 \u03ba\u03b1\u03b9 \u03ba\u03ac\u03b8\u03b5 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03bb\u03cd\u03c3\u03b1\u03bc\u03b5 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03b5\u03bb\u03b9\u03ba\u03ac \u03b5\u03cd\u03ba\u03bf\u03bb\u03b7!", - "Solution_4": "\u039a\u03c5\u03c1\u03b9\u03b5 \u03a3\u03c4\u03b5\u03c1\u03b3\u03b9\u03bf\u03c5 \u03b5\u03c7\u03b5\u03c4\u03b5 \u03b4\u03b9\u03ba\u03b9\u03bf!\u039a\u03b1\u03c0\u03bf\u03b9\u03b5\u03c2 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b5\u03b9\u03c2 \u03c3\u03bf\u03c5 \u03b4\u03b5\u03b9\u03c7\u03bd\u03bf\u03c5\u03bd \u03b1\u03c0\u03bf \u03bc\u03bf\u03bd\u03b5\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03bf\u03c4\u03b9 \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03c3\u03ba\u03b5\u03c6\u03c4\u03b5\u03b9\u03c2 \u03b1\u03c0\u03bb\u03b1,\u03c0\u03c1\u03b1\u03b3\u03bc\u03b1 \u03c0\u03bf\u03c5 \u03c0\u03bf\u03bb\u03bb\u03b5\u03c2 \u03c6\u03bf\u03c1\u03b5\u03c2 \u03c4\u03bf \u03b1\u03b3\u03bd\u03bf\u03bf\u03c5\u03bc\u03b5..\u039a\u03b1\u03b9 \u03c4\u03bf \u03b1\u03c0\u03bb\u03bf,\u03b5\u03b9\u03bd\u03b1\u03b9 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9 \u03b5\u03bd\u03c4\u03c5\u03c0\u03c9\u03c3\u03b9\u03b1\u03ba\u03bf...!", - "Solution_5": "\u03a3\u03c5\u03bc\u03c6\u03c9\u03bd\u03c9 \u03c6\u03b9\u03bb\u03b5!\u039a\u03b1\u03bb\u03bf\u03c3\u03b7\u03c1\u03b8\u03b5\u03c2! :lol:", - "Solution_6": "Alternative:\r\n$ \\sum\\frac{1}{1\\plus{}\\frac{xyz}{x^2}} \\ge 1$\r\n\u0388\u03c3\u03c4\u03c9 \u03b7 $ f(x)\\equal{}\\frac{1}{1\\plus{}\\frac{a}{x^2}}$, me $ a\\equal{}xyz$.\r\n$ f''(x) \\ge 0$\r\nAra \r\n$ LHS \\ge \\frac{3}{1\\plus{}\\frac{a}{\\left(\\frac{x\\plus{}y\\plus{}z}{3} \\right)^2}}$\r\nEtsi exoume ndo:\r\n$ (x\\plus{}y\\plus{}z)^2 \\ge \\frac{9xyz}{2}$\r\nAlla apo ton periorismo:\r\n$ xy\\plus{}yz\\plus{}zx \\equal{}\\frac{3}{2}xyz \\Leftrightarrow 3\\sum xy \\equal{}\\frac{9xyz}{2} \\le (x\\plus{}y\\plus{}z)^2$.", - "Solution_7": "[quote=\"Dimitris X\"]\n\u0388\u03c3\u03c4\u03c9 \u03b7 $ f(x) \\equal{} \\frac {1}{1 \\plus{} \\frac {a}{x^2}}$, me $ a \\equal{} xyz$.\n$ f''(x) \\ge 0$\nAra\n[...][/quote]\r\n\r\n\u0395\u03b4\u03ce \u03c7\u03ac\u03bd\u03b5\u03b9 \u03b7 \u03c0\u03c1\u03bf\u03c3\u03ad\u03b3\u03b3\u03b9\u03c3\u03ae \u03c3\u03bf\u03c5... \u03b4\u03b5 \u03c3\u03bf\u03c5 \u03b4\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b7 \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7 $ xyz \\equal{} \\text{const.}$ \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03c0\u03bf\u03c1\u03b5\u03af\u03c2 \u03bd\u03b1 \u03bc\u03b5\u03c4\u03b1\u03c7\u03b5\u03b9\u03c1\u03b9\u03c3\u03b8\u03b5\u03af\u03c2 \u03c4\u03bf $ a$ \u03c9\u03c2 \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03ac \u03c3\u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03b1\u03b3\u03ce\u03b3\u03b9\u03c3\u03b7.\r\n\r\nCheerio,\r\n\r\nDurandal 1707", - "Solution_8": "\u039d\u03bf\u03bc\u03b9\u03b6\u03c9 \u03bf\u03c4\u03b9 \u03b7 \u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03b1\u03c3\u03b7 \u03c3\u03c4\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03c4\u03bf\u03c5 \u0394\u03b7\u03bc\u03b7\u03c4\u03c1\u03b7 \u03c3\u03c9\u03b6\u03b5\u03c4\u03b1\u03b9.\r\n\r\n\u03a0\u03b9\u03c3\u03c4\u03b5\u03c5\u03c9 \u03c0\u03c9\u03c2 \u03b1\u03c5\u03c4\u03bf \u03b3\u03b9\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b4\u03bf\u03c5\u03bb\u03b5\u03c5\u03bf\u03bd\u03c4\u03b1\u03c2 \u03bc\u03b5 \u03bf\u03bb\u03b5\u03c2 \u03c4\u03b9\u03c2 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b5\u03b9\u03c2 $ f(x)\\equal{}\\frac{1}{1\\plus{}\\frac{a}{x^{2}}}$, \u03b3\u03b9\u03b1 \u03ba\u03b1\u03b8\u03b5 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03b7 \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03b1 $ a$\r\n\r\n\u0393\u03b9\u03b1 \u03bd\u03b1 \u03b3\u03b9\u03bd\u03b5\u03b9 \u03c0\u03b9\u03bf \u03b5\u03c5\u03ba\u03bf\u03bb\u03b1 \u03ba\u03b1\u03c4\u03b1\u03bd\u03bf\u03b7\u03c4\u03bf \u03b8\u03b1 \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b7\u03c3\u03c9 \u03bd\u03b1 \u03b4\u03c9\u03c3\u03c9 \u03bc\u03b9\u03b1 \u03b5\u03be\u03b7\u03b3\u03b7\u03c3\u03b7 \u03bc\u03b5 \u03b1\u03bd\u03c4\u03b9\u03b8\u03b5\u03c4\u03bf\u03b1\u03bd\u03c4\u03b9\u03c3\u03c4\u03c1\u03bf\u03c6\u03b7 \u03ba\u03b1\u03b9 \u03b1\u03c4\u03bf\u03c0\u03bf. \u0395\u03c3\u03c4\u03c9 \u03bc\u03b9\u03b1 \u03c3\u03c4\u03b1\u03b8\u03b5\u03c1\u03b7 \u03c4\u03c1\u03b9\u03b1\u03b4\u03b1 $ (x_0, y_0, z_0)$ \u03bc\u03b5 $ x_0y_0z_0 \\equal{} k \\equal{} const$ \u03c0\u03bf\u03c5 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03b5\u03b9 \u03c4\u03b7 \u03c3\u03c5\u03bd\u03b8\u03b7\u03ba\u03b7 \u03b1\u03bb\u03bb\u03b1 \u03bf\u03c7\u03b9 \u03c4\u03b7\u03bd \u03b1\u03bd\u03b9\u03c3\u03bf\u03c4\u03b7\u03c4\u03b1. \u03a4\u03bf\u03c4\u03b5 \u03b1\u03c0\u03bf Jensen \u03c3\u03c4\u03b7\u03bd $ f(x) \\equal{} \\frac{1}{1\\plus{}\\frac{k}{x^{2}}}$ \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03c0\u03c1\u03c9\u03c4\u03b7 \u03b1\u03bd\u03b9\u03c3\u03bf\u03c4\u03b7\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03b4\u03b9\u03c7\u03bd\u03b5\u03b9 \u03bf \u0394\u03b7\u03bc\u03b7\u03c4\u03c1\u03b7\u03c2 \u03b3\u03b9\u03b1 \u03bf\u03bb\u03b5\u03c2 \u03c4\u03b9\u03c2 \u03c4\u03c1\u03b9\u03b1\u03b4\u03b5\u03c2 \u03bf\u03c0\u03bf\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03c3\u03c5\u03b3\u03ba\u03b5\u03ba\u03c1\u03b9\u03bc\u03b5\u03bd\u03b7 \u03c4\u03c1\u03b9\u03b1\u03b4\u03b1 $ (x_0, y_0, z_0)$. \u039f\u03c0\u03bf\u03c4\u03b5 \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9 \u03b7 \u03b1\u03bd\u03b9\u03c3\u03bf\u03c4\u03b7\u03c4\u03b1 \u03b8\u03b1 \u03c0\u03c1\u03b5\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 $ 3(x_0y_0 \\plus{} y_0z_0 \\plus{} z_0x_0) > (x_0 \\plus{} y_0 \\plus{} z_0)^2$ \u03c4\u03bf \u03bf\u03c0\u03bf\u03b9\u03bf \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b1\u03b4\u03c5\u03bd\u03b1\u03c4\u03bf. \r\n\u03a0\u03b5\u03b9\u03c4\u03b5 \u03bc\u03bf\u03c5 \u03b1\u03bd \u03c7\u03b1\u03bd\u03b5\u03b9 \u03ba\u03b1\u03c0\u03bf\u03c5 \u03b1\u03c5\u03c4\u03bf \u03c0\u03bf\u03c5 \u03bb\u03b5\u03c9", - "Solution_9": "[quote=\"it's me\"]\u0391\u03bd \\[x,y,z>0\\] \u03bc\u03b5 \\[\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=\\frac{3}{2}\\] \u03bd\u03b1 \u03b4\u03b5\u03b9\u03be\u03b5\u03c4\u03b5 \u03bf\u03c4\u03b9 \\[\\frac{x}{x+yz}+\\frac{y}{y+zx}+\\frac{z}{z+xy}\\geq 1\\][/quote][quote=\"cretanman\"]H dosmeni sxesi ginetai $2(xy+yz+xz)=3xyz \\ \\ (1)$\n\n\n\nH pros apodeiksi anisotita grafetai $\\frac{x^{2}}{x^{2}+xyz}+\\frac{y^{2}}{y^{2}+xyz}+\\frac{z^{2}}{z^{2}+xyz}\\geq 1$\n\n\n\nOmws apo tin anisotita Andreescu exoume oti \n\n\n\n\\begin{eqnarray}\\frac{x^{2}}{x^{2}+xyz}+\\frac{y^{2}}{y^{2}+xyz}+\\frac{z^{2}}{z^{2}+xyz}&\\geq&\\frac{(x+y+z)^{2}}{x^{2}+y^{2}+z^{2}+3xyz}\\nonumber \\\\ &=&\\frac{x^{2}+y^{2}+z^{2}+2(xy+yz+xz)}{x^{2}+y^{2}+z^{2}+3xyz}\\nonumber \\\\ &\\stackrel{(1)}{=}& 1 \\nonumber \\end{eqnarray}\n[/quote] \u03bc\u03b9\u03b1 \u03c3\u03c5\u03bc\u03c0\u03bb\u03ae\u03c1\u03c9\u03c3\u03b7 \u03b1\u03c0\u03cc [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=780274#p780274]\u03b5\u03b4\u03c9[/url][quote=\"cretanman\"]Gia na isxiei i isotita stin anisotita Andreescu, tha prepei na isxiei \\[\\frac{x}{x^{2}+xyz}=\\frac{y}{y^{2}+xyz}=\\frac{z}{z^{2}+xyz}\\] 'h isodinama \\[x+yz=y+xz=z+xy\\] ap'opou linontas (fisika prepei to zeugos $(x,y,z)$ na epalitheuei kai tin $(1)$) pairnoume oti i isotita isxiei ean $x=y=z=2$[/quote]" -} -{ - "Problem": "show\r\n\r\n$\\; ( \\;\\mathbb{A}\\; ) \\;\\sum_{n=1}^{\\infty}\\;\\frac{1}{n\\cdot2^{n}}\\;=\\;\\ln{2}\\;$\r\n\r\n$\\; ( \\;\\mathbb{B}\\; ) \\;\\sum_{n=1}^{\\infty}\\;\\frac{1}{n^{2}\\cdot2^{n}}\\;=\\;\\frac{1}{2}\\; ( \\;\\frac{\\pi^{2}}{6}\\;-\\;\\ln^{2}{2}\\; ) \\;$\r\n\r\n :D", - "Solution_1": "[quote=\"misan\"]show\n\n$\\; ( \\;\\mathbb{A}\\; ) \\;\\sum_{n=1}^{\\infty}\\;\\frac{1}{n\\cdot2^{n}}\\;=\\;\\ln{2}\\;$\n[/quote]\r\n\r\nWe know that, for $-1p_0$\r\neither $ f(2p\\plus{}2)\\equal{}1$ and $ f(2p\\plus{}1)\\minus{}f(2p\\plus{}3)\\equal{}c$ from a given point \r\n\r\nCombining with the same analysis for second line, we conclude that, from a given point, we have :\r\neither $ a,b,a,b,a,b,....$\r\neither $ 1,k,1,k\\plus{}c,1,k\\plus{}2c,1,k\\plus{}3c, ...$\r\n\r\nUsing then the original equation, we can write :\r\n$ a\\plus{}b\\equal{}ab\\minus{}1996$ and so $ (a\\minus{}1)(b\\minus{}1)\\equal{}1997$ and so $ a\\equal{}2$ and $ b\\equal{}1998$ (or the reverse)\r\n$ 1\\plus{}k\\equal{}1(k\\plus{}c)\\minus{}1996$ and so $ c\\equal{}1997$\r\n\r\nSo the function ends with :\r\neither $ 2,1998,2,1998,2,1998,...$\r\neither $ 1,k,1,k\\plus{}1997,1,k\\plus{}2\\cdot 1997, ...$\r\n\r\nIt's then immediate to show that if this is true from a given point, it's true from the beginning. Hence the four solutions :\r\n\r\n$ \\{f(1),f(2),f(3),...\\}\\equal{}\\{2,1998,2,1998,2,1998, ...\\}$\r\n$ \\{f(1),f(2),f(3),...\\}\\equal{}\\{1998,2,1998,2,1998,2 ...\\}$\r\n$ \\{f(1),f(2),f(3),...\\}\\equal{}\\{1,a,1,a\\plus{}1997,1,a\\plus{}2\\cdot 1997,1,a\\plus{}3\\cdot 1997, ...\\}$\r\n$ \\{f(1),f(2),f(3),...\\}\\equal{}\\{a,1,a\\plus{}1997,1,a\\plus{}2\\cdot 1997,1,a\\plus{}3\\cdot 1997,1 ...\\}$", - "Solution_2": "thanks.did you think on my polynomials problems?", - "Solution_3": "[quote=\"MJ GEO\"]thanks.did you think on my polynomials problems?[/quote]\r\n\r\nSince you already have solutions, we are in no hurry to solve these problems ...", - "Solution_4": "yes.you are right no hurry :)", - "Solution_5": "Let $\\mathbb N$ be the set of positive integers. Find all functions defined on $\\mathbb N$ and taking values on $\\mathbb N$ satisfying, for all $n\\in\\mathbb N$,\n$$f(n)+f(n+1)=f(n+2)f(n+3)-1998.$$" -} -{ - "Problem": "Let $ \\{ f_k(z) \\}$ be a sequence of analytic functions on a domain $ D$ that converges normally to $ f(z)$. Suppose that $ f_k(z)$ attains each value $ w$ at most $ m$ times (counting multiplicity) in $ D$. Show that either $ f(z)$ is constant, or $ f(z)$ attains each value $ w$ at most $ m$ times in $ D$.\r\n\r\nWe have covered Hurwitz's Theorem and a theorem that says:\r\n\r\nSuppose $ \\{ f_k(z) \\}$ is a sequence of univalent functions on a domain $ D$ that converges normally on $ D$ to a function $ f(z)$. Then either $ f(z)$ is univalent or $ f(z)$ is constant. I am not sure if those help in this problem though. I need some hints on how to proceed here. Thank you.", - "Solution_1": "Use exactly the same proof, counting the number of times f attains a value in D using the argument principle" -} -{ - "Problem": "Let P is in the plane of \u0394$ DPQ$\uff0cand $ cot$ \u03b1$ \\plus{} cot$ \u03b2 $ \\plus{} cot$ \u03b3 $ \\equal{} cot m \\plus{} cot n \\plus{} cot p$. \r\nThen please give the locus of the point $ P$.\uff08Present now I know $ O,I,H,G,K,Ia,Ib,Ic$ is on the locus\uff09", - "Solution_1": "Testing a few points, I'm guessing the locus is the Thomson cubic.\r\n\r\nReferences:\r\n[url]http://mathworld.wolfram.com/ThomsonCubic.html[/url]\r\n[url]http://pagesperso-orange.fr/bernard.gibert/Exemples/k002.html[/url]", - "Solution_2": "Let me show you how I advance the problem\uff0cit is just that\uff1a\r\n\r\n$ cot$\u03b1$ \\minus{}cot$\u03b2$ \\equal{}cot$\u2220$ ABD\\minus{}cot$\u2220$ ADB$ \u21d4 $ quadrilateral ABCD$ is harmonious .\r\nSo\uff0cas for the point $ K$\uff0cwe have $ cot$\u03b1$ \\plus{}cot$\u03b2$ \\plus{}cot$\u03b3$ \\equal{}cotm\\plus{}cotn\\plus{}cotp$ .", - "Solution_3": "[quote=\"nsato\"]Testing a few points, I'm guessing the locus is the Thomson cubic.\n\nReferences:\n[url]http://mathworld.wolfram.com/ThomsonCubic.html[/url]\n[url]http://pagesperso-orange.fr/bernard.gibert/Exemples/k002.html[/url][/quote]\r\n\r\nMaybe\uff01But the $ Curve\\minus{}L$ I've given have a feature\uff1a\r\nIf $ P$ is on the $ Curve\\minus{}L$\uff0cand $ P$ have a Isogonal Conjugate point $ Q$\uff0cThen $ Q$ is also on the $ Curve\\minus{}L$ .\r\nIs the $ Thomson\\minus{}cubic$ have this feature\uff1f\r\n\r\nWho will have a consideration or a proof\uff1f", - "Solution_4": "Let the barycentric coordinates of $ P$ be $ (x,y,z)$. Let $ D$, $ E$, $ F$ be the projections of $ P$ onto side $ BC$, $ AC$, and $ AB$, respectively, and let the line through $ P$ parallel to $ BC$ intersect sides $ AB$ and $ AC$ at $ B'$ and $ C'$, respectively. Let the line through $ P$ parallel to $ AB$ and $ AC$ intersect $ BC$ at $ B''$ and $ C''$, respectively.\r\n\r\n[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/88647590aa1b1dd2dbda1d5a00d614ab.png[/img]\r\n\r\nThen $ PD \\equal{} x/a$, $ PE \\equal{} y/b$, and $ PF \\equal{} z/c$. (Actually, $ PD$ is proportional to $ x/a$. If you are really uncomfortable with this, let $ PD \\equal{} tx/a$, $ PE \\equal{} ty/b$, and $ PF \\equal{} tz/c$. In the end, all the variables $ t$ will cancel.) Then $ BB'' \\equal{} B'P \\equal{} z/(c \\sin B)$, and $ B''D \\equal{} (x \\cos B)/(a \\sin B)$, so\r\n\\[ BD \\equal{} \\frac{az \\plus{} cx \\cos B}{ac \\sin B},\\]\r\nand\r\n\\[ \\cot \\beta \\equal{} \\frac{BD}{PD} \\equal{} \\frac{az \\plus{} cx \\cos B}{cx \\sin B}.\\]\r\nSimilarly,\r\n\\[ CE \\equal{} \\frac{ay \\plus{} bx \\cos C}{ab \\sin C},\\]\r\nand\r\n\\[ \\cot p \\equal{} \\frac{ay \\plus{} bx \\cos C}{bx \\sin C}.\\]\r\n\r\nWe derive similar expressions for $ \\cot \\alpha$, $ \\cot \\gamma$, $ \\cot m$, and $ \\cot n$. Substituting into the equation $ \\cot \\alpha \\plus{} \\cot \\beta \\plus{} \\cot \\gamma \\equal{} \\cot m \\plus{} \\cot n \\plus{} \\cot p$, and expanding, it simplifies to\r\n\\[ x (c^2 y^2 \\minus{} b^2 z^2) \\plus{} y (a^2 z^2 \\minus{} c^2 x^2) \\plus{} z (b^2 x^2 \\minus{} a^2 y^2) \\equal{} 0,\\]\r\nwhich is the equation of the [url=http://pagesperso-orange.fr/bernard.gibert/Exemples/k002.html]Thomson cubic[/url].\r\n\r\nI wonder if any other [url=http://pagesperso-orange.fr/bernard.gibert/ctc.html]cubics[/url] have a similar property." -} -{ - "Problem": "find \r\n$ \\prod_{k \\equal{} 0}^{\\infty} \\frac {k^{k} \\minus{} 1}{k^{k} \\plus{} 1}$ - ?\r\n :huh: \r\nwhen I was solving this one I was playing with integrals like this $ \\int_{0}^{1}x^{ \\minus{} x}$ ..\r\nbut I can't get any closed answer.", - "Solution_1": "hello, i think it gives no closed answer.\r\nSonnhard." -} -{ - "Problem": "Let the numbers $ a,b,c$ be from interval $ [1;2]$. Prove the inequality:\r\n$ \\frac{a(b\\plus{}c)}{b^{2}\\plus{}c^{2}}\\plus{}\\frac{b(c\\plus{}a)}{c^{2}\\plus{}a^{2}}\\plus{}\\frac{c(a \\plus{} b)}{a^{2}\\plus{}b^{2}}<\\equal{}4.5$.", - "Solution_1": "[quote=\"lasha\"]Let the numbers $ a,b,c$ be from interval $ [1;2]$. Prove the inequality:\n$ \\frac{a(b \\plus{} c)}{b^{2} \\plus{} c^{2}} \\plus{} \\frac{b(c \\plus{} a)}{c^{2} \\plus{} a^{2}} \\plus{} \\frac{c(a \\plus{} b)}{a^{2} \\plus{} b^{2}} \\leq 4.5$.[/quote]" -} -{ - "Problem": "If $ a,b>0$, show that \r\n\r\n\\[ \\lim_{n\\rightarrow \\infty} \\sqrt{(n\\plus{}a)(n\\plus{}b)}\\minus{}n\\equal{}\\frac{(a\\plus{}b)}{2}\\]\r\n :)", - "Solution_1": "This is usually done by multiplying with $ \\frac {\\sqrt {(n \\plus{} a)(n \\plus{} b)} \\plus{} n}{\\sqrt {(n \\plus{} a)(n \\plus{} b)} \\plus{} n}$, but we can do this as well:\r\n\r\n$ \\text{L} \\equal{} \\lim _{n\\to 0} \\frac {\\sqrt {(1 \\plus{} a\\,n)(1 \\plus{} b \\, n)} \\minus{} 1}{n} \\equal{} ^{\\text{H}} \\lim_{n \\to 0} \\equal{} \\frac {2n \\plus{} a \\plus{} b}{2 \\sqrt {(1 \\plus{} a\\,n)(1 \\plus{} b \\, n)}} \\equal{} \\frac {a \\plus{} b}{2}$\r\n\r\n(The $ \\equal{} ^{\\text{H}}$ means the usage of L'Hospital's theorem)\r\n\r\nEDIT: How about this\r\n\r\n$ \\lim _{n \\to \\infty} \\left(\\sqrt [n]{\\prod^{n \\minus{} 1}_{i \\equal{} 0}(n \\plus{} c_i)} \\minus{} n\\right)$", - "Solution_2": "In fact, I need the solution for one of my friend in the $ \\epsilon$ language without using L'Hosopital's.\r\n\r\nthis question in [3.2, 12, pg 67 , Bartle and Sherbert , 3rd edition] ]Basic real analysis.\r\n\r\n---------\r\n\r\nIs the answer of your question is the mean on $ c_i$'s .?\r\n\r\n :)", - "Solution_3": "[quote=\"Hidden Scofield\"]If $ a,b > 0$, show that\n\\[ \\lim_{n\\rightarrow \\infty} \\sqrt {(n \\plus{} a)(n \\plus{} b)} \\minus{} n \\equal{} \\frac {(a \\plus{} b)}{2}\\]\n:)[/quote]\r\n$ T \\equal{} \\sqrt {(n \\plus{} a)(n \\plus{} b)} \\minus{} \\frac {a \\plus{} b \\plus{} 2n}{2} \\equal{} \\frac {4(n \\plus{} a)(n \\plus{} b) \\minus{} (a \\plus{} b \\plus{} 2n)^2}{4\\sqrt {(n \\plus{} a)(n \\plus{} b)} \\plus{} 2(a \\plus{} b \\plus{} 2n)}$\r\n$ \\equal{} \\minus{} \\frac {(a \\minus{} b)^2}{4\\sqrt {(n \\plus{} a)(n \\plus{} b)} \\plus{} 2(a \\plus{} b \\plus{} 2n)}$\r\n$ 4\\sqrt {(n \\plus{} a)(n \\plus{} b)} \\plus{} 2(a \\plus{} b \\plus{} 2n) \\ge 4n\\plus{}2a\\plus{}2b\\plus{}4n > n$\r\nSo: $ |T| < \\frac {(a \\minus{} b)^2}{n}$. And from here it is easy to see that $ T \\to 0$ as $ n \\to \\infty$", - "Solution_4": "[quote=\"Hidden Scofield\"]Is the answer of your question is the mean on $ c_i$'s .?[/quote]\r\n\r\nYes it is..", - "Solution_5": "Thank you :)", - "Solution_6": "hello, you can rewrite your term as\r\n$ \\frac{(n\\plus{}a)(n\\plus{}b)\\minus{}n^2}{\\sqrt{n^2\\plus{}na\\plus{}nb\\plus{}ab}\\plus{}n^2}\\equal{}$\r\n$ \\frac{a\\plus{}b\\plus{}\\frac{ab}{n}}{\\sqrt{1\\plus{}\\frac{a}{n}\\plus{}\\frac{b}{n}\\plus{}\\frac{ab}{n}}\\plus{}1}$\r\nand this tends to $ \\frac{a\\plus{}b}{2}$.\r\nSonnhard.", - "Solution_7": "In fact if $ p_1 \\plus{} ... \\plus{} p_m \\equal{}1$, then $ \\prod_{k\\equal{}1}^m (n\\plus{}a_k)^{p_k} \\minus{} n \\rightarrow p_1a_1 \\plus{} ...\\plus{} p_ma_m$ as $ n \\to \\infty$. You can allow any real numbers for $ a_k$ and $ p_k$ as long as $ p_1 \\plus{} ... \\plus{} p_m \\equal{}1$. To prove this, factor out an $ n^{p_k}$ from each term in the product to get $ n (\\prod_{k\\equal{}1}^m (1\\plus{}\\frac{a_k}{n})^{p_k}\\minus{}1)$, then use $ (1\\plus{}x)^p \\equal{} 1 \\plus{} px \\plus{} O(x^2)$ on each factor in the latter product. After multiplying this out the expression looks like $ n(1 \\plus{} (p_1a_1 \\plus{} ...\\plus{} p_ma_m)/n \\plus{} O(1/n^2) \\minus{} 1)$, which has the desired limit." -} -{ - "Problem": "Why is it not possible for a line in three-space to be represented by a scalar equation.", - "Solution_1": "That's, uh, not true, unless \"scalar equation\" means \"linear equation.\" For example, the graph of the solutions to the equation $ (x \\minus{} y)^2 \\plus{} (y \\minus{} z)^2 \\equal{} 0$ is of the line parametrized by $ x \\equal{} y \\equal{} z \\equal{} t$, $ t \\in \\mathbb{R}$, and probably you can figure out how to modify this trick to get a (quadratic) equation whose solutions are a given line of your choice. \r\n\r\nIf \"scalar equation\" does mean \"linear equation,\" then one way of phrasing the answer is to note that, generically speaking, the solution space of a system of $ k$ linear equations in $ n$ variables has dimension $ n \\minus{} k$ (each equation kills off one degree of freedom), and that a line is a 1-dimensional object in 3-dimensional space and so needs two linear equations in order to describe it." -} -{ - "Problem": "Solve in integers: \\[x(3y-5) = y^{2}+1\\]", - "Solution_1": "$y^{2}-3xy+5x+1=0\\iff y=\\frac{3x\\pm\\sqrt{9x^{2}-20x-4}}{2}$\r\n\r\n$9x^{2}-20x-4=t^{2}(t\\in\\mathbb{N}_{0})\\iff x=\\frac{20\\pm\\sqrt{544+36t^{2}}}{18}$\r\n\r\n$544+36t^{2}=u^{2}(u\\in\\mathbb{N}_{0})\\iff (u+6t)(u-6t)=2^{5}\\cdot 17$\r\n\r\nThe only pairs of divisors whose difference is divisible by $12$ (because $(u+6t)-(u-6t)=12t$) are $(68, 8)$ and $(136, 4)$. That gives $u_{1}=38, u_{2}=70$. Hence\r\n\r\n$x_{1,2}=\\frac{20\\pm 38}{18}, x_{3,4}=\\frac{20\\pm 70}{18}$. The only integer values of $x$ are $x=-1, x=5$. For those we get\r\n\r\n$y_{x=-1}=\\frac{-3\\pm\\sqrt{25}}{2}\\in\\{1,-4\\}$\r\n\r\n$y_{x=5}=\\frac{15\\pm\\sqrt{121}}{2}\\in\\{13,2\\}$\r\n\r\nThe solutions are $(-1,1),(-1,4),(5,13),(5,2)$", - "Solution_2": "$x(3y-5) = y^{2}+1\\ \\Longleftrightarrow\\ (3y-5)(9x-3y-5) = 34$. So $3y-5 \\in\\left\\{-34,-17,-2,-1,1,2,17,34\\right\\}$. And we're only looking for $y\\in\\mathbb{Z}$, so $y\\in\\{-4,1,2,13\\}$. We obtain $(x,y)\\in\\left\\{(-1,-4), (-1,1), (5,2), (5,13)\\right\\}$.\r\n\r\nJan, I think this was a question in Beersel, no? (IMO-game, so I don't know whether you got this question or not.)", - "Solution_3": "Indeed, this question is from the weekend in Beersel.\r\nI had the same reflex as Farenhajt, to solve it as a quadratic equation.\r\nBut Kurt, I really like your factorisation! Very nice and short solution...\r\n\r\nI know there is another solution, with divisibility. I tried to remember and reconstruct the proof, but I failed. \r\nI think the first trivial step was to notice that $x|y^{2}+1$, and then rewrite, and notice that $y|5x+1$, but I can't get any further than that.\r\n\r\nAnyone?", - "Solution_4": "[color=darkblue] Here is still another solution !\n\n$x=\\frac{y^{2}+1}{3y-5}\\in Z$ $\\Longrightarrow$ $3x=\\frac{3y^{2}+3}{3y-5}=y+\\frac{5y+3}{3y-5}\\in Z$ $\\Longrightarrow$ $\\frac{5y+3}{3y-5}\\in Z$ $\\Longrightarrow$ $\\frac{15y+9}{3y-5}=5+\\frac{34}{3y-5}\\in Z$ $\\Longrightarrow$\n\n$(3y-5)\\in\\{1,-1,2,-2,17,-17,34,-34\\}$ $\\Longleftrightarrow$ $3y\\in \\{6,4,7,3,22,-12,39,-29\\}$ $\\Longleftrightarrow$ $y\\in \\{2,1,-4,13\\}$,\n\ni.e. the solution is $\\boxed{\\ S\\ =\\ \\{\\ (5,2)\\ ;\\ (-1,1)\\ ;\\ (-1,-4)\\ ,\\ (5,13)\\ \\}\\ }\\ .$ But the \"Kurt Godel's solution \"broken\" me !\n\n[b]Remark.[/b] Between the my solution and the Kurt Godel's solution there is a connection :\n\n$x=\\frac{y^{2}+1}{3y-5}=$ $\\frac{1}{3}\\cdot \\frac{3y^{2}+3}{3y-5}=$ $\\frac{1}{3}\\cdot\\left(y+\\frac{5y+3}{3y-5}\\right)=$ $\\frac{1}{3}\\cdot\\left(y+\\frac{1}{3}\\cdot\\frac{15y+9}{3y-5}\\right)=$ $\\frac{1}{3}\\left[y+\\frac{1}{3}\\cdot \\left(5+\\frac{34}{3y-5}\\right)\\right]$ $\\Longrightarrow$\n\n$3\\cdot (3x-y)-5=$ $\\frac{34}{3y-5}$ $\\Longrightarrow$ $\\boxed{\\ (9x-3y-5)(3y-5)=34\\ }\\ !!$[/color]" -} -{ - "Problem": "I just got: \r\nAoPS Volume 1\r\nIntroduction to Counting & Probability\r\nIntroduction to Geometry\r\n\r\nI have a math contest similar to MathCounts on March 10th, and I wanted to know what is the best way to practice from these books (i.e. finish Geometry first, then Volume 1, or do a little of two).\r\n\r\nI am pretty confident in my basic Counting and Probability skills, so if there are some quick ways that are described in the C&P book, could you tell me about them?\r\n\r\nThanks for being my SparkNotes :P !", - "Solution_1": "actually do vol 1 first i think...it covers some of the topics in your other books (i think i just got aops vol 1 today :lol: ) then if u have any questions the other books might help" -} -{ - "Problem": "is there a number n, such that there is a number kn(k>1), where kn contains digits ONLY found in n.\r\n\r\nif so, what is the smallest possible value of n?", - "Solution_1": "um\r\n[hide]\n$ \\boxed{n\\equal{}0}$\n[/hide]", - "Solution_2": "Yeah, but I think there's another solution.", - "Solution_3": "If you are asking whether or not this is MATHCOUNTS level, then I think that it might be one of the last 5 of a chapter sprint.... as for the answer, I'm with alanchou", - "Solution_4": "It has more than one solution.\r\nExample, K is two and N is 7368421.\r\n$ 7368421\\times 2 \\equal{} 14736842$", - "Solution_5": "[quote=\"abacadaea\"]is there a number n, such that there is a number kn(k>1), where kn contains digits ONLY found in n.\n\nif so, what is the [u]smallest possible value[/u] of n?[/quote]", - "Solution_6": "Isnt 142857 and 2 work too? 142857*2 = 285714", - "Solution_7": "whoops i meant positive", - "Solution_8": "The smallest one is....\r\n\r\n10*10=100\r\n\r\nDo the solution and the multiple have to have the same number of each digit? :huh: \r\n\r\n :10:", - "Solution_9": "n is 1 and K is 11.\r\nHa.\r\nSmallest positive.", - "Solution_10": "um, possible, not positive :wink: \r\n\r\nI still think that this is mathcounts level, assuming $ 0$ is the answer", - "Solution_11": "21978*4=87912, I think that's the smallest non-trivial solution", - "Solution_12": "[quote=\"ZhangPeijin\"]n is 1 and K is 11.\nHa.\nSmallest positive.[/quote]\r\n\r\nThat is correct. +1 for you!\r\n\r\nWhen a question asks something about some smallest positive integer and the quickest method is guess and check, make sure you start at 1 (though most of the time, it's not 1). I'd say this is for Chapter Sprint Round maybe?" -} -{ - "Problem": "A few questions about the AMC12:\r\n\r\n1) After how much time are you usually done with the first 15 problems?\r\n\r\n2) After how much time are you done with the entire test?\r\n\r\n3) Do you check over every problem after you are done?", - "Solution_1": "For me, I can't finish the whole test in 75 minutes, so \r\n\r\n1) After how much time are you usually done with the first 15 problems? \r\n\r\nI'm generally done with the first 15 after 45 minutes. I probably got faster though.\r\n\r\n2) After how much time are you done with the entire test?\r\n\r\nLike I said, I'm usually not done with the entire test.\r\n\r\n3) Do you check over every problem after you are done?\r\n\r\nI usually check each problem after I finish it, and then I check it with about 3 minutes to go, regardless of how many problems are left. Double-checking helps prevent mistakes.", - "Solution_2": "I'll go by what I did last year. I promise I'll do better this year though. I didn't pace myself, so I barely finished the test on time. (i.e. didn't really, basically guessed/intuited answers to the last 5) and I didn't have time to check.", - "Solution_3": "Do you guys know [b]how[/b] to do every problem and spend the whole time trying to finish, or do you spend much of the time mulling over the questions?", - "Solution_4": "[quote]Do you guys know how to do every problem and spend the whole time trying to finish, or do you spend much of the time mulling over the questions?[/quote]\r\n\r\nEh, well, its been going weirdly for me lately. When I recently took the 1985 and 1986 AHSMEs, I managed to do all the problems with like 10 minutes left over to quickly run through and check my previous answers. However, I still made 3-4 careless errors on each of the tests. When I just did the AMC12B, I didn't manage to solve the last 4 problems within the 75min allotted. However, I didn't make a single careless mistakes on the ones I did solve though (even without going back to check any of them).", - "Solution_5": "I also cannot solve all problems within the allotted time. I'm a slow worker. As for the first 15 problems, I could probably get done in about 50 minutes. I don't really check my answers, but that doesn't affect me too much since I check as I go, which is one of the reasons for my slowness.\r\n\r\n-interesting_move", - "Solution_6": "when you guys check answers, do you do the problem, then immediately check it over (by doing it again), or do you do one problem, and continue to go on, then when there is x minutes left, you go back and cheeck?", - "Solution_7": "Well i'm not too sure about what sort of level this test is.. but it sounds like one with a reasonably large number of questions where you would \"expect\" (ok quite a few people wouldn't expect this but you know what I mean) to solve a lot of them. In that case, in those sort of exams I usually go the whole way through before checking, of course it depends on how quick you are.. however when checking, I don't just solve the problem again - you'll probably make the same mistake.. I find lots of other ways to solve problems, etc. Like working backwards, using approximations, I dunno all sorts of random stuff..", - "Solution_8": "MMM, it's 25 questions, 5 choices on each question, gets harder as the test goes on, scoring: 6 points for correct, 2.5 for omitted, 0 for wrong. The problems aren't very hard at all at first, but get significantly harder for the average highschool student by the end. Well, actually, I'm sure a lot or maybe even most highschool students would be struggling within the first five, but they aren't bad if you have any idea what you're doing. [url=http://www.math.uncc.edu/~hbreiter/AHSME/1997st.pdf]here's[/url] a sample test that you can look at. It's from 1997.", - "Solution_9": "i usually just go a little slower around the first time so im pretty confident in my answers", - "Solution_10": "[quote=\"confuted\"]MMM, it's 25 questions, 5 choices on each question, gets harder as the test goes on, scoring: 6 points for correct, 2.5 for omitted, 0 for wrong. The problems aren't very hard at all at first, but get significantly harder for the average highschool student by the end. Well, actually, I'm sure a lot or maybe even most highschool students would be struggling within the first five, but they aren't bad if you have any idea what you're doing. [url=http://www.math.uncc.edu/~hbreiter/AHSME/1997st.pdf]here's[/url] a sample test that you can look at. It's from 1997.[/quote]\r\n\r\nCool. We have a similar style exam over here, called the australian maths competition (not just for australians tho). Multichoice, lose points if you get some wrong instead of just leaving blank etc. Always been my favourite competition.." -} -{ - "Problem": "Let $m,n$ be positive integers and $S_{m}(n)=\\sum^{n}_{k=1}k^{m}$.\r\nProve that the number\r\n$(n+1)^{m+1}-(C^{2}_{m+1}S_{m-1}(n)+C^{3}_{m+1}S_{m-2}(n)+...+C^{m}_{m+1}S_{1}(n)+2)$\r\nis divisible by $m+1$.( where $C^{k}_{n}=\\frac{n!}{k!(n-k)!}$ )", - "Solution_1": "i give an example and after i'll generalize:\r\n\r\n$(1+1)^{3}=C^{0}_{3}1^{3}+C^{1}_{3}1^{2}*1+C^{2}_{3}1*1^{2}+C^{3}_{3}1^{3}$\r\n$...$\r\n$(n+1)^{3}=C^{0}_{3}n^{3}+C^{1}_{3}n^{2}*1+C^{2}_{3}n*1^{2}+C^{3}_{3}1^{3}$\r\naddictioning the relations we obtain\r\n$(n+1)^{3}=1+C^{1}_{3}S_{2}(n)+C^{2}_{3}S_{1}(n)+n$\r\nin general we have\r\n$(n+1)^{m+1}=1+C^{1}_{m+1}S_{m}(n)+...+C^{m}_{m+1}S_{1}(n)+n$\r\nand so\r\n$LHS=n-1+(m+1)S_{m}(n)$\r\nbut n-1 is not divisible by m+1.where is the mistake? thanks :(", - "Solution_2": "Oh sorry dude :oops: , i made a mistake, you are right, we have\r\nwe have:\r\n$\\sum^{n}_{k=1}\\left[(k+1)^{m+1}-k^{m+1}\\right]=\\sum^{n}_{k=1}\\left[C^{1}_{m+1}k^{m}+C^{2}_{m+1}k^{m-1}+...+C^{m}_{m+1}k+1\\right]$\r\nand so:\r\n$(n+1)^{m+1}-1=C^{1}_{m+1}S_{m}(n)+C^{2}_{m+1}S_{m-1}(n)+...+C^{m}_{m+1}S_{1}(n)+n$\r\n$\\Longrightarrow S_{m}(n)=\\frac{(n+1)^{m+1}-\\left[C^{2}_{m+1}S_{m-1}(n)+...+C^{m}_{m+1}S_{1}(n)+n+1\\right]}{m+1}$\r\nand because we know that $S_{m}(n)$ is integer, so \r\n$(n+1)^{m+1}-\\left[C^{2}_{m+1}S_{m-1}(n)+...+C^{m}_{m+1}S_{1}(n)+C^{m+1}_{m+1}S_{0}(n+1)\\right]$\r\nmust be divisible by $m+1$.( i put $n+1=S_{0}(n+1)$ )\r\nseems that I can't edit the problem now :( .", - "Solution_3": "i'm anyway satysfied. this is my first solution in this very difficult forum (all are very good solvers!). thanks sinajackson :lol:", - "Solution_4": "that's :coolspeak: I'm happy toooo :D", - "Solution_5": "good job girino!!\r\nyou are a very good mathlover!" -} -{ - "Problem": "$a+b+c=3,a,b,c \\ge 0$\r\nprove:\r\n$\\sqrt{3-ab}+\\sqrt{3-bc}+\\sqrt{3-ca} \\ge 3 \\sqrt 2$", - "Solution_1": "This problem is easily tackled using Jensen:\r\n\r\n$\\frac{\\sum_{\\text{sym}} \\sqrt{3 - ab}}{3} \\geq \\sqrt{\\frac{9 - ab - bc - ac}{3}}$\r\n\r\nNow note that by Cauchy-Schwartz, we have\r\n\r\n$(a^2 + b^2 + c^2)^2 \\geq(ab + bc + ac)^2 \\Rightarrow a^2 + b^2 + c^2 \\geq ab + bc + ac$\r\n$\\Rightarrow (a + b + c)^2 - 2(ab + bc + ac) \\geq ab + bc + ac \\Rightarrow 3 \\geq ab + bc + ac$\r\n\r\nUtilizing this in our original equation, we have\r\n\r\n$\\frac{\\sum_{\\text{cyc}} \\sqrt{3 - ab}}{3} \\geq \\sqrt{\\frac{9 - ab - bc - ac}{3}} \\geq \\sqrt{\\frac{9 - 3}{3}} = \\sqrt{2}$\r\n\r\ngiving us\r\n\r\n$\\sqrt{3-ab}+\\sqrt{3-bc}+\\sqrt{3-ca} \\ge 3 \\sqrt 2$", - "Solution_2": "[quote=\"zanttrang\"]This problem is easily tackled using Jensen:\n\n$\\frac{\\sum_{\\text{sym}} \\sqrt{3 - ab}}{3} \\geq \\sqrt{\\frac{9 - ab - bc - ac}{3}}$\n\n[/quote]\r\n\r\n\r\nYou're wrong because It's reveser.", - "Solution_3": "[quote=\"hungkhtn\"][quote=\"zanttrang\"]This problem is easily tackled using Jensen:\n\n$\\frac{\\sum_{\\text{sym}} \\sqrt{3 - ab}}{3} \\geq \\sqrt{\\frac{9 - ab - bc - ac}{3}}$\n\n[/quote]\n\n\nYou're wrong because It's reveser.[/quote]\r\nErgh, yeah, you're right - darn :P I guess I should be more careful when applying Jensen... it's funny how often applying Jensen the wrong way gives the desired result.", - "Solution_4": "Edit: nevermind, that method doesn't work either....", - "Solution_5": "OK. This's false.! Just take c=0.", - "Solution_6": "[quote=\"hungkhtn\"]OK. This's false.! Just take c=0.[/quote]\r\nHow does that provide a counter-example? I've failed miserably so far to prove this inequality, but I think it is true. I think AM-GM might help, but you have to prove that\r\n\r\n$(3 - ab)(3 - ac)(3 - bc) \\geq 8$, and I can't figure out how to do that, even after expanding and playing around with the numbers...", - "Solution_7": "[quote=\"zanttrang\"]\n\n$(3 - ab)(3 - ac)(3 - bc) \\geq 8$, and I can't figure out how to do that, even after expanding and playing around with the numbers...[/quote]\r\nIt's wrong.When $a=b=\\frac{3}{2},c=0$ then $(3-ab)(3-bc)(3-ca)=\\frac{27}{4}<8$", - "Solution_8": "[quote=\"Nameless\"][quote=\"zanttrang\"]\n\n$(3 - ab)(3 - ac)(3 - bc) \\geq 8$, and I can't figure out how to do that, even after expanding and playing around with the numbers...[/quote]\nIt's wrong.When $a=b=\\frac{3}{2},c=0$ then $(3-ab)(3-bc)(3-ca)=\\frac{27}{4}<8$[/quote]\r\nYeah, you're right - no wonder I couldn't prove it.", - "Solution_9": "For $0\\leq x\\leq 1,\\ \\sqrt{3-x}\\geq -(\\sqrt{3}-\\sqrt{2})x+\\sqrt{3}\\geq \\sqrt{2}.$\r\n\r\nFor $1\\leq x\\leq \\frac{9}{4},\\ \\sqrt{3-x}\\geq -\\frac{2\\sqrt{3}}{9}x+\\sqrt{3}.$", - "Solution_10": "I got it! :D\r\n\r\nWe are to prove\r\n\r\n\\[ x^2+y^2+z^2\\geq 6\\ (x\\geq 0,\\ y\\geq 0,\\ z\\geq 0)\\Longrightarrow x+y+z\\geq 3\\sqrt{2}. \\]", - "Solution_11": "[quote=\"kunny\"]I got it! :D\nWe are to prove\n\\[ x^2+y^2+z^2\\geq 6\\ (x\\geq 0,\\ y\\geq 0,\\ z\\geq 0)\\Longrightarrow x+y+z\\geq 3\\sqrt{2}. \\][/quote]\r\nTry $x=y=0,z=\\sqrt6$ ;)", - "Solution_12": "O.K. but, anyway, here is my solution.\r\n\r\nLet $x=\\sqrt{3-ab},\\ y=\\sqrt{3-bc}, z=\\sqrt{3-ca}$ we have $x^2+y^2+z^2=9-(ab+bc+ca).$\r\nThus $(a+b+c)^2\\geq 3(ab+bc+ca)\\Longleftrightarrow 3^2\\geq 3(ab+bc+ca),$ since $a\\geq 0,\\ b\\geq 0,\\ c\\geq 0,$ we obtain $0\\leq ab+bc+ca\\leq 3.$ \r\n\r\nThen since $c\\geq 0,$ we have $\\sqrt{ab}\\leq\\frac{a+b}{2}\\leq \\frac{3}{2},$ analogously $\\sqrt{bc}\\leq \\frac{3}{2}, \\sqrt{ca}\\leq \\frac{3}{2},$ thus \r\n\r\n$6\\leq x^2+y^2+z^2\\leq 9 \\left(x\\geq \\frac{\\sqrt{3}}{2}, y\\geq \\frac{\\sqrt{3}}{2}, z\\geq \\frac{\\sqrt{3}}{2}\\right)\\ \\cdots (*).$\r\n\r\nConsider for $xyz-space,$ two spheres $x^2+y^2+z^2=6\\ and\\ x^2+y^2+z^2=9\\ and\\ x\\geq \\frac{\\sqrt{3}}{2}, y\\geq \\frac{\\sqrt{3}}{2}, z\\geq \\frac{\\sqrt{3}}{2}.$\r\n\r\nAssumed that $x+y+z$ takes the value of $k$ for $(*).$ the plane $x+y+z=k$ are to have the intersection points with this domain,\r\n\r\nThus $\\sqrt{6}\\leq\\text{the distance between the center of the sphere and the plane}\\ x+y+z-k=0 \\leq 3.$\r\n\r\n\r\n$\\Longleftrightarrow\\sqrt{6}\\leq \\frac{|0+0+0-k|}{\\sqrt{1^2+1^2+1^2}}\\leq 3$, yielding $\\sqrt{6}\\leq \\frac{|-k|} {\\sqrt{3}}\\leq 3.$ Since $k\\geq 0,$ yielding $3\\sqrt{2}\\leq k \\leq 3\\sqrt{3}.$\r\n\r\nTherefore $3\\sqrt{2}\\leq \\sqrt{3-ab}+\\sqrt{3-bc}+\\sqrt{3-ca}\\leq 3\\sqrt{3}.$", - "Solution_13": "[quote=\"kunny\"]O.K. but, anyway, here is my solution.\n\nLet $x=\\sqrt{3-ab},\\ y=\\sqrt{3-bc}, z=\\sqrt{3-ca}$ we have $x^2+y^2+z^2=9-(ab+bc+ca).$\nThus $(a+b+c)^2\\geq 3(ab+bc+ca)\\Longleftrightarrow 3^2\\geq 3(ab+bc+ca),$ since $a\\geq 0,\\ b\\geq 0,\\ c\\geq 0,$ we obtain $0\\leq ab+bc+ca\\leq 3.$ \n\nThen since $c\\geq 0,$ we have $\\sqrt{ab}\\leq\\frac{a+b}{2}\\leq \\frac{3}{2},$ analogously $\\sqrt{bc}\\leq \\frac{3}{2}, \\sqrt{ca}\\leq \\frac{3}{2},$ thus \n\n$6\\leq x^2+y^2+z^2\\leq 9 \\left(x\\geq \\frac{\\sqrt{3}}{2}, y\\geq \\frac{\\sqrt{3}}{2}, z\\geq \\frac{\\sqrt{3}}{2}\\right)\\ \\cdots (*).$\n\nConsider for $xyz-space,$ two spheres $x^2+y^2+z^2=6\\ and\\ x^2+y^2+z^2=9\\ and\\ x\\geq \\frac{\\sqrt{3}}{2}, y\\geq \\frac{\\sqrt{3}}{2}, z\\geq \\frac{\\sqrt{3}}{2}.$\n\nAssumed that $x+y+z$ takes the value of $k$ for $(*).$ the plane $x+y+z=k$ are to have the intersection points with this domain,\n\nThus $\\sqrt{6}\\leq\\text{the distance between the center of the sphere and the plane}\\ x+y+z-k=0 \\leq 3.$\n\n\n$\\Longleftrightarrow\\sqrt{6}\\leq \\frac{|0+0+0-k|}{\\sqrt{1^2+1^2+1^2}}\\leq 3$, yielding $\\sqrt{6}\\leq \\frac{|-k|} {\\sqrt{3}}\\leq 3.$ Since $k\\geq 0,$ yielding $3\\sqrt{2}\\leq k \\leq 3\\sqrt{3}.$\n\nTherefore $3\\sqrt{2}\\leq \\sqrt{3-ab}+\\sqrt{3-bc}+\\sqrt{3-ca}\\leq 3\\sqrt{3}.$[/quote]\r\nkunny,can I ask why $x\\geq \\frac{\\sqrt{3}}{2}, y\\geq \\frac{\\sqrt{3}}{2}, z\\geq \\frac{\\sqrt{3}}{2}.$\r\nand $\\sqrt{6}\\leq\\text{the distance between the center of the sphere and the plane}\\ x+y+z-k=0 \\leq 3.$?\r\nthanks", - "Solution_14": "More general, the following inequality holds for $p \\geq \\frac{81}{32}$:\r\n$\\sqrt{p-ab}+\\sqrt{p-bc}+\\sqrt{p-ca} \\ge 3 \\sqrt{p-1}$\r\n\r\nA possible solution uses EV-Theorem.", - "Solution_15": "Long time ago,and anybody have a solution without n-1 principle theorem?", - "Solution_16": "Can you solve the first ineq:\r\n$\\sqrt{3-ab}+\\sqrt{3-bc}+\\sqrt{3-ca} \\ge 3 \\sqrt 2$\r\nI can't understand kunny's solution", - "Solution_17": "I can bet kunny's solution is certainly wrong ;)\r\nBut I also didn't have a solution.", - "Solution_18": "I find the solution today by accident,hope you can like it :D .\r\nBecause we have:\r\n$ \\sqrt {3 \\minus{} ab} \\equal{} \\frac {(6 \\minus{} 2ab)\\sqrt 2}{2\\sqrt {2(3 \\minus{} ab)}}\\ge \\frac {(6 \\minus{} 2ab)\\sqrt 2}{2 \\plus{} 3 \\minus{} ab} \\equal{} \\frac {(6 \\minus{} 2ab)\\sqrt 2}{5 \\minus{} ab}$\r\nSo we only need to prove:\r\n$ \\frac {3 \\minus{} ab}{5 \\minus{} ab} \\plus{} \\frac {3 \\minus{} bc}{5 \\minus{} bc} \\plus{} \\frac {3 \\minus{} ca}{5 \\minus{} ca} \\ge \\frac {3}{2}$\r\nWhich is also:\r\n$ \\frac {1}{5 \\minus{} ab} \\plus{} \\frac {1}{5 \\minus{} bc} \\plus{} \\frac {1}{5 \\minus{} ca} \\le \\frac {3}{4}$\r\nThis is the special case of Vasc's inequality,see \r\n[url]http://www.mathlinks.ro/viewtopic.php?t=18627&search_id=1516423496[/url]\r\n(Let d=1)Or,you can also prove it directly.\r\n\r\n\r\nBTW,Kunny's solution is wrong because:\r\nthe plane $ x \\plus{} y \\plus{} z \\equal{} k$ and the ball $ x^2 \\plus{} y^2 \\plus{} z^2 \\equal{} 6$ have the intersection points.\r\ndoesn't imply \r\n$ \\sqrt {6}\\leq\\text{the distance between the center of the sphere and the plane}\\ x \\plus{} y \\plus{} z \\minus{} k \\equal{} 0$", - "Solution_19": "[quote=\"zhaobin\"]I find the solution today by accident,hope you can like it :D .\nBecause we have:\n$ \\sqrt {3 \\minus{} ab} \\equal{} \\frac {(6 \\minus{} 2ab)\\sqrt 2}{2\\sqrt {2(3 \\minus{} ab)}}\\ge \\frac {(6 \\minus{} 2ab)\\sqrt 2}{2 \\plus{} 3 \\minus{} ab} \\equal{} \\frac {(6 \\minus{} 2ab)\\sqrt 2}{5 \\minus{} ab}$\nSo we only need to prove:\n$ \\frac {3 \\minus{} ab}{5 \\minus{} ab} \\plus{} \\frac {3 \\minus{} bc}{5 \\minus{} bc} \\plus{} \\frac {3 \\minus{} ca}{5 \\minus{} ca} \\ge \\frac {3}{2}$\nWhich is also:\n$ \\frac {1}{5 \\minus{} ab} \\plus{} \\frac {1}{5 \\minus{} bc} \\plus{} \\frac {1}{5 \\minus{} ca} \\le \\frac {3}{4}$\nThis is the special case of Vasc's inequality,see \n[url]http://www.mathlinks.ro/viewtopic.php?t=18627&search_id=1516423496[/url]\n(Let d=1)Or,you can also prove it directly.\n\n\nBTW,Kunny's solution is wrong because:\nthe plane $ x \\plus{} y \\plus{} z \\equal{} k$ and the ball $ x^2 \\plus{} y^2 \\plus{} z^2 \\equal{} 6$ have the intersection points.\ndoesn't imply \n$ \\sqrt {6}\\leq\\text{the distance between the center of the sphere and the plane}\\ x \\plus{} y \\plus{} z \\minus{} k \\equal{} 0$[/quote]\r\n\r\n\r\nThe inequality\r\n\r\n$ \\frac{1}{5\\minus{}ab}\\plus{} \\frac{1}{5\\minus{}bc}\\plus{} \\frac{1}{5\\minus{}ca} \\le \\frac{3}{4}$\r\n\r\nis wrong for $ a\\equal{}0,b\\equal{}c\\equal{}\\frac{3}{2}$ :wink:", - "Solution_20": "I have just deleted, because my reasoning was wrong, the deleted post, you will see shalex's post 2 posts forward.\r\n\r\nkunny", - "Solution_21": "[quote=\"zhaobin\"]$ a \\plus{} b \\plus{} c \\equal{} 3,a,b,c \\ge 0$\nprove:\n$ \\sqrt {3 \\minus{} ab} \\plus{} \\sqrt {3 \\minus{} bc} \\plus{} \\sqrt {3 \\minus{} ca} \\ge 3 \\sqrt 2$[/quote]\r\nWith the same conditions the following inequality is true too:\r\n\\[ \\sqrt {81 \\minus{} 32ab} \\plus{} \\sqrt {81 \\minus{} 32ac} \\plus{} \\sqrt {81 \\minus{} 32bc}\\geq21\r\n\\]", - "Solution_22": "[quote=\"kunny\"][quote=\"zhaobin\"]$ a \\plus{} b \\plus{} c \\equal{} 3,a,b,c \\ge 0$\nprove:\n$ \\sqrt {3 \\minus{} ab} \\plus{} \\sqrt {3 \\minus{} bc} \\plus{} \\sqrt {3 \\minus{} ca} \\ge 3 \\sqrt 2$[/quote]\n\nLet $ abc$ fix as $ abc \\equal{} k$, $ f(x): \\equal{} \\sqrt {3 \\minus{} \\frac {k}{x}} \\minus{} \\frac {k}{2\\sqrt {3 \\minus{} k}}\\ln x$ for $ x > 0$.\n\n$ f'(x) \\equal{} \\frac {k(x \\minus{} 1)(3x \\plus{} 3 \\minus{} k)}{2x\\sqrt {3 \\minus{} k}\\sqrt {3x^2 \\minus{} kx}(\\sqrt {3x^2 \\minus{} kx} \\plus{} \\sqrt {{3 \\minus{} k}})}$, yielding\n\n$ \\therefore f'(x) \\equal{} 0\\Longrightarrow x \\equal{} 1,\\ \\frac {k}{3} \\minus{} 1$, by A.M.-G.M. , $ 0\\leq abc\\leq \\left(\\frac {a \\plus{} b \\plus{} c}{3}\\right)^3\\Longleftrightarrow 0\\leq k\\leq 1$, yielding\n\n$ \\minus{} 1\\leq \\frac {k}{3} \\minus{} 1\\leq \\minus{} \\frac {2}{3}$. Thus we have $ f(x)\\geq f(1) \\equal{} \\sqrt {3 \\minus{} abc}\\geq \\sqrt {2}$.\n\n$ \\therefore f(a) \\plus{} f(b) \\plus{} f(c)\\geq 3\\sqrt {2}\\Longleftrightarrow \\sqrt {3 \\minus{} ab} \\plus{} \\sqrt {3 \\minus{} bc} \\plus{} \\sqrt {3 \\minus{} ca} \\ge 3 \\sqrt 2$. Q.E.D.[/quote]\r\n\r\nI don't understand why\r\n\r\n$ f(a) \\plus{} f(b) \\plus{} f(c) \\ge 3\\sqrt {2}\\Longleftrightarrow \\sqrt {3 \\minus{} ab} \\plus{} \\sqrt {3 \\minus{} bc} \\plus{} \\sqrt {3 \\minus{} ca}\\ge 3\\sqrt {2}$\r\n\r\nAnyway, I think\r\n\r\n$ f'(x) \\equal{} \\frac {k(1 \\minus{} x)(3x \\plus{} 3 \\minus{} k)}{2x\\sqrt {3 \\minus{} k}\\sqrt {3x^2 \\minus{} kx}(\\sqrt {3x^2 \\minus{} kx} \\plus{} \\sqrt {{3 \\minus{} k}})}$\r\n\r\nAnd $ f$ reach the maximum at the point $ x \\equal{} 1$, but not minimum.", - "Solution_23": "$ 0 < abc \\leq 1\\Longrightarrow \\ln abc < 0$, \r\n\r\n$ f(a) \\plus{} f(b) \\plus{} f(c) \\equal{} \\sqrt {3 \\minus{} ab} \\plus{} \\sqrt {3 \\minus{} bc} \\plus{} \\sqrt {3 \\minus{} ca} \\minus{} \\frac {abc}{2\\sqrt {3 \\minus{} abc}}\\ln abc$\r\n\r\n$ \\geq \\sqrt {3 \\minus{} ab} \\plus{} \\sqrt {3 \\minus{} bc} \\plus{} \\sqrt {3 \\minus{} ca}\\geq 3\\sqrt {2}$.\r\n\r\n[quote=\"shalex\"]\n$ f'(x) \\equal{} \\frac {k(1 \\minus{} x)(3x \\plus{} 3 \\minus{} k)}{2x\\sqrt {3 \\minus{} k}\\sqrt {3x^2 \\minus{} kx}(\\sqrt {3x^2 \\minus{} kx} \\plus{} \\sqrt {{3 \\minus{} k}})}$\n\nAnd $ f$ reach the maximum at the point $ x \\equal{} 1$, but not minimum.[/quote]\r\n\r\nWhose solution is this? :wink:", - "Solution_24": "[quote=\"kunny\"]$ 0 < abc \\leq 1\\Longrightarrow \\ln abc < 0$, \n\n$ f(a) \\plus{} f(b) \\plus{} f(c) \\equal{} \\sqrt {3 \\minus{} ab} \\plus{} \\sqrt {3 \\minus{} bc} \\plus{} \\sqrt {3 \\minus{} ca} \\minus{} \\frac {abc}{2\\sqrt {3 \\minus{} abc}}\\ln abc$\n\n$ \\geq \\sqrt {3 \\minus{} ab} \\plus{} \\sqrt {3 \\minus{} bc} \\plus{} \\sqrt {3 \\minus{} ca}\\geq 3\\sqrt {2}$.\n\n[quote=\"shalex\"]\n$ f'(x) \\equal{} \\frac {k(1 \\minus{} x)(3x \\plus{} 3 \\minus{} k)}{2x\\sqrt {3 \\minus{} k}\\sqrt {3x^2 \\minus{} kx}(\\sqrt {3x^2 \\minus{} kx} \\plus{} \\sqrt {{3 \\minus{} k}})}$\n\nAnd $ f$ reach the maximum at the point $ x \\equal{} 1$, but not minimum.[/quote]\n\nWhose solution is this? :wink:[/quote]\r\n\r\nDon't you see that\r\n\r\n$ f(a)\\plus{}f(b)\\plus{}f(c) \\ge 3\\sqrt{2}$\r\n\r\n(which your proved) is weaker than\r\n\r\n$ \\sqrt{3\\minus{}ab}\\plus{} \\sqrt{3\\minus{}bc}\\plus{} \\sqrt{3\\minus{}ca} \\ge 3\\sqrt{2}$?\r\n\r\nBecause $ f(a)\\plus{}f(b)\\plus{}f(c) \\ge 3\\sqrt{2}$ is equivalent to\r\n\r\n$ \\sqrt{3\\minus{}ab}\\plus{} \\sqrt{3\\minus{}bc}\\plus{} \\sqrt{3\\minus{}bc} \\ge 3\\sqrt{2}\\plus{}\\frac{abc}{2\\sqrt{3\\minus{}abc}}\\ln abc$\r\n\r\nBut the right side is less than or equals to $ 3\\sqrt{2}$\r\n\r\nAnyway I suggest you to check $ f'(x)$ by hand or with Mathematica or something else :wink:", - "Solution_25": "[b]shalex[/b], I'm very sorry, I would have to appologize to you, I admit that my calculation was wrong.\r\nPlease forgive my some critics to you.\r\n\r\nkunny", - "Solution_26": "That's alright :lol: \r\nNow we should concentrate on how to prove it in a good way...\r\nIt seems not a easy one.", - "Solution_27": "Thank you, :oops: I have tackled with several solution so far, not solved. Indeed how difficult the problem is.", - "Solution_28": "[quote=\"shalex\"]\n\n\nThe inequality\n\n$ \\frac {1}{5 \\minus{} ab} \\plus{} \\frac {1}{5 \\minus{} bc} \\plus{} \\frac {1}{5 \\minus{} ca} \\le \\frac {3}{4}$\n\nis wrong for $ a \\equal{} 0,b \\equal{} c \\equal{} \\frac {3}{2}$ :wink:[/quote]\r\nThank you,I am wrong :blush: \r\n\r\narqady,Do you have a solution?", - "Solution_29": "[quote=\"zhaobin\"]\n\narqady,Do you have a solution?[/quote]\r\nOf mine inequality? Yes, and very nice!", - "Solution_30": "Can you post it? \r\n\r\nBTW,\r\nDo you have some idea of mine?", - "Solution_31": "I'm eager to see your solution.\r\nCould you post it? Thanks a lot! :lol:", - "Solution_32": "[quote=\"zhaobin\"]Can you post it? [/quote]\nI have sent it to Crux a year ago. :wink: \nI am sorry, shalex and zhaobin.\n[quote=\"zhaobin\"]BTW,\nDo you have some idea of mine?[/quote]\r\nI think, other idea and without Vasc's EV-Theorem.:wink:" -} -{ - "Problem": "Let $f: N\\to R$ satify: $f(n+1)+f(n-1)=k.f(n)$ for all natural number $n$.\r\nFinl all values of $k$ so that $f$ ia period function", - "Solution_1": "it is problem in MOSP 1996", - "Solution_2": "what about $f(n)=0$??", - "Solution_3": "I think $f(n)=0$ is only special case for problem with all k belong to R. Problem will be harder if $f$ is general function that satified codition above.", - "Solution_4": "oh , it is $f: N^*\\to R$\r\nsolution $|k|\\leq 2$", - "Solution_5": "Dear Dieuhuynh, Can you post cpmletely solution?", - "Solution_6": "It's a sequense problem in fact.\r\n$a_{n+1}=ka_n-a_{n-1}$\r\nIf $|k|>2$,easy to find when $n$ infinite big,we can find a such sequence let $|a_{n+1}|>a_n$.So it's not period.\r\nWhen $|k|<2$ ,$k=2\\cos\\alpha$,\r\nThen we know the roots of $x^2-kx+1=0$ is $\\cos\\alpha+\\mbox{i}\\sin\\alpha,\\cos\\alpha-\\mbox{i}\\sin\\alpha$\r\nSo $a_n=A(\\cos\\alpha+\\mbox{i}\\sin\\alpha)^n+B(\\cos\\alpha-\\mbox{i}\\sin\\alpha)^n=A(\\cos n\\alpha+\\mbox{i}\\sin n\\alpha)+B(\\cos n\\alpha-\\mbox{i}\\sin n\\alpha)$ is real.\r\nSo $A=B$.$a_n=2A\\cos n\\alpha$.\r\nSo $\\{a_n\\}$ is period if and only if for some $m$,$\\cos (m+k)\\alpha =\\cos k\\alpha$ is hold for all $k$.\r\nThen we know $\\sin m\\alpha=0$.", - "Solution_7": "[quote=\"jin\"]It's a sequense problem in fact.\n$a_{n+1}=ka_{n}-a_{n-1}$\nIf $|k|>2$,easy to find when $n$ infinite big,we can find a such sequence let $|a_{n+1}|>a_{n}$.So it's not period.\n[/quote]\r\nWhy do you derive the fact?" -} -{ - "Problem": "If $\\theta=\\frac{\\pi}{2n+1}$ , $n\\in\\mathbb{N}$ , prove or disprove that \r\n\r\n$\\prod_{i=1}^n\\tan i\\theta=\\sqrt{2n+1}$", - "Solution_1": "At least it can be proved this way :\r\n\r\n$\\tan (\\pi - x) = - \\tan x$ so if we can compute $\\prod_{i=1}^{2n}\\tan i\\theta$ we are done.\r\n\r\n$(\\cos \\theta + i \\sin \\theta)^{2n+1} = 1$\r\nExpanding with binomial, dividing by $(cos \\theta)^{2n+1}$, we obtain an equation whose every roots are the $\\tan \\theta$.\r\nUsing Vi\u00e8ta relations gives $\\prod_{i=1}^{2n}\\tan i\\theta = \\binom{2n+1}{1} = 2n+1$", - "Solution_2": "can you further explain how binomial expansion and vieta formula work here ? I cant seem to get you :blush:", - "Solution_3": "[quote=\"shyong\"]If $\\theta=\\frac{\\pi}{2n+1}$ , $n\\in\\mathbb{N}$ , prove or disprove that \n\n$\\prod_{i=1}^n\\tan i\\theta=\\sqrt{2n+1}$[/quote]\r\n\r\nI would like to give you the source, where you can find the information required to solve this problem.\r\n\r\nYou can refer to the book \r\n\"Selected Problems and Theorems in Elementary Mathematics\" by\r\nD. O. Shklyarsky\r\nN. N. Chentsov\r\nI. M. Yaglom\r\n\r\nMir Publishers, Moscow.\r\n\r\nIn fact, This problem is solved in the book." -} -{ - "Problem": "I have never used Asymptote so it would be nice to see a complete example to draw the following type of diagrams:\r\n\r\nOn $ x$-axis, there are all integers $ 1,\\ldots, 10$ and on $ y$-axis, there are integers, say $ 50,\\ldots,300$. Then there are data points $ (i,y_i)$ where $ 50\\leq y_i\\leq 300$ and line segments from $ (i,y_i)$ to $ (i \\plus{} 1,y_{i \\plus{} 1})$ for all $ i\\in \\{ 1,\\ldots,9\\}$.\r\n\r\nAlso, I would like to know is the line segment diagram correct term for the diagram of that type.", - "Solution_1": "Is this what you were looking for?\r\n\r\n[asy]size(200, IgnoreAspect);\nint numberOfPoints = 10;\npair[] points;\npoints.push((1,77));\npoints.push((2,103));\npoints.push((3,94));\npoints.push((4,126));\npoints.push((5,155));\npoints.push((6,216));\npoints.push((7,195));\npoints.push((8,255));\npoints.push((9,240));\npoints.push((10,277));\n\nxaxis(0, 10, Ticks(true, true, 0, 0, 5, 1, 5, 3));\nyaxis(0, 300, Ticks(true, true, 0, 0, 50, 25, 5, 3));\n\nfor(int i = 0; i < numberOfPoints - 1; ++i) {\ndraw(points[i]--points[i+1],green+linewidth(1));\n}\nfor(int i = 0; i < numberOfPoints; ++i) {\ndot(points[i],blue+linewidth(3));\n}[/asy]\r\n\r\n[code]size(200, IgnoreAspect);[/code]\nThis code sets the size of the image, and allows the y-axis to be a different scale than the x-axis\n\n[code]int numberOfPoints = 10;\npair[] points;\npoints.push((1,77));\npoints.push((2,103));\npoints.push((3,94));\npoints.push((4,126));\npoints.push((5,155));\npoints.push((6,216));\npoints.push((7,195));\npoints.push((8,255));\npoints.push((9,240));\npoints.push((10,277));[/code]\nThis creates the an array of data points (like in a programming language). I just made up some values.\n\n[code]xaxis(0, 10, Ticks(true, true, 0, 0, 5, 1, 5, 3));\nyaxis(0, 300, Ticks(true, true, 0, 0, 50, 25, 5, 3));[/code]\nThis draws the x- and y-axes.\n\n[code]for(int i = 0; i < numberOfPoints - 1; ++i) {\ndraw(points[i]--points[i+1],green+linewidth(1));\n}\nfor(int i = 0; i < numberOfPoints; ++i) {\ndot(points[i],blue+linewidth(3));\n}[/code]\r\nFinally, this draws the line segments (first for loop) and creates dots on the data points (second for loop).", - "Solution_2": "[quote=\"MathWise\"]Is this what you were looking for?\n[/quote]\r\nAlmost. Thanks! But $ x$-axis should start from $ 1$ and $ y$-axis from $ 50$ instead of $ 0$. Also, can I name x-axis as \"days\" and y-axis as \"passed tests\"?", - "Solution_3": "Is this better?\r\n\r\n[asy]size(200, IgnoreAspect); \nint numberOfPoints = 10; \npair[] points; \npoints.push((1,77)); \npoints.push((2,103)); \npoints.push((3,94)); \npoints.push((4,126)); \npoints.push((5,155)); \npoints.push((6,216)); \npoints.push((7,195)); \npoints.push((8,255)); \npoints.push((9,240)); \npoints.push((10,277)); \n\nxaxis(Label(\"days\", MidPoint), YEquals(50), 1, 10, Ticks(true, true, 0, 0, 1, 0, 3, 0)); \nyaxis(Label(\"passed tests\", MidPoint), XEquals(1), 50, 300, Ticks(true, true, 0, 0, 50, 25, 5, 3)); \n\nfor(int i = 0; i < numberOfPoints - 1; ++i) { \ndraw(points[i]--points[i+1],green+linewidth(1)); \n} \nfor(int i = 0; i < numberOfPoints; ++i) { \ndot(points[i],blue+linewidth(3)); \n}[/asy]\r\n\r\nHere is the code:\r\n[code][asy]\nsize(200, IgnoreAspect); \nint numberOfPoints = 10; \npair[] points; \npoints.push((1,77)); \npoints.push((2,103)); \npoints.push((3,94)); \npoints.push((4,126)); \npoints.push((5,155)); \npoints.push((6,216)); \npoints.push((7,195)); \npoints.push((8,255)); \npoints.push((9,240)); \npoints.push((10,277)); \n\nxaxis(Label(\"days\", MidPoint), YEquals(50), 1, 10, Ticks(true, true, 0, 0, 1, 0, 3, 0)); \nyaxis(Label(\"passed tests\", MidPoint), XEquals(1), 50, 300, Ticks(true, true, 0, 0, 50, 25, 5, 3)); \n\nfor(int i = 0; i < numberOfPoints - 1; ++i) { \ndraw(points[i]--points[i+1],green+linewidth(1)); \n} \nfor(int i = 0; i < numberOfPoints; ++i) { \ndot(points[i],blue+linewidth(3)); \n}[/asy][/code]", - "Solution_4": "[quote=\"MathWise\"]Is this better?\n[/quote]\r\nAt the moment yes. I have to ask if this is what my boss wants." -} -{ - "Problem": "Viet Nam MO (group B)\r\n let a be a real number # 0 and the sequence of real numbers (x_n) is defined by : \r\n x1 = 0 , x_(n+1)(x_n+a)=a+1 , n=1,2,3...\r\n find the x_n ? \r\n \r\nThe solution from THTT (Mathematics and Youth of Vienamese) is very\r\n horrible !\r\nI solve it very easily and better than that way.\r\nDo you have nice solution ?", - "Solution_1": "At least it works by induction !\r\n\r\nx(2*n+1) = (a+1)a y(n) / z(n)\r\nand\r\nx(2*n) = (a+1)/a z(n) / y(n-1)\r\n\r\nwith z(n) = (a+1)^(2n) - ((a+1)^(2n)-1) / (a+2)\r\nand y(n) = ((a+1)^(2n)-1) / ((a+1)^2-1)\r\n\r\nit can be still be simplified (a bit) but it is \"convenient\" on this form to work by induction.\r\n\r\n\r\nBut probably it's smarter to set b=a+1, then it is immediate to see that\r\n\r\nx(n) = b (b^(n-2) - b^(n-3) + ... + (-1)^n) / (b^(n-1) - b^(n-2) + ... - (-1)^n) works :) \r\n\r\nIt can again be simplified a bit but it look nice this way.\r\n\r\n\r\nWhat's it your method A1lqdSchool ?", - "Solution_2": "Here is my solution :\r\nif a=-2 ,we have x_n=(n-1)/n;n=1,2,3...\r\nif a #-2\r\nConsider f(x)=(x+a+1)/(x-1)\r\nIt is easily to see that : f(x_(n+1))=-(a+1)*f(x_n)\r\nso f(x_n)=(-1)^n*(a+1)^n*f(0)=(-a-1)^(n+1)\r\nhence x_n=[(-1)^(n-2)+(a+1)^(n-1)](a+1)/[(-1)^(n-1)+(a+1)^n] , n=1,2,3.." -} -{ - "Problem": "I homeschool and want to go to Columbia. They require homeschoolers to take 4 SAT II's. Does anyone know what score I should aim for?", - "Solution_1": "Maybe you should head over to college confidential - they have a lot of information about this type of stuff. While it depends on the subject, you should probably aim for 750+, but that's not to say you can't get in with lower scores, because you can.", - "Solution_2": "750+ would be the safe zone. One below average score wouldn't necessarily harm you though.", - "Solution_3": "Depends on the subjects really. There's a table of subject test score percentiles lying somewhere on CB. That will give you a better idea. Or better- Check out the stats for Columbi and see the mid-ranges for subject test scores." -} -{ - "Problem": "If a and b are positive real numbers and each of the equations $x^{2}+ax+2b=0$ and $x^{2}+2bx+a=0$ has real roots, then find the smallest possible value of a+b.", - "Solution_1": "[hide]by taking the discriminant of each side, and knowing that the discriminant must be greater than or equal to 0, we get \n$a^{2}-8b\\ge0$\n$4b^{2}-4a\\ge0$\nWe then add $4a$ to each side of the second inequality to get $4b^{2}\\ge4a$, or $b^{2}\\ge a$. Square each side to get $b^{4}\\ge a^{2}$\nSubstitute this into the first inequality, so we have $b^{4}-8b\\ge a^{2}-8b\\ge0$.\nTherefore, we have $b^{4}-8b\\ge0$, and if we solve this for the positives, we get $b\\ge2$. When that is substituted into our first inequality, we can get $a\\ge4.$ We then check to see that the solutions work with our original inequalities (they do; we get $x^{2}+4x+4$, which of course has a double root of $x=-2$ Finally, we add $a$ and $b$ to get $6$.[/hide]", - "Solution_2": "couldn't $a$ and $b$ both be $0$, making the answer $0$?", - "Solution_3": "Positive implies nonzero.", - "Solution_4": ":oops_sign: \r\nman, am I stupid", - "Solution_5": "[quote=\"daermon\"]:oops_sign: \nman, am I stupid[/quote]\r\nnot stupid, just need to pay more attention. :) \r\nI do too." -} -{ - "Problem": "What wil i do to learn this so -called latex technique\r\nhelpful to one in expressing ideas,mathematically? Is there any\r\ncode require? If so,how will i use it? Or,do i've to download this\r\ntechnique?\r\n\r\n Respectfully,\r\n A.Eddington", - "Solution_1": "http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php\r\n\r\nYou don't need to download anything to use LaTeX in this forum. However, you can download it and create documents on your own computer as well. \r\n\r\nLaTeX is widely used by mathematicians, so you may be able to find a guide in your native language (I guess English is not your first :) ). \r\n\r\nYou can practice and ask about LaTeX here: http://www.artofproblemsolving.com/Forum/index.php?f=123" -} -{ - "Problem": "I am learning to distinguish between rational and polynomial functions. A questions reads:\r\n\r\n\"State whether the function is a polynomial, a rational function(but\r\nnot a polynomial), or neither a polynomial nor a rational function. If\r\nthe function is a polynomial, give the degree.\"\r\n\r\ng(x)= [x^2 - 2x - 8] / [x + 2]\r\n\r\n\r\nI know that the top tier is a polynomial of degree 2 and the bottom tier is a polynomial of degree 1. By definition, a rational function is a polynomial divided by a polynomial where the lower tier cannot equal zero, which is what I seem to have here.\r\n\r\nSince the function is rational, I do not have to give the degree, but can rational functions have a degree? Yes or no? If so, how would I calculate it.\r\n\r\nThanks,\r\nJack Daniels", - "Solution_1": "The difference between the degrees of numerator and denominator.", - "Solution_2": "So is this a rational function, or are you saying this is a polynomial?", - "Solution_3": "Rational, by all means :)", - "Solution_4": "It isn't a polynomial because a polynomial's domain is all real numbers, while the stated function is undefined at $ x \\equal{} \\minus{}2$." -} -{ - "Problem": "Numerele reale pozitive a,b verifica relatia a+b+ab=3.\r\nDemonstrati ca $a+b \\ge \\sqrt{a}+\\sqrt{b}$.\r\n*-macar asta e mai grea", - "Solution_1": "$\\sqrt{a}+\\sqrt{b}\\leq \\sqrt{2(a+b)}$, deci ne ramane $2\\leq a+b$.\r\n\r\nacum ne ducem in $a+b+ab=3$, deci $(a+b)+\\frac{(a+b)^{2}}{4}\\geq 3 \\Longleftrightarrow s^{2}+4s-12 \\geq 0$, cu $s=a+b$, acum rezolvand ecuatia asta pentru cand e $=0$, o sa obtinem $s = 2$ (celalalt $s$ da $-6$ , da nu se poate ca $a,b$ pozitive) , dar cum functia e crescatoare, rezulta concluzia.", - "Solution_2": "asta a fost misto. care a fost cel mai mare punctaj la voi?", - "Solution_3": "[quote=\"pohoatza\"]$\\sqrt{a}+\\sqrt{b}\\leq \\sqrt{2(a+b)}$, deci ne ramane $2\\leq a+b$.\n\nacum ne ducem in $a+b+ab=3$, deci $(a+b)+\\frac{(a+b)^{2}}{4}\\geq 3 \\Longleftrightarrow s^{2}+4s-12 \\geq 0$, cu $s=a+b$, acum rezolvand ecuatia asta pentru cand e $=0$, o sa obtinem $s = 2$ (celalalt $s$ da $-6$ , da nu se poate ca $a,b$ pozitive) , dar cum functia e crescatoare, rezulta concluzia.[/quote]\r\n\r\nsuper solutie, dar nu de locala.\r\nAplicam un Cauchy si ajungem la:\r\n$\\sqrt{a}+\\sqrt{b}\\leq 2$, adica la $a+b+2\\sqrt{ab}\\leq 1+a+b+ab$. Scadem $a+b$ si ramane inegalitatea mediilor.", - "Solution_4": "[u][b]O usoara generalizare :[/b][/u]\r\n\r\n$\\{a,b,c\\}\\subset (0,\\infty )\\ ,\\ a+b+ab=c(c+2)\\Longrightarrow \\frac{a+b}{\\sqrt a+\\sqrt b}\\ge \\sqrt c\\ .$", - "Solution_5": "asa am fkt-o si eu....k pohoatza", - "Solution_6": "[quote=\"Virgil Nicula\"][u][b]O usoara generalizare :[/b][/u]\n\n$\\{a,b,c\\}\\subset (0,\\infty )\\ ,\\ a+b+ab=c(c+2)\\Longrightarrow \\frac{a+b}{\\sqrt a+\\sqrt b}\\ge \\sqrt c\\ .$[/quote]\r\n\r\n\r\n:) metoda lui pohoatza ... ramane de demonstrat ca \r\n\r\n$a+b \\ge c$ , dar avem ca $(a+b)+\\frac{(a+b)^{2}}{4}\\ge c (c+2)$ si .. din semnul functiei de gradul II ( tinand cont ca $a+b > 0$ ) avem ca \r\n\r\n$a+b \\ge 2\\sqrt{1+c(c+2)}-2 = 2c$ .. ( cctd )" -} -{ - "Problem": "Now that the AIME II is over, we can discuss what we think the cutoff is. I'll start by saying 210. Thoughts?\r\n\r\nBy the way, when is the cutoff announced?\r\n\r\nYan Zhang", - "Solution_1": "There just doesn't seem to be a limit to the number of math geniuses on this site...\r\n\r\nIn any case, were you referring to the 1st 160 cutoff Yan?\r\n\r\nWe got an insider from TJ now...", - "Solution_2": "I don't think he is THE Yan Zhang, since that Yan has graduated high school and is at Harvard now.", - "Solution_3": "I'm pretty sure he's the Yan Zhang in \\mathbb{R}^- (i.e. he has also been to MOP).\r\n\r\nI'll guess 215.", - "Solution_4": "I believe Simon's guess to be slightly high. I will go with 200-210.", - "Solution_5": "a little off-topic, but do you get a certificate for participation in USAMO? btw I am hoping the cutoff is at 200", - "Solution_6": "Wait...are we talking about the cutoff for the first 160?", - "Solution_7": "Yes, they do give certificates of participation for the USAMO. Actually, I don't think I ever got one for 2002 for some reason.", - "Solution_8": "I keep on hearing about a 150 given to all sophomores and under who took the AMC 12 for the purposes of the USAMO index. Is this true?", - "Solution_9": "I am pretty sure that isn't true. What they may be refering to is that if you are a sophmore or less then getting a floor AIME gets you into USAMO (no matter how low AMC is, unless too many people qualify)", - "Solution_10": "Right. But what that translates to is basically giving everyone in grade 10 and below an automatic 150 since there will almost certainly be someone who did well on the AMC12 but poorly on the AIME and still made USAMO.", - "Solution_11": "Geez..that fact that people always ask the same old questions about the AMC exams really shows how we need a good FAQ about this sort of thing.", - "Solution_12": "[quote=\"Fierytycoon\"]Geez..that fact that people always ask the same old questions about the AMC exams really shows how we need a good FAQ about this sort of thing.[/quote]\r\n\r\nYou're quite right. I am trying to work on part of the FAQ, describing the AMC contests in general, but I am mystified myself by the selection rules for the USAMO, so I will simply say here that anyone who can put a FAQ together about that will have my thanks.", - "Solution_13": "[quote]I am mystified myself by the selection rules for the USAMO[/quote]\r\n\r\nHmm, what exactly are you confused about?", - "Solution_14": "[quote=\"ComplexZeta\"]Right. But what that translates to is basically giving everyone in grade 10 and below an automatic 150[/quote]\r\n\r\nYes! You've got it.", - "Solution_15": "Marma103, you are right about what the rules say, but the rules themselves are apparently incorrectly stated. We had a previous discussion of this topic in the following thread:\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1831[/url]\r\n\r\nThe consensus view was that by \"students who took AMC 10\", the AMC rules meant \"students in the 10th grade or below who took AMC 10 or 12\".", - "Solution_16": "Thank you Ravi, that's exactly what I meant... and thanks for the link, too.\r\n\r\nThey should've change the wording, though. I know kids who chose to take AMC-10 because of that wording.", - "Solution_17": "[quote=\"texas137\"]the first year that there were separate exams, there was a big advantage to taking the 10. Since then, the cutoffs have been sufficiently lower on the 12 than on the 10 that it is easier for students in grade 10 and below to qualify for the AIME by taking the 12. Once they take the AIME, they only have to make the AIME floor, not the USAMO index, so that removes any disadvantage they might have from a lower numerical score on the 12 than they would have had on the 10.[/quote]\r\n\r\nThis is correct, correct history, and correctly stated.\r\n\r\nSeveral of us in the office go separately through several stages of checking, double-checking and cross-checking to make sure that no deserving student, and in particular no 7th, 8th, 9th or 10th grade student misses a valid invitation to USAMO.\r\n\r\nSteve Dunbar\r\nAMC Director", - "Solution_18": "http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamoqualstate.html\r\n\r\n[quote]261 students from 190 schools \nThe USAMO index cutoff for participants was 210. \nIn addition, students in grades 7th through 10th achieving an AIME score of 7 or above\nwere also selected to write the USAMO. \n[/quote]", - "Solution_19": "Wow, 210, I called it right!\r\n\r\nI must be a prophet. :)\r\n\r\nYan P. Zhang", - "Solution_20": "Did it make a difference between AIME I and AIME II this year?", - "Solution_21": "[quote=\"compucomp\"]Wow, 210, I called it right!\n\nI must be a prophet. :)\n\nYan P. Zhang[/quote]\r\n\r\nI called it too, though I must admit to not being a prophet.", - "Solution_22": "YES!\r\n\r\nI squeezed past...barely...with a 7 on the AIME. (I'm in 8th grade.) I just got the email today (and I checked the website.)", - "Solution_23": "Would it be possible for an AIME qualifier but non-USAMO qualifier to take the USAMO unofficially, but on the official day. Also, could a person take the AIME I unofficially and the AIME II officially?", - "Solution_24": "Depends on what you mean by \"unofficial\". You can arrange for your school to put you in a closed room for the two days of 4.5 hrs that the USAMO takes. The problems become available on the webjust before the start, so you could certainly do this. Of course, you would have to wait for solutions before grading your own test. As for the AIME, same thing, just that if other students are officialy taking a test, you may not be able to sit in during it unofficialy.", - "Solution_25": "When I was a sophomore, I was the only USAMO qualifier at my school, and my teacher called the AMC office to ask if someone else could take it since he had missed it by two points. They said that he could take it unofficially, but he had to take it later. But questions weren't on the web back then.", - "Solution_26": "[quote=\"Pork_Chop8\"]Also, could a person take the AIME I unofficially and the AIME II officially?[/quote]\r\n\r\nI think this depends on whether you find a testing site with an extra exam that's willing to let you do this. Usually the questions from the AIME-I are on AoPS and probably other online places a day after the actual exam, so if you wanted to take it for practice, you could lock yourself in your room for three hours =) .. I'm pretty sure people (such as MithsApprentice) have sat for both exams, but had their AIME II score count - you need to decide this before taking either, though.", - "Solution_27": "official AMC rules state that the exam manager is only allowed to administer the AIME or USAMO to students who qualified for it. Now what you do on your own time, or what you work out informally with a teacher who is not administering the exam officially to anyone, is up to you.", - "Solution_28": "Do the AIME I and the AIME II had the same cutoff? This year's AIME II was much harder than AIME I.", - "Solution_29": "[quote=\"JP\"]Do the AIME I and the AIME II had the same cutoff? This year's AIME II was much harder than AIME I.[/quote]\r\n\r\nThe cutoffs were the same. While the AIME II did appear (at least statistically) to be harder than the AIME I, the deviation was not so large to expect more than a fraction of a problem difference for students between the exams. For that reason I doubt they spent very long considering a different cutoff.\r\n\r\nThe floor for 10th graders and below was a 7 regardless of which exam was taken." -} -{ - "Problem": "a) Suppose that three congruent disks with radius $R$ have their centers in the vertices of a given triangle which they cover. Find the smallest value of $R$ for a given triangle.\r\n\r\nb) (this problem just asks itself to generalize :P) Suppose that $n$ congruent disks with radius $R$ have their centers in the vertices of a given convex $n$-gon which they cover. Find the smallest value of $R$ for a given convex $n$-gon.", - "Solution_1": "for a), isn't the answer just the circumradius of the triangle itself? (at least for an acute triangle) because if R were to be smaller than the circumradius, then the circumcentre will never be covered by the disks....and i think for an obtuse triangle, min(R) is 1/2 of max{a,b,c}. :)", - "Solution_2": "Your answer is correct for acute triangle but wrong for obtuse where it is $\\frac{a}{2 \\cos {\\beta}}$ where $a$ is the second longest side and $\\beta$ is the smallest angle in the given triangle. :)", - "Solution_3": "ahh i see, it's because if BC is the largest side, have to make sure that R extended from A covers the midpoint of BC. okay, thanx! :D" -} -{ - "Problem": "Find the value of $ S$ if $ S\\equal{}\\sum_{n \\equal{} 1}^\\infty \\frac {3n \\minus{} 1}{5^n}$\r\n\r\nHere's what I did:\r\nS=2/5+5/5^2+8/5^3...\r\nS-S/5=2/5+(5-2)/5^2+(8-5)/5^3...\r\n=2/5+3(1/5+1/5^2...)\r\n=2/5+3(1/5 / ((1-1/5)))\r\n=23/20\r\nSo S=S=23/16? The correct answer is 11/16, and I don't think an answer greater than 1 would make sense anyway. I don't see what I did incorrectly at the moment", - "Solution_1": "[hide=\"Hint\"]\n$ S\\equal{}\\sum_{n\\equal{}1}^{\\infty}nr^{n\\minus{}1}\\equal{}1\\plus{}2r\\plus{}3r^{2}\\plus{}...$\n$ rS\\equal{}r\\plus{}2r^{2}\\plus{}3r^{3}\\plus{}...$\n$ S\\minus{}rS\\equal{}S(1\\minus{}r)\\equal{}\\frac{1}{1\\minus{}r} \\Rightarrow S\\equal{}\\frac{1}{(1\\minus{}r)^{2}}$\n[/hide]\n[hide=\"Solution\"]\n$ S\\equal{}\\sum_{n \\equal{} 1}^\\infty\\frac{3n\\minus{}1}{5^{n}}\\equal{}\\frac{3}{5}\\sum_{n\\equal{}1}^{\\infty}\\frac{n}{5^{n\\minus{}1}}\\minus{}\\sum_{n\\equal{}1}^{\\infty}\\left ( \\frac{1}{5} \\right )^{n}\\equal{}\\frac{3}{5}\\cdot \\frac{1}{(1\\minus{}\\frac{1}{5})^{2}}\\minus{}\\frac{\\frac{1}{5}}{1\\minus{}\\frac{1}{5}}\\equal{}\\boxed{\\frac{11}{16}}$\n[/hide]", - "Solution_2": "You decided $ \\frac{5 \\minus{} 2}{5^2} \\equal{} \\frac{3}{5}$ (and similarly down the line) ;)\r\n\r\n(BTW, that would have been much easier to notice if you'd done the entire post in LaTeX; this was a good place to use the align* environment.)", - "Solution_3": "[quote=\"calcc\"]Here's what I did:\nS=2/5+5/5^2+8/5^3...\nS-S/5=2/5+(5-2)/5^2+(8-5)/5^3...\n=2/5+3(1/5+1/5^2...)\n=2/5+3(1/5 / ((1-1/5)))\n=23/20\n[/quote]So to be clear about what JBL suggests, this is what you exactly typed (but in $ \\text{\\LaTeX}$ of course):\r\n\r\n$ \\begin{aligned}\r\nS&=\\frac25+\\frac5{5^2}+\\frac8{5^3}...\\\\\r\nS-\\frac{S}5&=\\frac25+\\frac{(5-2)}{5^2}+\\frac{(8-5)}{5^3}...\\\\\r\n&=\\frac25+3\\left(\\frac15+\\frac1{5^2}...\\right)\\\\\r\n&=\\frac25+3\\left(\\frac{\\frac15}{((1-\\frac15))}\\right)\\\\\r\n&=\\frac{23}{20}\r\n\\end{aligned}$\r\n\r\nHope you catch your mistakes.", - "Solution_4": "hello, $ \\sum_{n\\equal{}1}^{\\infty}\\frac{2n\\minus{}1}{5^n}\\equal{}\\frac{11}{16}$.\r\nSonnhard.", - "Solution_5": "[quote=\"Dr Sonnhard Graubner\"]hello, $ \\sum_{n \\equal{} 1}^{\\infty}\\frac {2n \\minus{} 1}{5^n} \\equal{} \\frac {11}{16}$.\nSonnhard.[/quote]Wrong. And there's already a solution given...", - "Solution_6": "[hide=\"just some arithmetico-geometric series\"]\nWe have\n$ S \\equal{} \\frac {2}{5} \\plus{} \\frac {5}{5^2} \\plus{} \\frac {8}{5^3} \\plus{} ...$\nSince this is an arithmetico-geomeric series, we divide the first equation by 5 to get\n$ \\frac {S}{5} \\equal{} \\frac {2}{5^2} \\plus{} \\frac {5}{5^3} \\plus{} \\frac {8}{5^4} \\plus{} ...$\nSubtracting the second equation from the first equation, we get\n$ \\frac {4S}{5} \\equal{} \\frac {2}{5} \\plus{} \\frac {3}{5^2} \\plus{} \\frac {3}{5^3} \\plus{} ... \\equal{} \\frac {2}{5} \\plus{} \\frac {3}{20} \\equal{} \\frac {11}{20}$\nMultiplying the equation gives us our desired result:\n$ S \\equal{} \\frac {11}{20}*\\frac {5}{4} \\equal{} \\boxed{\\frac {11}{16}}$\n[/hide]\r\n@10000th User: I think you made an error somewhere (its supposed to be $ 3(\\frac {1}{5^2} \\plus{} \\frac {1}{5^3} \\plus{} ...)$, not $ 3(\\frac {1}{5} \\plus{} \\frac {1}{5^2} \\plus{} ...)$). Fix that, and you'll get the right answer (I think).", - "Solution_7": "$ \\sum_{k\\equal{}1}^n \\frac{2k\\minus{}1}{5^k}\\equal{}\\frac 18 \\sum_{k\\equal{}1}^n \\left(\\frac{4k\\minus{}1}{5^{k\\minus{}1}}\\minus{}\\frac{4k\\plus{}3}{5^k}\\right)\\equal{}\\frac 38\\minus{}\\frac{4n\\plus{}3}{8\\cdot 5^n}\\longrightarrow \\boxed{\\frac{3}{8}}\\ (n\\rightarrow \\infty)$", - "Solution_8": "[quote=\"agentcx\"]@10000th User: I think you made an error somewhere (its supposed to be $ 3(\\frac {1}{5^2} \\plus{} \\frac {1}{5^3} \\plus{} ...)$, not $ 3(\\frac {1}{5} \\plus{} \\frac {1}{5^2} \\plus{} ...)$). Fix that, and you'll get the right answer (I think).[/quote]Correction: @calcc." -} -{ - "Problem": "can someone prove the equivalence of these two definitions of boiling point that it is the temp. at which vapour pressure of the liquid equals the atmospheric pressure and it is the temp. at which the liquid vapourizes", - "Solution_1": "The boiling of a liquid - when the molecules have enough energy to overcome the intermolecular forces existing in the liquid and transform in vapour - occurs not only at the surface of it but also (and predominantly) inside it. In the surface the molecules simply go directly to vapour after overcoming the forces that were exerted on them. Inside the liquid, a bubble of vapour has to form (and not colapse) and, because it is less dense than the liquid it rises to the surface and then breaks, with its vapour mixing with the already present vapour (and air) at the surface. However, for a recently formed bubble of vapour to survive inside the liquid, the pressure exerted by the vapour on the \"walls\" of the bubble must be greater or equal to the outside pressure: otherwise, it will colapse. And what is that outside pressure? From hydrostatics we know that in a given point of a liquid the total pressure is the sum of the atmosferic pressure with the hydrostatic pressure (the pressure due only to the liquid). As this last one is always neglegible, then for a bubble to be able to form and survive its vapour pressure (the pressure exerted by the vapour, not only at the surface of the liquid but also inside a bubble: the pressure exerted by the vapour form of the liquid) must be greater or equal to the atmosferic pressure. Of course, when this happens then the liquid is able to vapourize: its molecules (inside and at the surface) go to the vapour state.", - "Solution_2": "thanks :) carcul", - "Solution_3": "Another statement is the following: the boiling point of a substance is the maximum temperature for which the liquid state of that substance can exist.", - "Solution_4": "how do u imagine a 'bubble of vapour'", - "Solution_5": "also when u say the pressure inside the bubble must be graeater than or equal to the atmospheric pressure .then why isn't the vapour pressure due to already present vapour added here.i mean there are already some vapours so pressure at the top of the liquid equals this pressure and the atomspheric pressure(by the way i really don't in some sense understand why the atmospheric pressure gets added here. because i f u have to have vapours then u need to carry out your experiment in a close dcontainer otherwise these vapour molecules would escape.if it is close dthen what about the atm. pressure", - "Solution_6": "ok i f i am not wrong is this one of the assumptions in ur proof that the pressure due to the molecules inside the bubble is same as the vapour pressure", - "Solution_7": "[quote=\"Pardesi\"]how do u imagine a 'bubble of vapour'[/quote]\n\nA bubble of vapour is simple an aggregate of molecules with enough energy to be in the gaseous state, inside the liquid. The form of the bubble is spherical because the pressure inside a fluid acts equally in all directions.\n\n[quote=\"Pardesi\"]then why isn't the vapour pressure due to already present vapour added here.i mean there are already some vapours so pressure at the top of the liquid equals this pressure and the atomspheric pressure[/quote]\n\nBecause that vapour pressure contributes very little to the total pressure exerted on the (surface of the) liquid: that vapour pressure is a partial pressure (because the vapour mixes in the atmosphere), and because $ P_{i}= x_{i}P$, that partial pressure is always neglegible, as the atmosphere is such a \"big solution\".\n\n[quote=\"Pardesi\"]if it is close then what about the atm. pressure[/quote]\r\n\r\nIf a container is closed then the liquid will boil with greater difficulty because precisely the vapour accumulates above the liquid and now the total pressure to overcome is increasingly higher as the liquid is heated. That's why cooking food in a closed vessel (the appropriately called [i]pressure cooker[/i]) is much faster than in an open saucepan: the water can attain higher temperatures without boiling." -} -{ - "Problem": "Solve in the set of integers the equation :\r\n\r\n $ x^8\\plus{}2^{2^x\\plus{}2} \\equal{} p$ \r\n\r\n where $ p$ is a prime positive integer.", - "Solution_1": "well, x=1 and p=17 is a solution:\r\n\r\n1^8+2^(2^1+2) = 1+2^4 = 17", - "Solution_2": "we can easely prove that x^4+4y^4 is not prime except x=y=1 since it can be factorized. so x cannot be greater than 1." -} -{ - "Problem": "Suppose we have a triangle with vertices A, B, and C and an Orthocenter at D. Show that the locus of centers of the rectangular hyperbolas which pass through the four points (A, B, C, D) is a circle.", - "Solution_1": "(bump), and a first step:\r\n\r\nLet's assume we're dealing with an acute triangle. (This only eliminates the case of right triangles, since if angle ABC is obtuse and D is the orthocenter then triangle DBC is acute and A is the orthocenter -- the same set of 4 points.) Now, we cannot have all 4 points on one half of the hyperbola. We also can't have D on one half and the other three on the other half, and we can't even have D and A (or B or C) on one half and B and C (or A and C or A and B) on the other half, because then the secant AD would cross the secant BC on its interior and thus cross the other half of the hyperbola, but no secants of one half of a rectangular hyperbola intersect the other half. Thus, the only possibility is that A (or B or C) is on one half while B, C (or A, C or A, B) and D are on the other half.", - "Solution_2": "Very good, but how do we know that the locus of the centers makes a circle?" -} -{ - "Problem": "Prove that for every integer $ n>1$,\r\n\r\n$ n((n\\plus{}1)^{\\frac{2}{n}}\\minus{}1)<\\displaystyle\\sum_{i\\equal{}1}^{n}\\frac{2i\\plus{}1}{i^2}6$ then $ \\phi(n)>\\sqrt{n}.$\nii) $ \\forall{n}<10^{4},\\phi(n)\\not\\equal{}76.$ (Maple).\n\nMoreover, $37$ is the smallest prime number that answers the question." -} -{ - "Problem": "Prove the following statement: If $r_1$ and $r_2$ are real numbers whose quotient is irrational, then any real number $x$ can be approximated arbitrarily well by the numbers of the form $\\ z_{k_1,k_2} = k_1r_1 + k_2r_2$ integers, i.e. for every number $x$ and every positive real number $p$ two integers $k_1$ and $k_2$ can be found so that $|x - (k_1r_1 + k_2r_2)| < p$ holds.", - "Solution_1": "Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url] :)", - "Solution_2": "Okay...this solution may not be correct, please tell me where I'm wrong.\r\nWe are given that r1/r2 is irrational, which implies that r1 and r2 are both irrational. \r\n\r\nNow, put z=k1r1 +k2r2. \r\n\r\nChoosing r1, and r2 as a base, we can construct a coordinate system so that any number can be written in the form of z.\r\n\r\nSo, any number p can be written as the absolute value of the difference of two numbers. That number can be as small as we please. Hence the proof. Is this correct?! :) \r\n\r\nSorry for the bad presentation....how does this latex thing work?" -} -{ - "Problem": "im chinese. u guys are super good at math. just saying i am a fan.", - "Solution_1": "wat's up....", - "Solution_2": "im Chinese too\r\nis everyone hear living in china or just chinese and living somewhere including china", - "Solution_3": "Chinese in America here. :D", - "Solution_4": "I'm ABC (American Born Chinese). lol", - "Solution_5": "I am Chinese... How can you type in Chinese on Windows Vista?\r\n\r\nI just got a Windows Vista and it doesn't have a language bar :(", - "Solution_6": "Go to START -> CONTROL PANEL.\r\nThen click Clock, Language, Region.\r\nThere should be a link that says \"change keyboard or other input methods\".\r\nClick that, and a window should pop up.\r\nFind \"Keyboards and Languages\" in the top bar and click change keyboard.\r\nClick ADD, find Chinese, and add it." -} -{ - "Problem": "Let $ a,\\ b\\, c$ be positive real numbers. Find the minimum value of\r\n\r\n$ \\sqrt [3]{\\frac {(a^{2008} \\plus{} 2007b^{2008})(b^{2008} \\plus{} 2007c^{2008})(c^{2008} \\plus{} 2007a^{2008})}{a^{2008}b^{2008}c^{2008}}}$\r\n\r\n[color=red]Last Edited[/color]", - "Solution_1": "[quote=\"kunny\"]Let $ a,\\ b\\, c$ be positive real numbers. Find the minimum value of\n\n$ \\sqrt [3]{\\frac {(a^{2008} \\plus{} 2007b^{2007})(b^{2008} \\plus{} 2007c^{2007})(c^{2008} \\plus{} 2007a^{2007})}{a^{2008}b^{2008}c^{2008}}}$[/quote]\r\ni don't know this problem . \r\n $ LHS \\geq 1 \\plus{} \\frac{2007}{\\sqrt[3]{abc}} \\geq 1.$\r\n occur equal if only if : $ a, b,c \\to \\infty$", - "Solution_2": "Sorry for that I made a typo. I have just edited.\r\nThank you for your indication.\r\n\r\nkunny", - "Solution_3": "[quote=\"kunny\"]Let $ a,\\ b\\, c$ be positive real numbers. Find the minimum value of\n\n$ \\sqrt [3]{\\frac {(a^{2008} \\plus{} 2007b^{2008})(b^{2008} \\plus{} 2007c^{2008})(c^{2008} \\plus{} 2007a^{2008})}{a^{2008}b^{2008}c^{2008}}}$\n\n[color=red]Last Edited[/color][/quote]\r\n it's easy.\r\nsolution1:\r\nLemma : $ (x \\plus{} a)(y \\plus{} b)(z \\plus{} c) \\geq (\\sqrt [3]{xyz} \\plus{} \\sqrt [3]{abc})^3$\r\n$ \\sqrt [3]{\\frac {(a^{2008} \\plus{} 2007b^{2008})(b^{2008} \\plus{} 2007c^{2008})(c^{2008} \\plus{} 2007a^{2008})}{a^{2008}b^{2008}c^{2008}}} \\geq \\sqrt [3]{\\frac {(\\sqrt [3]{a^{2008}b^{2008}c^{2008}} \\plus{} 2007\\sqrt [3]{a^{2008}b^{2008}c^{2008})^3} }{a^{2008}b^{2008}c^{2008}}} \\equal{} 2008$\r\n solution2 : by cauchy's ineq for 2008 real positive numbers : \r\n$ a^{2008} \\plus{} b^{2008} \\plus{} ... \\plus{} b^{2008} \\geq 2008 ab^{2007}$\r\nthus we have : $ LHS \\geq 2008$", - "Solution_4": "Nice Solution. :)" -} -{ - "Problem": "find the no. of all integer sided isosceles obtuse angled triangles with perimeter 2008", - "Solution_1": "hi I've solved it in this way\r\nsuppose in triangle ABC ,C be the obtuse angle.\r\nnow 900$ - by making $ r$ sufficiently close to $ 1.$ Hence there is no such $ c$ and the norms are not equivalent.\r\n\r\nYou could get the same thing out of $ x_k\\equal{}\\begin{cases}\\frac1k,&1\\le k\\le n\\\\0,&k>n\\end{cases},$ for which $ \\|x\\|_2$ is bounded but $ \\|x\\|_1$ can be made arbitrarily large by choosing $ n$ large enough.", - "Solution_3": "Your answer is very useful. Thank u very much", - "Solution_4": "[quote=\"Kent Merryfield\"][quote=\"Ognjen09\"]...we have $ \\sqrt {\\sum_{k \\equal{} 1}^{n}x_k^2}\\leq\\sum_{k \\equal{} 1}^{n}|x_k|$ ...[/quote]\n\nWhile this step is true, it is far from obvious. In a textbook problem such as this, I would expect that the student would be expected to prove this as part of the overall proof.[/quote]\r\n\r\nsquare both sides. the expansion of the RHS contains the LHS, as well as a bunch of other clearly nonnegative terms..." -} -{ - "Problem": "Let $ a,b,c > 0$ and $ a\\plus{}b\\plus{}c\\equal{}1$ Prove that \\[ \\sum_{cyc}\\sqrt{\\frac{2b}{3\\plus{}a}} \\leq \\sqrt{\\sum_{cyc}\\frac{1}{1\\plus{}2\\sqrt{ab}}}\\]", - "Solution_1": "[quote=\"apollo\"]Let $ a,b,c > 0$ and $ a \\plus{} b \\plus{} c \\equal{} 1$ Prove that\n\\[ \\sum_{cyc}\\sqrt {\\frac {2b}{3 \\plus{} a}} \\leq \\sqrt {\\sum_{cyc}\\frac {1}{1 \\plus{} 2\\sqrt {ab}}}\n\\]\n[/quote]\r\n$ \\sum_{cyc}\\sqrt {\\frac {2b}{3 \\plus{} a}} \\leq \\sum_{cyc}b\\sqrt{\\frac{2}{b(3\\plus{}a)}}\\leq\\sqrt{\\sum_{cyc}\\frac{2}{3\\plus{}a}}$ and $ \\sqrt {\\sum_{cyc}\\frac {1}{1 \\plus{} 2\\sqrt {ab}}}\\geq\\sqrt {\\sum_{cyc}\\frac {1}{1 \\plus{} a\\plus{}b}}.$\r\nHence, it remains to prove that: $ \\sum_{cyc}\\frac{2}{3\\plus{}a}\\leq\\sum_{cyc}\\frac {1}{1 \\plus{} b\\plus{}c}.$ But $ \\sum_{cyc}\\frac{2}{3\\plus{}a}\\leq\\sum_{cyc}\\frac {1}{1 \\plus{} b\\plus{}c}\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}\\frac{1\\plus{}a\\minus{}2b\\minus{}2c}{(3\\plus{}a)(1\\plus{}b\\plus{}c)}\\geq0\\Leftrightarrow\\sum_{cyc}\\frac{a\\minus{}b\\minus{}(c\\minus{}a)}{(3\\plus{}a)(1\\plus{}b\\plus{}c)}\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(a\\minus{}b)\\left(\\frac{1}{(3\\plus{}a)(1\\plus{}b\\plus{}c)}\\minus{}\\frac{1}{(3\\plus{}b)(1\\plus{}a\\plus{}c)}\\right)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(a\\minus{}b)^2(2\\minus{}c)(3\\plus{}c)(1\\plus{}a\\plus{}b)\\geq0.$" -} -{ - "Problem": "The sequence $ \\{x_{n}\\}_{n \\ge 1}$ is defined by\r\n\\[ x_{1} \\equal{} 2, x_{n \\plus{} 1} \\equal{} \\frac {2 \\plus{} x_{n}}{1 \\minus{} 2x_{n}}\\;\\; (n \\in \\mathbb{N}).\r\n\\] Prove that \r\n\r\na) $ x_{n}\\not \\equal{} 0$ for all $ n \\in \\mathbb{N}$, \r\n\r\nb) $ \\{x_{n}\\}_{n \\ge 1}$ is not periodic.", - "Solution_1": "[quote=\"Peter\"]The sequence $ \\{x_{n}\\}_{n \\ge 1}$ is defined by\n\\[ x_{1} \\equal{} 2, x_{n \\plus{} 1} \\equal{} \\frac {2 \\plus{} x_{n}}{1 \\minus{} 2x_{n}}\\;\\; (n \\in \\mathbb{N}).\n\\]\nProve that [list=a][*] $ x_{n}\\not \\equal{} 0$ for all $ n \\in \\mathbb{N}$, [*] $ \\{x_{n}\\}_{n \\ge 1}$ is not periodic.[/list][/quote]\r\n$ a \\equal{} \\arctan {2}$\r\n$ x_1 \\equal{} \\tan{a}$\r\n$ x_{n \\plus{} 1} \\equal{} \\frac {x_n \\plus{} \\tan{a}}{1 \\minus{} x_n\\tan{a}}$\r\nWe prove by induction that:\r\n$ x_{n} \\equal{} \\tan{na}$\r\nSuppose $ x_n$ is period T.\r\n$ \\tan{(n \\plus{} T)a} \\equal{} \\tan{na}$\r\n$ \\rightarrow Ta\\equiv 0(\\mod \\pi)$\r\nSo $ a \\equal{} \\frac {mk\\pi}{T}$\r\nWe has a result :\r\n$ k\\in Q$ then $ \\tan{k\\pi}$ is not equal 2.", - "Solution_2": "[hide=\"Another approach\"]\nAs above let $ \\alpha \\equal{} \\arctan 2$ so that $ x_n \\equal{} \\tan n\\alpha$. For [b](a)[/b] suppose for the sake of contradiction that $ x_n \\equal{} 0$ for some $ n$. Let $ m \\equal{} n/2$ so that \\[ x_n \\equal{} x_{2m} \\equal{} \\frac{2\\tan m\\alpha}{1\\minus{}\\tan^2 m\\alpha} \\equal{} \\frac{2x_m}{1\\minus{}x_m^2}.\\] If $ n$ is even, $ x_n \\equal{} 0 \\iff x_m \\equal{} 0$. Otherwise let $ n \\equal{} 2^k (2j\\plus{}1)$ for some nonnegative integers $ j,k$. We can divide out powers of 2 until we are left with \\[ x_{n/2^k} \\equal{} x_{2j\\plus{}1} \\equal{} \\frac{2\\plus{}x_{2j}}{1\\minus{}x_{2j}} \\equal{} 0,\\] so $ x_{2j} \\equal{} \\minus{}2$. But \\[ x_{2j} \\equal{} \\frac{2x_j}{1\\minus{}x_j^2} \\equal{} \\minus{}2.\\] This equation has no rational roots so we have a contradiction (inductively, $ x_n$ must be rational since $ x_1 \\equal{} 2$ is rational). Therefore there are no zero values of $ \\{x_n\\}$.\n\nFor [b](b)[/b] suppose that $ x_{k\\plus{}t} \\equal{} x_{k}$ for some positive integers $ k,t$ with $ t\\ge 1$. Then \\[ x_{k\\plus{}t} \\minus{} x_{k} \\equal{} \\tan (k\\plus{}t)\\alpha \\minus{} \\tan k\\alpha \\equal{} \\frac{\\tan t\\alpha}{\\cos(k\\plus{}t)\\alpha} \\equal{} \\frac{x_t}{\\cos(k\\plus{}t)\\alpha} \\equal{} 0,\\] so $ x_t \\equal{} 0$. However, this is impossible by (a). Thus all terms of $ \\{x_n\\}$ are distinct.\n[/hide]" -} -{ - "Problem": "The problems states: \"Prove that there are infinite primes $ p$ such that $ \\minus{}3$ is a quadratic residue $ \\mod{p}$\".\r\n\r\nI feel it is very easy, but I can't solve it u_u ...", - "Solution_1": "This is equivalent to the claim that there are infinitely many primes dividing a value of the polynomial $ x^2 \\plus{} 3$ at an integer. There is a standard \"Euclidean\" proof of this; are you familiar with the Euclidean proof of the infinitude of the primes?" -} -{ - "Problem": "Thursday eve at Nationals\r\n\r\n8:00, during coaches meeting\r\nEVERYONE COME TO MY ROOM(wherever that may be)\r\n\r\nwe can play cards and watch SURVivor best show ever\r\nif anyone's interested", - "Solution_1": "Sorry, I can't make it...But can you imagine 227 mathletes stuck in the same room watching survivor?", - "Solution_2": "If Nationals included Wednesday, I'd have a LOST night. :D Good thing my flight's at 11:20, because I'm going to be up until 10 o'clock watching that. :D", - "Solution_3": "we're all in the same hotel so it couldn't be too hard...I'll come if I can convince any teammates to come with me", - "Solution_4": "I'll come... and ruin the whole thing\r\n\r\nit'll be fun", - "Solution_5": "How about a meeting of AoPSers only. Maybe at dinner on thursday we can all move into one particular corner, at one particular time :D.", - "Solution_6": "Yeah. that would greatly improve our scores...\r\n\r\n*can nobody feel the sarcasm*\r\n\r\nI'm pretty sure all the teams will want to meet.", - "Solution_7": "We dont practice the night before at all. last time all we did is stay up really late and drink beer", - "Solution_8": "[quote=\"biffanddoc\"]...and drink beer[/quote]\r\num...\r\nthat can't really help your score can it?", - "Solution_9": "[quote=\"k8reindeer\"][quote=\"biffanddoc\"]...and drink beer[/quote]\num...\nthat can't really help your score can it?[/quote]\r\n\r\nwell passing out early = more sleep :roll:", - "Solution_10": "I'll be busy cramming, obviously.\r\n\r\n\r\n[quote=\"biffanddoc\"]...and drink beer[/quote]\n\n...[/quote]", - "Solution_11": "[quote=\"Waxing Rhapsodic\"]I'll be busy cramming, obviously.[/quote]\r\n\r\nMe too. We should cram together... (I'll do it on the plane!)", - "Solution_12": "I wish I could go.", - "Solution_13": "Sorry to hear your disappointment tz, but Michigan is one of the hardest states.", - "Solution_14": "I. AM. IN. What state are you from ?\r\n\r\nName: Richard Ni\r\nPlace:[b] 5th\n[/b]\r\nI can spoil the show for you if you want... lol.\r\n\r\nHere's what happens if you want to know for the next few episodes\r\n\r\n[hide]Aras wins Immunity, Cirie goes home (possible tiebreaker required)\n\nThen I think Aras wins the immunity and takes Danielle, or vice versa. Terry goes home.\n\nAras is heavily favored to win.[/hide]", - "Solution_15": "[quote=\"Fanatic\"]I. AM. IN. What state are you from ?\n\nName: Richard Ni\nPlace:[b] 5th\n[/b]\nI can spoil the show for you if you want... lol.\n\nHere's what happens if you want to know for the next few episodes\n\n[hide]Aras wins Immunity, Cirie goes home (possible tiebreaker required)\n\nThen I think Aras wins the immunity and takes Danielle, or vice versa. Terry goes home.\n\nAras is heavily favored to win.[/hide][/quote]\n\n[hide=\"Off Topic\"]\nI doubt that's true unless Terry voted for Cirie, while Aras, Danielle, and Cirie all voted for Terry and he used the immunity idol.[/hide]", - "Solution_16": "Fanatic:\r\n\r\nEither Aras or Terry will win.\r\nThe next immunity is crucial. If Aras wins, then Danielle's gone and Terry is once again, without allies. Even if Terry wins final immunity, he'll lose to whoever he brings to the two.\r\n\r\nHowever, if Terry wins next immunity, he gives the idol to Danielle and Aras is gone. Then, either Danielle or Terry wins the final immunity, and will bring each other to the two (since Cirie can't win final immunity, but will most likely get the vote of the jury.\r\n\r\nIf Terry goes to the two with Danielle, he'll win, since Austin, Sally, Shane, and Courtney will all vote for him. Austin, Sally, and courtney were his former allies, and will vote for him. Shane won't vote for anyone else because everyone but Terry sold him out.\r\n\r\nMad analysis.", - "Solution_17": "Umm.. what I said wasn't analysis.\r\n\r\nIt was a spoiler. I said \"I can spoil the show for you if you want...\".\r\n\r\nTrust me. [hide]Cirie goes home, then Terry. DDL and Aras F2[/hide]", - "Solution_18": "[quote=\"XxHarryxX\"][quote=\"Waxing Rhapsodic\"]I'll be busy cramming, obviously.[/quote]\n\nMe too. We should cram together... (I'll do it on the plane!)[/quote]\r\n\r\nCramming is probably not a good idea. All the real preparation has been done months before the actual competition (like in sports), and you should be tapering the last week or so. Cramming is only useful for short term rote memorization of facts, formulas, etc, certainly not anything that will help you in MathCounts." -} -{ - "Problem": "Let P be an interior point of a triangle ABC. Cevians AP, BP, CP intersect sides \r\nin points A, B, C. Prove that:\r\n1) Lines that join midpoints of segments BC, AC, AB\r\nwith midpoints of sides BC, AC, AB respectively concur in a point T.\r\n2) T, P, G where G gravity center, are collinear.", - "Solution_1": "I will denote with M(XY) the midpoint of an arbitrary segment XY.\r\n\r\nLet T be the centroid of the four points A, B, C, P. Then, I will prove that this point T lies on the lines M(B'C')M(BC), M(C'A')M(CA) and M(A'B')M(AB), and that the points T, P, G are collinear. This will obviously solve the problem.\r\n\r\nBy the Gauss-Newton theorem, applied to the complete quadrilateral formed by the lines AB, CC', CA, BB', the midpoints of the segments BC, B'C' and AP are collinear. Hence, the line M(B'C')M(BC) coincides with the line M(BC)M(AP). But from the fundamental properties of centroids, the centroid T of the four points A, B, C, P lies on the line M(BC)M(AP). Hence, we have proven that the point T lies on the line M(B'C')M(BC). Similarly, we see that it also lies on the lines M(C'A')M(CA) and M(A'B')M(AB).\r\n\r\nNow it remains to show that the points T, P, G are collinear. But this is again a consequence of properties of centroids, since the point G is the centroid of the points A, B, C, and the point T is the centroid of the points A, B, C, P. Moreover, we have PT : TG = 3 (with directed segments).\r\n\r\n Darij", - "Solution_2": "Whe you say T is de centroid of ABCO Do you mean that M is the centroid of ABCO's Wittenbauer paralelogram?", - "Solution_3": "[quote=\"hucht\"]Whe you say T is de centroid of ABCO Do you mean that M is the centroid of ABCO's Wittenbauer paralelogram?[/quote]\r\n\r\nI don't say that T is the centroid of ABCP (not ABCO, by the way, there is no O in my post). I say that T is the centroid of the four points A, B, C, P, i. e. the midpoint of the segments $M_{AB}M_{CP}$, $M_{BC}M_{AP}$, $M_{CA}M_{BP}$, where $M_{XY}$ denotes the midpoint of any segment XY. This centroid is the center of the Varignon parallelogram (the parallelogram $M_{AB}M_{BC}M_{CD}M_{DA}$. The center of the Wittenbauer parallelogram is the \"area centroid\" of the quadrilateral ABCP and something completely different.\r\n\r\n Darij" -} -{ - "Problem": "In the plane we are given a set $ E$ of 1991 points, and certain pairs of these points are joined with a path. We suppose that for every point of $ E,$ there exist at least 1593 other points of $ E$ to which it is joined by a path. Show that there exist six points of $ E$ every pair of which are joined by a path.\r\n\r\n[i]Alternative version:[/i] Is it possible to find a set $ E$ of 1991 points in the plane and paths joining certain pairs of the points in $ E$ such that every point of $ E$ is joined with a path to at least 1592 other points of $ E,$ and in every subset of six points of $ E$ there exist at least two points that are not joined?", - "Solution_1": "The wording is a bit iffy for those who know graph theory. Here, by \"path\" they mean the graph-theoretic edge - not path.\r\n\r\nLet's say we have a point $ Q_{n\\plus{}1}$ in a set $ S_n$ of points connected by an edge to points $ Q_1,Q_2,...,Q_n$ and we want to determine if $ Q_{n\\plus{}1}$ is connected to another vertex in $ S_n$. Then we simply have to check whether $ 1991\\minus{}|S_n|\\minus{}n< 1593$. Then we are assured that there exists a set $ S_{n\\plus{}1}\\in S_n$ with $ |S_{n\\plus{}1}|\\ge|S_n|\\plus{}n\\plus{}1593\\minus{}1991\\equal{}|S_n|\\plus{}n\\minus{}398$ and for all $ P\\in S_{n\\plus{}1}$, $ P$ is connected by an edge to points $ Q_1,Q_2,...,Q_{n\\plus{}1}$.\r\n\r\nSince the sum of the degrees of a graph must be even, we must have one vertex with degree at least $ 1594$. It follows that $ |S_1|\\ge 1594$. Then we have $ |S_2|\\ge1197,|S_3|\\ge801,|S_4|\\ge406,|S_5|\\ge12$. Since $ 12>5$, there exists $ Q_6\\in S_5$ so that $ Q_1,Q_2,Q_3,Q_4,Q_5,Q_6$ are our desired points.", - "Solution_2": "We use Zarankiewicz Lemma. \n\nAssume the graph is $6$-free. Then by Zarankiewicz lemma, there exists a vertex with at most degree $[\\frac{4}{5}*1991] = 1592$. However, this is a contradiction as all vertices have degree strictly greater than $1592$. ", - "Solution_3": " [quote=va2010]We use Zarankiewicz Lemma. \n\nAssume the graph is $6$-free. Then by Zarankiewicz lemma, there exists a vertex with at most degree $[\\frac{4}{5}*1991] = 1592$. However, this is a contradiction as all vertices have degree strictly greater than $1592$.[/quote]\nIn the actual contest though, (Indian National MO (INMO) in my case), would we need to prove the Zarankiewicz Lemma, or would we get full marks for just this much?\n" -} -{ - "Problem": "Let x is the the maximum natural number.\r\n\r\nx^2<=x ( because x greater than or equal to any natural number)\r\nx^2-x<=0\r\nx(x-1)<=0\r\n\r\nx is natural number x>0 ,x-1<=0 , x<=1\r\n\r\nx is natural number, so x=1.\r\n\r\nWhat is wrong?", - "Solution_1": "You know what is wrong :) \r\nThere is no maximal natural number.", - "Solution_2": "What you've proven is true: if there is a maximal natural number, then it is equal to 1. However, it's true for the not-very-interesting reason that the premise is false." -} -{ - "Problem": "If $t = \\frac{1}{1 - \\sqrt[4]{2}}$, then $t$ equals\r\n\\[ \\text{(A)}\\ (1 - \\sqrt[4]{2})(2 - \\sqrt{2}) \\qquad \\text{(B)}\\ (1 - \\sqrt[4]{2})(1 + \\sqrt{2}) \\qquad \\text{(C)}\\ (1 + \\sqrt[4]{2})(1 - \\sqrt{2}) \\]\r\n\\[ \\text{(D)}\\ (1 + \\sqrt[4]{2})(1 + \\sqrt{2}) \\qquad \\text{(E)} -(1 + \\sqrt[4]{2})(1 + \\sqrt{2}) \\]", - "Solution_1": "[hide=\"My solution\"] let $x=\\sqrt{2}$ then $\\sqrt{x}=\\sqrt[4]{2}$ \nand $t = \\frac{1}{1 - \\sqrt[4]{2}}$=$\\frac{1}{1 -\\sqrt{x}}=\\frac{1+\\sqrt{x}}{1 - x}=\\frac{(1+\\sqrt{x)}(1+x)}{1 - x^2}=-(1 + \\sqrt[4]{2})(1 + \\sqrt{2})$[/hide]", - "Solution_2": "[hide=\"Answer\"]$t=\\frac{1}{1-\\sqrt[4]{2}}=\\frac{1+\\sqrt[4]{2}}{1-\\sqrt{2}}=\\frac{(1+\\sqrt[4]{2})(1+\\sqrt{2})}{-1}\\Rightarrow \\boxed{E}$[/hide]", - "Solution_3": "[hide]$t=\\frac{1}{1-\\sqrt[4]{2}}$\n\n$t=\\frac{1+\\sqrt[4]{2}}{1-\\sqrt{2}}$\n\n$t=\\frac{(1+\\sqrt[4]{2})(1+\\sqrt{2})}{1-2}$\n\n$t=-(1+\\sqrt[4]{2})(1+\\sqrt{2})$\n\n$\\boxed{E}$[/hide]" -} -{ - "Problem": "What is the shortest distance from the point(5,10) to the curve $ x^2\\equal{}12y$?", - "Solution_1": "Any point on the curve would be of the form $ (x,\\frac{x^2}{12})$. By the distance formula, we want to determine the minimum value of\r\n\r\n$ \\sqrt{(5\\minus{}x)^2 \\plus{} (10\\minus{}\\frac{x^2}{12})^2}$\r\n\r\nSince the square root function is increasing, we then wish to minimize\r\n\r\n$ (5\\minus{}x)^2 \\plus{} (10\\minus{}\\frac{x^2}{12})^2$\r\n\r\nThen, we differentiate and take the critical points. I used Mathematica and got a horrible answer. :|\r\n\r\nhttp://www.wolframalpha.com/input/?i=minimum+of+sqrt%28%285-x%29^2+%2B+%2810-x^2%2F12%29^2%29", - "Solution_2": "oh... hahaha!!! thanks for your answer... ^^" -} -{ - "Problem": "hey guys ,try this\r\na^5+b^5+c^5 >= 5abc(b2-ac)\r\nwith a,b,c being positive reals", - "Solution_1": "Let [tex] f(a,b,c)=a^5+b^5+c^5-5abc(b^2-ac)[/tex]\r\nEasily see that:[tex] f(a,b,c)\\geq f( \\sqrt{ac},b,\\sqrt{ac}) [/tex].\r\nSo take $a=c$ ,and we need to prove that:\r\n[tex] 2a^5+b^5-5a^2b(b^2-a^2)\\geq 0[/tex]\r\nTake a=1 then we have:\r\n[tex] b^5-5b^3+5b+2 \\geq0 [/tex]\r\nAnd it is easily now. :)", - "Solution_2": "Please, a solution for \r\n\r\n$ b^5 + 5b +2 \\geq 5b^3 $\r\n\r\nYou can assume $ b > 1 $", - "Solution_3": "$b^5-5b^3+5b+2=(b+2)(b^2-b-1)^2\\geq 0$", - "Solution_4": "Yes,I am really stupid :blush:" -} -{ - "Problem": "Assume $\\phi: \\mathbb R\\to\\mathbb R$ is a differentiable function such that ($L$ and $K$ are two real numbers):\r\n\r\n$\\int_1^{\\infty}\\frac{\\phi(x)}{x^2}dx=L$ and $\\phi'(x)\\leq K$, $\\forall x\\geq 0$\r\n\r\nThen $\\lim_{x\\to+\\infty}\\frac{\\phi(x)}{x}=0$.", - "Solution_1": "One can take $K=1$:\r\nsuppose that $\\frac{\\phi(x)}{x}$ does not tends towards $0$. WLOG, then there exists a sequence of real numbers $x_n>1$ such that $x_{n}$ tends towards $+\\infty$ and $\\frac{\\phi(x_n)}{x_n} > a$, where $a>0$ is a certain real. \r\n\r\n1/first one show that the function stays positive during a long time after $x_n$\r\n\r\nTake such a real number $x_n$:\r\nLet $y$ the smallest real number such that $y>x_n$ and $\\phi(y)=0$, and $y=+ \\infty$ if it does not exists.\r\n$\\frac{\\phi(x)}{x} > \\frac{\\phi(x)}{y}$ for all $x \\in [x_n; y]$.\r\nBut $\\frac{\\phi(x_n)}{y}>\\frac{ax_n}{y}$, and because $| \\phi'(x) | < 1$, $\\frac{\\phi(x)}{x}>\\frac{\\phi(x)}{y}>0$ for all $x \\in [x_n;(1+a)x_n)[$. \r\nThis is why $y \\geq (1+a)x_n$\r\n\r\n2/from here one can obtain a contradiction:\r\n\r\n$\\frac{\\phi(x)}{x} > \\frac{\\phi(x)}{(1+a)x_n}$ for all $x \\in [x_n; (1+a)x_n]$.\r\nBut $\\frac{\\phi(x_n)}{(1+a)x_n}>\\frac{a}{1+a}$, and because $| \\phi'(x) | < 1$, $\\frac{\\phi(x)}{(1+a)x_n}>\\frac{a}{2(1+a)}$ for all $x \\in [x_n;(1+C)x_n]$ where $C=\\frac{a}{2}$ .\r\n\r\n\r\nThis is why $\\int_{x_n}^{(1+C)x_n} \\frac{\\phi(x)}{x^2}dx>\\frac{1}{(1+C)x_n}\\int_{x_n}^{(1+C)x_n} \\frac{\\phi(x)}{x}dx>\\frac{1} {(1+C)x_n} C x_n \\frac{a} {2(1+a)} = E > 0$, where $E$ is a constant independant of $x_n$.\r\nFrom this we obtain that $\\int_{1}^{+\\infty} \\frac{\\phi(x)}{x^2}dx$ does not exist, contradiction.", - "Solution_2": "it is not entirely correct. you use $|\\phi'(x)|\\leq 1$ yet only $\\phi'(x)\\leq 1$ is given, and that actually makes a difference. \r\n\r\nalso, you say at some point that $\\frac{\\phi(x)}{(1+a)x_n}>\\frac{a}{2(1+a)}$, which is incorrect since $x_n$ itself is large.", - "Solution_3": "actually it is almost correct:\r\n\r\n*$\\frac{\\phi(x_n)}{(1+a)x_n}>\\frac{a}{1+a}$. And because the derivative of $x\\to \\frac{\\phi(x)}{(1+a)x_n}$ is superior to $\\frac{-1}{(1+a)x_n}$ if we assume $|\\phi'(x)|<1$, then $\\frac{\\phi(x)}{(1+a)x_n}>\\frac{a}{2(1+a)}$ for all $x \\in [x_n; (1+\\frac{a}{2})x_n]$\r\n\r\n*but I just used the fact that $\\phi'(x)>-1$. The same idea works : it suffices to work on $[(1-\\frac{a}{2})x_n;x_n]$ instead of $[x_n; (1+\\frac{a}{2})x_n]$.", - "Solution_4": "this is indeed the sort of analysis that is needed here.\r\nbut you need more than that under the assumption that $\\phi'$ is bounded from above (below). as it is the argument doesnt stand. you can convince yourself of that." -} -{ - "Problem": "For any positive real numbers $a,\\ b$ and $c$.\r\n\\[ \\frac{c}{a(b+c)}+\\frac{a}{b(c+a)}+\\frac{b}{c(a+b)} \\ge \\frac{9}{2(a+b+c)} \\]", - "Solution_1": "Cauchy + AM-GM?\r\n\r\n$((b+c)+(c+a)+(a+b))\\left( \\frac{c}{a(b+c)}+\\frac{a}{b(c+a)}+\\frac{b}{c(a+b)} \\right ) \\geq \\left(\\sqrt{\\frac ca} + \\sqrt{\\frac ab} + \\sqrt{\\frac bc} \\right)^2 \\geq 9$" -} -{ - "Problem": "Source: ACoPS\r\n(Hungary, 1899) Let $r$ and $s$ be the roots of \r\n$x^{2}-(a+d)x+(ad-bc)=0$.\r\nProve that $r^{3}$ and $s^{3}$ are the roots of \r\n$y^{2}-(a^{3}+d^{3}+3abc+3bcd)y+(ad-bc)^{3}=0$.", - "Solution_1": "From ?vietta's? sums $r+s=a+d$ and $rs=ad-bc$. Thus $r^{3}+s^{3}=(r+s)^{3}-3rs(r+s)=(a+d)^{3}-3(ad-bc)(a+d)=a^{3}+3a^{2}d+3ad^{2}+d^{3}-3a^{2}d+3abc-3ad^{2}+3bcd=a^{3}+d^{3}+3abc+3bcd$. Also, $(rs)^{3}=(ad-bc)^{3}$. Therefore the roots of $y^{2}-(a^{3}+d^{3}+3abc+3bcd)y+(ad-bc)^{3}$ are $r^{3}$ and $s^{3}$.", - "Solution_2": "[quote=\"drunner2007\"]From ?vietta's? sums $r+s=a+d$ and $rs=ad-bc$. Thus $r^{3}+s^{3}=(r+s)^{3}-3rs(r+s)=(a+d)^{3}-3(ad-bc)(a+d)=a^{3}+3a^{2}d+3ad^{2}+d^{3}-3a^{2}d+3abc-3ad^{2}+3bcd=a^{3}+d^{3}+3abc+3bcd$. Also, $(rs)^{3}=(ad-bc)^{3}$. Therefore the roots of $y^{2}-(a^{3}+d^{3}+3abc+3bcd)y+(ad-bc)^{3}$ are $r^{3}$ and $s^{3}$.[/quote]\r\nYes, that's how i did..this was pretty easy, since all u have to do is just applying Vieta's theorem.", - "Solution_3": "yeah its a little too easy for acops" -} -{ - "Problem": "Day 1\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=246409]Problem 1.[/url] Let $a_{1},..., a_{n},...$ be positive real numbers such that $\\frac{a_{n}+a_{n+2}}{2}< a_{n+1}$ for every $n \\geq 1$. Show that $a_{n}< a_{n+1}$ for every $n \\geq 1$.\r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=246410]Problem 2.[/url] An acute triangle $ABC$ is given. Find the locus of points $M$ in the interior of $ABC$ such that $AB-FG = \\frac{MF \\cdot AG+MG \\cdot BF}{CM}$ where $F$ and $G$ are the feet of the perpendiculars from $M$ to $BC$ and $AC$ respectively. \r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=246413]Problem 3.[/url] Find all positive intergers $n$ such that $2^{\\phi(n)}-1$ divides $n^{n}$, where $\\phi$ is the Euler $\\phi$-function. \r\n\r\nDay 2\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=247004]Problem 4.[/url] Let $n \\geq 2$ be a natural number. A pyramid $P$ has base $A_{1}A_{2}...A_{2n}$ and apex $O$. The polygon $A_{1}A_{2}...A_{2n}$ is regular and the point $C$ is its centre. The line $OC$ is perpendicular to the plane of the base of $P$. A sphere passes through $O$ and meets each of the line segments $OA_{i}$ internally. For each $i = 1, 2,..., 2n$, let $X_{i}$ be the point (other than $O$) where the sphere meets $OA_{i}$. Prove that $OX_{1}+OX_{3}+...+OX_{2n-1}= OX_{2}+OX_{4}+...+OX_{2n}$ \r\n \r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=247003]Problem 5.[/url] Find the number of subsets $B$ of $\\{1,2,3,...,2005\\}$ such that the sum of the elements of $B$ is congruent to $2006 \\mod 2048$. \r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=247002]Problem 6.[/url] Let $n \\geq 3$ be an integer. Consider positive real numbers $a_{1},..., a_{n}$ such that their product is 1. Show that: $\\sum_{i=1}^{n}\\frac{a_{i}+3}{(a_{i}+1)^{2}}\\geq 3$ \r\n\r\nPS: Can this be a sticky and/or be listed in the front?", - "Solution_1": "Disastrous, I solved 3 on the first day and none on the next.", - "Solution_2": "[quote=\"kueh\"]PS: Can this be a sticky and/or be listed in the front?[/quote]Yes it can :D When will be the results available? :)", - "Solution_3": "The team for Mexico is\r\n\r\nSaul Glasman (myself)\r\nNathan Kettle\r\nAndre Kueh (kueh)\r\nMatthew Lee\r\nMartin Orr\r\nJack Shotton", - "Solution_4": "Yet two more MathLinkers present at the IMO :D Congrats guys! :)", - "Solution_5": "It seems we must provide all the ML-shirts at the beginning of the IMO opening ceremony so that the MLer have enough time to get to know each other. :)", - "Solution_6": "Well, I have now joined, to make it three (I am Jack)", - "Solution_7": "Yay it's Jack!", - "Solution_8": "Congratulatinos to all UK students on the team", - "Solution_9": "congratulations to you! thanks valentin (=", - "Solution_10": "All except for the solid geometry (which was plucked from the air) was from the Balkan Olympiad Shortlists.", - "Solution_11": "You all from England, me too :D :D :D :D for a moment I thought.... nevermind. :roll:" -} -{ - "Problem": "There is 6 points on a plane.Every 2 point is connected by a segment.From 6 points perpendiculars are drawen to this line segments.Find the maximum value of\r\nintersection points of these perpendiculars.(perpendiculars dont have length they are unlimited from both points)", - "Solution_1": "max no of poit of itersection =3780", - "Solution_2": "can you post whole solution i think there are less than 1500 points", - "Solution_3": "still nobody solved :S its not hard question \r\nhint[hide]3 heights in a triangle intersects on same point[/hide]" -} -{ - "Problem": "A hand of cards consists of all 13 hearts from an ordinary pack. The cards are arranged in random order. Find the probability that the ace is somewhere in front of the king.\r\nIf anybody is able to answer this, please could you show how you worked it all out. Thanks", - "Solution_1": "[quote=\"nuerofuzz\"]A hand of cards consists of all 13 hearts from an ordinary pack. The cards are arranged in random order. Find the probability that the ace is somewhere in front of the king.\nIf anybody is able to answer this, please could you show how you worked it all out. Thanks[/quote]\r\n\r\nWhat do you mean with \"in front of\" ?\r\n\r\nIf it is \"before\", then the probability is the same as the probability of having the king before the ace, so $ \\frac{1}{2}$\r\n\r\nIf it is \"beside\", then :\r\nYou have $ 13!$ ordered positions.\r\nYou have $ 12!$ ordered positions of the $ 13$ cards less ace and then $ 2$ possibilities to place the remaining king (left or right of ace)\r\nSo the requested probability is $ \\frac{12!*2}{13!}=\\frac{2}{13}$\r\n\r\nIf it is anything else, please give some precision on the meaning of the question.\r\nThanks.", - "Solution_2": "the answer is 1/2. Thanks for such a clear answer." -} -{ - "Problem": "For [i]n [/i]distinct points chosen randomly and independently in the unit interval, what is the expected length of the largest sub-interval.\r\n\r\nWhere \"randomly\" means uniform distribution and the \"largest subinterval\" is the maximum of {$x_1,x_2-x_1,x_3-x_2,\\dots,x_n-x_{n-1},1-x_n$} where $0\\leq x_1\\leq x_2\\leq \\dots\\leq x_n\\leq 1$.", - "Solution_1": "The differences are distributed uniformly on the simplex $u_i>0,\\sum u_i=1$. This is not obvious, but it can be routinely proven by induction.\r\n\r\nNow, the largest coordinate- that's ugly. It comes down to finding the center of mass of an irregular box-like region (quadrilateral when $n=3$, hexahedral with no two faces parallel when $n=4$, etc.)", - "Solution_2": "[quote=\"jmerry\"]The differences are distributed uniformly on the simplex $u_i>0,\\sum u_i=1$. This is not obvious, but it can be routinely proven by induction. [/quote]\r\n\r\nAll that is needed is that the probability distribution on the differences is cyclically invariant.\r\nThus, one can assume the longest interval is the last one, $[x,1]$. The probability density of such configurations (among all $n$-point configurations) is proportional to $x^{n-1}$. The integral of that density is $1/n$ so that the correctly normalized probability density is $n x^{n-1}$. The answer to the problem is therefore $\\int_0^1 (1-x) n x^{n-1} dx = \\frac{1}{n+1}$.", - "Solution_3": "[quote=\"fleeting_guest\"]\nAll that is needed is that the probability distribution on the differences is cyclically invariant.\n[/quote]\r\nYou mean that for example $1-x_n$ and $x_{n}-x_{n-1}$ has the same distribution??? ($x_{n-1} 1$ for all $ x\\in \\mathbb{N}.$", - "Solution_1": "This doesn't feel like the \"right\" proof for a) (mainly on aesthetic grounds...more on this after the argument), but here goes anyways:\r\n\r\nWe will actually prove the following:\r\n$ \\sum_{k\\equal{}0}^n (\\minus{}1)^k 2^{2n\\plus{}1\\minus{}2k} \\binom{2n\\plus{}1\\minus{}k}{k}\\equal{}2(n\\plus{}1)$\r\n\r\nWe view the left hand side as follows: How many ways are there to tile a $ 2n\\plus{}1 \\times 1$ strip by red $ 1 \\times 1$ squares, blue $ 1 \\times 1$ squares, and uncolored $ 2 \\times 1$ dominoes? To do so, we first need to pick some number $ k$ of dominoes (at least 0, at most $ n$). There will be $ 2n\\plus{}1\\minus{}2k$ $ 1 \\times 1$ squares, so $ \\binom{2n\\plus{}1\\minus{}k}{k}$ ways to place the dominoes relative to the squares. We then color the squares in one of the $ 2^{2n\\plus{}1\\minus{}2k}$ ways. \r\n\r\nIn other words, the left side would exactly count these arrangements, except for that $ (\\minus{}1)^k$ term. So now what we aim to do is to \"pair off\" the arrangements with an even number of dominoes with those with an odd number. We do so as follows: Start from the left of the row of blocks and move right until we encounter either\r\n\r\na) a domino with a $ 1 \\times 1$ square immediately to its left. We replace it by two identical squares of the opposite color of that $ 1 \\times 1$ square. Or \r\n\r\nb) a pair of identically colored squares with a $ 1 \\times 1$ square of the opposite color immediately to their left. We replace them by a domino.\r\n\r\nThis operation is its own inverse, and changes the parity of the number of dominoes, so the sum cancels everywhere the operation is defined (as it cancels on each pair $ (x, f(x))$). However, there are some configurations where it is not defined. \r\n\r\nWhat are those configurations? No domino can appear to the right of a colored square (else we could apply a) ), so they consist of some number of dominoes, then colored squares thereafter. To avoid being able to apply b), the colored squares must consist of some number of identically colored squares, followed by squares alternating in color.\r\n\r\nIn other words, we can define our remaining configurations by picking some number $ k$ of dominoes to start with, picking the color of the next square (2 choices), and choosing how long the next stretch of that color is ($ 2n\\plus{}1\\minus{}2k$ choices). There are therefore $ 2(2n\\plus{}1\\minus{}2k)$ choices with $ k$ dominoes remaining, so the whole sum is $ 2( (2n\\plus{}1)\\minus{}(2n\\minus{}1)\\plus{}(2n\\minus{}3)\\minus{}...\\plus{}1)\\equal{}2(n\\plus{}1)$.\r\n\r\n----------------------------\r\nNow, at the beginning I said I thought this was the wrong argument. There are two reasons for this.\r\n\r\n1. It doesn't generalize to part $ b$. But I don't think any pure combinatorial proof would do that anyways, so I don't mind that too much.\r\n\r\n2. I dislike that I still had more counting and cancellation even after defining my involution. The \"right\" involution would have left exactly $ 2(n\\plus{}1)$ elements unpaired, all of which would have an even number of dominoes (0, perhaps?)\r\n\r\nTherefore my challenge is\r\n\r\n[b]Define a better pairing on the configurations I defined in my proof, in the sense of objection 2.[/b]", - "Solution_2": "[quote=\"kevinatcausa\"]\n\n1. It doesn't generalize to part $ b$. But I don't think any pure combinatorial proof would do that anyways, so I don't mind that too much.\n\n[/quote]\r\nAre you sure ? :lol:", - "Solution_3": "Not at all actually. I guess I was misled somewhat by the notation $ p(n)$...there's no real reason to think of it as a function of $ n$ at all, at least for my proof method. Here's a reworked proof, with what I consider to be the \"right\" bijection:\r\n\r\nAs before, we view the left hand side as counting colorings of a strip of length $ p(n)$ with red and blue $ 1 \\times 1$ blocks and colorless $ 2 \\times 1$ dominoes, with signs determined by the number of dominoes. Given a configuration $ A$, we define $ f(A)$ as follows.\r\n\r\nRead from left to right until the first time we encounter either a domino or a red block immediately to the left of a blue block. We replace the domino by Red/Blue, and Red/Blue by a domino. As $ f(f(A)) \\equal{} A$, the sum over all pairs $ (A,f(A))$ is 0.\r\n\r\nWhat's left is all the colorings containing no dominoes, and no red blocks to the left of blue blocks. In other words, there is some number $ j$ of blue blocks ($ 0 \\leq j \\leq p(n)$), followed by red blocks. There are $ p(n) \\plus{} 1$ choices for $ j$, so the sum is $ p(n) \\plus{} 1$." -} -{ - "Problem": "$ L_k(x) $ a sequence of polynomials defined by $L_0(x) = 1 $, $L_1(x) = 1+x, L_k (x) = L_{k-1}(x) + xL_{k-2}(x) $for $ k>=2 $. Give an expression of $ L_k(x)$\r\nTHanx A lOT..\r\nI'm WaiTin' for an Answer..!!", - "Solution_1": "Not exactly a calculus problem ...\r\n\r\nJust running the recursion for a while (a spreadsheet helps) and looking for patterns, we see that\r\n\r\n$L_k(x)=\\sum_{j=0}^{\\lfloor\\frac{k+1}2\\rfloor}{{k+1-j}\\choose j}x^j$\r\n\r\nNote: $L_k(1)=F_{k+1},$ where $F_n$ is the $n$th Fibonacci number.", - "Solution_2": "[quote=\"Kent Merryfield\"]Not exactly a calculus problem ...\n\nJust running the recursion for a while (a spreadsheet helps) and looking for patterns, we see that\n\n$L_k(x)=\\sum_{j=0}^{\\lfloor\\frac{k+1}2\\rfloor}{{k+1-j}\\choose j}x^j$\n\nNote: $L_k(1)=F_{k+1},$ where $F_n$ is the $n$th Fibonacci number.[/quote]\r\nthanx a alot H :roll: OW CAN I thank u!!!", - "Solution_3": "[quote=\"various\"]thanx a alot H :roll: OW CAN I thank u!!![/quote]\r\n\r\nBy contributing nontrivial solutions/ideas/information in these bulletin boards.\r\n\r\n\"From each according to his abilities, to each according to his needs.\"" -} -{ - "Problem": "The sum of two numbers $ x$ and $ y$ is 399, and the value of the fraction $ \\frac{x}{y}$ is 0.9. What is the value of $ y \\minus{} x$?", - "Solution_1": "hello, with $ x\\equal{}\\frac{9}{10}y$ we get $ \\frac{19}{10}y\\equal{}399$, from here we have $ x\\equal{}189$ and $ y\\equal{}210$, so\r\n$ y\\minus{}x\\equal{}21$.\r\nSonnhard." -} -{ - "Problem": "an irregular quadrangle ABCD,E,G and F,H trisect AD and BC,then prove\r\nthat the sguare measure of quadrangle EFGH equals 1/3 of the sguare measure of quadrangle ABCD.", - "Solution_1": "Is \"sguare measure\" area?", - "Solution_2": "Yes,I'm not good at this", - "Solution_3": "[quote]Let ABCD be a quadrilateral. Points E and H trisect AD and points F and G trisect BC. Show that the area of quadrilateral EFGH is one third the area of ABCD.[/quote]\r\n\r\n[hide=\"Steps to take\"]First, you want some casework: if $AD \\parallel BC$, you have trapezoids and the result follows easily from the area of a trapezoid. Otherwise, extend $AD$ and $BC$ until they meet at a point $J$ and then use what you know about triangle areas (most importantly, that if you fix side $XY$ and angle $\\angle X$ of $\\triangle XYZ$, then the area of $\\triangle XYZ$ is proportional to the length $XZ$. After that, it should all fall out with a little algebra.[/hide]", - "Solution_4": "Thank you, I have solved it." -} -{ - "Problem": "Ellen has five different jobs to be done. She assigns all five jobs to her four kids. Each kid will have at least one job. How many ways can Ellen assign the jobs?", - "Solution_1": "[hide=\"solution\"]well, there are 5 jobs for the first kid, 4 for the second, 3 for the 3rd, 2 for the fourth, last job can go to any of them, so x4. Also multiply by 4 for each of the four kids for the jobs being switched between them. this is $5\\cdot4\\cdot3\\cdot2\\cdot4\\cdot4 = \\boxed{1920}?$\n[/hide]", - "Solution_2": "This has been answered already, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=jobs&t=135461]here.[/url]\r\n\r\n[hide]One kid will get two jobs. There are $\\binom{5}{2}$ ways to choose those two jobs. If we consider those to be one single job altogether, we can assign it and the other three jobs to the kids in $4!$ ways. $\\binom{5}{2}\\cdot4!=\\boxed{240}.$[/hide]", - "Solution_3": "oops, i was kinda way off. No wonder i didn't make nats..." -} -{ - "Problem": "$ f: [0,1] \\rightarrow R$ continous and $ F: [0,1]\\rightarrow R$, $ F(x)\\equal{}\\int_{0}^{1}min(x,t)f(t)dt$\r\na)Prove that $ F^{\\prime\\prime}(x)\\plus{}f(x)\\equal{}0$\r\nb) find $ \\lim_{x\\to1}\\frac{F(x)}{x\\minus{}1}$", - "Solution_1": "you just need to write it as :\r\n$ F(x)\\equal{}\\int_{0}^{1} \\min(x,t) f(t)dt \\equal{} \\int_{0}^{x}tf(t) dt \\plus{}x\\int_{x}^{1}f(t)dt$\r\nNow is easy to answer on your question :)", - "Solution_2": "a) $ F(x)\\equal{}\\int_{0}^{x} tf(t) dt \\plus{} x\\int_{x}^{1} f(t)dt$.\r\n So $ F'(x)\\equal{}xf(x)\\minus{}xf(x)\\plus{}\\int_{x}^{1}f(t)dt \\Rightarrow F''(x)\\equal{}\\minus{}f(x)$\r\n\r\nb) by using l'Hospital if $ \\int_{0}^{1}tf(t)dt \\equal{}0$ we get $ \\lim_{x \\rightarrow 1} \\frac{F(x)}{x\\minus{}1}\\equal{}\\lim_{x \\rightarrow 1} F'(x) \\equal{} \\lim_{x \\rightarrow 1} \\int_{x}^{1}f(t)dt \\equal{}0$; if $ \\int_{0}^{1} tf(t) dt >0$ then the limit is $ \\minus{}\\infty$ or if $ \\int_{0}^{1} tf(t)dt<0$ then limit is $ \\plus{}\\infty$." -} -{ - "Problem": "prove that\r\n\r\n\r\n$ \\frac{f(a)\\plus{}f(b)}{2}\\equal{}f ( \\frac{a\\plus{}b}{2} )\\plus{} \\frac{(b\\minus{}a)^2}{8}\\cdot f {''}(c)$", - "Solution_1": "You forget many thigs $ f\\in D^{(2)}[a,b],$ then exist $ c\\in (a,b)$", - "Solution_2": "yes :blush:" -} -{ - "Problem": "Sorry, but I haven't been able to find this on the wiki...\r\n\r\nHow do you create a large brace for systems of equations? Thanks!", - "Solution_1": "Well actually, I dunno if you want something like\r\n\r\n[code]\\begin{tabular}{rcl|}\n$$ & = & $$ \\\\\n$$ & = & $$ \\\\ \\hline\n\\end{tabular}[/code]", - "Solution_2": "Something like this:", - "Solution_3": "Ohhh, that's easy\r\n\r\n$ x_{k}\\equal{}\\left\\{\\begin{array}{cr}0 &\\text{if }x_{k\\minus{}1}\\equal{} 0,\\\\ \\left\\{\\frac{p_{k}}{x_{k\\minus{}1}}\\right\\}&\\text{if }x_{k\\minus{}1}\\ne0,\\end{array}\\right.$\r\n\r\n[color=red]x_k=\\left\\{\\begin{array}{cr}0&\\text{if }x_{k-1}=0,\\\\\n\\left\\{\\frac{p_k}{x_{k-1}}\\right\\}&\\text{if }x_{k-1}\\ne0,\\end{array}\\right.[/color]\r\n\r\nDoes that make sense?", - "Solution_4": "The cases environment is simpler\r\n$ x_{k}\\equal{}\\begin{cases}0 &\\text{if }x_{k\\minus{}1}\\equal{}0,\\\\ \\left\\{\\frac{p_{k}}{x_{k\\minus{}1}}\\right\\}&\\text{if }x_{k\\minus{}1}\\ne 0\\end{cases}$\r\n\r\n[code]x_k=\n\\begin{cases}\n0 & \\text{if } x_{k-1}=0,\\\\ \n\\left\\{\\frac{p_k}{x_{k-1}}\\right\\} & \\text{if } x_{k-1} \\ne 0\n\\end{cases}[/code]", - "Solution_5": "Thanks for your help! And what package is the cases environment included in?", - "Solution_6": "It's in amsmath which you should use in every document along with amssymb (which means you can write $ x\\in\\mathbb{R}$ for example)." -} -{ - "Problem": "Sorry, if this question has been posted before. Is there a formula to find even positive divisors of a number. I know there's formula to determine number of divisors of a number. \r\n\r\nI've to find number of even positive divisors of 22680\r\n$22680 = 2^{3}* 3^{4}* 5 * 7$ and the number of divisors are $(3+1)*(4+1)*(1+1)*(1+1) = 80$\r\n\r\nCould someone help me to find the even positive divisors of 22680, thanks in advance!", - "Solution_1": "It's very similar; what is special about an even number? It has at least one power of $2$. So the exponent on $2$ is $\\ge 1$. Hence there are $3 \\cdot 5 \\cdot 2 \\cdot 2 = 60$ even divisors.", - "Solution_2": "Alternately, you could figure out what the odd divisors are, which means there can't be any powers of $2$. This is equivalent to finding the divisors of\r\n\r\n$2835 = 3^{4}\\cdot 5 \\cdot 7$\r\n\r\nAnd subtracting this number from your original number.\r\n\r\nEdit: This is a great example of an occasion where it helps you more to understand the concepts behind a rule (in this case, finding the number of divisors of an integer) than to remember the rule itself. ;)", - "Solution_3": "[quote=\"t0rajir0u\"]Alternately, you could figure out what the odd divisors are, which means there can't be any powers of $2$. This is equivalent to finding the divisors of\n\n$2835 = 3^{4}\\cdot 5 \\cdot 7$\n\nAnd subtracting this number from your original number.[/quote]\r\n\r\nyep that's what i did, thanks for the help guys", - "Solution_4": "Or you count divisors of $\\frac{n}{2}$.", - "Solution_5": "[b]Ok..Lemme twist it a lil bit...Find all the divisors of 22680 of the form 1)4k 2)4k+1 3)4k+2 4)4k+3 and 5)5k 5)5k+1 6)5k+2 7)5k+3 8)5k+4 ???????? [/b]" -} -{ - "Problem": "John and a group of friends took a bus trip. Each person paid the bus driver with the same 9 coins. If the bus driver recieved $8.41 from the group, how many dimes did he recieve?", - "Solution_1": "[hide]\n$ 841 \\equal{} 29^2$\n\nSo there are 29 people and each paid 29 cents. If we have a quarter, it's impossible to have 9 coins, since there are only 4 cents left, so assume there are only pennies, nickels and dimes:\n$ p \\plus{} n \\plus{} d \\equal{} 9$\n$ p \\plus{} 5n \\plus{} 10d \\equal{} 29$\nSubtract the first from the second, \n$ 4n \\plus{} 9d \\equal{} 20$\nIt's easy to find a solution $ (4,5,0)$, and we can verify there's no other positive integer solutions.\n\nTherefore, there are no dimes.[/hide]\n[/hide]" -} -{ - "Problem": "Let G be a tree, and the average of the degrees of its vertices is 1:5. How\r\nmany vertices are in G?", - "Solution_1": "Use the identities $ \\sum_{v\\in V(G)}d(v)\\equal{}2e(G)$ and $ e(G)\\equal{}|V|\\minus{}1$.", - "Solution_2": "Thank you! Appreciate it. I knew there were identities, but I couldn't find them in my notes." -} -{ - "Problem": "[b]Let be a,b and c non-negative reals numbers. Show that[/b]\r\n\r\n$\\frac{b+c}{a^{2}}+\\frac{a+c}{b^{2}}+\\frac{a+b}{c^{2}}\\geq2(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})$\r\n\r\n[hide]We can suppose $a\\geq b\\geq c$, so $\\frac{1}{c}\\geq\\frac{1}{b}\\geq\\frac{1}{a}$ and $\\frac{a+b}{c}\\geq\\frac{a+c}{b}\\geq\\frac{b+c}{a}$ , then, we can use Chebychev's inequality :\n\n$\\frac{b+c}{a^{2}}+\\frac{a+c}{b^{2}}+\\frac{a+b}{c^{2}}\\geq\\frac{1}{3}(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})(\\frac{b+c}{a}+\\frac{a+c}{b}+\\frac{a+b}{c})$\n\nBut, always from Chebychev's inequality, we have,\n\n$\\frac{b+c}{a}+\\frac{a+c}{b}+\\frac{a+b}{c}\\geq\\frac{2}{3}(a+b+c)(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})$\n\n\nNext, we will use this inequality, defined for all x > 0\n\n$x+\\frac{1}{x}\\geq2$\n\n$(a+b+c)(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})=3+(\\frac{a}{b}+\\frac{b}{a})+(\\frac{a}{c}+\\frac{c}{a})+(\\frac{b}{c}+\\frac{c}{b})\\geq3+2+2+2=9$\n\nSo,\n\n$\\frac{b+c}{a}+\\frac{a+c}{b}+\\frac{a+b}{c}\\geq\\frac{2}{3}*9=6$\n\nAnd finally,\n\n$\\frac{b+c}{a^{2}}+\\frac{a+c}{b^{2}}+\\frac{a+b}{c^{2}}\\geq\\frac{1}{3}*6*(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})=2(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c})$\n\nwith egality iff a = b= c[/hide]", - "Solution_1": "[hide=\"much simpler\"]\nit's equivalent to\n\\[(a+b+c)\\left( \\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}\\right) \\geq 3\\left( \\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)\\]\nafter adding $\\frac{1}{a}$ to $\\frac{b+c}{a^{2}}$, etc. Now the above inequality is evident by Chebyshev.\n[/hide]", - "Solution_2": "[hide=\"Another solution\"] \n$ \\frac {b}{a^2}\\plus{}\\frac 1b\\geq \\frac2a$\n$ \\frac c{a^2}\\plus{}\\frac 1c\\geq \\frac 2a$\nSo summing up;\n$ \\frac{c\\plus{}b}{a^2}\\geq \\frac 4a\\minus{}\\frac 1b\\minus{}\\frac 1c$\nSimilarly we get three other relations, and summing up again gives the desired result. :) [/hide]", - "Solution_3": "[hide=\"Solution\"]$ b(\\frac {1}{a^2} \\plus{} \\frac {1}{c^2}) \\ge b(\\frac {2}{ac})$ by AM-GM, so we have to show\n$ \\frac {b}{ac} \\plus{} \\frac {c}{ab} \\plus{} \\frac {a}{bc} \\ge \\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c} \\iff a^2 \\plus{} b^2 \\plus{} c^2 \\ge$ $ ab \\plus{} bc \\plus{} ca \\iff (a \\minus{} b)^2 \\plus{} (b \\minus{} c)^2 \\plus{} (c \\minus{} a)^2 \\ge 0$ which is obvious.[/hide]", - "Solution_4": "My solution:\r\nNotice that:\r\n \r\n\r\n $ \\frac{b\\plus{}c}{a^2}\\plus{}\\frac{a\\plus{}c}{b^2}\\plus{}\\frac{a\\plus{}b}{c^2}\\equal{}(\\frac{a}{b^2}\\plus{}\\frac{b}{a^2})\\plus{}(\\frac{b}{c^2}\\plus{}\\frac{c}{b^2})\\plus{}(\\frac{c}{a^2}\\plus{}\\frac{a}{c^2}).$\r\n\r\nBut the inequality: $ \\frac{a}{b^2}\\plus{}\\frac{b}{a^2} \\geq \\frac{1}{a}\\plus{}\\frac{1}{b}$ is true since it's equivalent to $ a\\plus{}b \\geq a\\plus{}b$\r\n\r\nSimilarly we get the other two inequalities like above and by adding all three inequalities we get the desired result.\r\n\r\n~Kostas", - "Solution_5": "expanding , we\u00b4ll have to prove that that $ b^2c^2(b \\plus{} c) \\plus{} a^2b^2(a \\plus{} b) \\plus{} a^2c^2(a \\plus{} c)\\geq2abc(ab \\plus{} bc \\plus{} ca)$, which is straight forward from muirhead (the sequence (3,2,0)majorizes(2,2,1) ).", - "Solution_6": "[quote=\"new_member\"]My solution:\nNotice that:\n \n\n $ \\frac {b \\plus{} c}{a^2} \\plus{} \\frac {a \\plus{} c}{b^2} \\plus{} \\frac {a \\plus{} b}{c^2} \\equal{} (\\frac {a}{b^2} \\plus{} \\frac {b}{a^2}) \\plus{} (\\frac {b}{c^2} \\plus{} \\frac {c}{b^2}) \\plus{} (\\frac {c}{a^2} \\plus{} \\frac {a}{c^2}).$\n\nBut the inequality: $ \\frac {a}{b^2} \\plus{} \\frac {b}{a^2} \\geq \\frac {1}{a} \\plus{} \\frac {1}{b}$ is true since it's equivalent to $ a \\plus{} b \\geq a \\plus{} b$\n\nSimilarly we get the other two inequalities like above and by adding all three inequalities we get the desired result.\n\n~Kostas[/quote]\r\n\r\ni\u00b4m sorry, but how did you arrive that it is equivalent to $ a \\plus{} b \\geq a \\plus{} b$? i actually think that it is equivalentt to $ (a\\minus{}b)^2\\geq0$.\r\ncould clarify this , please?", - "Solution_7": "Hey v235711,\r\nI was in a hurry so yeah it's not equivalent to $ a\\plus{}b \\geq a\\plus{}b$.\r\nBut the inequality after multiplying with $ a^2,b^2$, gives:\r\n$ a^3\\plus{}b^3 \\geq ab(a\\plus{}b)$ which is a well-known inequality which is equivalent to:\r\n$ (a\\minus{}b)^2(a\\plus{}b) \\geq 0$ which is true since $ a,b>0$. So yeah,I am sorry about that.\r\n\r\n~Kostas" -} -{ - "Problem": "Find $ \\int \\frac{\\sqrt{1\\plus{}x^4}}{x^3}\\,dx$.", - "Solution_1": "$ {\\int\\frac{\\sqrt{x^4\\plus{}1}}{x^3}dx}$\r\n\r\nUse the supstitution:\r\n\r\n$ {x^2\\equal{}sinh(t)}$\r\n$ {2xdx\\equal{}cosh(t)dt}$\r\n$ {dx\\equal{}\\frac{cosh(t)}{2x}dt}$\r\n\r\n$ {\\int\\frac{\\sqrt{1\\plus{}sinh(t)^2}}{x^3}\\frac{cosh(t)}{2x}dt\\equal{}}$\r\n$ {\\int\\frac{cosh(t)^2}{x^4}dt\\equal{}}$\r\n$ {\\int\\frac{cosh(t)^2}{sinh(t)^2}dt\\equal{}}$\r\n$ {\\int{coth(t)^2}dt\\equal{}\\frac{1}{2}(t\\minus{}coth(t))\\equal{}\\frac{arcsinh(x^2)}{2}\\minus{}\\frac{cosh(t)}{2sinh(t)}\\equal{}\\frac{arcsinh(x^2)}{2}\\minus{}\\frac{\\sqrt{1\\plus{}x^4}}{2x^2}}$" -} -{ - "Problem": "Let $a,b,c,d,e,f$ be six real numbers with sum 10, such that \\[ (a-1)^2+(b-1)^2+(c-1)^2+(d-1)^2+(e-1)^2+(f-1)^2 = 6. \\] Find the maximum possible value of $f$. \r\n\r\n[i]Cyprus[/i]", - "Solution_1": "Let the a,b,c,d, and e expressions be all treated as (x-1)^2\r\nbecause they're not being asked to be maximized.\r\n\r\n5(x-1)^2 + (f-1)^2 = 6\r\n\r\nAnd because we have the variables adding up to 10, then we have:\r\n\r\n5x + f = 10\r\n\r\nThen f = 10 - 5x\r\n\r\nSubstituting,\r\n\r\n5(x-1)^2 +(9 - 5x)^2 = 6\r\n\r\n>\r\n>\r\n> Steps left out here\r\n>\r\n>\r\n\r\n30x^2 - 100x + 86 = 6\r\n\r\n30x^2 - 100x + 80 = 0\r\n\r\n3x^2 - 10x + 8 = 0\r\n\r\n(3x - 4)(x - 2) = 0\r\n\r\nx = 4/3, x = 2\r\n\r\nf = 10/3, f = 0\r\n\r\nThe maximum f = 10/3.", - "Solution_2": "You can do it the same way as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47256]1978 USAMO, #1[/url], but the numbers do not seem to come out as nice.", - "Solution_3": "Sorry to revive, but here is a solution, similar to what nsato was probably describing.\n\n[hide=\"Solution\"]Let $A=a-1$, $B=b-1$, $C=c-1$, $D=d-1$, $E=e-1$, $F=f-1$. Then $A^2+B^2+C^2+D^2+E^2+F^2=6$ and $A+B+C+D+E+F=4$.\n\nThen $A+B+C+D+E=4-F$ AND $A^2+B^2+C^2+D^2+E^2=6-F^2$. By Cauchy-Schwarz, we have $5(A^2+B^2+C^2+D^2+E^2) \\ge (A+B+C+D+E)^2 \\Rightarrow 5(6-F^2) \\ge (4-F)^2$. Expanding gives $30-5F^2 \\ge F^2-8F+16$ so $6F^2+8F-14 \\le 0 \\Rightarrow 3F^2+4F-7=(F-1)(3F+7) \\le 0$. Therefore $-\\frac{7}{3} \\le F \\le 1$ so $F$ is at most $1$. Therefore, $f$ is at most $1+1=\\boxed{2}$.[/hide]", - "Solution_4": "[quote=\"AwesomeToad\"]Let $A=a-1$, $B=b-1$, $C=c-1$, $D=d-1$, $E=e-1$, $F=f-1$. Then $A^2+B^2+C^2+D^2+E^2+F^2=6$ and $A+B+C+D+E+F=4$.\n\nThen ... so $F$ is at most $1$.[/quote]\nHow about $(A,B,C,D,E,F) = (0,0,0,1,1,2)$ ?", - "Solution_5": "[quote=\"AwesomeToad\"]... Expanding gives $30-5F^2 \\ge F^2-8F+16$ so $6F^2+8F-14 \\le 0 \\Rightarrow 3F^2+4F-7=(F-1)(3F+7) \\le 0$. Therefore $-\\frac{7}{3} \\le F \\le 1$ so $F$ is at most $1$. Therefore, $f$ is at most $1+1=\\boxed{2}$.[/quote]\nThe error stems from the fact that from $30-5F^2 \\ge F^2-8F+16$ follows $6F^2-8F-14 \\le 0$, i.e. $3F^2-4F-7=(F+1)(3F-7) \\le 0$. Therefore $-1 \\le F \\le \\frac{7}{3}$.\n\nEquality occurs when $A=B=C=D=E = \\frac {1} {3}$ and $F = \\frac {7} {3}$, thus for $a=b=c=d=e=\\frac {4} {3}$ and $f = \\frac {1} {3}$." -} -{ - "Problem": "I am looking for help. Does anyone know how to convert GSP files to Latex?", - "Solution_1": "I would use jPicedit. But for the best results this topic should go to the [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]LaTeX forum[/url]", - "Solution_2": "[quote=\"mastermathman\"]I am looking for help. Does anyone know how to convert GSP files to Latex?[/quote]You can't directly. Either save to a format LaTeX can read, such as eps, pdf, png, gif, jpg or take a screen shot of the picture, copy into Paint and save in one of these formats. Then include the picture into your tex document - see [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Pictures]LaTeX:Pictures[/url]." -} -{ - "Problem": "Suppose $ f$ is a bounded entire function on $ \\mathbb{C}$. Since all holomorphic functions on the compact Riemann surface $ C_{\\infty}$ (the Riemann sphere) are constant, can we find an extension of $ f$ on $ C_{\\infty}$ to prove Liouville's Theorem? If so, would it make sense to use the stereographic projection from the Riemann sphere to the complex plane?", - "Solution_1": "To find that extension, you'll have to prove that the singularity at $ \\infty$ is actually removable. Stereographic projection won't get you that argument, but looking at $ f\\left(\\frac1z\\right)$ will." -} -{ - "Problem": "We take $ x,y,z>0$ such that $ \\sqrt{xy}\\plus{}\\sqrt{yz}\\plus{}\\sqrt{zx}\\equal{}1$.Prove that $ 8xyz\\geq (1\\minus{}x)(1\\minus{}y)(1\\minus{}z)$.", - "Solution_1": "[quote=\"skylover\"]We take $ x,y,z > 0$ such that $ \\sqrt {xy} \\plus{} \\sqrt {yz} \\plus{} \\sqrt {zx} \\equal{} 1$.Prove that $ 8xyz\\geq (1 \\minus{} x)(1 \\minus{} y)(1 \\minus{} z)$.[/quote]\r\nLet $ x\\equal{}a^2,$ $ y\\equal{}b^2$ and $ z\\equal{}c^2,$ where $ a,$ $ b$ and $ c$ are non-negative. Then your inequality is equivalent to $ \\sum_{sym}(a^4b^2\\minus{}a^3b^3\\plus{}a^4bc\\minus{}2a^3b^2c\\plus{}a^2b^2c^2)\\geq0,$ which true by Schur." -} -{ - "Problem": "Calculate\r\n\\[ \\int_{0}^{\\pi}\\frac{\\cos (rx)}{1\\minus{}2a\\cos x\\plus{}a^2}\\;dx\\]", - "Solution_1": "hello, what do we know about $ a$ and $ r$?\r\nSonnhard.", - "Solution_2": "For any real numbers $ r$ and $ a$. :)", - "Solution_3": "But it diverges for $ |a|\\ge 1.$ And it's a nicer problem for $ r$ an integer than it is for $ r$ not an integer.\r\n\r\nIf we assume that $ 0\\le a<1,$ the following may be a useful formula:\r\n\r\n$ \\sum_{k\\equal{}\\minus{}\\infty}^{\\infty}a^{|k|}e^{ikx}\\equal{}1\\plus{}2\\sum_{k\\equal{}1}^{\\infty}a^k\\cos kx\\equal{}\\frac{1\\minus{}a^2}{1\\minus{}2a\\cos x\\plus{}a^2}.$", - "Solution_4": "Could you please elaborate why the integral is diverges when $ |a|\\geq1$ ?", - "Solution_5": "Oops, sorry about that. It certainly diverges for $ a\\equal{}1,$ since in that case the denominator would be $ 2\\minus{}2\\cos x\\equal{}4\\sin^2\\left(\\frac x2\\right),$ so the integral would be comparable to $ \\frac1{x^2}$ near zero.\r\n\r\nBut it turns out that $ a>1$ is OK. We can write $ 1\\minus{}2a\\cos x\\plus{}a^2\\equal{}|1\\minus{}ae^{ix}|^2,$ which is never zero for either $ a>1$ or $ a<1.$\r\n\r\nCan you find a different geometric sum, similar to the one I posted, but valid for $ a>1?$", - "Solution_6": "[quote=\"Kent Merryfield\"]\n$ \\sum_{k \\equal{} \\minus{} \\infty}^{\\infty}a^{|k|}e^{ikx} \\equal{} 1 \\plus{} 2\\sum_{k \\equal{} 1}^{\\infty}a^k\\cos kx \\equal{} \\frac {1 \\minus{} a^2}{1 \\minus{} 2a\\cos x \\plus{} a^2}.$[/quote]\r\n\r\nAt first, I confused with the second equality but now I understand. :lol: \r\n\r\nFor $ a>1$, $ \\sum_{k\\equal{}\\minus{}\\infty}^{\\infty}a^{\\minus{}|k|}e^{ikx}\\equal{}1\\plus{}2\\sum_{k\\equal{}1}^{\\infty}a^{\\minus{}k}\\cos kx$\r\n\r\n$ \\equal{}\\frac{1\\minus{}\\left(\\frac{1}{a}\\right)^{2}}{1\\minus{}2\\left(\\frac{1}{a}\\right)\\cos x\\plus{}\\left(\\frac{1}{a}\\right)^{2}}\\equal{}\\frac{a^{2}\\minus{}1}{1\\minus{}2a\\cos x\\plus{}a^{2}}$." -} -{ - "Problem": "Find the following derivative.\r\n\r\n[1] $\\frac{d}{dx} \\cos x$\r\n\r\n[2] $\\frac{d}{dx} \\sin 4x$\r\n\r\n[3] $\\frac{d}{dx} \\frac{(x-1)(x+2)^2}{(x+1)^3}$\r\n\r\n[4] $\\frac{d}{dx} \\sqrt{x^2+2x+3}$\r\n\r\n[5] $\\frac{d}{dx} \\frac{\\sin x}{x+1}$\r\n\r\n[6] $\\frac{d}{dx} x^{\\sin x}$\r\n\r\n[7] $\\frac{d}{dx} \\cos x\\sin ^ 2 x$\r\n\r\n[8] $\\frac{d}{dx} \\sqrt[4]{\\frac{x}{x+1}}$", - "Solution_1": "1\r\n$-\\sin{x}$\r\n2\r\n$\\frac{-\\cos{4x}}{4}$\r\n3\r\n$\\frac{[(x+2)^2+2(x+2)(x-1)](x+3)^3-3(x+3)^2(x-1)(x+2)^2}{(x+3)^6}$ dont want to simplify this :) \r\n4\r\n$\\frac{x+1}{\\sqrt{x^2+2x+3}}$\r\n5\r\n$\\frac{\\cos{x}(x+1)-\\sin{x}}{(x+1)^2}$\r\n7\r\n$2\\cos^2{x}\\sin{x}-\\sin^3{x}$\r\n8\r\n$\\frac{-1}{4(x+1)^2\\sqrt[4]{(\\frac{x}{x+1})^3}}$", - "Solution_2": "not sure for 6:\r\n$x^{\\sin{x}}(\\cos{x}\\ln{x}+\\frac{\\sin{x}}{x})$", - "Solution_3": "[quote=\"amirhtlusa\"]1\n$-\\sin{x}$ :) \n2\n$\\frac{-\\cos{4x}}{4}$ :( \n3\n$\\frac{[(x+2)^2+2(x+2)(x-1)](x+3)^3-3(x+3)^2(x-1)(x+2)^2}{(x+3)^6}$ dont want to simplify this :) Yes. :) \n4\n$\\frac{x+1}{\\sqrt{x^2+2x+3}}$ :) \n5\n$\\frac{\\cos (x+1){x}-\\sin{x}}{(x+1)^2}$ :) \n7\n$2\\cos^2{x}\\sin{x}-\\sin^3{x}$ :) \n8\n$\\frac{-1}{4(x+1)^2\\sqrt[4]{(\\frac{x}{x+1})^3}}$ :( [/quote]", - "Solution_4": "[quote=\"amirhtlusa\"]not sure for 6:\n$x^{\\sin{x}}(\\cos{x}\\ln{x}+\\frac{\\sin{x}}{x})$[/quote]\r\n\r\n :) :)", - "Solution_5": "what? 2 is wrong? i thing $\\frac{-\\cos{(4x)}}{4}$ IS the correct answer...\r\n8 typo\r\n$\\displaystyle \\frac{1}{4(x+1)^2\\sqrt[4]{(\\frac{x}{x+1})^3}}$", - "Solution_6": "[quote=\"amirhtlusa\"]what? 2 is wrong? i thing $\\frac{-\\cos{(4x)}}{4}$ IS the correct answer... :? you made intgeral\n8 typo\n$\\displaystyle \\frac{1}{4(x+1)^2\\sqrt[4]{(\\frac{x}{x+1})^3}}$ :) [/quote]", - "Solution_7": "yeah, that's right,\r\ntoo much \"TODAYS INTEGRAL\" lol :) \r\n\r\n$4\\cos{(4x)}$", - "Solution_8": ":)" -} -{ - "Problem": "$ f(x)$ is a differentiable function and $ g(x)$ is a double differentiable function such that $ |f(x)|\\leqslant 1$ and $ f'(x)\\equal{}g(x)$. If $ (f(0))^2\\plus{}(g(0))^2\\equal{}9$. Prove that there exists some $ c\\in (\\minus{}3,3)$ such that $ g(c).g''(c)<0$", - "Solution_1": "Let $ I$ be one of the intervals $ [0,3]$ or $ [\\minus{}3,0]$.\r\n\r\nHere is a sketch, which I think will work (I have left out the details): \r\n\r\nWe consider 3 cases.\r\n\r\ni) $ g(x)$ is of the the same sign on $ I$. This implies that $ f(x)$ is monotonic and the fact that $ |f'(0)| \\ge 2\\sqrt(2)$ would imply that one of $ |f(3)|$ or $ |f(\\minus{}3)|$ is $ > 1$. \r\n\r\n\r\nii) $ g'(x)$ is never $ 0$. Then we have that $ g'(x) > 0 \\forall x$ or $ g'(x) < 0 \\forall x$. Thus $ g(x)$ is monotonic and would fall into case i).\r\n\r\niii) $ g'(a) \\equal{} 0$ for some $ a$. Also assume that $ g(x)$ is not a function of the kind describe in case i). If $ g(a)<0$ and $ g''(a)<0$ then $ a$ is a local maxima. If $ g$ changes sign, there must a local minima c in I such that $ g(c) < g(a)$. Similary consider the case when $ g(a) > 0$. $ g(a) \\equal{} 0$ is probably the trickiest case.\r\n\r\nBy drawing a graph, we should get an idea of how to prove it rigorously I think." -} -{ - "Problem": "Does there exist a sequence $ a_{1}, a_{2}, ... ,a_{2010}$ of positive reals, such that for all integer $ k$, $ 1\\le$ $ k$ $ \\le 2010$ all the roots $ z$ of the polynomial $ p_{k}(x) \\equal{} a_{k \\plus{} 2009}x^{2009} \\plus{} a_{k \\plus{} 2008}x^{2008} \\plus{} \\cdots \\plus{} a_{k \\plus{} 1}x \\plus{} a_k$ satisfy $ |Im(z)| \\le |Re(z)|$? Here $ a_{i \\plus{} 2010} \\equal{} a_{i}$, for $ i \\ge 1$.", - "Solution_1": "The answer is NO!\r\n\r\nLet $ n \\equal{} 2010$. Assume there exists such polynomials.\r\nFix $ k$ and let $ x_i \\equal{} \\alpha_i \\plus{} \\beta_i\\sqrt { \\minus{} 1},1\\le i\\le n \\minus{} 1$ be all the roots to $ p_k(x)$.\r\nNotice $ \\sum_ix_i^2$ is a real number. \r\nHence, by assumption, we have\r\n$ \\sum_ix_i^2 \\equal{} \\sum_i(\\alpha_i^2 \\minus{} \\beta_i^2)\\ge0$, or\r\n$ \\sum_ix_i^2 \\equal{} (\\sum_ix_i)^2 \\minus{} 2\\sum_{i < j}x_ix_j \\equal{} (\\frac {a_{k \\plus{} n \\minus{} 2}}{a_{k \\plus{} n \\minus{} 1}})^2 \\minus{} 2\\frac {a_{k \\plus{} n \\minus{} 3}}{a_{k \\plus{} n \\minus{} 1}}\\ge0$, or\r\n$ 2\\frac {a_{k \\plus{} n \\minus{} 1}}{a_{k \\plus{} n \\minus{} 2}}\\le \\frac {a_{k \\plus{} n \\minus{} 2}}{a_{k \\plus{} n \\minus{} 3}}$.\r\n\r\nLet $ k$ go through $ 1$ to $ n$. We will have\r\n$ 2^{n \\minus{} 1}\\frac {a_2}{a_1}\\le \\frac {a_2}{a_1}$. Contradiction!" -} -{ - "Problem": "\u0388\u03c3\u03c4\u03c9 $ ABCD$ \u03ba\u03c5\u03c1\u03c4\u03cc \u03c4\u03b5\u03c4\u03c1\u03ac\u03c0\u03bb\u03b5\u03c5\u03c1\u03bf \u03ba\u03b1\u03b9 $ A_1, A_2 \\in AB$, $ B_1, B_2 \\in BC$, $ C_1, C_2 \\in CD$ \u03ba\u03b1\u03b9 $ D_1, D_2 \\in DA$ \u03ba\u03b1\u03b9 $ A_0, B_0, C_0, D_0$ \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03b5\u03c3\u03c9\u03c4\u03b5\u03c1\u03b9\u03ba\u03ac \u03c4\u03bf\u03c5 \u03c4\u03b5\u03c4\u03c1\u03b1\u03c0\u03bb\u03b5\u03cd\u03c1\u03bf\u03c5. \u039d\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03b1\u03bd \u03c4\u03b1 $ A_0B_0C_0D_0, AA_1A_0D_2, BB_1B_0A_2, CC_1C_0B_2, DD_1D_0C_2$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c1\u03b9\u03b3\u03c1\u03ac\u03c8\u03b9\u03bc\u03b1, \u03c4\u03cc\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03c4\u03bf $ ABCD$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c1\u03b9\u03b3\u03c1\u03ac\u03c8\u03b9\u03bc\u03bf.", - "Solution_1": "\u03a4\u03bf \u03bc\u03cc\u03bd\u03bf \u03c0\u03bf\u03c5 \u03c7\u03c1\u03b5\u03b9\u03ac\u03b6\u03b5\u03c4\u03b1\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b8\u03b5\u03c9\u03c1\u03af\u03b1 \u03c0\u03b5\u03c1\u03b9\u03b3\u03c1\u03ac\u03c8\u03b9\u03bc\u03bf\u03c5 \u03c4\u03b5\u03c4\u03c1\u03b1\u03c0\u03bb\u03b5\u03cd\u03c1\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03b2\u03b3\u03b1\u03af\u03bd\u03b5\u03b9 . \u039c\u03cc\u03bb\u03b9\u03c2 \u03b2\u03c1\u03ce \u03c7\u03c1\u03cc\u03bd\u03bf \u03bd\u03b1 \u03ba\u03ac\u03bd\u03c9 \u03c4\u03bf \u03c3\u03c7\u03ae\u03bc\u03b1 \u03c3\u03c4\u03bf Euclidraw \u03b8\u03b1 \u03b2\u03ac\u03bb\u03c9 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 .", - "Solution_2": "\u0391\u03c1\u03c7\u03b9\u03ba\u03ac \u03be\u03ad\u03c7\u03b1\u03c3\u03b5\u03c2 \u03bd\u03b1 \u03b1\u03bd\u03b1\u03c6\u03ad\u03c1\u03b5\u03b9\u03c2 '\u03bf\u03c4\u03b9 \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 $ D_2 , A_0 , B_0 , B_1$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03b5 \u03b5\u03c5\u03b8\u03b5\u03af\u03b1 \u03bf\u03bc\u03bf\u03af\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b1 $ D_1 , D_0 , C_0 , B_2$ . \u039b\u03cd\u03c3\u03b7 : \r\n\r\n\u0395\u03bc\u03b5\u03af\u03c2 \u03b8\u03ad\u03bb\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03bf\u03c5\u03bc\u03b5 \u03bf\u03c4\u03b9 \u0391\u0392+DC=\u0391D+\u0392C \u03b1\u03bb\u03bb\u03ac \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03bc\u03b5\u03c4\u03b1\u03c4\u03c1\u03ad\u03c0\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b5 $ A_3B_3\\plus{}D_4C_4\\equal{}A_5D_5\\plus{}B_5C_5$ . \u039b\u03cc\u03b3\u03c9 \u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03c1\u03af\u03b1\u03c2 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 $ A_3B_3\\equal{}A_4B_4 , D_4C_4\\equal{}D_3C_3 , A_5D_5\\equal{}LM , B_5C_5\\equal{}KN$. \u0393\u03bd\u03c9\u03c1\u03b9\u03b6\u03bf\u03c5\u03bc\u03b5 \u03cc\u03bc\u03c9\u03c2 \u03cc\u03c4\u03b9 $ A_4A_0\\equal{}A_0L , KB_0\\equal{}B_0B_4 , C_0C_3\\equal{}C_0N , D_0D_3\\equal{}D_0M$ \u03b1\u03c6\u03bf\u03cd \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03bc\u03ae\u03bc\u03b1\u03c4\u03b1 \u03b5\u03c6\u03b1\u03c0\u03c4\u03cc\u03bc\u03b5\u03bd\u03b1 \u03b1\u03c0\u03cc \u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf . \u0388\u03c4\u03c3\u03b9 \u03b7 \u03c3\u03c7\u03ad\u03c3\u03b7 \u03bc\u03b5\u03c4\u03b1\u03c4\u03c1\u03ad\u03c0\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b5 $ A_0B_0\\plus{}D_0C_0\\equal{}A_0C_0\\plus{}B_0D_0$ \u03c0\u03bf\u03c5 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b1\u03c6\u03bf\u03cd \u03b3\u03bd\u03c9\u03c1\u03af\u03b6\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd \u03c5\u03c0\u03cc\u03b8\u03b5\u03c3\u03b7 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03c0\u03bb\u03b5\u03c5\u03c1\u03bf $ A_0B_0C_0D_0$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c1\u03b9\u03b3\u03c1\u03ac\u03c8\u03b9\u03bc\u03bf . \u0386\u03c1\u03b1 \u03c4\u03bf \u03c4\u03b5\u03c4\u03c1\u03ac\u03c0\u03bb\u03b5\u03c5\u03c1\u03bf ABCD \u03b5\u03af\u03bd\u03b1\u03b9 \u03c0\u03b5\u03c1\u03b9\u03b3\u03c1\u03ac\u03c8\u03b9\u03bc\u03bf . \u03a3\u03c5\u03b3\u03b3\u03bd\u03ce\u03bc\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03b7 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03ac \u03b3\u03bb\u03ce\u03c3\u03c3\u03b1\u03c2 \u03b1\u03bd\u03ac\u03bc\u03b5\u03c3\u03b1 \u03c3\u03b5 \u03c3\u03c7\u03ae\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03c4\u03cd\u03c0\u03bf\u03c5\u03c2.", - "Solution_3": "\u0394\u03af\u03ba\u03b9\u03bf \u03ad\u03c7\u03b5\u03b9\u03c2. \u0395\u03af\u03bd\u03b1\u03b9 \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac \u03ba\u03b1\u03b9 \u03c4\u03bf \u03be\u03ad\u03c7\u03b1\u03c3\u03b1." -} -{ - "Problem": "Let ABC be a triangle with fixed perimeter. Let the coordinates of A be (0,0) and B be (c,0). Show that the third vertex C lies on a common ellipse.", - "Solution_1": "Hint: [hide]Let the two vertices of the triangle be the foci of the ellipse.[/hide]", - "Solution_2": "[hide=\"Solution\"]As darkdieuguerre said, $ A$ and $ B$ are the foci of the ellipse. Now, we can show that the sum of the distance between $ C$ and $ A$ and the distance between $ C$ and $ B$ is a constant by noting that, given a perimeter $ P$, $ AC\\plus{}BC\\plus{}AB\\equal{}P$, or $ AC\\plus{}BC\\equal{}P\\minus{}c$, which is a constant.[/hide](Possibly wrong)", - "Solution_3": "Well, I guess that's correct. I have my proof but I just want to have other solutions.", - "Solution_4": "+) Let p is perimeter of $ \\triangle ABC$, we have $ AB \\plus{} BC \\plus{} CA \\equal{} p \\Longrightarrow CA \\plus{} CB \\equal{} p \\minus{} c \\equal{} const \\Longrightarrow$\r\n$ C$ lie on a elipse.\r\n\r\n+) Let $ C(x;y)$, We have $ AC \\plus{} CB \\plus{} AB \\equal{} p$\r\n$ \\Longleftrightarrow \\sqrt {x^2 \\plus{} y^2} \\plus{} \\sqrt {(x \\minus{} c)^2 \\plus{} y^2}\\plus{}c \\equal{} p$\r\n\r\n$ \\Longleftrightarrow \\sqrt {(x \\minus{} c)^2 \\plus{} y^2} \\equal{} p\\minus{}c \\minus{} \\sqrt {x^2 \\plus{} y^2}$\r\n\r\n$ \\Longleftrightarrow x^2 \\plus{} y^2 \\plus{} c^2 \\minus{} 2cx \\equal{} (p \\minus{} c)^2 \\minus{} 2(p \\minus{} c)\\sqrt {x^2 \\plus{} y^2} \\plus{} x^2 \\plus{} y^2$\r\n\r\n$ \\Longleftrightarrow 2(p \\minus{} c)\\sqrt {x^2 \\plus{} y^2} \\equal{} (p \\minus{} c)^2 \\minus{} 2(p \\minus{} c)\\sqrt {x^2 \\plus{} y^2}$\r\n\r\n............\r\n$ \\Longleftrightarrow \\frac {(x \\minus{} \\frac {c}{2})^2}{\\frac {(p \\minus{} c)^2}{4}} \\plus{} \\frac {y^2}{\\frac {p(p \\minus{} 2c)}{4}} \\equal{} 1$", - "Solution_5": "I've tried laboriously but I can't arrive at \r\n\r\n$ \\frac {(x \\minus{} \\frac {c}{2})^2}{\\frac {(p \\minus{} c)^2}{4}} \\plus{} \\frac {y^2}{\\frac {p(p \\minus{} 2c)}{4}} \\equal{} 1$\r\n\r\nThere must be something wrong with my solution. Can you fill up the gap?", - "Solution_6": "[quote=\"guile\"]I've tried laboriously but I can't arrive at \n\n$ \\frac {(x \\minus{} \\frac {c}{2})^2}{\\frac {(p \\minus{} c)^2}{4}} \\plus{} \\frac {y^2}{\\frac {p(p \\minus{} 2c)}{4}} \\equal{} 1$\n\nThere must be something wrong with my solution. Can you fill up the gap?[/quote]\r\n\r\n[size=150][color=darkblue]Yes! It's easy. I will help you:[/color][/size]\r\n\r\n\r\n[color=darkblue]Let $ C(x;y)$, We have $ AC \\plus{} CB \\plus{} AB \\equal{} p$[/color]\r\n\r\n$ \\Longleftrightarrow \\sqrt {x^2 \\plus{} y^2} \\plus{} \\sqrt {(x \\minus{} c)^2 \\plus{} y^2} \\plus{} c \\equal{} p$\r\n\r\n$ \\Longleftrightarrow \\sqrt {(x \\minus{} c)^2 \\plus{} y^2} \\equal{} p \\minus{} c \\minus{} \\sqrt {x^2 \\plus{} y^2}$\r\n\r\n$ \\Longleftrightarrow x^2 \\plus{} y^2 \\plus{} c^2 \\minus{} 2cx \\equal{} (p \\minus{} c)^2 \\minus{} 2(p \\minus{} c)\\sqrt {x^2 \\plus{} y^2} \\plus{} x^2 \\plus{} y^2$\r\n\r\n$ \\Longleftrightarrow 2(p \\minus{} c)\\sqrt {x^2 \\plus{} y^2} \\equal{} (p \\minus{} c)^2 \\minus{} c^2 \\plus{} 2cx$\r\n\r\n$ \\Longleftrightarrow 2(p \\minus{} c)\\sqrt {x^2 \\plus{} y^2} \\equal{} p(p \\minus{} 2c) \\plus{} 2cx$\r\n\r\n$ \\Longleftrightarrow 4(p \\minus{} c)^2(x^2 \\plus{} y^2) \\equal{} p^2(p \\minus{} 2c)^2 \\plus{} 4c^2x^2 \\plus{} 4cxp(p \\minus{} 2c)$\r\n\r\n$ \\Longleftrightarrow 4(p \\minus{} c)^2x^2 \\plus{} 4(p \\minus{} c)^2y^2 \\minus{} 4c^2x^2 \\minus{} 4cxp(p \\minus{} 2c) \\equal{} p^2(p \\minus{} 2c)^2$\r\n\r\n$ \\Longleftrightarrow \\left[ 4(p \\minus{} c)^2x^2 \\minus{} 4c^2x^2 \\minus{} 4cxp(p \\minus{} 2c)\\right] \\plus{} 4(p \\minus{} c)^2y^2 \\equal{} p^2(p \\minus{} 2c)^2$\r\n\r\n$ \\Longleftrightarrow \\left[ 4x^2p(p \\minus{} 2c) \\minus{} 4cxp(p \\minus{} 2c)\\right] \\plus{} 4(p \\minus{} c)^2y^2 \\equal{} p^2(p \\minus{} 2c)^2$\r\n\r\n$ \\Longleftrightarrow 4p(p \\minus{} 2c)\\left( x^2 \\minus{} cx \\right) \\plus{} 4(p \\minus{} c)^2y^2 \\equal{} p^2(p \\minus{} 2c)^2$\r\n\r\n$ \\Longleftrightarrow 4p(p \\minus{} 2c)\\left( x^2 \\minus{} cx \\plus{} \\frac {c^2}{4} \\minus{} \\frac {c^2}{4}\\right) \\plus{} 4(p \\minus{} c)^2y^2 \\equal{} p^2(p \\minus{} 2c)^2$\r\n\r\n$ \\Longleftrightarrow 4p(p \\minus{} 2c)\\left( x^2 \\minus{} cx \\plus{} \\frac {c^2}{4}\\right) \\plus{} 4(p \\minus{} c)^2y^2 \\equal{} p^2(p \\minus{} 2c)^2 \\plus{} 4p(p \\minus{} 2c).\\frac {c^2}{4}$\r\n\r\n$ \\Longleftrightarrow 4p(p \\minus{} 2c)\\left( x \\minus{} \\frac {c}{2}\\right)^2 \\plus{} 4(p \\minus{} c)^2y^2 \\equal{} p(p \\minus{} 2c)(p \\minus{} c)^2$\r\n\r\n$ \\Longleftrightarrow \\frac {(x \\minus{} \\frac {c}{2})^2}{\\frac {(p \\minus{} c)^2}{4}} \\plus{} \\frac {y^2}{\\frac {p(p \\minus{} 2c)}{4}} \\equal{} 1$\r\n\r\n[color=darkblue]Thanhnam2902[/color] :P", - "Solution_7": "Wow, Thank you!", - "Solution_8": "Question: In line 5, we have\r\n$ \\Longleftrightarrow 2(p \\minus{} c)\\sqrt {x^2 \\plus{} y^2} \\equal{} p(p \\minus{} 2c) \\plus{} 2cx$\r\nThen in line 6:\r\n\r\n$ \\Longleftrightarrow 4(p \\minus{} c)^2(x^2 \\plus{} y^2) \\equal{} p^2(p \\minus{} 2c)^2 \\plus{} 4c^2x^2 \\plus{} 4cxp(p \\minus{} 2c)$\r\n\r\nBoth sides were squared right? where did $ 4cxp(p \\minus{} 2c)$ come from?", - "Solution_9": "He square both sides. and $ (p(p\\minus{}2c)\\plus{}2cx)^2\\equal{}p^2(p\\minus{}2c)^2\\plus{}2*2cx*p(p\\minus{}2c)\\plus{}4c^2x^2$ .", - "Solution_10": "Something to notice: This is just the definition (or one of them) of an ellipse. As the second poster said, let $ A$ and $ B$ be the foci of the ellipse.\r\n\r\n$ AB$ is clearly constant. So we want $ AC\\plus{}CB$ to be constant...", - "Solution_11": "[quote=\"doabli\"]He square both sides. and $ (p(p \\minus{} 2c) \\plus{} 2cx)^2 \\equal{} p^2(p \\minus{} 2c)^2 \\plus{} 2*2cx*p(p \\minus{} 2c) \\plus{} 4c^2x^2$ .[/quote]\r\n\r\nOh, yeah i got it. my paper was so messy that i didn't see it right away. thanks!", - "Solution_12": "[quote=\"Yongyi781\"]Something to notice: This is just the definition (or one of them) of an ellipse. As the second poster said, let $ A$ and $ B$ be the foci of the ellipse.\n\n$ AB$ is clearly constant. So we want $ AC \\plus{} CB$ to be constant...[/quote]\r\n\r\nyeah. actually, the locus of points of $ C$ will be an ellipse since $ c$ is constant and therefore the sum of the remaining two sides is constant." -} -{ - "Problem": "What is the algebra generated by the nilpotent matrices in $M_n(C)$?", - "Solution_1": "Isn't it the entire $\\mathcal M_n(\\mathbb C)$?\r\n\r\nFirst of all, all the matrices having $1$ in the position $(i,j)$ for some $i\\ne j$ and $0$ everywhere else are nilpotent, so they belong to the algebra. Next, given $1$, we can consider the matrix faving $1$ in positions $(i,i)$ and $(i,j)$ and $-1$ in positions $(j,i),(j,j)$. It's nilpotent, and from the first observation it follows that the matrix having $1$ on $(i,i)$ and $-1$ on $(j,j)$ belongs to the algebra. This shows that it's square, which has $1$ on $(i,i),(j,j)$ also belongs to the algebra, so their sum divided by $2$, which has $1$ on $(i,i)$ and $0$ everywhere else, is in the algebra.\r\n\r\nWe have just shown that for all $i,j$, the matrix having $1$ on $(i,j)$ and $0$ everywhere else belongs to the algebra, so the algebra must be the entire $\\mathcal M_n(\\mathbb C)$.\r\n\r\nMaybe I misunderstood something? :?", - "Solution_2": "and what are those matrices that are a product of nilpotent matrices ?", - "Solution_3": "I think we have already discussed that, alekk... but I'm sure you are not satisfied with the hints of killer... ;)\r\n To grobber: no, I did not say it is hard problem, just nice. My idea was to use the fact that any matrix of trace 0 is linear combination of nilpotent matrices and thus the dimension is at least $n^2-1$. But I've shown that a hyperplan cannot be an algebra." -} -{ - "Problem": "If anyone knows about PDA's(push-down automata) and CFG's(Context-free grammars), I need help with these following problems:\r\n\r\n1. Consider the following language:\r\na*b*c* \u2212 {(a^i)(b^i)(c^i) | i >= 0}\r\n\r\n\u2022 Give a CFG that generates this set\r\n\u2022 Give a PDA that recognizes this set\r\n\r\n2.Give a PDA for the following language:\r\n{x \u2208 {0, 1}* | count(0, x) <= 2 * count(1, x)}", - "Solution_1": "Also, does anybody know what Chomsky Normal Form is? I also need to do the 2 problems:\r\n\r\nGive a grammar in Chomsky Normal Form for the language\r\n{a^(i)b^(2i)c^(j) | i, j >=1 }\r\n\r\nGive a PDA that recognizes the language of the following grammar:\r\nA -> aA | aE | bAA\r\nE -> e | aEb | EE\r\nwhere e is the empty string", - "Solution_2": "Context-free grammar is in Chomsky Normal Form if(f) all the rules are in form\r\n\r\nA -> BC (one nonterminal to two nonterminals) or\r\nA -> a (one nonterminal to one terminal) or\r\nS -> e (starting nonterminal to empty string)." -} -{ - "Problem": "If x is 150% of y, what percent of 3x is 4y? Express your answer to the nearest\r\nwhole number.", - "Solution_1": "[hide]89[/hide]", - "Solution_2": "[hide] Isn't it $\\boxed{89}$, or the reciprocal of what you had? :? [/hide]", - "Solution_3": "[hide]set x to be 15 and y to be 10...[/hide]", - "Solution_4": "Umm... [hide] Set x=3, y=2. 3x=9, 4y=8. What percent of 9 is 8? Answer: $\\frac{8}{9}$, which is $88.8888...$%. So, the answer is $\\boxed{89}$. [/hide]", - "Solution_5": "you are right. Need to edit because I misread it.", - "Solution_6": "[hide]\nI think the easiest way is plug in numbers. Let x be 30 and y be 20.\n$\\frac{80}{90}=.88888888888\\approx\\boxed{89}$[/hide]" -} -{ - "Problem": "What is the largest possible area, in square centimeters, of a right\ntriangle with one side of length 12 cm and another side of length\n20 cm?", - "Solution_1": "We want the largest possible area, so we will assume that 20 is the length of one the legs, meaning that the hypotenuse is longer and therefore the triangle is bigger. So we use the formula, $ \\frac{1}{2}bh$, and get $ \\frac{1}{2} \\cdot 12 \\cdot 20 \\equal{} \\boxed{120}$.", - "Solution_2": "Let these be the two smalles sides, so 12*20=240 divide by 2=120\r\n\r\nps i got this one in about 3 sec.", - "Solution_3": "[quote=\"AIME15\"]Let these be the two smalles sides, so 12*20=240 divide by 2=120\n\nps i got this one in about 3 sec.[/quote]\r\n\r\nDUDE! STOP SPAMMING!\r\n\r\n[size=59][color=white]Sorry for spamming.[/color][/size]", - "Solution_4": "[quote=\"isabella2296\"]We want the largest possible area, so we will assume that 20 is the length of one the legs, meaning that the hypotenuse is longer and therefore the triangle is bigger. So we use the formula, $ \\frac {1}{2}bh$, and get $ \\frac {1}{2} \\cdot 12 \\cdot 20 \\equal{} \\boxed{120}$.[/quote]\r\nis it possible to find the smallest possible area", - "Solution_5": "It is, but that's not what the question asks.", - "Solution_6": "[quote=\"AIME15\"]Let these be the two smalles sides, so 12*20=240 divide by 2=120\n\nps i got this one in about 3 sec.[/quote]\r\n\r\nI answered this a day ago. If you bothered to actually read posts, you would realize that I wrote basically the EXACT SAME THING. Do I need to recite the spiel again?\r\n\r\nOnly make another post if:\r\n\r\n1. You found another way.\r\n2. You got a different answer.\r\n3. You noticed a mistake.", - "Solution_7": "[quote=\"myyellowducky82\"][quote=\"isabella2296\"]We want the largest possible area, so we will assume that 20 is the length of one the legs, meaning that the hypotenuse is longer and therefore the triangle is bigger. So we use the formula, $ \\frac {1}{2}bh$, and get $ \\frac {1}{2} \\cdot 12 \\cdot 20 \\equal{} \\boxed{120}$.[/quote]\nis it possible to find the smallest possible area[/quote]\r\n\r\nThe smallest possible answer would be really tiny.", - "Solution_8": "SMallest possible would be 0 but then the trianlge would be degenerate. Smallest possible for right triangle is 96. I'll leave you to figure out why", - "Solution_9": "[quote=\"mewto55555\"]SMallest possible would be 0 but then the trianlge would be degenerate. Smallest possible for right triangle is 96. I'll leave you to figure out why[/quote]\r\n\r\nBut wouldn't that triangle not exist...\r\n\r\nBut maybe...", - "Solution_10": "Actually if you want to be technical, there is no minimum value!", - "Solution_11": "Wouldn't it be 0? You can't have negative area, can you?", - "Solution_12": "Yes but if it's 0, it's no longer a triangle. As the largest angle approaches 180 degrees, the area approaches 0, but if the largest angle is 180, then it has area of 0, but isnt triangle." -} -{ - "Problem": "How many positive three-digit integers with each digit greater than 4 are divisible by 6?", - "Solution_1": "If the number is divisible by six, then the units digit must be even (6 or 8) and the numbers must add up to a multiple of 3. So, first lets list the possible numbers.\n\n$ 0 \\mod 3 \\rightarrow \\{6,9\\}$\n$ 1 \\mod 3 \\rightarrow \\{7 \\}$\n$ 2 \\mod 3 \\rightarrow \\{5, 8 \\}$\n\n[b]Case 1:[/b]\n$ a \\plus{} b \\plus{} 6 \\equiv a \\plus{} b \\equiv 0 \\mod 3$\n\n$ (0,0), (1,2), (2,1) \\mod 3$\n$ 2 \\cdot 2 \\plus{} 1 \\cdot 2 \\plus{} 2 \\cdot 1 \\equal{} 8$\n\n[b]Case 2:[/b]\n$ a \\plus{} b \\plus{} 8 \\equiv a \\plus{} b \\plus{} 2 \\rightarrow a \\plus{} b \\equiv 1 \\mod 3$\n\n$ (1,0),(0,1),(2,2)$\n$ 1 \\cdot 2 \\plus{} 2 \\cdot 1 \\plus{} 2 \\cdot 2 \\equal{} 8$\n\nIn total, there are $8+8= \\boxed{16 \\ numbers}$.\n\nNote: I don't know if this is the fastest way to do this problem, but it is the fastest I can think of right now. If you are comfortable with modular arithmetic, it shouldn't take too long.", - "Solution_2": "This isn't right. Can someone post a correct solution?", - "Solution_3": "I edited it. It should be 16, not 8, since I forgot to add the results from each case.", - "Solution_4": "The official solution is much simpler; I used a bash method involving selecting the hundreds digit first, but selecting the ones digit first yields an excellent solution!", - "Solution_5": "why is this wrong? \n\nThere are $5$ options for the first digit, $6$ options for the second digit, and since we want it to be divisible by $6$ the last digit can only take evens which means there are $3$ options. Now by symmetry exactly a third of these are divisible by $3$. So our answer is $5 \\cdot 6 = 30$?", - "Solution_6": "Take this with a grain of salt as I haven't seen this problem in a while: it's actually not symmetric; if you count out all integers, the number of non-multiples would [b]not[/b] be two times the number of multiples.\n\nIt's kind of like if I asked you \"How many multiples of 3 are between 0 and 10,\" you can't use symmetry.", - "Solution_7": "Ok, but how is this problem supposed to be state sprint if it takes this long :(", - "Solution_8": "ummm who else just literally listed? bad problem imo", - "Solution_9": "[quote=618173]ummm who else just literally listed? bad problem imo[/quote]\n\nyeah same i just made a list", - "Solution_10": "Don\u2019t revive posts from 2009", - "Solution_11": "[quote=jason.ca]Don\u2019t revive posts from 2009[/quote]\n\nLast post was from 2021 also I clicked the discuss button\nIf you can\u2019t revive threads started in 2009 then am I not allowed to discuss the problem??" -} -{ - "Problem": "Hey, I got the prize today.\r\n360 Problems .. and T-shirt !\r\n\r\nThey were really awesome! The book was hardcover too!\r\n\r\nBTW, what is the level of difficulty of 360 Problems M C??", - "Solution_1": "It ranges from medium AIME - hard USAMO, I think (probably with a few outliers). There are some old USAMO problems (a couple of 3/6's) but the beginning problems of each chapter are decently easy, too. Really good book in my opinion, though.", - "Solution_2": "No wonder it looked hard...\r\nLast time I read a post someone posted in this forum describing the USAMTS prizes, and it says the level of difficulty of this book is 9/10. I thought it meant that the book is for 9th and 10th graders... :D Now I think it is based on the scale 1(the easiest) to 10(the hardest).\r\n\r\nAnyway, I think it's really good book too!", - "Solution_3": "I didn't get mine yet; did anyone else get/not get their books?", - "Solution_4": "[quote=\"d343seven\"]I didn't get mine yet; did anyone else get/not get their books?[/quote]\r\n\r\nI didn't receive it either. I was wondering to post thread with title \"SHOULD I GET SCARED!?\" :P", - "Solution_5": "I just got mine today. I live in Utah. I think if you live in the east you should get it later this week.", - "Solution_6": "Hmmm...360 Problems book was REALLY hard. I thought I would use it to practice for AIME. Didn't work out too well. I thought the problems had a different feel than most contest problems.", - "Solution_7": "[quote=\"d343seven\"]I didn't get mine yet; did anyone else get/not get their books?[/quote]\r\nme neither. is it time to panic? :D", - "Solution_8": "I didn't get mine yet either.\r\n\r\nDon't worry yet, guys. They should come tomorrow or Friday.\r\n\r\nJB", - "Solution_9": "Don't get scared - many packages were mailed out at the end of last week, so it's too early to worry.", - "Solution_10": "how were the book and t-shirt shipped? UPS like the mathematica?", - "Solution_11": "[quote=\"tongchen1226\"]how were the book and t-shirt shipped? UPS like the mathematica?[/quote]\r\n\r\nUS mail.", - "Solution_12": "I got mine today. YAY\r\n\r\nJust wondering, mine didn't have certificate. I got shirt and 360 problem book but not the certificate.\r\n\r\n :? :?", - "Solution_13": "No certificates this year; we sent medals (as keychains) instead.", - "Solution_14": "[quote=\"rrusczyk\"]No certificates this year; we sent medals (as keychains) instead.[/quote]\r\n\r\nDid everyone suppose to get it?", - "Solution_15": "I got a gold one, it's nice :lol:", - "Solution_16": "Should I be worried now? My stuff hasn't arrived. Were they shipped out in alphabetical order or something? :D", - "Solution_17": "I just got my package today.\r\n\r\nFor any USAMTS graders and organizers, I just wanted to say thank you, again, for everything this year. This competition has been absolutely amazing. The problems were amazing, the grading was fair and efficient, and, of course, the prizes were great to receive. I can only begin to understand all the work, time, and effort you all put in to making this competition run smoothly. I really appreciate everything you all have done, and am really looking forward to next year.\r\n\r\nJB", - "Solution_18": "I didn't get the email from amazon yet, so I'm a little worried because i leave for india tomorrow, then as soon as i get back, I leave for PROMYS, so i'm worried i won't have it by then...\r\n\r\nBut yeah I got that book. Sounds quite good.", - "Solution_19": "I didn't participate in the USAMTS this year. I'm curious, is it possible to get a copy of 360 Problems through Amazon or somewhere other than as a prize for USAMTS?", - "Solution_20": "[quote=\"mna851\"]I didn't participate in the USAMTS this year. I'm curious, is it possible to get a copy of 360 Problems through Amazon or somewhere other than as a prize for USAMTS?[/quote]Probably. Try searching on Amazon.", - "Solution_21": "Arrived today, Eastern MA. So expect it to arrive soon if you haven't gotten it yet.", - "Solution_22": "[quote=\"mna851\"]I didn't participate in the USAMTS this year. I'm curious, is it possible to get a copy of 360 Problems through Amazon or somewhere other than as a prize for USAMTS?[/quote]\r\nGo to this web: http://www.gil.ro/en/index.html\r\nThis is the GIL publishing site... It says on the left column that for orders e-mail: sales@gil.ro", - "Solution_23": "I got mine yesterday: t-shirt, silver medal, 360 problems, and the geometry book.", - "Solution_24": "Got my t-shirt, key chain, and [i]The Elegant Universe[/i] yesterday :) Thanks USAMTS!!!", - "Solution_25": "I don't think admins answered my question.\r\n\r\nAre all contestants receiving key chain? I'm asking this because I'm Honorable Mention.. :?", - "Solution_26": "I think the key chain thing is the medals. in other words, gold people get gold key chain, silver people get silver key chains, etc.", - "Solution_27": "They would have sent it with all of your other stuff.", - "Solution_28": "Has anyone else looked at the Diophantine EQ's book yet?? Cuz it's really good!\r\n\r\nJB", - "Solution_29": "Hm, so it seems like just about everyone's gotten their stuff already... I'm in NYC, and I haven't gotten it yet, should I be worried now?", - "Solution_30": "[quote=\"joml88\"]They would have sent it with all of your other stuff.[/quote]\r\n\r\nThey didn't so that's why I'm worrying. I would be OK if I didn't receive anything.. But I got all things except key chain (unless HM isn't suppose to get key chain?)", - "Solution_31": "Indeed, the keychains only went to medalists.", - "Solution_32": "Oh ok. Thanks.\r\n\r\nDarn, I could've got keychain! :ninja:" -} -{ - "Problem": "Is it true that \r\n\r\n$ a^2\\plus{}b^2\\plus{}c^2\\plus{}d^2\\minus{}ab\\minus{}ac\\minus{}ad\\minus{}bc\\minus{}bd\\minus{}cd\\geq0$\r\n\r\ngiven the condition that $ a\\leq 0\\leq b\\leq c\\leq d$ and the fact that $ a\\plus{}b\\geq 0$", - "Solution_1": "It is possile due to the fact that\r\n$ a^2 \\minus{} a(b\\plus{}c\\plus{}d) \\ge 0$\r\nand $ b^2 \\plus{}c^2 \\plus{}d^2 \\ge bc\\plus{}cd\\plus{}bd$\r\nWe can judge from above statements easily!", - "Solution_2": "how is $ a^2\\minus{}ab\\minus{}ac\\minus{}ad\\geq 0$?", - "Solution_3": "It follows from $ a \\le 0$.", - "Solution_4": "I mean, can you show for this condition as well?\r\n\r\n$ 0\\leq a\\leq b\\leq c\\leq d$?\r\n\r\nthanks.", - "Solution_5": "[quote=\"macrokim\"]I mean, can you show for this condition as well?\n\n$ 0\\leq a\\leq b\\leq c\\leq d$?\n\nthanks.[/quote]\r\nWith your conditions the original inequality is wrong. Try $ a\\equal{}b\\equal{}c\\equal{}d\\equal{}1.$", - "Solution_6": "sorry, how about $ 00$ and $a+b+c=3$ prove that \r\n\r\n$\\sqrt{\\frac{a+b}{c+1}}+\\sqrt{\\frac{b+c}{a+1}}+\\sqrt{\\frac{c+a}{b+1}}\\geq 3$", - "Solution_1": "Oh ! if your problem is true then it's easy.it's old. :P :D", - "Solution_2": "[quote=\"silouan\"]If $a,b,c>0$ and $a+b+c=3$ prove that \n\n$\\sqrt{\\frac{a+b}{c+1}}+\\sqrt{\\frac{b+c}{a+1}}+\\sqrt{\\frac{c+a}{b+1}}\\geq 3$[/quote]\r\n$\\sum_{cyc}\\sqrt{\\frac{a+b}{c+1}}\\geq3\\Leftrightarrow\\sum_{cyc}(\\sqrt{\\frac{3-a}{1+a}}+\\frac{1}{2}(a-1)-1)\\geq0\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}\\frac{(a-1)^2(3-a)}{\\sqrt{\\frac{3-a}{1+a}}+\\frac{1}{2}(3-a)}\\geq0.$ :)", - "Solution_3": "My equivalent solution:\r\n\r\n${\\sqrt{\\frac{a+b}{c+1}}} \\ge \\frac{a+b}{2}$." -} -{ - "Problem": "Prove for every a,b,c,d>0, that\r\n\r\n(ac+bd)^5+(ad+bc)^5<=(a+b)^5(c^5+d^5)", - "Solution_1": "[quote=\"rumtreiber\"]Prove for every $a,b,c,d>0,$ that\n\n$(ac+bd)^{5}+(ad+bc)^{5}\\leq(a+b)^{5}(c^{5}+d^{5})$[/quote]\r\n$f(x)=x^{5}$ is convex function in $(0,+\\infty).$\r\nHence, $\\frac{a}{a+b}\\cdot c^{5}+\\frac{b}{a+b}\\cdot d^{5}\\geq\\left(\\frac{ac}{a+b}+\\frac{bd}{a+b}\\right)^{5}$ and $\\frac{b}{a+b}\\cdot c^{5}+\\frac{a}{a+b}\\cdot d^{5}\\geq\\left(\\frac{ad}{a+b}+\\frac{bc}{a+b}\\right)^{5}.$\r\nThus, $c^{5}+d^{5}\\geq\\left(\\frac{ac}{a+b}+\\frac{bd}{a+b}\\right)^{5}+\\left(\\frac{ad}{a+b}+\\frac{bc}{a+b}\\right)^{5}\\Leftrightarrow$\r\n$\\Leftrightarrow(ac+bd)^{5}+(ad+bc)^{5}\\leq(a+b)^{5}(c^{5}+d^{5}).$ :)", - "Solution_2": "Nice , arqady ! :D :lol:", - "Solution_3": "hello, your inequality is aquivalent to\r\n$5ab(a+b)(c-d)^{2}(c+d)(a^{2}c^{2}+abc^{2}+b^{2}c^{2}+2abcd+a^{2}d^{2}+abd^{2}+b^{2}d^{2}) \\ge 0$,\r\nwhich is true.\r\nSonnhard." -} -{ - "Problem": "The problem and solution:\r\n\r\nhttp://www.artofproblemsolving.com/Wiki/index.php/1983_AIME_Problems/Problem_3\r\n\r\nThe solution claims that the second solution for y gives non-real roots, though I calculated it to be otherwise (final answer: 720?). I'm not sure if I have a misunderstanding of non-real roots. Can anyone correct me where I'm wrong or clarify the solution?", - "Solution_1": "[quote=\"TCAtect\"]The problem and solution:\n\nhttp://www.artofproblemsolving.com/Wiki/index.php/1983_AIME_Problems/Problem_3\n\nThe solution claims that the second solution for y gives non-real roots, though I calculated it to be otherwise (final answer: 720?). I'm not sure if I have a misunderstanding of non-real roots. Can anyone correct me where I'm wrong or clarify the solution?[/quote]\r\n\r\nYes non-real roots in the [u]orignial[/u] equation, not the $ x^2\\plus{}18x\\plus{}30\\equal{}\\minus{}6$ equation.\r\n\r\nPlugging the two roots you get from that equation back into the main equation you get the RHS to be sqrt of a negative number", - "Solution_2": "Ahhh...I see it. Thanks!" -} -{ - "Problem": "is this true? it is probably well know if it is known, but could somebody prove it anyway?\r\n\r\n\\[\\frac{3\\sqrt3}{4}\\le\\sin^3A+\\sin^3B+\\sin^3C\\le2\\]\r\n\r\nwith $A,B,C$ being the angles of a triangle", - "Solution_1": "Actually, we have\r\n\r\n\\[\r\n0 < \\sin^3(A) + \\sin^3(B) + \\sin^3(C) < 2\r\n\\]\r\n\r\nThe lower bound is obvious - take two small angles and a large obtuse angle. We seek the upper bound.\r\n\r\nFirst, observe that $f(x) = \\sin^3(x)$ has $f'(x) = 3\\sin^2(x)\\cos(x)$ and $f''(x) = 6\\sin(x)\\cos^2(x) - 3\\sin^3(x) = 3\\sin(x)\\left(2\\cos^2(x)-\\sin^2(x)\\right) = 3\\sin(x)\\left(2-3\\sin^2(x)\\right)$. Hence, $f$ has inflexion points at $x_0 = \\sin^{-1}\\left(\\sqrt{\\frac{2}{3}}\\right)$ and $\\pi - x_0$, where $f$ is convex over $(0,x_0]$ and $[\\pi-x_0,\\pi)$, and concave between the points of inflexion. (We restrict the domain to possible angles of a triangle.) We remark that a triangle with an obtuse angle, since that obtuse angle could be decreased while the any acute angle could be increased, [is not optimal.] Therefore, we may assume $A,B,C \\in \\left(0,\\frac{\\pi}{2}\\right)$. Since there is a single inflexion point in the interval, WLOG $A=B$; hence, we consider only sums of the form\r\n\r\n\\[\r\nS(A) = f(A) + f(A) + f(2A)\r\n\\]\r\n\r\nwhere $0 < A < \\frac{\\pi}{2}$ since $\\sin(x) = \\sin(\\pi-x)$. We have $S'(A) = 6\\sin(2A)\\sin(A)\\left(1 + 2\\cos(A)\\cos(2A)\\right)$. The bondary limits are $S(0) = 0$ and $S(\\pi/2)=2$. The $A$ derivative is 0 at $0, \\pi/2$. ($1 + 2\\cos(A)\\cos(2A)$ is never zero; use $\\cos(2A) = 2\\cos^2(A)-1$ and check the derivative of $g(x) = 4x^3-2x+1$ over $0 < x < 1$.)\r\n\r\nEdit - Added a few missing words.", - "Solution_2": "For the upper bound, see also http://www.mathlinks.ro/Forum/viewtopic.php?t=33518 .\r\n\r\n darij", - "Solution_3": "o yea, that makes sense ... i was doing a usamo problem from 1981 and misscalculated somewhere and had this weird result and just wanted to ask if it is not maybe true after all, but i was stuck thinking of acute angles so one extreme would be (90,90,0) but i just got messed up :blush: \r\nthanks for help\r\n\r\npeeta\r\n\r\nps: couldnt quite follow ur prove the end and could quite understand the proofs of the site darij gave either :(\r\ndoes anybody has a more simple prove?" -} -{ - "Problem": "\"Failed to parse (Can't write to or create math output directory): \\frac{y} {20} = \\frac{1} {k}\"\r\n\r\nhttp://www.artofproblemsolving.com/Wiki/index.php/User:Temperal/Introductory_Proportion\r\n\r\nI've gotten this several times. Is it a bug?", - "Solution_1": "Might be the math directory has been corrupted? Anyway I'm currently upgrading the software, so by Monday the Wiki will be using the Forum latex parser (and asy parser for that matter).", - "Solution_2": "It\u2019s a very common problem with LaTeX on the AoPSWiki. On average when I add a new AIME problem/solution, I get about 3 ~ 8 of these [color=red]Can't write to or create math output directory errors[/color]. The best way to get rid of them is to just randomly toss in some symbols to change the content in the math tags (white spaces don\u2019t change anything); my favorite to add in is \\displaystyle. \r\n\r\nIf adding \\displaystyle to the beginning of the equation doesn\u2019t work, I just add another one until eventually the math will be parse-able.\r\n\r\nHopefully the upgrade will remove this issue :)", - "Solution_3": "[quote=\"Valentin Vornicu\"]Anyway I'm currently upgrading the software, so by Monday the Wiki will be using the Forum latex parser (and asy parser for that matter).[/quote]\r\n :thumbup: \r\nI hate those \"can't write to math directory\" messages. I think another way of getting rid of them is to put random curly brackets around the LaTeX.", - "Solution_4": "[quote=\"archimedes1\"][quote=\"Valentin Vornicu\"]Anyway I'm currently upgrading the software, so by Monday the Wiki will be using the Forum latex parser (and asy parser for that matter).[/quote]\n :thumbup: \nI hate those \"can't write to math directory\" messages. I think another way of getting rid of them is to put random curly brackets around the LaTeX.[/quote]That is a bad idea. I am currently upgrading the software (probably will be released tomorrow) and normal forum $ \\text{\\LaTeX}$ and Asymptote rendering will be available on Wiki.", - "Solution_5": "[quote=\"Valentin Vornicu\"]... normal forum $ \\text{\\LaTeX}$ and Asymptote rendering will be available on Wiki.[/quote]\r\nAll hail the mighty site admin! :D\r\n\r\nThis will make creating images and typing up solutions much easier.", - "Solution_6": "[quote=\"Valentin Vornicu\"][quote=\"archimedes1\"][quote=\"Valentin Vornicu\"]Anyway I'm currently upgrading the software, so by Monday the Wiki will be using the Forum latex parser (and asy parser for that matter).[/quote]\n :thumbup: \nI hate those \"can't write to math directory\" messages. I think another way of getting rid of them is to put random curly brackets around the LaTeX.[/quote]That is a bad idea. I am currently upgrading the software (probably will be released tomorrow) and normal forum $ \\text{\\LaTeX}$ and Asymptote rendering will be available on Wiki.[/quote]\r\nOh, I see what you mean.... So that people won't have to edit out the extra brackets when the new parser comes, right?", - "Solution_7": "[quote=\"archimedes1\"]Oh, I see what you mean.... So that people won't have to edit out the extra brackets when the new parser comes, right?[/quote]Yes." -} -{ - "Problem": "I\u2019m throwing 8 dice up in the air. What\u2019s the probability that half land on an even number, while the other half land on 1?", - "Solution_1": "[hide]$ \\binom{8}{4}$ ways to choose the 4 that land on 1, then $ 3^4$ ways that the other 4 can land on even numbers. the probability is then $ \\frac{3^4\\binom{8}{4}}{6^8}$.[/hide]", - "Solution_2": "If they are 6 sided die:\r\n[hide=\"Solution\"]\nYou need to choose 4 of the 8 die to land on one, then figure out the possible ways to get the remaining to land on an even number.\n\n$ \\frac{\\binom{8}{4} \\cdot 3^4}{6^8} \\equal{} \\frac{8 \\cdot 7 \\cdot 6 \\cdot 5}{4 \\cdot 3 \\cdot 2 \\cdot 2^4 \\cdot 6^4} \\equal{} \\frac{8 \\cdot 7 \\cdot 6 \\cdot 5}{12^5} \\equal{} \\frac{35}{5184}$\n[/hide]" -} -{ - "Problem": "Prove that the triangle $ABC$ is equilateral if and only if the following system has at least two solutions and in this case interpret geometrically:\r\n\\[ x+y+z=1 \\] \\[ bc\\cdot x+ca\\cdot y+ab\\cdot z=2R(R+r) \\] \\[ \\left(b^2+c^2\\right)\\cdot x+\\left(c^2+a^2\\right)\\cdot y+\\left(a^2+b^2\\right)\\cdot z=6R^2 \\]", - "Solution_1": "The system of equations\r\n\r\n\r\n$\\left(\\begin{array}{ccc} 1&1&1\\\\\\ bc&ca&ab\\\\\\ b^2+c^2&a^2+c^2&b^2+a^2\\end{array}\\right) \\cdot\\left(\\begin{array}{c} x&y&z\\end{array}\\right)=\\left(\\begin{array}{c} 1&2R(r+r)&6R^{2}\\end{array}\\right)$\r\n\r\nhas two or more solutions if $det\\left(\\begin{array}{ccc} 1&1&1\\\\\\ bc&ca&ab\\\\\\ b^2+c^2&a^2+c^2&b^2+a^2\\end{array}\\right) =0$\r\n\r\nBut $det\\left(\\begin{array}{ccc} 1&1&1\\\\\\ bc&ca&ab\\\\\\ b^2+c^2&a^2+c^2&b^2+a^2\\end{array}\\right) =0 \\Longrightarrow$\r\n\r\n$(a-b)(b-c)(c-a)(a+b+c)=0 \\Longrightarrow\\, a=b=c$\r\n\r\nWhat do you mean by interpret geometrically ?? :)", - "Solution_2": "[quote=\"Ramanujan\"]The system of equations\n$\\left(\\begin{array}{ccc} 1&1&1\\\\\\ bc&ca&ab\\\\\\ b^2+c^2&a^2+c^2&b^2+a^2\\end{array}\\right) \\cdot\\left(\\begin{array}{c} x&y&z\\end{array}\\right)=\\left(\\begin{array}{c} 1&2R(r+r)&6R^{2}\\end{array}\\right)$\n\nhas two or more solutions if $det\\left(\\begin{array}{ccc} 1&1&1\\\\\\ bc&ca&ab\\\\\\ b^2+c^2&a^2+c^2&b^2+a^2\\end{array}\\right) =0$\n........................................................................................\n$(a-b)(b-c)(c-a)(a+b+c)=0 \\Longrightarrow\\, a=b=c.$[/quote] \r\nIn my opinion, you made two mistakes:\r\n$1.\\ \\ (a-b)(b-c)(c-a)=0\\Longrightarrow a=b$ [b]or[/b] $b=c$ [b]or[/b] $c=a$;\r\n$2.$ the system $x+y+z=3$, $x+2x+3z=6$, $3x+4y+5z=10$ has the determinant of the coefficients equal with $0$ but the system hasn't solutions !", - "Solution_3": "I know that\r\n[quote]the system $x+y+z=3, x+2x+3z=6, 3x+4y+5z=10$ has the determinant of the coefficients equal with $0$ but the system hasn't solutions[/quote]\nbut we dont need that(we need that for the converse). :) \nAlso i know that\n[quote]$(a-b)(b-c)(c-a)=0\\Longrightarrow a=b$ or $b=c$ or $c=a$[/quote]\r\nbut i just wrote the idea not the complete solution because i was in hurry. :)\r\nOk .I'll continue the solution.\r\n\r\nFrom $(a-b)(b-c)(c-a)(a+b+c)=0\\Longrightarrow a=b$ or $b=c$ or $c=a$.\r\nWe take $b=c$ WLOG.\r\nFor the triangle $ABC$ since $b=c$ we get $R=\\frac{b^2}{\\sqrt{2b-a}\\,\\sqrt{2b+a}}$ and $r=\\frac{a}{2}\\frac{\\sqrt{2b-a}}{\\sqrt{2b+a}}$.\r\n\r\nNow the system $\\left(\\begin{array}{ccc} 1&1&1\\\\\\ bc&ca&ab\\\\\\ b^2+c^2&a^2+c^2&b^2+a^2\\end{array}\\right) \\cdot\\left(\\begin{array}{c} x&y&z\\end{array}\\right)=\\left(\\begin{array}{c} 1&2R(R+r)&6R^{2}\\end{array}\\right)$ becomes\r\n\r\n$\\left(\\begin{array}{ccc} 1&1&1\\\\\\ b^2&ba&ab\\\\\\ 2b^2&a^2+b^2&b^2+a^2\\end{array}\\right) \\cdot\\left(\\begin{array}{c} x&y&z\\end{array}\\right)=\\left(\\begin{array}{c} 1&2R(R+r)&6R^{2}\\end{array}\\right)$\r\n\r\nor equivalently after eliminating\r\n\r\n$\\left(\\begin{array}{ccc} 1&1&1\\\\\\ b-a&0&0\\\\\\ 0&0&0\\end{array}\\right) \\cdot\\left(\\begin{array}{c} x&y&z\\end{array}\\right)=\\left(\\begin{array}{c} 1&\\frac{(b-a)(2b^2-a^2)}{4b^2-a^2}&0\\end{array}\\right)$.\r\n\r\nBut there are at least two different solutions of the equation ${(b-a)x=\\frac{(b-a)(2b^2-a^2)}{4b^2-a^2}}$ therefore $b-a=0$ or $b=a$.\r\n\r\nFor the converse we assume that $a=b=c$\r\nThen the system becomes (in this case $R=\\frac{a}{\\sqrt3}$ and $r=\\frac{a}{2\\sqrt3}$)\r\n\r\n$\\left(\\begin{array}{ccc} 1&1&1\\\\\\ a^2&a^2&a^2\\\\\\ 2a^2&2a^2&2a^2\\end{array}\\right) \\cdot\\left(\\begin{array}{c} x&y&z\\end{array}\\right)=\\left(\\begin{array}{c} 1&a^2&2a^2\\end{array}\\right)$\r\n\r\nor equivalently\r\n\r\n$\\left(\\begin{array}{ccc} 1&1&1\\\\\\ 0&0&0\\\\\\ 0&0&0\\end{array}\\right) \\cdot\\left(\\begin{array}{c} x&y&z\\end{array}\\right)=\\left(\\begin{array}{c} 1&0&0\\end{array}\\right)$\r\nand this system has more than one solutions.\r\n\r\nIt seems correct now :) :roll:" -} -{ - "Problem": "I was doing Mathcounts Chapter Sprint 2002, and could understand how to solve all questions but this one.\r\n\r\n28. The length of a year on planet Mars is exactly 697 days. If Mars has a calendar with a 12-day week, and year 0 begins on the first day of the week, what is the next year that will begin on the first day of the week?\r\n\r\nThanks in advance.", - "Solution_1": "[hide]Since $ 697\\equiv 1 \\pmod {12}$, the first day of every year will be the day after the day that the previous year started on.\nSo, the pattern is like this.\nYear 0-Day 1\nYear 1-Day 2\nYear 2-Day 3\netc.\nSo, the year that you will get the 12th day of the week first is Year 11. The next year, Year 12, will be the year where the first day of the week is the first day of the year, so the answer is Year 12. [/hide]", - "Solution_2": "I believe you need to find the lcm. Let $ n$ be a non-negative integer. Then $ 697n$ represents the number of days that have passed after $ n$ years. In order for a year to start on the first day of the week, some multiple of 12 days must have passed. So find the lcm, which is $ 8364$. This gives a value of $ n\\equal{}12$, which would mean 12 years.", - "Solution_3": "697/12= blah remainder 1\r\nEvery year the day the year starts off with will be moved forward one. Since there are 12 days in a week, that takes 12 years until you can come back on the same day. The year 12 would be the answer" -} -{ - "Problem": "Let $ABC$ be a triangle with circumcentre $O$. A point $P$ is moving on the extension of $OA$ towards $A$. Let $g$ be the reflection of the line $PB$ with respect to $BA$, $h$ be the reflection of the line $PC$ with respect to $CA$ and $Q$ be the point of intersection of $g$ and $h$.\r\nFind the locus of $Q$ as $P$ moves on the extension of $OA$ towards $A$.", - "Solution_1": "Answer: Locus of poits $Q$ is a ray. \r\nLet be $Q,Q'$ intersections of ${AB,g}{AC,h}$ ,respectively.we will prove $Q=Q'$.Easily, we check that: $\\angle OBP=\\angle BQO$.so, $\\triangle OBP\\cong \\triangle OQB$.Therefore, $OB^{2}=OP.OQ....(1)$.\r\nSimilarly, \r\n$\\angle OCP=\\angle CQ'O$.so, $\\triangle OCP\\cong \\triangle OQ'C$.Therefore, $OC^{2}=OP.OQ'....(2)$.\r\nfrom $(1)$ and $(2)$, $OP.OQ=OP.OQ'\\Longrightarrow OQ=OQ'$.That is $Q=Q'$.The lines $BQ,CQ,PA$ are concurrent at point $Q$.Point $Q$ is on the extension of direction of $A$ of ray $[PA$ .\r\nLokman G\u00d6K\u00c7E.", - "Solution_2": "[quote=\"scarface\"]Answer: Locus of poits $Q$ is a ray. [/quote] Well, not exactly...the locus of $Q$ is only the segment $OA$.\r\n\r\nThere is also a very nice solution using the circle of Apollonius :)", - "Solution_3": "sorry, If I'm mistaken.in my figure, point $Q\\in [PA$, but $Q$ is not elemenet of $(AP]$.I solved the problem according to the below figure.I checked the my solution with sketchpad program.", - "Solution_4": "Ok, sorry...bad wording :blush: \r\n$P$ is on the ray $OA$ but not on the segment $OA$, ie on the extension of $OA$ beyond $A$. (Just swap $P$ and $Q$ in your figure :D)\r\n\r\nBut this doesn't change much in your solution :)", - "Solution_5": "We denote as $K,$ $L,$ the orthogonal projections of $P,$ on the sidelines $AC,$ $AB,$ of the given triangle $\\bigtriangleup ABC,$ respectively.\r\n\r\nAlso, we denote as $D,$ $E,$ the reflections of $P,$ with respect to $K,$ $L$ respectively, lie on the lines $(h),$ $(g)$ and as $A',$ the antidiametric ( = diametric opposite ) point of $A,$ with respect to the circumcircle $(O),$ of $\\bigtriangleup ABC.$\r\n\r\nIt is easy to show that $KL\\parallel BC,$ because of $\\frac{AK}{AC}= \\frac{AP}{AA'}= \\frac{AL}{AB}$ $($ from $A'C\\parallel PK$ and $A'B\\parallel PL$ $).$\r\n\r\nWe have also that $DE\\parallel KL$ ( because of the points $K,$ $L,$ are the midpoints of the segments $PD,$ $PE$ respectively ) and so, we conclude that $DE\\parallel BC.$\r\n\r\nBecause of now, $\\angle ABO = \\angle BAO = \\angle PAL = \\angle LAE,$ we conclude that $BO\\parallel AE$ and similarly $CO\\parallel AD.$\r\n\r\nWe see that the triangles $\\bigtriangleup OBC,$ $\\bigtriangleup AED,$ have their sidelines parallel one per one and so, they are perspective ( the intersection points of the pairs of their homologous sidelines at infinity, are collinear ).\r\n\r\nHence, by [b][size=100]Desarques\u2019s theorem,[/size][/b] we conclude that the segment lines $BE,$ $CD,$ $OA,$ ( connecting the pairs of their homologous vertices ), are concurrent at one point.\r\n\r\nThat is the point $Q,$ as the intersection point of the segment lines $BE\\equiv (g)$ and $CD\\equiv (h),$ lies on the segment line $OA.$\r\n\r\nIt is easy to show that when the point $P,$ is variable on the extension of $OA$ towards $A,$ then the locus of $Q$ is the segment $OA$ and the solution is completed.\r\n\r\nKostas Vittas." -} -{ - "Problem": "Let $ P(x)$ be the real polynomial function, $ P(x) \\equal{} ax^3 \\plus{} bx^2 \\plus{} cx \\plus{} d.$ Prove that if $ |P(x)| \\leq 1$ for all $ x$ such that $ |x| \\leq 1,$ then\r\n\r\n\\[ |a| \\plus{} |b| \\plus{} |c| \\plus{} |d| \\leq 7.\\]", - "Solution_1": "Let $ P( \\minus{} 1) \\equal{} y_1,\\ P( \\minus{} 1/2) \\equal{} y_2,\\ P(1/2) \\equal{} y_3,\\ P(1) \\equal{} y_4$. Then from Lagrange's interpolation formula (or from a linear system of equations) we obtain:\r\n$ P(x) \\equal{} \\frac 1 3 (( \\minus{} 2y_1 \\plus{} 4y_2 \\minus{} 4y_3 \\plus{} 2y_4)x^3 \\plus{} (2y_1 \\minus{} 2y_2 \\minus{} 2y_3 \\plus{} 2y_4)x^2 \\plus{}$\r\n$ (\\frac 1 2 y_1 \\minus{} 4y_2 \\plus{} 4y_3 \\minus{} \\frac 1 2 y_4)x \\plus{} ( \\minus{} \\frac 1 2y_1 \\plus{} 2y_2 \\plus{} 2y_3 \\minus{} \\frac 1 2 y_4)$.\r\nIt is easy to check, that if $ y_i \\in [ \\minus{} 1,1]$, then $ \\pm a \\pm b \\pm c \\pm d \\le 7$ for any combination of pluses and minuses. So, $ |a| \\plus{} |b| \\plus{} |c| \\plus{} |d| \\le 7$.\r\n\r\n[b]Remark.[/b] This solution applies the technics, which is common for polynomials, bounded on [-1, 1]. Actually we consider Chebyshev's polynomial $ T_n(x)\\equal{}\\cos(n\\arccos(x))$ of the same degree and compare it with our polynomial in points, where $ T_n(x)\\equal{} \\pm 1$. In our case it will be a polynomial $ 4x^3\\minus{}3x$ and the corresponding points are $ \\minus{}1,\\ \\minus{}1/2,\\ 1/2,\\ 1.$", - "Solution_2": "[quote=\"Sasha Rybak\"]It is easy to check, that if $ y_i \\in [ \\minus{} 1,1]$, then $ \\pm a \\pm b \\pm c \\pm d \\le 7$ for any combination of pluses and minuses. So, $ |a| \\plus{} |b| \\plus{} |c| \\plus{} |d| \\le 7$.[/quote]\r\nCan you explain this further? Why is it \"easy to check,\" and how do you check it?", - "Solution_3": "Let $a \\le 0, b \\le 0$.Then set $Q(x)=-P(x)$\nLet $a \\le 0,b \\ge 0$.Then set $Q(x)=P(-x)$\nLet $a \\ge 0,b \\le 0$.Then set $Q(x)=-P(-x)$\n\nIn each of these cases we see that $Q(x)$ satisfies all the conditions of the problem and the coefficient of $x^3$ and $x^2$ are nonnegative.So wlog let $a \\ge 0,b \\ge 0$.\n\n$c \\ge 0,d \\ge 0 \\Rightarrow |a|+|b|+|c|+|d|=a+b+c+d=P(1) \\le 1$\n$c \\ge 0,d < 0 \\Rightarrow |a|+|b|+|c|+|d|=a+b+c-d=P(1)-2P(0) \\le 3$\n$c < 0,d \\ge 0 \\Rightarrow |a|+|b|+|c|+|d|=a+b-c+d=\\frac{4}{3}P(1)-\\frac{1}{3}P(-1)-\\frac{8}{3}P(\\frac{1}{2})+\\frac{8}{3}P(-\\frac{1}{2}) \\le 7$\n$c < 0,d < 0 \\Rightarrow |a|+|b|+|c|+|d|=a+b-c-d=\\frac{5}{3}P(1)-4P(\\frac{1}{2})+\\frac{4}{3}P(-\\frac{1}{2}) \\le 7$\n\nSo we are done!!", - "Solution_4": "This problem is very reminiscent of Chebyshev bounds on polynomials. What I have so far is that $|f(x)| \\le |C_3(x)|$ for $|x| \\ge 1$, where $C_3(x)$ is the 3rd degree Chebyshev polynomial. However, I cannot seem to isolate into $|a| + |b| + |c| + |d|$? Does anyone have a completion?", - "Solution_5": "[quote=orl]Let $ P(x)$ be the real polynomial function, $ P(x) \\equal{} ax^3 \\plus{} bx^2 \\plus{} cx \\plus{} d.$ Prove that if $ |P(x)| \\leq 1$ for all $ x$ such that $ |x| \\leq 1,$ then\n\n\\[ |a| \\plus{} |b| \\plus{} |c| \\plus{} |d| \\leq 7.\\][/quote]\nLet $f(x)=a+bx+cx^2+dx^3$ and $|f(0)|\\leq 1,|f'(0)|\\leq 1,|f''(0)|\\leq 1,|f(1)|\\leq \\frac{1}{2}.$\nThen for all $x\\in [0,1]$ [url=https://artofproblemsolving.com/community/c6h1425490p8030153] we have [/url]$|f(x)|\\leq \\frac{11}{8}.$\n", - "Solution_6": "[quote=sayantanchakraborty]Let $a \\le 0, b \\le 0$.Then set $Q(x)=-P(x)$\nLet $a \\le 0,b \\ge 0$.Then set $Q(x)=P(-x)$\nLet $a \\ge 0,b \\le 0$.Then set $Q(x)=-P(-x)$\n\nIn each of these cases we see that $Q(x)$ satisfies all the conditions of the problem and the coefficient of $x^3$ and $x^2$ are nonnegative.So wlog let $a \\ge 0,b \\ge 0$.\n\n$c \\ge 0,d \\ge 0 \\Rightarrow |a|+|b|+|c|+|d|=a+b+c+d=P(1) \\le 1$\n$c \\ge 0,d < 0 \\Rightarrow |a|+|b|+|c|+|d|=a+b+c-d=P(1)-2P(0) \\le 3$\n$c < 0,d \\ge 0 \\Rightarrow |a|+|b|+|c|+|d|=a+b-c+d=\\frac{4}{3}P(1)-\\frac{1}{3}P(-1)-\\frac{8}{3}P(\\frac{1}{2})+\\frac{8}{3}P(-\\frac{1}{2}) \\le 7$\n$c < 0,d < 0 \\Rightarrow |a|+|b|+|c|+|d|=a+b-c-d=\\frac{5}{3}P(1)-4P(\\frac{1}{2})+\\frac{4}{3}P(-\\frac{1}{2}) \\le 7$\n\nSo we are done!![/quote]\n\nWhat is Q(x) here\n\n", - "Solution_7": "Also if |X| \u2264 1 then isn't X only 1and -1", - "Solution_8": "Let $x_1=-1,x_2=-\\frac 12,x_3=\\frac 12,x_4=1.$ From Lagrange's interpolation formula$,$\n$$\\begin{aligned}P(x)&=\\sum\\limits_{k=1}^4P(x_k)\\prod\\limits_{j\\neq k}\\frac {x-x_j}{x_k-x_j}\\\\&=-\\frac 23P(-1)\\left(x^3-x^2-\\frac x4+\\frac 14\\right)+\\frac 43P\\left(-\\frac 12\\right)\\left(x^3-x^2-\\frac x4+\\frac 14\\right)\\\\&\\quad\\text{ }-\\frac 43P\\left(\\frac 12\\right)\\left(x^3+\\frac {x^2}2-x-\\frac 12\\right)+\\frac 23P(1)\\left(x^3+x^2-\\frac x4-\\frac 14\\right).\\end{aligned}$$\nFor $p,q,r,s\\in\\mathbb R,$ define $(p,q,r,s)=pP(x_1)+qP(x_2)+rP(x_3)+sP(x_4).$\nThen $a=\\left(-\\frac 23,\\frac 43,-\\frac 43,\\frac 23\\right),b=\\left(\\frac 23,-\\frac 23,-\\frac 23,\\frac 23\\right),c=\\left(\\frac 16,-\\frac 43,\\frac 43,-\\frac 16\\right),d=\\left(-\\frac 16,\\frac 23,\\frac 23,-\\frac 16\\right).$\nTherefore\n$a+b=\\left(0,\\frac 23,-2,\\frac 43\\right),a-b=\\left(-\\frac 43,2,-\\frac 23,0\\right)\\Rightarrow |a|+|b|=\\max\\{|a+b|,|a-b|\\}\\leqslant 4.$\n$c+d=\\left(0,-\\frac 23,2,-\\frac 13\\right),c-d=\\left(\\frac 13,-2,\\frac 23,0\\right)\\Rightarrow |c|+|d|=\\max\\{ |c+d|,|c-d|\\}\\leqslant 3.$\n$\\therefore |a| + |b| + |c| + |d|\\leqslant 4+3=7.\\blacksquare$", - "Solution_9": "[img]https://i.ibb.co/x6SRRwK/1996-ISL-A5.png[/img]\n[hide=P(x) is a real polynomial function, why do you subtitute i?]In [url=https://mathworld.wolfram.com/RealPolynomial.html]Wolfram Mathworld[/url], I saw this definition of [b]Real Polynomial[/b]:\n[img]https://i.ibb.co/wQ7rSNz/def.png[/img]Also, it's given that $ |P(x)| \\le 1$ $\\boxed{\\text{for all} \\; x\\; \\text{such that}\\; |x| \\le 1},$ not $\\text{for all real}\\; x\\; \\text{such that}\\; |x|\\le 1.$[/hide]" -} -{ - "Problem": "If $[0,1]$ is a subset of the $\\text{Range}$ of function $y=\\frac{x+1}{a+x^2}$, find $a$.", - "Solution_1": "Differentiate?", - "Solution_2": "Differentiation throws away all information about the range, because to go backwards you can only keep your function up to a constant. [hide=\"solution\"]0 being in the range doesn't tell us anything except that a isn't -1. For 1 to be in the range, we know that the equation $x^2 + a = x + 1$ has a real solution, so $1 - a \\geq -\\frac{1}{4}$, $a \\leq \\frac{5}{4}$. So, the question is, for what a less than or equal to 5/4 do we get the whole interval? Clearly, for positive a, this function exists everywhere and so everything works and we get what we want. So our question is reduced to $a \\leq 0$. So, we need to solve $t = \\frac{x+1}{x^2+a}$ for x for any t in [0, 1]. At t = 0, we again have solutions except for a = -1. Otherwise, we find solutions $x = \\frac{1 \\pm \\sqrt{1-4t(at-1)}}{2t}$. So, we need to know for what combinations $(a, t)$ with $a \\leq 0, 0\\leq t \\leq 1$ have $1-4at^2 + 4t < 0$. However, this never occurs, so the answer is all a less than or equal to 5/4 except for -1.[/hide]" -} -{ - "Problem": "Let $ABC$ be a triangle with orthocentre $H$. Prove that the circumradii of triangles $ABC$, $HBC$, $HAB$, and $HAC$ are equal.\r\n\r\nMasoud Zargar", - "Solution_1": "[hide=\"Hint.\"]The reflections of the orthocenter $H$ w.r.t. the sides of the triangle $ABC$ belong to the its circumcircle.[/hide]", - "Solution_2": "[hide=\"outline of solution\"]Simple angle chasing and then extended Law of Sines. Pretty much a two-liner.[/hide]" -} -{ - "Problem": "Let $F_n$ be the $n$-th Fibonacci number. Prove that if $p>5$ is a prime, then\r\n\\[ F_p\\equiv \\left(\\frac{p}5\\right) \\mod p \\]", - "Solution_1": "Sorry if this has been posted before but I don't want to search it since searching for 'Fibonacci' is a bit depressing (too many results obviously)", - "Solution_2": "Because of the Law of Quadratic Reciprocity: $(\\frac{p}{5}) = (\\frac{5}{p})$.\r\n\r\nSuppose that $5$ is quadratic residue mod $p$.\r\nThen, $\\sqrt{5}$ is an element of $Z_p$, so:\r\n$F_p=\\frac{(\\frac{1+\\sqrt{5}}{2})^p - (\\frac{1-\\sqrt{5}}{2})^p}{\\sqrt{5}} \\equiv \\frac{\\frac{1+\\sqrt{5}}{2}-\\frac{1-\\sqrt{5}}{2}}{\\sqrt{5}} = \\frac{\\sqrt{5}}{\\sqrt{5}}=1=(\\frac{5}{p})$\r\nas desired\r\n\r\nNow, suppose that $5$ is not quadratic residue mod $p$.\r\nWe extend field $\\mathbb{Z}_p$ by the new element $\\sqrt{5}$. We see that:\r\n$(1+\\sqrt{5})^p \\equiv 1 + (\\sqrt{5})^p$\r\n(other terms cancel $\\mod{p}$)\r\nAlso:\r\n$1+(\\sqrt{5})^p=1+5^\\frac{p-1}{2}*\\sqrt{5}=1-\\sqrt{5}$\r\nbecause $5$ is not a quadratic residue.\r\nAnalogously: $(1-\\sqrt{5})^p \\equiv 1 + \\sqrt{5}$\r\nSo:\r\n$F_p=\\frac{(\\frac{1+\\sqrt{5}}{2})^p - (\\frac{1-\\sqrt{5}}{2})^p}{\\sqrt{5}} \\equiv \\frac{\\frac{1-\\sqrt{5}}{2} - \\frac{1+\\sqrt{5}}{2}}{\\sqrt{5}}=\\frac{-\\sqrt{5}}{\\sqrt{5}}=-1=(\\frac{5}{p})$\r\nas desired...", - "Solution_3": "[quote=\"Yimin Ge\"]Sorry if this has been posted before but I don't want to search it since searching for 'Fibonacci' is a bit depressing (too many results obviously)[/quote]\r\n\r\nTry searching for Fibonacci and Legendre ;)", - "Solution_4": "I don't know more about the number theory in $Z_n$,where can we find a nice book or article or post which introduce it?", - "Solution_5": "Who can help me?" -} -{ - "Problem": "Given $ f(0;1)\\minus{}\\minus{}>(0;1)$ function and number $ M$ from the interval $ (0;1)$. Assume that for any numbers $ x,y$ from $ (0;1)$, the following inequality holds: $ |f(x)\\minus{}f(y)|<\\equal{}M|x\\minus{}y|$. \r\n Show that the equation $ f(x)\\equal{}x$ has exactly one root.", - "Solution_1": "hint:\r\n[hide]prove $ f$ is continious and consider a sequence $ (u_n)$ such that $ u_0 \\equal{} a \\in (0,1)$ and $ u_{n \\plus{} 1} \\equal{} f(u_n)$, prove $ (u_n)$ converge and then.. and it's clear that f has at most one fixed point.[/hide]", - "Solution_2": "gamarjoba lasha :wink: \r\nI want to show you more general theorem:\r\nsuppose $ X$ is [b] complete [/b] metric space and $ f: X \\to X$ also $ f(X) \\equal{} X$\r\nwith the property:\r\n$ d(f(x),f(y))\\leq M d(x,y)$\r\nfor all $ x,y \\in X$ and some $ 0 < M < 1$ \r\nthen $ f$ have exactly one \r\nmotionless point or $ f(x_{0}) \\equal{} x_{0}$\r\n\r\nsuppose we have two such points say $ f(x) \\equal{} x$ and $ f(y) \\equal{} y$ hence :\r\n$ (1 \\minus{} M)d(x,y)\\leq 0$ therefore $ x \\equal{} y$\r\n\r\nNow, Let's show that for all $ x_{0}$ sequance $ x_{n} \\equal{} f^{n}(x_{0})$ is Cauchy.\r\n(by $ f^{n}$ I mean composition of $ f$ -$ n$-th times at point $ x_{0}$)\r\n$ d(x_{n},x_{n \\plus{} p})\\leq M^{n} d(x_{0},x_{p}) \\leq M^{n} \\sum_{i \\equal{} 1}^{p}d(x_{i \\minus{} 1},x_{i}) \\leq M^{n} d(x_{0},x_{1}) \\sum_{i \\equal{} 1}^{p} M^{i \\minus{} 1} \\leq \\frac {M^{n} d(x_{0},x_{1})}{1 \\minus{} M} \\to 0$ as $ n \\to \\infty$\r\nthus we have $ x \\equal{} \\lim x_{n}$ \r\nnote that $ f$ is continuous , hence $ f(x) \\equal{} x$", - "Solution_3": "What about $ f(x)\\equal{}\\frac{x}{2}$? ;)", - "Solution_4": "then by theorem $ f$ have exactly one\r\nmotionless point or there exist $ x_{0}$ such that $ f(x_{0}) \\equal{} x_{0}$ in your case $ x_{0} \\equal{} 0$ :)", - "Solution_5": "Yes you're right. I just made a mistake while posting the condition. The correct condition would be $ f: [0;1) \\minus{} \\minus{} > [0;1)$. :)", - "Solution_6": "you can't choose $ f(x) \\equal{} x/2$ since theorem holds only when $ f(X) \\equal{} X$ in your case $ X \\equal{} [0,1)$ and if $ f \\equal{} x/2$ hence :\r\n$ f([0,1)) \\equal{} [0,1/2) \\neq [0,1)$.\r\nalso your problem (original) is true if $ f(0,1) \\equal{} (0,1)$", - "Solution_7": "This result is usually referred to as the [url=http://en.wikipedia.org/wiki/Banach_fixed_point_theorem]Banach fixed point theorem[/url] or as the contraction principle." -} -{ - "Problem": "A set $ P$ consists of $ 2005$ distinct prime numbers. Let $ A$ be the set of all possible products of $ 1002$ elements of $ P$, and $ B$ be the set of all products of $ 1003$ elements of $ P$. Find a one-to-one correspondance $ f$ from $ A$ to $ B$ with the property that $ a$ divides $ f(a)$ for all $ a \\in A$.", - "Solution_1": "In the language of graph theory, we're looking for a matching in the graph whose vertices are subsets of size $ n$ and $ n\\plus{}1$ in a set $ S$ of size $ 2n\\plus{}1$, and whose edges correspond to set inclusion (so edges connect a set $ A$ of size $ n$ with a set $ B$ of size $ n\\plus{}1$ iff $ A \\subset B$). Since $ \\binom{2n\\plus{}1}{n} \\equal{} \\binom{2n\\plus{}1}{n\\plus{}1}$, the number of vertices of each kind is the same, and a matching is at least feasible. In what follows, an \"$ n$-set\" is a set of size $ n$. \r\n\r\nTo prove that a matching exists, we use the standard \"marriage theorem\" criterion: every set of $ k$ vertices is connected to at least $ k$ others. If this fails for some set, then a matching is clearly impossible; Hall's theorem guarantees that this necessary condition is also sufficient. \r\n\r\nTo check the condition, let $ X$ be a set of $ n$-element subsets of $ S$, with $ |X|\\equal{}k$. Let $ Y$ be the collection of all $ (n\\plus{}1)$-sets which contain at least one member of $ X$, and let $ Y\\equal{}l$. Let us count how many pairs $ (A,B)$ are there with $ A \\in X$ and $ B \\in Y$ such that $ A \\subset B$. \r\n\r\nOn the one hand, for every $ A \\in X$ there are obviously $ n\\plus{}1$ ways of extending it to subset $ B$ of size $ n\\plus{}1$, since there are $ n\\plus{}1$ elements of $ S$ outside of $ A$. Thus the number of pairs is $ k(n\\plus{}1)$.\r\n\r\nOn the other hand, for every $ B \\in Y$ there are $ n\\plus{}1$ ways of removing an element from $ B$ to obtain an element of $ X$, so the number of such pairs is $ l(n\\plus{}1)$. So in fact, $ l\\equal{}k$, and by Hall's theorem there is a matching as required.\r\n\r\nIncidentally, the graph we just analyzed is conjectured to be Hamiltonian, but this is still an open problem (sometimes called \"the mid-level conjecture\", \"[url=http://arxiv.org/abs/math/0608485]the middle level problem[/url]\" or \"Erd\u0151s' revolving doors conjecture\".)" -} -{ - "Problem": "Hey guys.. I just came up with this problem which - once you know the method - isn't that difficult :)\r\n\r\nShow that:\r\n$ k| \\left[\\left( \\frac {2k \\plus{} 1 \\plus{} \\sqrt {4k^2 \\plus{} 1}}{2} \\right)^n \\right], k \\in \\mathbb{N}$\r\n\r\nAny solutions and/or suggestions on how to make this problem harder is appreciated:)", - "Solution_1": "[hide=\"hint\"] Let $ r_1 \\equal{} \\frac {2k \\plus{} 1 \\plus{} \\sqrt {4k^2 \\plus{} 1}}{2}$ and $ r_2 \\equal{} \\frac {2k \\plus{} 1 \\minus{} \\sqrt {4k^2 \\plus{} 1}}{2}$. $ r_1,r_2$ are roots of the quadratic equation\n\\[ x^2 \\minus{} (2k \\plus{} 1)x \\plus{} k \\equal{} 0\n\\]\nand therefore for all $ c_1,c_2\\in\\mathbb{R}$, we have that $ a_n \\equal{} c_1 r_1^n \\plus{} c_2 r_2^n$ is a solution of the recurrence equation\n\\[ a_n \\equal{} (2k \\plus{} 1)a_{n \\minus{} 1} \\minus{} ka_{n \\minus{} 2}\n\\]\nNotice also that for each $ k$, $ 0 < r_2 < 1$ and so for sufficiently large $ n$ we have that $ a_n \\equal{} \\left \\lfloor c_1r_1^n\\right \\rfloor \\pm 1$. Thus to prove the original statement it suffices to show $ k \\mid a_k \\pm 1$ (which sign depends on the choice of $ c_2$). \n\nThough it remains to choose convenient integral initial conditions $ a_0,a_1$ so that $ c_1 \\equal{} 1$, and $ c_2$ is small enough so that you can bound $ 0 < c_2r_2^k < 1$. I think $ a_0 \\equal{} 0$ and $ a_1 \\equal{} 2k \\plus{} 1$ would work.\n[/hide]", - "Solution_2": "Maybe it was a little more than just a hint? :):):)\r\n\r\nCool you solved it that way - I was actually using induction. Do you have a suggestion on how to make this harder?\r\n\r\nEdit: Just have $ c_1 \\equal{} c_2 \\equal{} 1$", - "Solution_3": "$ r_1r_2 \\equal{} k$, because $ r_2 > k \\plus{} 0.5$, $ r_2 < 1$. Therefore \r\n$ a_n \\equal{} r_1^n \\plus{} r_2^n \\minus{} 1 \\equal{} (r_1 \\plus{} r_2)(r_1^{n \\minus{} 1} \\plus{} r_2^{n \\minus{} 1}) \\minus{} r_1r_2(r_1^{n \\minus{} 2} \\plus{} r_2^{n \\minus{} 2}) \\minus{} 1 \\equal{} (2k \\plus{} 1)(a_{n \\minus{} 1} \\plus{} 1) \\minus{} k(a_{n \\minus{} 2} \\plus{} 1) \\minus{} 1 \\equal{} (2k \\plus{} 1)a_{n \\minus{} 1} \\minus{} ka_{n \\minus{} 2} \\plus{} k$.\r\nBecause $ a_1 \\equal{} k,a_2 \\equal{} 4k^2 \\plus{} 2k \\plus{} 1$, we get $ a_n \\equal{} a_{n \\minus{} 1}\\mod k \\equal{} 1\\mod k$ for $ n > 2$." -} -{ - "Problem": "Let $f(x)$ be a polynomial with integer coefficients, satisfying $f(0)=11$, $f(x_{1})=f(x_{2})=......=f(x_{n})= 2002$, for some distinct integers $x_{1},x_{2},.....,x_{n}$. Find the maximal value of $n$.", - "Solution_1": "[hide=\"Solution\"] $a-b | f(a)-f(b) \\implies x_{i}| (2002-11) = 1991, i = 1, 2, ... n$. We have $1991 = 11 \\cdot 181$ which has $\\boxed{8}$ distinct integral factors. [/hide]", - "Solution_2": "[quote=\"t0rajir0u\"][hide=\"Solution\"] $a-b | f(a)-f(b) \\implies x_{i}| (2002-11) = 1991, i = 1, 2, ... n$. We have $1991 = 11 \\cdot 181$ which has $\\boxed{8}$ distinct integral factors. [/hide][/quote]\r\n\r\nLook more closely at your factorisation.", - "Solution_3": "Do you have to double the number of factors to include for negative integers?", - "Solution_4": "[quote=\"BanishedTraitor\"][quote=\"t0rajir0u\"][hide=\"Solution\"] $a-b | f(a)-f(b) \\implies x_{i}| (2002-11) = 1991, i = 1, 2, ... n$. We have $1991 = 11 \\cdot 181$ which has $\\boxed{8}$ distinct integral factors. [/hide][/quote]\n\nLook more closely at your factorisation.[/quote]\r\n\r\nI fail to see the problem.", - "Solution_5": "Well your answer is not correct. And when you do find a maximum value for $n$ you should provide an example just to make sure the maximum works.\r\n\r\n[hide=\"Outline Of What I Did\"] Let $g(x)= f(x)-2002$ and analyse the case when $x=0$. Then notice for all values of $x$ such that $f(x)=2002$, these values become the root of $g(x)$.[/hide]", - "Solution_6": "Oh. Point taken.\r\n\r\n[hide=\"Then...\"] $g(x) = f(x) (x-x_{1})(x-x_{2})... (x-x_{n})$\n$g(0) = f(0) (-x_{1})(-x_{2})...(-x_{n}) =-1991$\n\nWe want to write $-1991$ as the product of a maximal number of distinct integers. Well,\n\n$-1991 = 1 \\cdot-1 \\cdot 11 \\cdot 181$\n\nIs the best we can do, so the maximal value of $n$ is $\\boxed{4}$. [/hide]" -} -{ - "Problem": "If $x,y,z$ are positive reals such that $x^2+y^2+z^2\\ge 3$, then\r\n\r\n$\\frac{x^5-x^2}{x^5+y^2+z^2}+\\frac{y^5-y^2}{x^2+y^5+z^2}+\\frac{z^5-z^2}{x^2+y^2+z^5} \\ge 0$.\r\n\r\n(Notice that $xyz\\ge 1$ implies $x^2+y^2+z^2\\ge 3$).", - "Solution_1": "No solution ?\r\n?????? :(", - "Solution_2": "Have you found a solution to it by now?", - "Solution_3": "[quote=\"Vasc\"]If $x,y,z$ are positive reals such that $x^2+y^2+z^2\\ge 3$, then\n$\\frac{x^5-x^2}{x^5+y^2+z^2}+\\frac{y^5-y^2}{x^2+y^5+z^2}+\\frac{z^5-z^2}{x^2+y^2+z^5} \\ge 0$.\n[/quote]\r\nProof:\r\nLet $x^2+y^2+z^2=3.$\r\nHence, $\\frac{x^5-x^2}{x^5+y^2+z^2}+\\frac{y^5-y^2}{x^2+y^5+z^2}+\\frac{z^5-z^2}{x^2+y^2+z^5} \\ge 0\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum\\frac{-x^2-y^2-z^2}{x^5+y^2+z^2}\\geq-3\\Leftrightarrow\\sum\\frac{1}{x^5-x^2+3}\\leq1\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum(\\frac{1}{x^5-x^2+3}+\\frac{1}{6}(x^2-1)-\\frac{1}{3})\\leq0\\Leftrightarrow$\r\n$\\sum\\frac{(x-1)^2(x^5+2x^4-3x^2-6x-3)}{x^5-x^2+3}\\leq0.$\r\nLet $x\\geq y\\geq z$ and $f(x)=x^5+2x^4-3x^2-6x-3,$ $x_{0}$ is root of $f$, where $x\\geq1.$\r\n$f'(x)=5x^4+8x^3-6x-6\\geq0,$ where $x\\geq1.$\r\nHence, $f(x)\\leq f(x_0)=0,$ where $1\\leq x\\leq x_0=1.5263...$\r\nIf $00$ and $a^2+b^2+c^2=3$ then\r\n$(ka)^2+(kb)^2+(kc)^2\\geq3.$ Hence $k\\geq1.$\r\nHence, $\\sum\\frac{x^2+y^2+z^2}{x^5+y^2+z^2}=\\sum\\frac{a^2+b^2+c^2}{k^3a^5+b^2+c^2}\\leq\\sum\\frac{a^2+b^2+c^2}{a^5+b^2+c^2}.$\r\nHence, enough to prove that $\\sum\\frac{1}{a^5+b^2+c^2}\\leq1,$ for $a^2+b^2+c^2=3.$", - "Solution_6": "arqady, what inspired you in putting a $\\frac{1}{6}(x^2-1)$ term?\r\n\r\nSecondly, can the inequality be generalized?", - "Solution_7": "[quote=\"perfect_radio\"]arqady, what inspired you in putting a $\\frac{1}{6}(x^2-1)$ term?[/quote]\nLet $f(x)=\\frac{1}{x^5-x^2+3}+\\lambda(x^2-1)-\\frac{1}{3}.$\nThen $f'(1)=0$ and $\\lambda=\\frac{1}{6}.$ :)\n[quote=\"perfect_radio\"]\nSecondly, can the inequality be generalized?[/quote]\r\nI would not do it. ;)", - "Solution_8": "[quote=\"arqady\"][quote=\"Vasc\"]$\\sum\\frac{1}{x^5-x^2+3}\\leq1$:)[/quote][/quote]\r\nit is will be easier\r\nif we can prove:$(x^5-x^2+1)(y^5-y^2+1)(z^5-z^2+1) \\ge 1$\r\n(which is a sepical case I mentioned [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=62775]here[/url],nobody proved or disproved it.)\r\nAnd then use the famous inequality:\r\n$abc=1,a,b,c>0$\r\n$\\sum \\frac 1{2+a} \\le 1$", - "Solution_9": "[quote=\"perfect_radio\"] Secondly, can the inequality be generalized?[/quote]\r\nLong time ago I conjectured that:\r\n[i]If $x_1,x_2,...,x_n$ are positive numbers such that $x_1+x_2+...+x_n \\geq n$, then for $p \\geq 1$ the inequality holds:\n$\\frac{1}{x_1^p+x_2+...+x_n}+\\frac{1}{x_1+x_2^p+...+x_n}...+\\frac{1}{x_1+...+x_{n-1}+x_n^p} \\leq 1$.[/i]", - "Solution_10": "I proved this conjecture only for $p=2$ and $p=3$.\r\nOn the other hand, it is very nice and interesting the zhaobin's inequality:\r\n$(1-a+a^p)(1-b+b^p)(1-c+c^p) \\ge 1$\r\nfor $a,b,c$ positive numbers such that $a+b+c=3$.\r\nI proved the similar inequality\r\n$(1-a_1+a_1^2)(1-a_2+a_2^2)...(1-a_n+a_n^2) \\geq 1$\r\nfor $a_i$ positive numbers such that $a_1+a_2+...+a_n=n$, and $n \\leq 10$.\r\nHowever, it seems that this ineq. is valid for $n \\leq 13$.", - "Solution_11": "[quote=\"Vasc\"]I proved the similar inequality\n$(1-a_1+a_1^2)(1-a_2+a_2^2)...(1-a_n+a_n^2) \\geq 1$\nfor $a_i$ positive numbers such that $a_1+a_2+...+a_n=n$, and $n \\leq 10$.\nHowever, it seems that this ineq. is valid for $n \\leq 13$.[/quote]\r\n\r\n[b]What follows has a critical sign error, as pointed out by Vasc in the next post.[/b]\r\n\r\nIf you have proved it for the cases $n \\le 10$, then I believe it is true for all positive integers $n$. I'll outline the solution here.\r\n\r\nNote that $\\frac{d^2}{dx^2}\\left[\\ln(x^2-x+1)\\right] = \\frac{1+2x-2x^2}{(1-x+x^2)^2}$ has a single positive zero, i.e. $f(x) = \\ln(x^2-x+1)$ has a single inflexion point. By my version of the $n-1$ equal value principle (illustrated in example 21, page 16 of http://web.mit.edu/~tmildorf/www/Inequalities.pdf ) we can write $x_1 = \\cdots = x_{n-1} = a$ and $x_n = n-an+a$. It is then left to see that $g(a) = (n-1)\\ln(a^2-a+1) + \\ln((n-an+a)^2-(n-an+a)+1) \\ge 0$. For $n \\ge 10$, this single variable inequality in $a$ falls to the same lines of argumentation illustrated in http://web.mit.edu/~tmildorf/www/ASharpBound.pdf . $g(1)=0, g'(1)=0, g''(1) > 0$, the inequality holds as $a \\rightarrow \\pm \\infty$, so if false there must be a root of $g'(a) = 0$ other than $a=1$. Removing nonzero junk factors of $g'(a)$, it must be that \r\n\r\n$h(a) = g'(a) * (a^2-a+1)*(1-a+a^2-n+3an-2a^2n+n^2-2an^2+a^2n^2)/(n(n-1))$\r\n$= (a-1)(1+2a-2a^2+n-3an+2a^2n)$ \r\n\r\nhas root $h(a)=0$ other than 1. Then by Rolle's theorem there must be some root of $h'(a) = -1+8a-6a^2+4n-10an+6a^2n = 0$. The minimum of $h'(a)$, a quadratic in $a$, occurs where $h''(a) = 8-12a-10n+12an=0$ or $a = (5n-4)/(6n-6)$, but we have\r\n\r\n$h'((5n-4)/(6n-6)) = (n^2-10n+10)/(6(n-1)) > 0$\r\n\r\nfor $n\\ge 10$. Thus there can be no solution to $g(a) = 0$ other than $a=1$.\r\n\r\n[b]In the above work, I dropped a minus sign at the very last step. It is actually:\n\n$h'((5n-4)/(6n-6)) = -(n^2-10n+10)/(6(n-1)) > 0$ for $n < 9$.[/b]", - "Solution_12": "Mildorf, I found\r\n\r\n$h'((5n-4)/(6n-6)) = -(n^2-10n+10)/(6(n-1)) < 0$ for $n \\geq 9$. :mad:", - "Solution_13": "Darn!\r\n\r\nYou are right, I have dropped a minus sign in my computations.\r\n\r\nAn even more straightforward, brutally computational solution is possible. First a clarification. $\\ln(x^2-x+1)$ has two inflexion points over the reals, convex on the bounded interval. By Karamata, we can spread values on the concave intervals so that $n-2$ values are on the convex interval. But since $x^2-x+1$ is minimized in the convex interval, we can push the two values on the concave intervals together so that at least one can be combined with the $n-2$ in the convex interval, all the while decreasing the logarithmic sum in question. Having equated $n-1$ variables, the derivative with respect to $a$ is 0 only when $a=1$ or $a = (-2+3n\\pm\\sqrt{12-12n+n^2})/(2(-2+2n))$. $a=1$ is the trivial equality case. The two nontrivial critical points arise when $n \\ge 11$.\r\n\r\n\\[ g\\left((-2+3n-\\sqrt{12-12n+n^2})/(2(-2+2n))\\right) \\qquad (*) \\]\r\n\r\nis the interesting critical value. One easily checks that it $g$ positive at this value for $11 \\le n \\le 13$. $g$ at the other critical value is also positive for these values of $n$. What seems apparent from the graph of (*) is that it is negative for $n \\ge 14$, which would mean 13 is the best possible. I will check for this later.", - "Solution_14": "Of course 13 is the best bound! The same computations lead to the counterexample $a_1=\\cdots=a_{13} = 0.6476047054\\dots = \\frac{20-\\sqrt{10}}{26}$ and $a_{14} = 5.581138830\\dots = 4+\\sqrt{5/2}$. In this case,\r\n\r\n\\[ \\prod_{i=1}^{14} (a_i^2-a_i+1) = 0.9158544491\\dots \\]\r\n\r\nFinally, take $a_{15} = \\cdots = a_{n} = 1$, completing the problem.", - "Solution_15": "When $a_1=\\cdots =a_{13}=0.6,a_{14}=6.2,$\n $$(a^2_1-a_1+1)(a^2_2-a_2+1)\\cdots (a^2_{14}-a_{14}+1) < 1.$$\n[url=https://artofproblemsolving.com/community/c6h2041314p14482425]Thank arqady[/url]" -} -{ - "Problem": "A box has height $h$. A chain of length $L$ is placed such that $L-h$ of it is on (the top) of the box and the remaining $h$ of it is hanging down the box. When the end of the chain runs through the $L-h$ length and passes the edge of the box, what will be its velocity?", - "Solution_1": "i'm thinking... use energy\r\n\r\n\r\nedit.\r\n$\\sqrt{2(L-h)(g)(h)/L}$ ?", - "Solution_2": "It also somewhat depends on what happens to the chain after it reaches the bottom of the box. Does it keep falling or is there some table / floor on which the chain stops?", - "Solution_3": "I was thinking of energy as well but after the chain is off the box,it will fall under gravity which we can find but if the question is asking the velovity right after the chain which off the box,then also if there is no friction wouldn't it all still be under free gravity?", - "Solution_4": "well, I'm assuming that the box is resting on a table...\r\nunless its a *floating* box", - "Solution_5": "well, let's say the box is floating.. :)", - "Solution_6": "If the box is floating you may still use the energy, only with different outcome. If you set zero potential at the top of the box then at the beginning, the chain has\r\n\r\n$h\\rho\\times g\\times -h/2$\r\n\r\npotential energy and at the end it is\r\n\r\n$L\\rho\\times g\\times -L/2$.\r\n\r\nDifference must equal the kinetic energy:\r\n\r\n$1/2\\times L\\rho\\times v^2=(-h^2-(-L^2))\\rho g/2$\r\n$v=\\sqrt{\\frac{(L^2-h^2)g}{L}}$\r\n\r\nDid I get it right?\r\n\r\nOh - and for some reason, my solution for the other case is $\\sqrt{2(L-h)g}$. Perhaps you forgot that the L-h length of the chain has zero kinetic when it's at the end lying on the table?", - "Solution_7": "Hey gyus, you're playing with energy as if it was an 'untouchable' quantity. While I think bus's solution to the first case is OK (even though I still have some doubts), the second one can not be! Imagine a situation such that h=1 cm and set L, say, 101 cm. And then set L 1001 cm. It doesn't seem right to me.\r\nWhen the end of the chain hits the ground, it DOES lose some (a bit)(in fact, all) of its kinetic energy. I'd rather use the force approach. The hanging part of the chain has a constant length of $h$, therefore the force on the chain is $F = \\rho .h.g$. Now, I think the outcome of this problem depends on how the chain is resting on the top of the box in the beginning. If it's all just at the egde of the box, it's different than if it's laid in a line... I'll try to do the math but it seems quite complicated.\r\n\r\nAnother thing that bothers me. Suppose you have the chain laid out in a line. Then, when it starts to slide down, the sliding part has some momentum in the horisontal direction. Where it disappears when the whole chain comes down? There is no force pushing to stop the chain going behind the edge. Therefore I think that even bus's first solution using energies is not quite kosher.", - "Solution_8": "OK. I've worked through it, and I don't guarrantee it's right, but if all the chain is rolled up on the edge of the table, the answer is:\r\n\\[ v = \\sqrt{gh}\\sqrt{1-e^{-\\frac{2(L-h)}{h}}} \\]\r\nnotice that the velocity of the chain approaches the critical value $\\sqrt{gh}$, when all the pulling force is used to accelerate new pieces of chain which are initially at rest on the table.", - "Solution_9": "[quote=\"kubus\"]Imagine a situation such that h=1 cm and set L, say, 101 cm. And then set L 1001 cm. It doesn't seem right to me.\nWhen the end of the chain hits the ground, it DOES lose some (a bit)(in fact, all) of its kinetic energy. \n\nAnother thing that bothers me. Suppose you have the chain laid out in a line. Then, when it starts to slide down, the sliding part has some momentum in the horisontal direction. Where it disappears when the whole chain comes down? There is no force pushing to stop the chain going behind the edge. Therefore I think that even bus's first solution using energies is not quite kosher.[/quote]\r\n\r\nI think your example works ok with the equation I got. \r\nThe thing that I'm doing is this. We're looking at the energy difference from the beginning and the end. That will give the KE of the whole chain(if part of it were not stopped). However, because a chain doesn't exert compressive forces, the front of the chain(which has been stopped by the ground) doesn't slow down the rest of the chain any, so we can use the KE expression to find the velocity. The rest of the chain should be moving at the same velocity as it would be if there was a smooth, frictionless ramp on the bottom that diverted all the bottom so that it would still move when it hits bottom.\r\n\r\nAbout the conservation of momentum- I think that if the surface the box is resting on is frictionless, the box would slide the other way, such that the center of mass doesn't move. But because of friction, the box doesn't move- the whole system kind of does, in a small way.\r\n\r\nKubus, how did you derive your result? What do you mean by \"rolled\"?", - "Solution_10": "I see what you guys are saying... and when I realized that the chain loses some of its energy, daveed's solution with the frictionless ramp seemed to be all right... until I read the second part of your argument kubus. Daveed, you can see what kubus means by \"rolled\" and \"laid out straight\" on my picture - is that right kubus? So, the difference is that in the \"rolled\" case the gravitational force always accelerates only $h$ length of the chain and in the other case, the gravitational force accelerates the whole chain at the same time (although the force remains the same). This could possibly lead to different results. So what's wrong with the energy + frictionless ramp approach? The problem is that the parts of chain that were already deflected by the ramp at the bottom of the box all move with different velocities. It seems like you, daveed, thought that all of the chain will be at the end moving with the same speed $v$ and you used this to calculate your $v=\\sqrt{2(L-h)gh/L}$. (But I admit that it's one step closer to the right solution than what I said :).) On the other hand, kubus took the acceleration of the chain and integrated it. I checked your solution kubus and it's correct - although I have no idea how you found it :).\r\n\r\nOh, and I wouldn't worry about the momentum - suppose you have a sloped surface and a ball rolling down this surface. You would surely calculate it's velocity using energy and never worry about the fact that it has gained some horizontal momentum which came from nowhere.", - "Solution_11": "Bus, thanks for doing all the dirty work for me;) & a nice picture! And thanks for checking the solution .. I guess you've worked out the differential equation of motion if you have checked my solution. Anyway, it was a long way to the solution but all very, very interesting! I suggest you go and try it! Being lazy and hiding it behind the fact that I don't want to take away from you the feeling of discovering the solution yourself, only a few guidelines:\r\n\r\n$F = dp/dt$\r\n\r\nF is constant, work out the derivative (formally, you'd say using the product rule;), get to\r\n\r\n[hide]\n$a = (F - v^2 \\rho)/(h \\rho)$\n\n[hide=\"solving\"]\n$2\\sqrt{g/h}T = \\ln{\\frac{\\sqrt{gh}+v}{\\sqrt{gh}-v}}$\n\nwhich forces you to integrate once more, 'cause you know the distance not the time after which the end of the chain leaves the table, with a little of fussy math and a handy substitution when integrating,\n[hide]\n... you come to the desired result and plug everything back in, sorry, I cannot figure it out from my messy work from yesterday, the next figure in my notebook that I'm able to read is the result;/ If you have any problems, I'll try to remember or do it again:)\n[/hide]\n[/hide]\n[/hide]", - "Solution_12": "Hey, thanks, any time. I only got to\r\n\r\n$\\frac{\\text{d}v}{\\text{d}l}=-\\frac{v}{h}+\\frac{g}{v}$\r\n\r\nand didn't know what to do next so I checked your solution :). So, that's it.\r\n\r\nAnd don't forget - beware of flying spiders!", - "Solution_13": "heyy thanks for the explanation :D" -} -{ - "Problem": "Consider $ \\triangle ABC$ with vertices $ A(4,8),B( \\minus{} 1,2),$ and $ C(0, \\minus{} 3).$ Find the point $ D$ such that $ \\triangle ABD,\\triangle ACD$ and $ \\triangle BCD$ all have the same area.", - "Solution_1": "The centroid is such a point...", - "Solution_2": "How would you prove it?", - "Solution_3": "Let $ G$ be the centroid; let $ A'\\in AG\\cap BC$. Then $ A'$ is the midpoint of $ BC$. It is well known that $ \\frac{AA'}{GA'}\\equal{}\\frac{1}{3}\\equal{}\\frac{[ABC]}{[BCG]}$ since we can drop altitudes from $ A$; $ G$ to $ BC$ the the ratio of the heights is $ 1: 3$ and the bases are the same so the area ratio is also $ 1: 3$. By symmetry of argument, all these triangles have an area that is a third of the original triangle so they are equal.", - "Solution_4": "I didn't know that, I used matrices instead. But I didn't know whether the absolute values of the expressions were positive or negative so I had to try out possibilities.", - "Solution_5": "I didn't prove that I found all solutions. In fact, there are three other solutions!\r\n\r\nIn fact, one can show that the triple ratio $ u: v: w\\equal{}[PBC]: [PCA]: [PAB]$ uniquely defines a point $ P$ in the plane of $ ABC$. (The areas are signed, if $ P$ is in the same half plane as $ A$ wrt line $ BC$, $ [PBC]$ is positive. If $ P$ and $ A$ are on opposite sides of $ BC$, then $ [PBC]$ is negative.)\r\n\r\nNoting that $ [PBC]\\plus{}[PCA]\\plus{}[PAB]\\equal{}[ABC]$ (using directed areas), the problem postulates that\r\n\r\n$ |[PBC]|\\equal{}|[PCA]|\\equal{}|[PAB]|$\r\n\r\nWe can normalize to $ u\\plus{}v\\plus{}w\\equal{}1$. So we either have the solutions $ (u,v,w)\\equal{}\\left(\\frac{1}{3},\\frac{1}{3},\\frac{1}{3}\\right),(1,1,\\minus{}1)$ and permutations.\r\n\r\nThe first solution corresponds to the centroid. The second corresponds to the reflection of a vertex in the midpoint of the opposite side." -} -{ - "Problem": "Find all polynomial functions $f: \\mathbb{R}\\rightarrow \\mathbb{R}$ with real coefficient that they are bijective functions and satisfy the condition:\r\n$\\forall x\\in \\mathbb{R} , f^2(x)=f(x^2)-2f(x)+a$ ,(where $a$ is a constant)", - "Solution_1": "[quote]bebakhshd ke saal ro engilisi ferestadam[/quote] \r\n\r\n :P", - "Solution_2": "$p(x)=f(x)+1$\r\n\r\n$p^2(x)-p(x)^2=a$ :rotfl:" -} -{ - "Problem": "In a large crate of oranges, 100[i]p[/i]% are bad. A random sample of 10 oranges produces 2 bad ones. On the assumption that this is the most likely number of bad oranges to find in the sample, what can you say about the possible values of [i]p[/i]?", - "Solution_1": "[hide]Let $ P(X\\equal{}k)$ be the probability of getting k bad oranges out of a group of 10.\n\nIn order for 2 to be the most likely choice, you want $ P(X\\equal{}2) > P(X\\equal{}1)$ and $ P(X\\equal{}2) > P(X\\equal{}3)$ [think about what a histogram of binomial probability density looks like].\n\nSince $ P(X\\equal{}k) \\equal{} \\binom{10}{k}p^k(1\\minus{}p)^{10\\minus{}k}$, we want:\n\n$ \\binom{10}{2}p^2(1\\minus{}p)^8 > \\binom{10}{1}p(1\\minus{}p)^9$\n$ \\Rightarrow 45p^2(1\\minus{}p)^8 > 10p(1\\minus{}p)^9$\n$ \\Rightarrow p > \\frac{2}{11}$\n\nand\n\n$ \\binom{10}{2}p^2(1\\minus{}p)^8 > \\binom{10}{3}p^3(1\\minus{}p)^7$\n$ \\Rightarrow 45p^2(1\\minus{}p)^8 > 120p^3(1\\minus{}p)^7$\n$ \\Rightarrow p < \\frac{3}{11}$\n\nTherefore, $ \\boxed{\\frac{2}{11} < p < \\frac{3}{11}}$.[/hide]", - "Solution_2": "I get that. Thanks gauss.", - "Solution_3": "gauss202 where did you learn that stuff?", - "Solution_4": "I'm a high school teacher. But ideally, one would learn about binomial probability distributions in either an AP/Introductory Statistics class, or possibly in an Algebra II, Precalculus, or Data Analysis class." -} -{ - "Problem": "Let $ A_1A_2A_3A_4A_5A_6A_7, B_1B_2B_3B_4B_5B_6B_7, C_1C_2C_3C_4C_5C_6C_7$ be regular heptagons with areas $ S_A, S_B$ and $ S_C$, respectively. Let $ A_1A_2 \\equal{} B_1B_3 \\equal{} C_1C_4$. Prove that\r\n\r\n$ \\frac{1}{2} < \\frac{S_B \\plus{} S_C}{S_A} < 2 \\minus{} \\sqrt{2}$", - "Solution_1": "$ a_2 = A_1A_2 < a_3 = A_1A_3 < a_4 = A_1A_4$\r\n$ b_2 = B_1B_2 < b_3 = B_1B_3 < b_4 = B_1B_4$\r\n$ c_2 = C_1C_2 < c_3 = C_1C_3 < c_4 = C_1C_4$\r\n$ b_3 = a_2\\ \\Longrightarrow\\ b_2 = \\frac {b_2b_3}{b_3} = \\frac {a_2^2}{a_3},\\ \\ \\ c_4 = a_2\\ \\Longrightarrow\\ c_2 = \\frac {c_2c_4}{c_4} = \\frac {a_2^2}{a_4}$\r\n\r\n$ q(7) = \\frac {S_B + S_C}{S_A} = \\frac {b_2^2 + c_2^2}{a_2^2} = \\frac {a_2^2}{a_3^2} + \\frac {a_2^2}{a_4^2}$\r\nfor regular heptagons, similarly $ q(6)$ for regular hexagons. Sides/areas of hexagons $ (A_i)_1^6 \\longrightarrow (B_i)_1^6 \\longrightarrow (C_i)_1^6$ shrink slower then for heptagons $ (A_i)_1^7 \\longrightarrow (B_i)_1^7 \\longrightarrow (C_i)_1^7$ $ \\Longrightarrow$\r\n$ q(7) < q(6) = \\frac {1}{3} + \\frac {1}{4} = 2 - \\frac {17}{12} = 2 - \\sqrt {2} \\sqrt {\\frac {289}{288}} < 2 - \\sqrt {2}$\r\n\r\n$ a_2a_3 + a_2a_4 = a_3a_4$ by Ptolemy theorem for the cyclic quadrilateral $ A_1A_2A_4A_7 \\Longrightarrow$ $ \\frac {a_2}{a_3} + \\frac {a_2}{a_4} = 1 \\Longrightarrow$ $ q(7) = 1 - \\frac {2a_2^2}{a_3a_4}.$ By the sine theorem for the $ \\triangle A_1A_3A_4$ with $ a_2 = A_3A_4$ and easily calulated angles $ \\angle A_1A_3A_4 = \\frac {4\\pi}{7},\\ \\angle A_3A_4A_1 = \\frac {2\\pi}{7},\\ \\angle A_4A_1A_3 = \\frac {\\pi}{7},$\r\n$ q(7) = 1 - \\frac {2\\sin^2 \\frac {\\pi}{7}}{\\sin \\frac {2\\pi}{7} \\sin \\frac {4\\pi}{7}} = 1 - \\frac {1 - \\cos \\frac {2\\pi}{7}}{2\\left(1 - \\cos^2 \\frac {2\\pi}{7}\\right) \\cos \\frac {2\\pi}{7}} = 1 - \\frac {1}{2(x_0 + 1)x_0}$\r\nwhere $ x_0 = \\cos \\frac {2\\pi}{7}$ is the only positive root of $ Q(x) = 8x^3 + 4x^2 - 4x - 1 = 0$ (the other roots are $ \\cos \\frac {4\\pi}{7},\\ \\cos \\frac {6\\pi}{7},$ both negative). Q(x) is obtained similarly as in [url=http://www.mathlinks.ro/viewtopic.php?t=173185]Regular heptagon[/url], by substitution ${ \\frac {(2k + 1)\\pi}{7}\\longrightarrow \\frac {2l\\pi}{7}} = \\pi - \\frac {(2k + 1)\\pi}{7},\\ x \\longrightarrow - x,$ where $ k = 0, 1, 2$ and $ l = 3 - k = 1, 2, 3.$ We have to show $ (x_0 + 1)x_0 > 1.$\r\n\r\n[color=blue][b](1)[/b][/color] $ P(x) = (x + 1)x - 1 = 0$ has roots $ x_{12} = - G,\\ 1/G$ where G is the golden ratio. Using $ G^2 = G + 1$ repeatedly,\r\n$ Q(x_2) = Q(1/G) = \\frac {3 - 2G}{G^3} < 0 \\Longrightarrow$ $ 1/G = x_2 < x_0 = \\cos \\frac {2\\pi}{7} \\Longrightarrow$ $ P(x_0) > 0,\\ (x_0 + 1)x_0 > 1.$\r\n\r\n[color=blue][b](2)[/b][/color] $ Q(x) = 4(2x - 1)(x + 1)x - 1,$ $ Q(5/8) = 1/64 > 0 \\Longrightarrow x_0 < 5/8,\\ 4(2x_0 - 1) < 1 \\Longrightarrow$ $ (x_0 + 1)x_0 = \\frac {1}{4(2x_0 - 1)} > 1.$\r\n\r\nAs a result, $ q(7) = 1 - \\frac {1}{2(x_0 + 1)x_0} > \\frac {1}{2}.$ $ 1/G < x_0 < 5/8$ allows to improve the right inequality to $ \\frac{1}{2} < q(7) < \\frac{1}{2} + \\frac{1}{130}.$" -} -{ - "Problem": "The maximum speed of a 2.7 kg mass attached to a spring is 0.73 m/s, and the maximum force exerted on the mass is 13 N. \r\n(a) What is the amplitude of motion for this mass? (m)\r\n(b) What is the force constant of the spring? (N/m)\r\n(c) What is the frequency of this system? (Hz)", - "Solution_1": "why don't u write down the relevant formulaes.These are direct questions", - "Solution_2": "because i do not know the formula for amplitude that uses the given information.", - "Solution_3": "$ v_{max}\\equal{}A\\omega$\r\n$ \\omega\\equal{}\\sqrt{\\frac{k}{m}}$\r\n$ F_{max}\\equal{}kA$", - "Solution_4": "[hide=\"My Solution for (a)\"]\n(a)\nConservation of energy: [Subcript m - max]\n$ \\frac {1}{2} mv_m^2 = \\frac {1}{2} kA^2 $ \n$ mv_m^2 = F_mA $\n$ A = \\frac {mv_m^2}{F_m} $\n$ A = 0.11 m $\n[/hide]\n[hide=\"My Solution for (b)\"]\n$ F_m = kA $\n$ k = \\frac {F_m}{A} $ \n$ k = 120 N/m $\n[/hide]\n[hide=\"My Solution for (c)\"]\n$ f = \\frac {1}{2\\pi} \\sqrt {\\frac {k}{m}} $\n$ f = 1.0 Hz $\n[/hide]", - "Solution_5": "[quote=\"pardesi\"]$ v_{max}\\equal{}A\\omega$\n$ \\omega\\equal{}\\sqrt{\\frac{k}{m}}$\n$ F_{max}\\equal{}kA$[/quote]\n\nIs $ \\omega $ equal to $ 2*\\pi*f $?", - "Solution_6": "Yes, the angular speed $\\omega=\\frac{2\\pi}{T}=2\\pi f$.", - "Solution_7": "[quote=\"gaussintraining\"]Yes, the angular speed $\\omega=\\frac{2\\pi}{T}=2\\pi f$.[/quote]\nThanks", - "Solution_8": "[quote=\"gaussintraining\"]Yes, the angular speed $\\omega=\\frac{2\\pi}{T}=2\\pi f$.[/quote]\n\nYou might be interested in: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=636&t=351053", - "Solution_9": "[quote=\"gaussintraining\"]Yes, the angular [b]frequency[/b] $\\omega=\\frac{2\\pi}{T}=2\\pi f$.[/quote]\n\nCorrected a mistake :blush: (I can no longer edit).", - "Solution_10": "[quote=\"gaussintraining\"][quote=\"gaussintraining\"]Yes, the angular [b]frequency[/b] $\\omega=\\frac{2\\pi}{T}=2\\pi f$.[/quote]\n\nCorrected a mistake :blush: (I can no longer edit).[/quote]\n\nYou made a mistake? :)" -} -{ - "Problem": "count the number of pairs (a,b) in N, for which 1/a + 1/b = 1/2004", - "Solution_1": "[hide]We can simplify to. This means that \n\n\n\n\n\n\n\nI'm not exactly sure, but wouldn't that mean that ?[/hide]", - "Solution_2": "This should work: [tex](a-2004)(b-2004)=2004^2[/tex]", - "Solution_3": "hehe that's getting pretty close yet :)", - "Solution_4": "Yes the classic problem resolves into just finidng the factors of (2004)^2.\r\n\r\nNow an extension of the problem. (Hopefully it is not too difficult for the level here)\r\n\r\n[b]given (1/a)+(1/b) = (1/c)\n\nif a,b,c all are positive integers, and have NO common factors (for all the three numbers) (In other words GCD of a,b.c is 1) \n\nprove a+b is a square number.[/b]", - "Solution_5": "1/a + 1/b = 1/c\r\n\r\nab = ac + bc\r\n\r\nab - c(a+b) = 0\r\n\r\n(a-c)(b-c) = ab - c(a+b) + c 2 = c 2\r\n\r\n(a-c)(b-c) = c 2\r\n\r\nI think I'm stuck...", - "Solution_6": "Attempt :\r\n\r\n$\\frac{1}{a} + \\frac{1}{b} = \\frac{1}{c}$\r\n$\\Rightarrow a + b = \\frac{ab}{c}$\r\n\r\nOne can see that any prime divisor $p$ of $c$ divides either $a$ or $b$, but not both else $p$ is a common divisor of $a$, $b$, $c$.\r\n $p | c$ $\\Rightarrow$ $p|a$ or $p|b$ but not both --------------- (1) \r\n\t\t\t\t\t\t\r\nNow, assume that some prime divisor $p$ of $c$ also divides $ab/c$, then it also divides $a + b$. \r\n $p | (a + b)$ --------------- (2)\r\n\r\nBy (1) and (2), $p$ divides both $a$ and $b$, and hence $a$, $b$ and $c$ have a common divisor $p$.\r\n\r\nThus, no prime divisor of $c$ divides $ab/c$.\r\n\r\nNow, consider any prime divisor $q$ of $ab/c$. It divides either $a$ or $b$. This prime $q$ cannot be a divisor of $c$ as argued before. But as $q$ divides \r\n$ab/c = a + b$, so it divides both $a$ and $b$, and hence, $q^2 | (ab/c)$, as $q$ does not divide $c$. \r\n\r\nThus, if a prime divides $ab/c$, then its square also divides the latter.\r\nAs, $a + b = ab/c$, we conclude that $a + b$ is a square.\r\n\r\n*******************************************************************************************************\r\nSambit", - "Solution_7": "Very nice solution!! :)" -} -{ - "Problem": "Prove or disprove that there is an $ n$ such that $ 1988n\\equiv\\minus{}1\\pmod{3^{100}}$.", - "Solution_1": "In the title, don't you mean $ 3^{100}$ divides 1988n+1?", - "Solution_2": "Yeah, I'm sorry. Just ignore the title. :D", - "Solution_3": "[hide]Since $ 1988$ is relatively prime to $ 3^100$, there will be a multiplicative inverse of $ 1988 \\pmod {3^100}$. Take the negative of this inverse and put that as $ n$.[/hide]", - "Solution_4": "I would guess $ n\\equal{}1988^{3^{99} \\minus{}1}$", - "Solution_5": "[quote=\"Johan Gunardi\"]Prove or disprove that there is an $ n$ such that $ 1988n\\equiv\\minus{}1\\pmod{3^{100}}$.[/quote]\n\nWe have $\\gcd(1988,3^{100})=1$, so $1988^{\\phi( 3^{100})} \\equiv 1 \\pmod{3^{100}}$.\n\nAlso $\\phi(p^k)=p^{k-1}(p-1)$ for every prime $p$ and positive integer $k$.\n\nSo $\\phi( 3^{100})=2 \\times 3^{99}$ and you can easily find $n=1988^{2 \\times 3^{ 99} -1}$.", - "Solution_6": "[quote=\"amparvardi\"]...\nSo $\\phi( 3^{100})=2 \\times 3^{99}$ and you can easily find $n=1988^{2 \\times 3^{ 99} -1}$.[/quote]\n(Not that it matters - as it is easy to adjust n) but wouldn't $1988 n$ would be $\\equiv +1 \\pmod {3^{100}}$\n\n(Question is little more interesting if it was to find $n$ such that $1988^n \\equiv -1 \\pmod {3^{100}}$" -} -{ - "Problem": "has anyone here taken the moems 2007 division M contests 1-4? if so, can you check if my answers are right through private message? Thanks", - "Solution_1": "is division M the middle school division?\r\n\r\nif so, i never took MOEMS. How do you enter it?", - "Solution_2": "hmm...i would first like to check your location. As we know, the Moems is now an international math test, so every country has a different way of registration...yeah, it's middle school (division M)...and by the way, from what country are you?" -} -{ - "Problem": "What exactly does this ring look like: Z[x]/(x^2+1, 10x-10).\r\n\r\nI realize this is really the Gaussian integers, Z[i]/(10i-10), but I really don't know how this looks since the linear term is not monic.\r\n\r\nthanks", - "Solution_1": "By the Chinese remainder thm\r\n$\\mathbb Z[i]/(10i-10)\\simeq \\mathbb Z[i]/(1+i)^{3}\\times\\mathbb Z[i]/(1+2i)\\times\\mathbb Z[i]/(1-2i)$. The later two rings each have $5$ elements, so it is not too hard to figure out who they are. The other is a ring with $8$ elements and an element $x$ such that $x^{3}=0$ and $3$ is the least such number.\r\n\r\n[edit] I don't know how to accurately describe $\\mathbb Z[i]/(1+i)^{3}$.\r\nWe could say that it is $\\mathbb Z[X]/(X^{2}+1,2X-2)$.\r\nWe can see that it has $8$ elements i.e. the norm of $(1+i)^{3}$.\r\nIt has characteristic $4$ because $4~(1+i)^{4}$ in $\\mathbb Z[i]$.\r\nIt is not $\\mathbb Z/4\\times\\mathbb Z/2$ because it has a nilpotent of order $3$ nor is it $\\mathbb Z/8$ or $(\\mathbb Z/2)^{2}$ because it has characteristic $4$. Yet, it is a commutative ring.\r\nWe will have to live with it as it is.\r\nAdditively, it is $\\mathbb Z/4\\times\\mathbb Z/2$.\r\nOne can give a multiplication table, but I won't." -} -{ - "Problem": "Show that if A+B+C=180 then\r\n$(\\sqrt{3}-sinA)(\\sqrt{3}-sinB)(\\sqrt{3}-sinC) \\ge sinAsinBsinC$\r\n\r\n :) :)", - "Solution_1": "Is it so hard? :(" -} -{ - "Problem": "Find all the positive real number $a$ such that there exist a positive integer $n$ and $n$ pairwisely disjoint sets $A_1, A_2, ..., A_n$ with infinite elements respectively, satisfying\r\n(1) $A_1 \\cup A_2 \\cup ... \\cup A_n = Z$, $Z$ is the set of integer.\r\n(2) For any two elements $b > c$ in any set $A_i$, $b - c \\geq a^i$, $i = 1, 2, ..., n$.", - "Solution_1": "i guess the answer should be all the numbers $a\\in (0,2)$.\r\nfirstly note that if we denote the required set by $A$ and that $a\\notin A$, then $b\\notin A\\ \\forall b\\geq a$.\r\nwe shall see the claim in two steps:if $a<2$ then $a\\in A$.indeed, if $a<2$ then there exists an $n$ such that $a\\leq 2^{1-1/n}$.(that is so since $\\lim_{n\\rightarrow\\infty} 2^{1-1/n}=2$). \r\nin this case we claim that a \"good\" partitioning of $\\mathbb{Z}$ is possible into $n$ such sets.(we shall only partition the positive integers.but the partitions can be extended symmetrically,w.r.t $0$ and will still be \"good\"):\r\n Indeed, $A_1=\\{1,3,\\ldots\\}$;\r\n $A_2=\\{0,2,6,10,14,\\ldots\\}$\r\n $A_3=\\{4,12,20,\\ldots\\}$\r\n\r\n $A_{n-1}=\\{2^{n-2},2^{n-2}+2^{n-1},\\ldots\\}$\r\n $A_n=\\{2^{n-1},\\ldots\\}$\r\n(i hope the idea is clear;in general what we need is that any even number uniquely exactly one of $2\\mod 4,4\\mod 8,\\ldots,2^{n-1}\\mod 2^n$ or finally $0\\mod 2^n$).\r\nnow suppose such a partition were possible for $a=2$ for some $n$.consider the partial partition induced on $S=\\{1,2,\\ldots,2^n\\}$.since for any two consecutive elements of $A_n$, the difference is atleast $2^n$ $A_n$ contains atmost one element from $S$.likewise, $A_{n-1}$ contains atmost $2$ elements of $S$ and in general $A_i$ contains atmost $2^{n-i}$ elements of $S$.but then the $A_i's$ only cover $1+2+\\cdots+2^{n-1}=2^n-1$ elements of $S$,a contradiction.", - "Solution_2": "Your answer is correct!" -} -{ - "Problem": "Calculate:\r\n$S=\\cos\\frac{\\pi}{2n+1}-\\cos\\frac{2\\pi}{2n+1}+...+(-1)^{n+1}\\cos\\frac{n\\pi}{2n+1}$", - "Solution_1": "[quote=\"hien\"]Calculate:\n$S=\\cos\\frac{\\pi}{2n+1}-\\cos\\frac{2\\pi}{2n+1}+...+(-1)^{n+1}\\cos\\frac{n\\pi}{2n+1}$[/quote]\r\ntake \r\n$C=\\sin\\frac{\\pi}{2n+1}-\\sin\\frac{2\\pi}{2n+1}+...+(-1)^{n+1}\\sin\\frac{n\\pi}{2n+1}$\r\nand calcuate\r\n$S+iC$\r\n \r\n\r\n$S=Re(S+iC)$", - "Solution_2": "Yes, it's too my solution. But i post it here in order to hope somebody can kill :) it by a elementary solution not using complex number" -} -{ - "Problem": "Determine all the strictly increasing functions $ f\\colon \\mathbb{R}\\to \\mathbb{R}$ which satisfy\r\n\r\n$ f(x\\plus{}f(y))\\equal{}f(x\\plus{}y)\\plus{}2009$,\r\n for any reals $ x,y$.", - "Solution_1": "put in x such that x = -f(y), we get f(0) - 2009 = const = f(-f(y) + y), f is injection since f if strictly monotonic, so there exist a constant a such that a = -f(y) + y (*)\r\n\r\nnow we get f(y) = y - a\r\n\r\nputting in x = y = 0, we get f(f(0)) = f(0) + 2009, now using (*) , -2a = -a + 2009, a = -2009,\r\nso f(x) = x + 2009, putting back in starting relation we see that this function is the soluton", - "Solution_2": "Sorry, can you write this in LATEX?", - "Solution_3": "[quote=\"felixx\"]put in x such that x = -f(y) [/quote]\r\n\r\nThis step throws out any potentially non-surjective solutions. In other words you've only proven that $ f(x) \\equal{} x \\plus{} 2009$ for $ x\\in f(\\mathbb{R})$. In order to make the argument solid, you'd need to show $ f$ is surjective.", - "Solution_4": "[quote=\"JoeBlow\"][quote=\"felixx\"]put in x such that x = -f(y) [/quote]\n\nThis step throws out any potentially non-surjective solutions.[/quote]\r\n\r\nCan you explain it? I think putting that into an equation does not need surjectivity, because here x is arbitrary. :huh:", - "Solution_5": "[quote=\"Inequalities Master\"]Determine all the strictly increasing functions $ f\\colon \\mathbb{R}\\to \\mathbb{R}$ which satisfy\n\n$ f(x \\plus{} f(y)) \\equal{} f(x \\plus{} y) \\plus{} 2009$,\n for any reals $ x,y$.[/quote]\r\nFrom hypothesis we have $ f(x\\plus{}f(y))\\equal{}f(y\\plus{}f(x))\\forall x,y\\in\\mathbb{R}$ therefore $ x\\plus{}f(y)\\equal{}y\\plus{}f(x)\\forall x,y\\in\\mathbb{R}$ because $ f$ is strictly increasing, so $ f(x)\\equal{}x\\plus{}C\\forall x\\in\\mathbb{R}$. Now, check original identity. :)", - "Solution_6": "[quote=\"polskimisiek\"][quote=\"JoeBlow\"][quote=\"felixx\"]put in x such that x = -f(y) [/quote]\n\nThis step throws out any potentially non-surjective solutions.[/quote]\n\nCan you explain it? I think putting that into an equation does not need surjectivity, because here x is arbitrary. :huh:[/quote]\r\nYes, he is wrong! I think so, too. :lol:" -} -{ - "Problem": "Let $ p_1,p_2,\\dots,p_r$ be distinct primes, and let $ n_1,n_2,\\dots,n_r$ be arbitrary natural numbers. Prove that the number of pairs of integers $ (x, y)$ such that\r\n\\[ x^3 \\plus{} y^3 \\equal{} p_1^{n_1}p_2^{n_2}\\cdots p_r^{n_r}\\]\r\ndoes not exceed $ 2^{r\\plus{}1}$.", - "Solution_1": "This is the only problem of Korea 1997 National Olympiad which happen to have no solution on MathLinks... :?", - "Solution_2": "I get what is wrong about it!! :P\ni searched the original problem on web; and there's another condition: [b]x,y are coprimes.[/b]\n\nnow it leaves a bit easy, since $gcd(x+y,x^2-xy+y^2)=1 or 3$", - "Solution_3": "Let try together to solve this hard problem.This problem shames us", - "Solution_4": "If $(x+y,x^2-xy+y^2)=1$, we have to split up $r$ primes into two groups, and set $$x+y=p^{n_{i_1}}_{i_1} \\cdots p^{n_{i_k}}_{i_k}$$ $$x^2-xy+y^2 = p^{n_{j_1}}_{j_1} \\cdots p^{n_{j_{k'}}}_{j_{k'}}$$ where $\\{i_1, \\cdots i_k \\}$ and $\\{ j_1, \\cdots j_k' \\}$ is a partition of $\\{1,2, \\cdots r\\}$. \nThis pair of equations have at most two solutions, and there are $2^r$ ways to make the partition, so at most $2^{r+1}$ solutions.\n\nIf $(x+y,x^2-xy+y^2)=3$, set $x+y=3^a \\cdot t$, where $t$ is coprime with $3$. This will lead us to $$ x^3 + y^3 = 3^{a+1}t (3^{2a-1}t^2-xy)$$\nOf course, $t$ and $3^{2a-1}t^2 - xy$ are coprime, so we are left with a similar situation with above.\nWe have $r-1$ primes now because we erased the prime $3$. \nAnyway, the number of solutions, in this case, is no more than $2^{r+1}$. (isw the first case) Done.", - "Solution_5": "[quote=rkm0959]If $(x+y,x^2-xy+y^2)=1$, we have to split up $r$ primes into two groups, and set $$x+y=p^{n_{i_1}}_{i_1} \\cdots p^{n_{i_k}}_{i_k}$$ $$x^2-xy+y^2 = p^{n_{j_1}}_{j_1} \\cdots p^{n_{j_{k'}}}_{j_{k'}}$$ where $\\{i_1, \\cdots i_k \\}$ and $\\{ j_1, \\cdots j_k' \\}$ is a partition of $\\{1,2, \\cdots r\\}$. \nThis pair of equations have at most two solutions, and there are $2^r$ ways to make the partition, so at most $2^{r+1}$ solutions.\n\nIf $(x+y,x^2-xy+y^2)=3$, set $x+y=3^a \\cdot t$, where $t$ is coprime with $3$. This will lead us to $$ x^3 + y^3 = 3^{a+1}t (3^{2a-1}t^2-xy)$$\nOf course, $t$ and $3^{2a-1}t^2 - xy$ are coprime, so we are left with a similar situation with above.\nWe have $r-1$ primes now because we erased the prime $3$. \nAnyway, the number of solutions, in this case, is no more than $2^{r+1}$. (isw the first case) Done.[/quote]\n\nBut your estimation estimates $2^r$ but why in problem asked $2^{r+1}$. I think problem also true for not relavitely primes $x,y$" -} -{ - "Problem": "Twenty students bought tickets for a school party. All of the money recieved for these 20 tickets was used to purchase beverages. Then, and additional 10 students bought tickets. Rater than use this additional money to buy more refreshments, all 30 students recieved a $3.00$(three dollar) refund. How many dollars were used to buy beverages?", - "Solution_1": "[hide]\nLet's say each ticket costs x dollars. Then a total of 20x dollars was bought on refreshments. When 10 more students bought tickets, they spent 10x dollars. All 30 students also got 3 dollars back so the 10x dollars that they spent is equal to 90, so each ticket costs 9 dollars. We're looking for the cost of the beverages so multiply that by twenty. The answer is $180$.[/hide]", - "Solution_2": "[hide]\nwe know that one person's cost is 9.00\n\nwe multiply that by 20 and get 180\n\nthe answer is $\\$$180[/hide]\r\n-jorian" -} -{ - "Problem": "Let P be a regular 20-gon which is inscribed in a circle of radius 1. Find the sum of the squares of the lengths of all of the sides and diagonals of P.", - "Solution_1": "hmm...\n\n\n\n[hide]\n\n\n\n5*(5-2sqrt(5)+1)\n\n\n\n[/hide]\n\n\n\n?", - "Solution_2": "Sorry. That's incorrect. Check your work, and remember you're looking for the sum of the [i]squares[/i] of the sides and diagonals.", - "Solution_3": "EEK! AND diagonals... i totally missed that part of the question. I gave you only the sides :)", - "Solution_4": "LoL.\r\n\r\nBy the way ComplexZeta, I'm sure there is a better way to do it, but would getting the diagonals require Ptolemy's, or just taking small n-gons and looking for a pattern?", - "Solution_5": "I wouldn't suggest finding the lengths of diagonals that you don't just know instantly (such as diameters). Don't go finding cos(18) or anything like that. You don't need Ptolemy, and it would be better for you not to look at small cases. You get a better insight into the nature of the problem if you can do it without relying on small cases. Having said that, when you are actually taking a competition, solve it in any way you can. But right now it's a challenge problem on the forum, so try not to look for patterns.", - "Solution_6": "hmm this looked interesting.... \n\n[hide]\n\nwe can pair off every diagonal/side except for the main ones going through the center of the circle. pythagoras' theorem yields the sum of squares for each pair is equal to the main diagonal squared = 4. \n\n\n\ntaking the main diagonals (all 10 of them) separate:\n\n\n\nmains: 10 * 4 = 40\n\nothers: (20 C 2) - 10 sides/diags = 180 => 90 pairs => 360\n\n\n\n40 + 360 = 400.\n\n[/hide]", - "Solution_7": "Very good.", - "Solution_8": "ooooo\r\nniceness\r\ni shulda thought of that one", - "Solution_9": "Sorry for being slow to pick up on this new concept... Can you specify how you are pairing these diagonals and how you get the inequalities?", - "Solution_10": "Those aren't inequalities. => means implies. The thing you use here is that if an angle is inscribed in a semicircle, then it is a right angle. That's how you pair them.", - "Solution_11": "And he's not actually pairing diagonals. He's pairing each diagonal with the side adjaccent to it such that they form a right triangle. with the hypotenuse being the diameter of the circle.", - "Solution_12": "two diagonals can be paired together...\r\n\r\nbut i do emphasize can. its not always the case; however it is irrelevant to the problem as long as the pairings of line segments do exist, which is fairly clear as any angle with vertex on a circle subtending half the circle is 90 degrees in measure." -} -{ - "Problem": "Bow to solition the number of equation on $ [\\minus{}2\\pi;2\\pi]$\r\n$ \\sin(1\\cdot x)\\cdot \\sin(1\\cdot 3x) \\cdot \\sin(1\\cdot 3\\cdot 5x)...\\sin(1\\cdot3\\cdot ...\\cdot 2009x)\\equal{}1$\r\n\r\n\r\n\r\n\r\n\r\n\r\n_______________________________________\r\nAzerbaijan Land of Fire \r\n$ \\Rightarrow\\bigstar\\Leftarrow$\r\n :ninja:", - "Solution_1": "[quote=\"Pirkuliyev Rovsen\"]Bow to solution the number of equation ... [/quote]\r\n\r\nWhat does this mean ?", - "Solution_2": "Bow to the number of :!:", - "Solution_3": "[quote=\"Pirkuliyev Rovsen\"]Bow to solition the number of equation on $ [ - 2\\pi;2\\pi]$\n$ \\sin(1\\cdot x)\\cdot \\sin(1\\cdot 3x) \\cdot \\sin(1\\cdot 3\\cdot 5x)...\\sin(1\\cdot3\\cdot ...\\cdot 2009x) = 1$\n[/quote]\r\n\r\nWe need all these factors be $ 1$ or $ -1$ and so $ x\\in\\{-\\frac{3\\pi}2,{-\\frac{\\pi}2,{\\frac{\\pi}2,{\\frac{3\\pi}2\\}}}}$\r\n\r\nFor $ x\\in\\{-\\frac{3\\pi}2,{\\frac{\\pi}2\\}}$, we have $ 502$ factors equal to $ -1$ and $ 503$ equal to $ 1$, and so these values are solution.\r\n\r\nFor $ x\\in\\{-\\frac{\\pi}2,{\\frac{3\\pi}2\\}}$, we have $ 503$ factors equal to $ -1$ and $ 502$ equal to $ 1$, and so these values are not solution.\r\n\r\nAnd so only two solutions in $ [-2\\pi,2\\pi]$ : $ -\\frac{3\\pi}2$ and $ \\frac{\\pi}2$" -} -{ - "Problem": "given that $(0,0)$ and $(-4,3)$ are the coordinates of two points of a triangle, find a third point that makes this an equilateral triangle.", - "Solution_1": "[hide=\"Messy solution 1\"] The third point must lie on the perpendicular bisector of the line segment connecting those initial two points. This bisector has slope $\\frac{4}{3}$ and passes through the point $(-2, \\frac{3}{2})$, so it is given by\n\n$y-\\frac{3}{2}= \\frac{4}{3}(x+2) \\Leftrightarrow$\n$y = \\frac{4}{3}x+\\frac{7}{6}$\n\nThe third point must have a distance of $\\sqrt{3^{2}+4^{2}}= 5$ from the origin, so we solve the quadratic equation\n\n$x^{2}+\\left( \\frac{4}{3}x+\\frac{7}{6}\\right)^{2}= 25$\n\nTo find our messy answer. [/hide]\n[hide=\"Messy solution 2\"] The side length of the triangle is $5$, so we know the third point must be at some $(5 \\cos \\theta, 5 \\sin \\theta)$.\n\nMoreover, letting $\\phi = \\arctan \\frac{3}{-4}$ (so that $\\sin \\phi = \\frac{3}{5}, \\cos \\phi = \\frac{-4}{5}$) we have $\\theta = \\phi-60^{\\circ}$, so\n\n$x = 5 \\cos \\theta = 5 \\cos (\\phi-60^{\\circ}) = 5 \\cos \\phi \\cos 60^{\\circ}+5 \\sin \\phi \\sin 60^{\\circ}=-4 \\cdot \\frac{1}{2}+3 \\cdot \\frac{ \\sqrt{3}}{2}$\n\nAnd\n\n$y = 5 \\sin \\theta = 5 \\sin (\\phi-60^{\\circ}) = 5 \\sin \\phi \\cos 60^{\\circ}-5 \\cos \\phi \\sin 60^{\\circ}= 3 \\cdot \\frac{1}{2}-4 \\cdot \\frac{ \\sqrt{3}}{2}$ [/hide]", - "Solution_2": "[quote=\"t0rajir0u\"][hide=\"Messy solution 1\"] The third point must lie on the perpendicular bisector of the line segment connecting those initial two points. This bisector has slope $\\frac{4}{3}$ and passes through the point $(-2, \\frac{3}{2})$, so it is given by\n\n$y-\\frac{3}{2}= \\frac{4}{3}(x+2) \\Leftrightarrow$\n$y = \\frac{4}{3}x+\\frac{7}{6}$\n\nThe third point must have a distance of $\\sqrt{3^{2}+4^{2}}= 5$ from the origin, so we solve the quadratic equation\n\n$x^{2}+\\left( \\frac{4}{3}x+\\frac{7}{6}\\right)^{2}= 25$\n\nTo find our messy answer. [/hide]\n[hide=\"Messy solution 2\"] The side length of the triangle is $5$, so we know the third point must be at some $(5 \\cos \\theta, 5 \\sin \\theta)$.\n\nMoreover, letting $\\phi = \\arctan \\frac{3}{-4}$ (so that $\\sin \\phi = \\frac{3}{5}, \\cos \\phi = \\frac{-4}{5}$) we have $\\theta = \\phi-60^{\\circ}$, so\n\n$x = 5 \\cos \\theta = 5 \\cos (\\phi-60^{\\circ}) = 5 \\cos \\phi \\cos 60^{\\circ}+5 \\sin \\phi \\sin 60^{\\circ}=-4 \\cdot \\frac{1}{2}+3 \\cdot \\frac{ \\sqrt{3}}{2}$\n\nAnd\n\n$y = 5 \\sin \\theta = 5 \\sin (\\phi-60^{\\circ}) = 5 \\sin \\phi \\cos 60^{\\circ}-5 \\cos \\phi \\sin 60^{\\circ}= 3 \\cdot \\frac{1}{2}-4 \\cdot \\frac{ \\sqrt{3}}{2}$ [/hide][/quote]\nI like the first messy way better because there is no trig involved. :) \nI solved your way for the first way because I like solving these kinds of equations. I found x to be \n[hide]$x=\\frac{\\sqrt{739}}{10}$.[/hide]\r\nI'm sure theres some mistake but I'm too tired." -} -{ - "Problem": "Let $ \\omega$ be the circumcircle of $ ABC$ and let $ D$ be a fixed point on $ BC$, $ D\\neq B$, $ D\\neq C$. Let $ X$ be a variable point on $ (BC)$, $ X\\neq D$. Let $ Y$ be the second intersection point of $ AX$ and $ \\omega$. Prove that the circumcircle of $ XYD$ passes through a fixed point.", - "Solution_1": "w.l.o.g. $ \\beta \\ge \\gamma$. Let E the second intersection between the circumcircle of XYD and $ \\omega$. DE meets $ \\omega$ on F. So 'cause DXYE is cyclic $ \\angle FEY \\equal{} \\angle YAC \\plus{} \\angle BCA$, so $ \\angle FEC \\equal{} \\angle BCA$ and then F is the symmetric of A w.r.t. the axis of BC. Then 'cause E is the intersection between FD and $ \\omega$ that are all fixed it's also fixed.", - "Solution_2": "[quote=\"freemind\"]Let $ \\omega$ be the circumcircle of $ ABC$ and let $ D$ be a fixed point on $ BC$, $ D\\neq B$, $ D\\neq C$. Let $ X$ be a variable point on $ (BC)$, $ X\\neq D$. Let $ Y$ be the second intersection point of $ AX$ and $ \\omega$. Prove that the circumcircle of $ XYD$ passes through a fixed point.[/quote]\r\n[hide=\"Solution\"]\nConsider two points on $ BC$ not equal to $ D$, $ B$, or $ C$. Call them $ X$ and $ X'$. Let $ AX\\perp BC$ and $ X'$ be a random point. Let $ AX$ and $ AX'$ meet the circle at $ Y$ and $ Y'$ respectively. Allow the circumcircle of $ \\triangle DXY$ to meet $ \\omega$ at $ Z$. Let $ \\angle Y'ZY \\equal{} \\alpha$, so since $ \\angle DXY \\equal{} 90\\implies \\angle DZY \\equal{} 180 \\minus{} \\angle DXY \\equal{} 90$, which means that $ \\angle Y'ZD \\equal{} 90 \\minus{} \\alpha$. Also, $ \\angle Y'AY \\equal{} \\alpha$ and $ \\angle AXB \\equal{} 90$, so $ \\angle DX'Y' \\equal{} \\angle AXX' \\plus{} \\angle XAX' \\equal{} 90 \\plus{} \\alpha$. This results in $ \\angle DX'Y' \\plus{} \\angle DZY' \\equal{} 90 \\minus{} \\alpha \\plus{} 90 \\plus{} \\alpha\\equal{}180$, so $ X'DZY'$ is cyclic, which means that the circumcircle of $ X'Y'D$ passes through $ Z$. Yet, $ Z$ is fixed because it is the intersection between the circumcircle of $ XDY$ (which is fixed because $ X$, $ Y$, and $ D$ are fixed points) and $ \\omega$, which is also fixed. Hence, the circumcircle of $ \\triangle X'DY'$ passes through a fixed point, as desired. [/hide]", - "Solution_3": "Let $ X_1$ be a fixed point on $ BC$ and $ X_{2}$ be any point on $ BC$ ($ X_{1}, X_{2}$ are distinct from D). Let $ AX_{i}$ cut the circumcenter of the triangle $ ABC$ at $ Y_{i}$.\r\nLet the circumcenter of the triangle $ DX_{1}Y_1$ touch the circumcenter of the triangle $ ABC$ at $ T$.\r\n\r\nThen $ \\angle TDX_{1}\\equal{} 180^{\\circ}\\minus{}\\angle{TY_1A}\\equal{}\\angle{TBA}\\equal{}180^{\\circ}\\minus{}\\angle{TY_{2}A}$, therefore the point $ T$ belongs to the circumcenter of the triangle $ DX_{2}Y_{2}$, and since the points $ X_{2}, Y_{2}$ are mobile, the problem is solved.", - "Solution_4": "Suppose that the diagram is as given (the other situations are the same). Let $ AD\\cap \\omega \\equal{} A,F$. Let $ \\Gamma$ be the circle through $ D$, $ F$ that is tangent to $ BC$. Clearly $ \\Gamma$ is fixed. Let $ \\Gamma\\cap \\omega \\equal{} E,F$ (so $ E$ is fixed). Now \\[ \\angle XYE \\equal{} \\angle AYE \\equal{} \\pi \\minus{} \\angle AFE \\equal{} \\pi \\minus{} \\angle AFD \\equal{} \\pi \\minus{} \\angle EDX\\] so $ DXYE$ is cyclic. Hence the circumcircle of $ \\triangle DXY$ passes through a fixed point, $ E$.", - "Solution_5": "For the record, the circumcircle of DXY passes through D, a fixed point ;)\r\n\r\nBut we know what you mean.", - "Solution_6": "If $E=(DXY)\\cap \\omega$ and $X'$ is another variable point in $BC$ and $Y'$ the intersection of $AX'$ with $\\omega$ we have,\n$\\angle X'DE=AYE=180-AY'E=180-X'Y'E$ \nSo $X'Y'DE$ is cyclic and hence we are done" -} -{ - "Problem": "Let $R$ be a noetherian local ring. Prove that if there exists a nonzero, finitely generated, injective $R$-module, then $R$ is artinian.", - "Solution_1": "First, let's make some notations and some simplifications:\r\n\r\nLet $\\frak m$ be the maximal ideal of $R$. Also, denote the non-zero, finitely generated, injective module by $M$. Given an element $x\\in R,\\ \\mbox{Ann}(x)$ is its annihilator, i.e. the set $\\{y\\in R\\ |\\ yx=0\\}$. Similarly, we can define the annihilator $\\mbox{Ann}(m)$ of an element $m\\in M$. If $x\\in R$, then $(x)$ denotes the ideal of $R$ generated by $x$. \r\n\r\nConsider the set $\\frak a$ of elements in $R$ which are nilpotent and annihilate every element of $M$. Clearly, $\\frak a$ is an ideal of $R$, and nothing changes if we replace $R$ with $R/\\frak a$. This is because on the one hand, $M$ can be naturally considered to be an $R/\\frak a$-module, with all its properties intact (non-triviality, finite generation, and injectivity), and on the other hand, all prime ideals of $R$ contain $\\frak a$, so the conclusion we want to reach (namely that $R$ has precisely one prime ideal: Artinian rings = Noetherian rings of dimension zero) is not affected by passing to $R/\\frak a$. Now, instead of writing $R/\\frak a$, I'll keep $R$ and I'll simply make the additional assumption that $(*)$ there are no non-zero nilpotent elements of $R$ which annihilate all of $M$. \r\n\r\nOne observation that will be used repeatedly: there is an $R$-module homomorphism from $(x)$ into $M$ sending $x$ onto $m\\in M$ iff $\\mbox{Ann}(x)\\le\\mbox{Ann}(m)$. Whenever this happens, by injectivity, the homomorphism can be extended to $R\\to M$. If $1\\mapsto n$, then $x\\mapsto xn=m$, so $m\\in xM$. \r\n\r\n\r\nNow for the actual proof:\r\n\r\nAssume there is a non-zero-divisor $x\\in\\frak m$. Then for any $m\\in M$ we can find a homomorphism $(x)\\to M$ with $x\\mapsto m$. By the remarks made above, $m\\in xM$.Since $m$ was arbitrary, this says that $M=xM$. $x$ belongs to the radical of $R$ and $M$ is finitely generated, so Nakayama's Lemma gives $M=0$, contradiction. Hence, all elements of $\\frak m$ are zero-divisors. \r\n\r\nAssume now that the conclusion we want to reach is false, i.e. assume $\\frak m$ is not the only prime ideal. Equivalently, assume not all elements in $\\frak m$ are nilpotent. \r\n\r\nThe set of zero-divisors is the union of the finitely many prime ideals belonging to $(0)$ in $R$ (the radicals of the members of a primary decomposition of $(0)$, that is), so by teh above, $\\frak m$ must coincide with one of these prime ideals. Since the mentioned prime ideals are precisely the prime annihilators of elements of $R$, we can find $x\\in R$ with $\\mbox{Ann}(x)=\\frak m$. If $x$ happens to be invertible, we're done: the annihilator of an invertible element is $(0)$, so $\\frak m=(0)$, contradicting our assumption. Otherwise, $x$ is nilpotent. By $(*)$, it does not annihilate $M$, so we can find $xm\\ne 0$ for some $m\\in M$. The annihilator of $xm$ is $\\frak m$, so if $a$ is any non-zero element of $R$, then there is a homomorphism $(a)\\to M,\\ a\\mapsto xm$. In turn, this implies $xm\\in aM$. Now take $a\\in\\frak m$ to be non-nilpotent (there is one such element, by assumption), and repeat the procedure for $a^{n},\\ n\\ge 1$. We get $xm\\in\\bigcap_{n\\ge 1}a^{n}M=\\{0\\}$ (this last equality is a particular case of a result of Krull, I believe), contradicting $xm\\ne 0$." -} -{ - "Problem": "prove or disprove:\r\nWe have triangle ABC and CL is the bisector. Points M and N are on CL such that angles CAM and BAN are equal. Prove that angles ABN and CBM are equal, too.", - "Solution_1": "[quote=\"Beat\"]We have triangle ABC and CL is the bisector. Points M and N are on CL such that angles CAM and BAN are equal. Prove that angles ABN and CBM are equal, too.[/quote]\r\n\r\nThis is a consequence of [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=359003#p359003]http://www.mathlinks.ro/Forum/viewtopic.php?t=58872 post #1[/url] Theorem 2. In fact, that theorem yields that the reflections of the lines AM, BM, CM in the angle bisectors of the angles CAB, ABC, BCA concur at one point. Denote this point by Q. Now, since the line AQ is the reflection of the line AM in the angle bisector of the angle CAB, while the line AB is the reflection of the line AC in this angle bisector, the angle between the lines AB and AQ must equal to the angle between the lines AC and AM (since reflections preserve angles); in other words, < BAQ = < CAM. Comparing this with < CAM = < BAN, we get < BAQ = < BAN, and thus the point Q lies on the line AN. On the other hand, since the line CM is the line CL, i. e. the angle bisector of the angle BCA, the line CQ, being the reflection of the line CM in the angle bisector of the angle BCA, must be the angle bisector of the angle BCA, as well; thus, the point Q lies on the angle bisector of the angle BCA, i. e. on the line CL, i. e. on the line CN. So, altogether, we see that the point Q lies on the lines AN and CN. Thus, Q = N. Now, similarly to how we showed that < BAQ = < CAM, we can find that < ABQ = < CBM; since Q = N, this becomes < ABN = < CBM, and the proof is complete.\r\n\r\nSorry for the messy explanation, I hope it was, at least, understandable.\r\n\r\n Darij", - "Solution_2": "Thanks a lot. But is there a simple proof only for this fact? Just asking. If there is please post it.. Thank you", - "Solution_3": "[b]A interesting Lemma.[/b] $\\triangle ABC,\\ M\\in BC\\Longrightarrow\\frac{MB}{MC}=\\frac{AB}{AC}\\cdot \\frac{\\sin \\widehat {MAB}}{\\sin \\widehat {MAC}}\\ \\ (1)\\ .$\r\n[b]Proof.[/b] Denote $m(\\widehat {MAB})=x$, $m(\\widehat {MAC})=y$. From the product of the relations\r\n$\\frac{MB}{MA}=\\frac{\\sin x}{\\sin B},\\ \\frac{MA}{MC}=\\frac{\\sin C}{\\sin y},\\ \\frac{\\sin C}{\\sin B}=\\frac{AB}{AC}$ we obtain the relation $(1)\\ .$\r\n\r\n[b]The Steiner's theorem.[/b] Let $ABC$ be a triangle and let $M,N\\in BC$ be two points\r\nso that the rays $[AM$ and $[AN$ are isogonals in the angle $\\widehat {BAC}$,i.e. the angles\r\n$\\widehat {BAC}$ and $\\widehat {MAN}$ have same bisector. Then $\\frac{MB}{MC}\\cdot \\frac{NB}{NC}=\\left(\\frac{AB}{AC}\\right)^2\\ \\ (2)\\ .$\r\n[b]Proof.[/b] Denote $m(\\widehat {MAB})=m(\\widehat {NAC})=x$, $m(\\widehat {MAC})=m(\\widehat {NAB})=y$.\r\nUsing the above Lemma, from the product of the relations\r\n$\\frac{MB}{MC}=\\frac{AB}{AC}\\cdot \\frac{\\sin x}{\\sin y},\\ \\frac{NB}{NC}=\\frac{AB}{AC}\\cdot \\frac{\\sin y}{\\sin x}$\r\nresults the [u]Steiner's relation[/u] $(2)\\ .$\r\n\r\n[b]A solution of the proposed problem.[/b] Denote $m(\\widehat {ABN})=x$, $m(\\widehat {MBN})=z$,\r\n\r\n$m(\\widehat {CBM})=y.$ From the product of the relations\r\n\r\n$\\left(\\frac{AL}{AC}\\right)^2=\\frac{NL}{NC}\\cdot \\frac{ML}{MC}\\ \\ \\ ;\\ \\ \\ \\frac{NL}{NC}=\\frac{BL}{BC}\\cdot \\frac{\\sin x}{\\sin (y+z)}$\r\n\r\n$\\frac{ML}{MC}=\\frac{BL}{BC}\\cdot \\frac{\\sin (x+z)}{\\sin y}\\ \\ \\ ;\\ \\ \\ \\left(\\frac{BL}{BC}\\right)^2=\\left(\\frac{AL}{AC}\\right)^2$\r\n\r\nwe obtain the relation $\\sin x\\sin (x+z)=\\sin y\\sin (y+z)\\Longleftrightarrow$\r\n\r\n$\\cos z-\\cos (2x+z)=\\cos z-\\cos (2y+z)\\Longleftrightarrow2x+z=2y+z\\Longleftrightarrow x=y\\ .$", - "Solution_4": "Virgil Nicula Thanks ! :omighty: \r\nI`m sure that Darij`s solution is nice, too but i don`t have time for it. this problem is my homework for tomorrow :) \r\nThanks again both of you" -} -{ - "Problem": "At a shotgun factory in Ouagadougou, Burkina Faso, the $N$ employees don't feel like working. Unfortunately, they ran out of luck since the new manager, Oleg Protasov, wants work to get done. He gave each of the workers a task and he computed how long each task lasts. The task of employee $i$ lasts for $T_i$ days.\r\n\r\nTo waste as much time as possible, the employees started to make pertinent comments: \"I can't start working until some other employees of my choice finish their work... Off with Protasov's head.\" Thus, every employee $i$ has a list of employees on which he depends, i.e. those workers must finish their tasks before $i$ can start his own. Obviously, some of them do not depend on anybody, so they can start right away.\r\n\r\nProtasov, in order to mantain his head on his shoulders, accepted the sindicate's proposal, but he also computed the minimum time required to complete the tasks. Furthermore, it was commonly agreed that every employee will start his/her work as late as possible, but so that the time required is the minimum time computed a sentence ago.\r\n\r\n\r\nInput:\r\n\"protasov.in\": on the first line, the number $N$; on the next $N$ lines, the $i$th line describes the task of the $i$th employee by $T_i \\ \\ D_i \\ \\ S_{i 1} \\ \\ S_{i 2} \\ \\ \\ldots \\ \\ S_{i D_i}$, where $T_i$ is the time required to fulfill the task, $D_i$ is the number of employees on which he/she depends and $S_{i1},\\ldots,S_{i D_i}$ are the respective employees.\r\n\r\n\r\nOutput:\r\n\"protasov.out\": $N+1$ lines; on the first line is the minimum time required to complete the job; on the next $N$ lines are the maximum delay time of each employee.\r\n\r\n\r\nRestrictions:\r\n$0 < N , T_i < 101$.\r\n\r\n\r\nObservations:\r\nThere are no circular dependencies.", - "Solution_1": "Construct an oriented graph and use greedy to calculate the minimum total time - keep removing leaves and set the starting time for each removed vertex to the maximum of ending times of its dependencies.\r\n\r\nAfter you calculate the ending time of the last vertex, take again the original graph but this time remove vertices from the top (i.e. the ones on who nobody depends first). Set the ending times as late as possible. And that's it!" -} -{ - "Problem": "Prove that, for any positive real numbers $a,b$ and $c$,\r\n\\[ \\sum_{cycl}\\frac{a^2+bc}{a^2+(b+c)^2}\\le\\frac{18}{5}\\cdot\\frac{a^2+b^2+c^2}{(a+b+c)^2} \\]", - "Solution_1": "My ugly proof:\r\n$\\sum_{cyc}\\frac{a^2+bc}{a^2+(b+c)^2}\\le\\frac{18}{5}\\cdot\\frac{a^2+b^2+c^2}{(a+b+c)^2}\\Leftrightarrow\\sum\\left(\\frac{a^2+bc}{a^2+(b+c)^2}-\\frac{2}{5}\\right)\\leq\\frac{6}{5}\\left(\\frac{3(a^2+b^2+c^2)}{(a+b+c)^2}-1\\right)\\Leftrightarrow$\r\n$\\sum_{cyc}\\frac{3a^2-3bc-2(b-c)^2}{a^2+(b+c)^2}\\leq6\\cdot\\sum_{cyc}\\frac{(a-b)^2}{(a+b+c)^2}\\Leftrightarrow$\r\n$\\Leftrightarrow3\\cdot\\sum_{cyc}\\frac{2a^2-2bc}{ a^2+(b+c)^2}\\leq\\sum_{cyc}(a-b)^2\\left(\\frac{12}{(a+b+c)^2}+\\frac{4}{c^2+(a+b)^2}\\right)\\Leftrightarrow$\r\n$\\Leftrightarrow3\\cdot\\sum_{cyc}\\left(\\frac{(a-b)(a+c)}{a^2+(b+c)^2}-\\frac{(c-a)(a+b)}{a^2+(b+c)^2}\\right)\\leq\\sum_{cyc}(a-b)^2\\left(\\frac{12}{(a+b+c)^2}+\\frac{4}{c^2+(a+b)^2}\\right)\\Leftrightarrow$\r\n$\\Leftrightarrow3\\cdot\\sum_{cyc}(a-b)\\left(\\frac{a+c}{a^2+(b+c)^2}-\\frac{b+c}{b^2+(a+c)^2}\\right)\\leq\\sum_{cyc}(a-b)^2\\left(\\frac{12}{(a+b+c)^2}+\\frac{4}{c^2+(a+b)^2}\\right)\\Leftrightarrow$\r\n$\\Leftrightarrow3\\cdot\\sum_{cyc}\\frac{(a-b)^2(a^2+b^2+3c^2+2ac+2bc)}{(a^2+(b+c)^2)(b^2+(a+c)^2)}\\leq\\sum_{cyc}(a-b)^2\\left(\\frac{12}{(a+b+c)^2}+\\frac{4}{c^2+(a+b)^2}\\right)\\Leftrightarrow$\r\n$\\Leftrightarrow\\sum_{cyc}(a-b)^2S_c\\geq0,$ where\r\n$S_c=\\frac{12}{(a+b+c)^2}+\\frac{4}{c^2+(a+b)^2}-\\frac{3((a+c)^2+(b+c)^2+c^2)}{(a^2+(b+c)^2)(b^2+(a+c)^2)}=$ \r\n$=\\frac{12}{(a+b+c)^2}+\\frac{4}{c^2+(a+b)^2}-\\frac{3(b^2+(a+c)^2+a^2+(b+c)^2+c^2-a^2-b^2)}{(a^2+(b+c)^2)(b^2+(a+c)^2)}=$\r\n$=\\frac{12}{(a+b+c)^2}-\\frac{3}{a^2+(b+c)^2}-\\frac{3}{b^2+(a+c)^2}+\\frac{4}{c^2+(a+b)^2}+\\frac{3(a^2+b^2-c^2)}{(a^2+(b+c)^2)(b^2+(a+c)^2)}.$\r\n$\\frac{6}{(a+b+c)^2}-\\frac{3}{a^2+(b+c)^2}\\geq0\\Leftrightarrow2(a^2+(b+c)^2)\\geq(a+b+c)^2,$\r\n$\\frac{6}{(a+b+c)^2}-\\frac{3}{b^2+(a+c)^2}\\geq0\\Leftrightarrow2(b^2+(a+c)^2)\\geq(b+a+c)^2$ \r\n and\r\n$\\frac{4}{c^2+(a+b)^2}+\\frac{3(a^2+b^2-c^2)}{(a^2+(b+c)^2)(b^2+(a+c)^2)}\\geq0,$ which obviously true. \r\nId est, our inequality is proved. :)", - "Solution_2": "(My friend & I solved it)\r\nWLOG,assume that $a+b+c=3$.We have to prove that $\\frac{2}{5}(a^2+b^2+c^2)\\geq\\sum\\frac{4a^2+(3-a)^2}{4(a^2+(3-a)^2)}$\r\nThis is an easy problem because:\r\n$\\frac{2}{5}a^2-\\frac{4a^2+(3-a)^2}{4(a^2+(3-a)^2)}\\geq\\frac{11}{25}a-\\frac{11}{25}\\Leftrightarrow(a-1)^2(80a^2-168a+171)\\geq 0\\ldots$ :lol:", - "Solution_3": "Nice proof! Hiep.\r\n\r\nI found this inequality the same time as this one\r\n[url]http://www.artofproblemsolving.com/Forum/topic-79611.html[/url]", - "Solution_4": "[quote=\"ductrung\"]Prove that, for any positive real numbers $a,b$ and $c$,\n\\[\\sum_{cycl}\\frac{a^{2}+bc}{a^{2}+(b+c)^{2}}\\le\\frac{18}{5}\\cdot\\frac{a^{2}+b^{2}+c^{2}}{(a+b+c)^{2}}\\]\n[/quote]\r\nMy solution:\r\nThe inequality is equivalent to\r\n\\[\\sum\\frac{(b+c)^{2}-bc}{a^{2}+(b+c)^{2}}+\\frac{18}{5}\\cdot\\frac{a^{2}+b^{2}+c^{2}}{(a+b+c)^{2}}\\ge 3 \\]\r\nSince $(b+c)^{2}\\ge 4bc$ hence it suffices to prove\r\n\\[\\sum\\frac{(b+c)^{2}}{4(a^{2}+(b+c)^{2})}+\\frac{6}{5}\\cdot\\frac{a^{2}+b^{2}+c^{2}}{(a+b+c)^{2}}\\ge 1 \\]\r\nBy Cauchy Inequality we have\r\n\\[\\sum\\frac{(b+c)^{2}}{4(a^{2}+(b+c)^{2})}\\ge \\frac{(a+b+c)^{2}}{\\sum(a^{2}+(b+c)^{2})}=\\frac{(a+b+c)^{2}}{2(a^{2}+b^{2}+c^{2})+(a+b+c)^{2}}\\]\r\nWLOG, we may assume $a+b+c=1$. Setting $x=a^{2}+b^{2}+c^{2}$ then $3x \\ge 1$, it remains to prove that \r\n\\[\\frac{1}{2x+1}+\\frac{6x}{5}\\ge 1 \\]\r\n\\[\\Leftrightarrow x(3x-1) \\ge 0 \\]\r\nWhich is true. \r\nWe are done. :)", - "Solution_5": "A similar inequality that I just thought of:\r\n\r\nProve that $\\sum \\frac{a^{2}+bc}{2a^{2}+(b+c)^{2}}\\le \\frac{7}{4}-\\frac{1}{4}\\cdot \\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}$", - "Solution_6": "[quote=\"probability1.01\"]A similar inequality that I just thought of:\n\nProve that $\\sum \\frac{a^{2}+bc}{2a^{2}+(b+c)^{2}}\\le \\frac{7}{4}-\\frac{1}{4}\\cdot \\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}$[/quote]\r\nWe can prove it by the same method. :)" -} -{ - "Problem": "Show that: $(-1)^{(p-1)/2}\\binom{p-1}{(p-1)/2}-4^{p-1}$ is divided by $p^3$.\netc. where $\\binom{x}{y}$ is a binomial coefficient.\nIt is a strong theorem.(at least for me!).\nI haven't known its solution. Can anybody help me?", - "Solution_1": "I played with it and got to this:\r\n\r\n$\\displaystyle (\\sum _{i=1}^{\\frac {p-1}2}\\frac 1i)^2-\\sum_{1\\le j3 by p 2 ) has its names:\"Eisteiner's theorem\".Consequently,it is easy to prove.I got some results,but not much as you... :D", - "Solution_10": "I'll prove my last observation:\r\n\r\nWe want to show that $\\displaystyle \\sum \\frac 1{ij}=0$, where the ordered pairs $i,j$ are such that $i+j\\ge \\frac {p+1}2$ and $i,j\\in [1,\\frac {p-1}2]$. We can write this sum as $S=\\frac 11\\cdot \\frac 1{\\frac{p-1}2}+\\frac 12\\cdot(\\frac 1{\\frac {p-1}2}+\\frac1{\\frac {p-3}2})+\\ldots+\\frac 1{\\frac {p-1}2}\\cdot (\\frac 1{\\frac {p-1}2}+\\ldots+\\frac 11)$.\r\n\r\nWe now order this sum after $i+j$, and each time we write $\\frac 1{ij}=\\frac 1{i+j}\\cdot(\\frac 1i+\\frac 1j)$. $i+j$ ranges from $\\frac {p+1}2$ to $p-1$, so we have the sum $\\frac 1{\\frac{p+1}2}\\cdot 2(\\frac 1{\\frac {p-1}2}+\\frac 1{\\frac {p-3}2}+\\ldots+\\frac 11)+\\ldots+\\frac 1{p-2}\\cdot 2(\\frac 1{\\frac {p-1}2}+\\frac 1{\\frac {p-3}2})+\\frac 1{p-1}\\cdot 2\\frac 1{\\frac {p-1}2}=-2S$ because $\\frac {p+1}2=-\\frac {p-1}2,\\ldots,p-2=-2,p-1=-1$, where everything is computed in $\\mathbb Z_p$.\r\n\r\nAs for the fact that this implies the theorem we were initially trying to prove, I'm sorry, but I really don't want to start going through so many notebook pages and filter the necessary steps from those which are redundant and so on. \r\n\r\nSorry to disappoint you.. :)" -} -{ - "Problem": "Hello everyone,\r\n\r\nMy question is this and is very specific:\r\n\r\n100000 euros deposited on 1 January in a bank at 8% annually.\r\nWhat is the value at the end of each quarter of the year?\r\n\r\nIs this correct?:\r\n\r\n$ Amount \\equal{} C(1 \\plus{} i*n)$.\r\n\r\n$ n$ taking the value of $ 3$ to calculate the value of the first quarter.\r\n\r\nBut I would not know if $ i \\equal{} 0.08.$\r\n\r\nA greeting", - "Solution_1": "Unfortunately, your question is incomplete. You also should specify the interest compounding method (which can range from \"annually\" to \"continuously\" with plenty of options in between...). You seem to assume monthly compounding. If so, there are two wrong things with your solution:\r\n1) The monthly interest rate is not 0.08 but __________\r\n2) After the first month, you'll also gain interest on interest, and so on, so the money do not increase by the same amount each month but are _______ by the same number, which is _________" -} -{ - "Problem": "Let $ S$ be a set of all positive integers which can be represented as $ a^2 \\plus{} 5b^2$ for some integers $ a,b$ such that $ a\\bot b$. Let $ p$ be a prime number such that $ p \\equal{} 4n \\plus{} 3$ for some integer $ n$. Show that if for some positive integer $ k$ the number $ kp$ is in $ S$, then $ 2p$ is in $ S$ as well.\r\n\r\nHere, the notation $ a\\bot b$ means that the integers $ a$ and $ b$ are coprime.", - "Solution_1": "What is the meaning of $ a\\bot b$ :blush:", - "Solution_2": "It means: the numbers are coprime.\r\n\r\n[Moderator edit: Added to the first post. I love this notation, but I wouldn't use it without explanation. ;) - darij]", - "Solution_3": "Here is my solution : \r\nFrom the condition we have property ,there exist $ x,y\\in N$ such that \r\n$ p|x^2+5y^2$ \r\nExist $ (\\frac{-5}{p})=1$ \r\nIt mean that there exist a number $ a\\in N$ such that $ p|a^2+5$\r\nLet $ m=[\\sqrt {p}]$ \r\nConsider ${ A=\\{x+ay}$ where $ x,y=0,..,m$ \r\nBecause $ |A|=(m+1)^2>p$ so exist $ (x_1,y_1),(x_2,y_2)$ such that\r\n$ x_1+ay_2\\equiv x_2+ay_2 (\\mod p )$\r\n$ \\Leftrightarrow x_1-x_2\\equiv a(y_2-y_1) (\\mod p )$\r\nTake $ x=|x_1-x_2| ,y=|y_1-y_2|$ then \r\n$ p|x^2+5y^2$ \r\nBut $ 00$, then $ X_k\\to\\plus{}\\infty$ almost surely. Of course, this is rather useless without an efficient way to compute $ E\\log V$ in terms of $ a$ and $ b$ and I haven't figured out how to do it yet. Any ideas?\r\n\r\nNeedless to say, the non-positive case is much worse. Currently I do not know how to get anything reasonably nice for it (OK, you, probably, can get an answer similar to the positive case one, but it will be completely useless because the involved quantities are next to impossible to compute).", - "Solution_8": "general theorems on continuous state Markov state that under mild conditions (kind of irreducibility and aperiodicity), if this Markov chain $ (Y_k)_k$ has an invariant probability $ \\mu$ then this Markov chain is also Ergodic. The Markov chain of interest is the one defined on $ E \\equal{} [a; \\plus{} \\infty[$ by:\r\n\\[ Y_{k \\plus{} 1} \\equal{} U_{k} \\plus{} \\frac {\\alpha}{Y_k}\\]\r\nwhere $ U_k$ are iid and uniformly distributed on $ [a;b]$. Irreducibility (for Lebesgue measure for example) and aperiodicity are not hard to prove, but how did you show that there exists an invariant probability $ \\mu$. If $ \\mu$ exists (and $ \\ln$ is $ \\mu$-integrable), then $ \\mu$-almost everywhere one would have $ \\frac {1}{n} \\sum_1^n \\ln(Y_k) \\to \\int_E \\ln(x) \\, d \\mu(x)$, which gives the condition you mentioned, but I am still struggling to show that this $ \\mu$ exists :blush:", - "Solution_9": "For the positive case, just assume that $ U_k$ have density bounded by $ M$ and supported on finite interval. Write the convolution equation for the density of $ Y$. The right hand side will give you a compact linear operator from the convex closed set $ \\{p\\in L^1: 0\\le p\\le M,\\textstyle{\\int_0^\\infty p=1\\}\\subset L^1}$ to itself (you can even add the $ p(x)=O(x^{-2})$ condition at infinity, so summability of logarithm will be automatic). The Shauder lemma finishes the story. Another way is just to show directly that the continuous fraction in question converges with probability $ 1$ ;)", - "Solution_10": "thank you Fedja :)", - "Solution_11": "Let's talk about the positive case a bit more. What bothers me a lot is that we have this intractable sharp condition $ E\\log V<0$. But, if you just look at $ EX_k$, you'll get a simple sufficient condition $ EU_k<1\\minus{}\\alpha$. And I have no idea how to deduce the former from the latter. :wacko:", - "Solution_12": "a side question: how would you do to compute a good approximation of this invariant measure (ie. the distribution of the continuous fraction) A few million iterations (computation time ~ 10sec) of a naive Monte Carlo estimate of $ Q_{k\\plus{}1} \\equal{} U_k \\plus{} \\frac{ \\alpha}{ Q_{k\\minus{}1}}$ gives a pretty good approximation (but of course the approximation is horrible near the points where the density is very small), but I am curious about other available methods. (I do not know anything about computational methods for integral equation etc...) This should not be so hard to compute that invariant distribution this it is included in $ [m;M]$ with $ m\\equal{}a \\plus{} \\frac{\\alpha}{M}$ and $ M\\equal{}b\\plus{} \\frac{\\alpha}{m}$.", - "Solution_13": "[quote=\"alekk\"]This should not be so hard to compute that invariant distribution this it is included in $ [m;M]$ with $ m \\equal{} a \\plus{} \\frac {\\alpha}{M}$ and $ M \\equal{} b \\plus{} \\frac {\\alpha}{m}$.[/quote]\r\nWell, I'm not so sure about this: if you have the equation $ Y_k\\equal{}U\\plus{}\\alpha Y_{k\\minus{}1}$, you already have a big headache computing the corresponding infinite convolution and our problem is hardly any better. Of course, you can always just experiment, but I would prefer some more explicit answer. The problem is that I do not see one yet. :wallbash:", - "Solution_14": "Here's a random, perhaps naive, thought. We can write the recurrence $ X_k \\equal{} U_k X_{k\\minus{}1} \\plus{} \\alpha X_{k\\minus{}2}$ in the matrix form\r\n\\[ \\begin{bmatrix} x_k \\\\ x_{k \\minus{} 1} \\end{bmatrix} \r\n \\equal{} \\begin{bmatrix} U_k & \\alpha \\\\ 1 & 0 \\end{bmatrix}\r\n \\begin{bmatrix} x_{k\\minus{}1} \\\\ x_{k\\minus{}2} \\end{bmatrix}.\\]\r\nSo our question boils down to the convergence of a product of random iid 2x2 matrices. Does that make life easier? I know that there is a ton of work on the product of random matrices, but I don't know if any of it is useful to us.", - "Solution_15": "Alas, it seems pretty much the same to me: we still need that invariant distribution to reduce the question to a pure computation. I looked at some of the relevant papers starting with the famous 1973 paper of Kesten. They have great results [b]under[/b] the assumption that our limit is $ 0$ (or is not $ 0$) but do not provide the slightest clue as to how to check that assumption in practice except for the brute force random simulation (they have some equivalent conditions, but they are equally useless. :( )\r\n\r\nIf you find anything, let me know :).", - "Solution_16": "@Fedja: I do not understand why this is not true that the invariant measure of $ Q_{n\\plus{}1} \\equal{} U \\plus{} \\frac{\\alpha}{Q_n}$ is not included in $ E \\equal{} [m;M]$. If one starts in $ E$ one stays in it and the invariant distribution is unique isn't it. Where am I wrong ?", - "Solution_17": "You aren't wrong. The invariant measure is compactly supported if $ a>0$. I just said that it may be rather hard to find a function even if you know that it has compact support, is infinitely smooth, etc. \r\n\r\nActually, when I looked at the phrase again, I realized that I probably just misunderstood your broken English. The intended meaning apparently was \"It is not hard to show that the invariant measure is supported on $ [m,M]$\". The meaning I got was \"The invariant measure should not be so hard to compute since it is supported on $ [m,M]$\". :)", - "Solution_18": "thanks Fedja. Yes, I meant \"The invariant measure should not be so hard to compute since it is supported on $ [m,M]$\", and by \"compute\" I meant: find a good numerical approximation. Basically, if we are given $ a,b,\\alpha$, how can we efficiently compute $ \\int_{E} \\ln(x) d \\mu(x)$ ?", - "Solution_19": "That's exactly the question that gives me a headache... I even went as far as asking a few probabilists I know. No result so far :(.\r\n\r\nOf course, you can try to solve the integral equation by projecting it onto a finite dimensional space and solving the corresponding linear system, but this method can hardly be called \"efficient\"." -} -{ - "Problem": "Let $ a,b,c,d$ be positive numbers sach that $ a\\plus{}b\\plus{}c\\plus{}d\\equal{}4$. Prove that\r\n\r\n$ \\frac 1{ab}\\plus{}\\frac 1{bc}\\plus{}\\frac 1{cd}\\plus{}\\frac 1{da}\\ge a^2\\plus{}b^2\\plus{}c^2\\plus{}d^2$.", - "Solution_1": "The problem is equivalent to \\[ (a\\plus{}c)(b\\plus{}d)\\geq abcd\\cdot\\sum a^{2}.\\]\r\nDenote $ \\sqrt{abcd}\\equal{}t$, and also $ x\\equal{}a\\plus{}c, y\\equal{}b\\plus{}d$. Then $ x\\plus{}y\\equal{}4$ and $ a^{2}\\plus{}b^{2}\\plus{}c^{2}\\plus{}d^{2}\\equal{}x^{2}\\plus{}y^{2}\\minus{}2(ac\\plus{}bd)\\leq x^{2}\\plus{}y^{2}\\minus{}4t$.\r\nTherefore it will suffice to prove that \\[ xy\\geq t^{2}(x^{2}\\plus{}y^{2})\\minus{}4t.\\]But $ x\\plus{}y\\equal{}4$, therefore $ x^{2}\\plus{}y^{2}\\equal{}16\\minus{}2xy$.We denote $ a\\equal{}xy$ and therefore it suffices to prove that \\[ t^{2}(16\\minus{}2a)\\minus{}4t\\minus{}a\\leq 0.\\]\r\nIt`s discriminant is $ \\triangle\\equal{}8(\\minus{}a^{2}\\plus{}8a\\plus{}2)>0$, because $ a\\leq 4$. The two roots of the above equation are\r\n\\[ t_{1}\\equal{}\\frac{4\\minus{}2\\sqrt{2(\\minus{}a^{2}\\plus{}8a\\plus{}2)}}{4(8\\minus{}a)}<0\\leq t\\]\r\n\\[ t_{2}\\equal{}\\frac{4\\plus{}2\\sqrt{2(\\minus{}a^{2}\\plus{}8a\\plus{}2)}}{4(8\\minus{}a)}\\geq 1\\geq t\\]\r\n\r\nTherefore \\[ t^{2}(16\\minus{}2a)\\minus{}4t\\minus{}a\\equal{}(16\\minus{}2a)(t\\minus{}t_{1})(t\\minus{}t_{2})\\leq 0,\\] so the problem is solved!\r\n\r\nThe last two inequalities are easily proven, keeping in mind that $ a\\leq 4.$\r\n\r\nI hope I haven`t made any mistake :)", - "Solution_2": "[quote=\"andyciup\"]\n\\[ t_{2} \\equal{} \\frac {4 \\plus{} 2\\sqrt {2( \\minus{} a^{2} \\plus{} 8a \\plus{} 2)}}{4(8 \\minus{} a)}\\geq 1\\geq t\n\\][/quote]\r\nIt is wrong. See $ a\\equal{}1$.", - "Solution_3": "That`s right :blush: sorry..Maybe I can fix it in some way.. Anyway nice inequality!", - "Solution_4": "Using AM-GM inequality we have: $ 4 \\equal{} a \\plus{} b \\plus{} c \\plus{} d\\ge 4\\sqrt [4]{abcd}\\to abcd\\le 1$\r\nOn the other hand\r\n\\[ f(a,b,c,d) \\equal{} \\left(\\frac {1}{a} \\plus{} \\frac {1}{c}\\right)\\left(\\frac {1}{b} \\plus{} \\frac {1}{d}\\right) \\minus{} a^2 \\minus{} b^2 \\minus{} c^2 \\minus{} d^2\r\n\\]\r\n\r\n\\[ f(a,b,c,d) \\minus{} f\\left(a,\\frac {b \\plus{} d}{2},c,\\frac {b \\plus{} d}{2}\\right) \\equal{} (b \\minus{} d)^2\\left(\\frac {(a \\plus{} c)}{abcd(b \\plus{} d)} \\minus{} \\frac {1}{2} \\right)\r\n\\]\r\n\r\n\\[ \\ge (b \\minus{} d)^2\\left(\\frac {(a \\plus{} c)}{(b \\plus{} d)} \\minus{} \\frac {1}{2} \\right) \\ge 0\r\n\\]\r\nThe rest part is not hard to prove:\r\n\\[ f(a,t,t,t) \\equal{} \\minus{} a^2 \\minus{} 3t^2 \\plus{} \\frac {2}{at} \\plus{} \\frac {2}{t^2} \\equal{} \\frac {(t \\minus{} 1)^2(9t^3 \\minus{} 12t^2 \\plus{} 3t \\plus{} 2)}{(4 \\minus{} 3t)t^2}\\ge 0\r\n\\]\r\nThe equality hold when $ a \\equal{} b \\equal{} c \\equal{} d \\equal{} 1$", - "Solution_5": "[quote=\"zaizai-hoang\"]\\[ f(a,b,c,d) \\equal{} \\left(\\frac {1}{a} \\plus{} \\frac {1}{c}\\right)\\left(\\frac {1}{b} \\plus{} \\frac {1}{d}\\right) \\minus{} a^2 \\minus{} b^2 \\minus{} c^2 \\minus{} d^2\n\\]\n\n\\[ f(a,b,c,d) \\minus{} f\\left(a,\\frac {b \\plus{} d}{2},c,\\frac {b \\plus{} d}{2}\\right) \\equal{} (b \\minus{} d)^2\\left(\\frac {(a \\plus{} c)}{abcd(b \\plus{} d)} \\minus{} \\frac {1}{2} \\right)\n\\][/quote]\r\n\r\nI do not understand :maybe: \r\nWhat do you achieve by replacing b and d by (b+d)/2?\r\nI mean is it some standard method of proving??", - "Solution_6": "I used Mixing Variable method to proving above inequality, isomorphism :wink:", - "Solution_7": "[quote=\"zaizai-hoang\"]Using AM-GM inequality we have: $ 4 \\equal{} a \\plus{} b \\plus{} c \\plus{} d\\ge 4\\sqrt [4]{abcd}\\to abcd\\le 1$\nOn the other hand\n\\[ f(a,b,c,d) \\equal{} \\left(\\frac {1}{a} \\plus{} \\frac {1}{c}\\right)\\left(\\frac {1}{b} \\plus{} \\frac {1}{d}\\right) \\minus{} a^2 \\minus{} b^2 \\minus{} c^2 \\minus{} d^2\n\\]\n\n\\[ f(a,b,c,d) \\minus{} f\\left(a,\\frac {b \\plus{} d}{2},c,\\frac {b \\plus{} d}{2}\\right) \\equal{} (b \\minus{} d)^2\\left(\\frac {(a \\plus{} c)}{abcd(b \\plus{} d)} \\minus{} \\frac {1}{2} \\right)\n\\]\n\n\\[ \\ge (b \\minus{} d)^2\\left(\\frac {(a \\plus{} c)}{(b \\plus{} d)} \\minus{} \\frac {1}{2} \\right) \\ge 0\n\\]\nThe rest part is not hard to prove:\n\\[ f(a,t,t,t) \\equal{} \\minus{} a^2 \\minus{} 3t^2 \\plus{} \\frac {2}{at} \\plus{} \\frac {2}{t^2} \\equal{} \\frac {(t \\minus{} 1)^2(9t^3 \\minus{} 12t^2 \\plus{} 3t \\plus{} 2)}{(4 \\minus{} 3t)t^2}\\ge 0\n\\]t,\nThe equality hold when $ a \\equal{} b \\equal{} c \\equal{} d \\equal{} 1$[/quote]\r\nI think you can't conclude that f(a,b,c,d)>=f(a,t,t,t) instead of f(a,b,c,d)>=f(a,t,c,t) and prove similar to conclude that f(a,b,c,d)>=f(h,h,t,t)! Anyway, your solution is fine!", - "Solution_8": "[quote=\"ghjk\"]\nI think you can't conclude that f(a,b,c,d)>=f(a,t,t,t) instead of f(a,b,c,d)>=f(a,t,c,t) [/quote]\nYou are right.\n\n[quote=\"ghjk\"]\nand prove similar to conclude that f(a,b,c,d)>=f(h,h,t,t)! [/quote]\r\nI think you mean f(a,b,c,d)>=f(h,t,h,t).\r\nWe can also prove directly f(a,b,c,d)>=f(h,t,h,t).\r\n\r\nAn elegant solution is to add the inequalities\r\n$ ac(b \\plus{} d)^4 \\ge 8abcd(b^2 \\plus{} d^2)$,\r\n$ bd(a \\plus{} c)^4 \\ge 8abcd(a^2 \\plus{} c^2)$.", - "Solution_9": "of course, I have known how to use Mixing Variable method :wink: but I am a lazy person, I think the last part is easy to prove because it has only two-variable. So I used to S.M.V for the remain part. Anyway, above inequality is really nice, quite easy and I am interested in solve it :) \r\nCould you solve the general case with $ n$ variable ? Maybe, (n-1)EV will kill it easily :)", - "Solution_10": "This is a generalization of a known similar one.\r\nIf $ a_1,a_2,...a_n$ are positive numbers such that $ a_1\\plus{}a_2\\plus{}...\\plus{}a_n\\equal{}n$, $ n\\ge 4$, then\r\n\r\n$ \\frac 1{a_1^2}\\plus{}\\frac 1{a_2^2}\\plus{}...\\plus{}\\frac 1{a_n^2} \\plus{}8\\minus{}n \\ge \\frac 8{n}(a_1^2\\plus{}a_2^2\\plus{}...\\plus{}a_n^2)$.", - "Solution_11": "[quote=\"Vasc\"]\nIf $ a_1,a_2,...a_n$ are positive numbers such that $ a_1+a_2+...+a_n=n$, $ n\\ge 4$, then\n\n$ \\frac 1{a_1^2}+\\frac 1{a_2^2}+...+\\frac 1{a_n^2} +8-n \\ge \\frac 8{n}(a_1^2+a_2^2+...+a_n^2)$.[/quote]\nLet $f(x)=8x^2-\\frac{n}{x^2}+n-8$.\nHence, we need to prove that $\\sum_{i=1}^nf\\left(a_i\\right)\\leq nf\\left(\\frac{a_1+a_2+...+a_n}{n}\\right)$.\nBut $f''(x)=\\frac{16\\left(x^4-\\frac{3n}{8}\\right)}{x^4}$. Hence, $f$ is a concave function on $(0, 1]$ for $n\\geq4$.\nHence, by LCF-Theorem we need to prove that\n$\\frac{n-1}{x^2}+\\frac{1}{(n-(n-1)x)^2}+8-n\\geq\\frac{8}{n}\\left((n-1)x^2+(n-(n-1)x)^2\\right)$ \nfor $00$ and $16x^3-16x^2+x+2>0$ for $00$. \nId est, the inequality is proven.", - "Solution_12": "The case $n=8$ is nice. :wink:", - "Solution_13": "[quote=\"Vasc\"]The case $n=8$ is nice. :wink:[/quote]\nA form of the inequality in this case is nice. But the proof is the same. No?", - "Solution_14": "[quote=zaizai-hoang]Using AM-GM inequality we have: $ 4 \\equal{} a \\plus{} b \\plus{} c \\plus{} d\\ge 4\\sqrt [4]{abcd}\\to abcd\\le 1$\nOn the other hand\n\\[ f(a,b,c,d) \\equal{} \\left(\\frac {1}{a} \\plus{} \\frac {1}{c}\\right)\\left(\\frac {1}{b} \\plus{} \\frac {1}{d}\\right) \\minus{} a^2 \\minus{} b^2 \\minus{} c^2 \\minus{} d^2\n\\]\n\n\\[ f(a,b,c,d) \\minus{} f\\left(a,\\frac {b \\plus{} d}{2},c,\\frac {b \\plus{} d}{2}\\right) \\equal{} (b \\minus{} d)^2\\left(\\frac {(a \\plus{} c)}{abcd(b \\plus{} d)} \\minus{} \\frac {1}{2} \\right)\n\\]\n\n\\[ \\ge (b \\minus{} d)^2\\left(\\frac {(a \\plus{} c)}{(b \\plus{} d)} \\minus{} \\frac {1}{2} \\right) \\ge 0\n\\]\nThe rest part is not hard to prove:\n\\[ f(a,t,t,t) \\equal{} \\minus{} a^2 \\minus{} 3t^2 \\plus{} \\frac {2}{at} \\plus{} \\frac {2}{t^2} \\equal{} \\frac {(t \\minus{} 1)^2(9t^3 \\minus{} 12t^2 \\plus{} 3t \\plus{} 2)}{(4 \\minus{} 3t)t^2}\\ge 0\n\\]\nThe equality hold when $ a \\equal{} b \\equal{} c \\equal{} d \\equal{} 1$[/quote]\n\n$ \\[ \\ge (b \\minus{} d)^2\\left(\\frac {(a \\plus{} c)}{(b \\plus{} d)} \\minus{} \\frac {1}{2} \\right) \\ge 0\n\\]$I do not think it is always true...so how could you conclude $f\\ge f(a,b,a,b)$?", - "Solution_15": "How to give an upper bound to $\\sum a^2$?? Most of the ineq.s I am concerned of, says $\\sum a^2 \\geq $(something).... :(", - "Solution_16": "[quote=lwwwww][quote=zaizai-hoang]Using AM-GM inequality we have: $ 4 \\equal{} a \\plus{} b \\plus{} c \\plus{} d\\ge 4\\sqrt [4]{abcd}\\to abcd\\le 1$\nOn the other hand\n\\[ f(a,b,c,d) \\equal{} \\left(\\frac {1}{a} \\plus{} \\frac {1}{c}\\right)\\left(\\frac {1}{b} \\plus{} \\frac {1}{d}\\right) \\minus{} a^2 \\minus{} b^2 \\minus{} c^2 \\minus{} d^2\n\\]\n\n\\[ f(a,b,c,d) \\minus{} f\\left(a,\\frac {b \\plus{} d}{2},c,\\frac {b \\plus{} d}{2}\\right) \\equal{} (b \\minus{} d)^2\\left(\\frac {(a \\plus{} c)}{abcd(b \\plus{} d)} \\minus{} \\frac {1}{2} \\right)\n\\]\n\n\\[ \\ge (b \\minus{} d)^2\\left(\\frac {(a \\plus{} c)}{(b \\plus{} d)} \\minus{} \\frac {1}{2} \\right) \\ge 0\n\\]\nThe rest part is not hard to prove:\n\\[ f(a,t,t,t) \\equal{} \\minus{} a^2 \\minus{} 3t^2 \\plus{} \\frac {2}{at} \\plus{} \\frac {2}{t^2} \\equal{} \\frac {(t \\minus{} 1)^2(9t^3 \\minus{} 12t^2 \\plus{} 3t \\plus{} 2)}{(4 \\minus{} 3t)t^2}\\ge 0\n\\]\nThe equality hold when $ a \\equal{} b \\equal{} c \\equal{} d \\equal{} 1$[/quote]\n\n$ \\[ \\ge (b \\minus{} d)^2\\left(\\frac {(a \\plus{} c)}{(b \\plus{} d)} \\minus{} \\frac {1}{2} \\right) \\ge 0\n\\]$I do not think it is always true...so how could you conclude $f\\ge f(a,b,a,b)$?[/quote]\n\nAnyone understand?", - "Solution_17": "[quote=andyciup]The problem is equivalent to \\[ (a\\plus{}c)(b\\plus{}d)\\geq abcd\\cdot\\sum a^{2}.\\]\nDenote $ \\sqrt{abcd}\\equal{}t$, and also $ x\\equal{}a\\plus{}c, y\\equal{}b\\plus{}d$. Then $ x\\plus{}y\\equal{}4$ and $ a^{2}\\plus{}b^{2}\\plus{}c^{2}\\plus{}d^{2}\\equal{}x^{2}\\plus{}y^{2}\\minus{}2(ac\\plus{}bd)\\leq x^{2}\\plus{}y^{2}\\minus{}4t$.\n[color=#f00]Therefore it will suffice to prove that \\[ xy\\geq t^{2}(x^{2}\\plus{}y^{2})\\minus{}4t.\\][/color][/quote]\n\nDear [b]andyciup[/b]\nIt will suffice to prove this\n\\[ xy\\geq t^{2}(x^{2}\\plus{}y^{2})\\minus{}4t^3.\\]\n", - "Solution_18": "[quote=Vasc]Let $ a,b,c,d$ be positive numbers sach that $ a\\plus{}b\\plus{}c\\plus{}d\\equal{}4$. Prove that\n\n$ \\frac 1{ab}\\plus{}\\frac 1{bc}\\plus{}\\frac 1{cd}\\plus{}\\frac 1{da}\\ge a^2\\plus{}b^2\\plus{}c^2\\plus{}d^2$.[/quote]\n\nBY Lagrange method it's easy to prove." -} -{ - "Problem": "The diagonol of a cube is 5 inches. How many square inches are in the total surface area of the cube? :?:", - "Solution_1": "hello, let $ a$ be the side length of the given cube, then we have after the theorem of Pythagorean\r\n$ 3a^2 \\equal{} 5$ and from here get $ a \\equal{} \\sqrt{\\frac {5}{3}} \\text{inches}$, so the surface of the cube is\r\n$ a^2 \\equal{} 6\\cdot \\frac{5}{3}\\equal{}10 \\text{inches}^2$\r\nSonnhard.", - "Solution_2": "hello, do you mean the space diagonal?\r\nSonnhard.", - "Solution_3": "In MATHCOUNTS problems, the diagonal of a cube is the space diagonal of the cube.\r\n\r\nWe can write an equation for the side length $ s$:\r\n\r\n$ s\\sqrt {3} \\equal{} 5 \\implies s \\equal{} \\frac {5}{\\sqrt {3}} \\implies s \\equal{} \\frac {5\\sqrt {3}}{3}$.\r\n\r\nThe surface area of a cube is $ 6s^2 \\implies \\text{SA} \\equal{} 6 \\times \\left (\\frac {25 \\times 3}{9} \\right ) \\equal{} \\boxed{50}$ as nick42 said.", - "Solution_4": "AIME, you screwed your latex :wink: .", - "Solution_5": "[quote=\"AIME15\"]In MATHCOUNTS problems, the diagonal of a cube is the space diagonal of the cube.\n\nWe can write an equation for the side length $ s$:\n\n$ s\\sqrt {3} = 5 \\implies s = \\frac {5}{\\sqrt {3}} \\implies s = \\frac {5\\sqrt {3}}{3}$.\n\nThe surface area of a cube is $ 6s^2 \\implies \\text{SA} = 6 \\times \\left (\\frac {25 \\times 3}{9 \\right ) = \\boxed{50}}$ as nick42 said.[/quote]\r\n\r\nWow... I think he messed up on the left and right parts. it's creepy how people are so advanced in LaTeX. :(", - "Solution_6": "Forgot to close my fraction.", - "Solution_7": "if the diagonal of a cube is 5 inches, then each side would equal 5/root 3. The surface area of a cube is 6s^2, where s= the side length. Therefore, the surface area would be 6*(5/root3)^2=6*25/3=25*2=50.\r\nTherefore, 50 is the answer!" -} -{ - "Problem": "Hi, I am trying to solve this problem where a storage tank ruptures at time t= 0 and oil leaks from the tank at a rate of r(t) = 100e^(-0.01t). I am trying to find out how mmuch oil leaks during the first hour.\r\n\r\nSo far I found r '(t) = e^(-0.01t)\r\nI plugged in 1 for t and got 0.99.\r\n\r\nI am just wondering if this problem is any more complicated then the two steps I just did. \r\n\r\nThank you so much for your help!", - "Solution_1": "Yes, the problem is more complicated than what you just did. What exactly do you think $ r'$ measures?", - "Solution_2": "the rate of oil pumped in?", - "Solution_3": "I don't see oil being pumped in anywhere. I do see oil leaking out, but the rate at which it's leaking is (given as) $ r(t)$, not $ r'(t)$. And if $ r'(t)$ [i]were[/i] the rate at which oil was leaking out, the value $ r'(1)$ would be the rate at which it was leaking after one hour, not the total amount that had leaked over that time. (As it is, $ r'(1)$ is the rate at which the rate at which oil is leaking out is changing.) So, the first thing you should do is to sit and think about what (first in words, then in symbols) the quantity you're interested in is.\r\n\r\nI didn't notice this before, but you also calculated $ r'$ incorrectly." -} -{ - "Problem": "Title is self explanatory.\r\n\r\nThe link is [url]http://www.octmtournament.org/octmattach%5Ctestresults%5C000000000073.pdf[/url]", - "Solution_1": "You get more money for being tied for 4th in the OCTM than being 4th overall in the OHMIO. And the difference between 4th and 3rd overall in the OHMIO is only 50 dollars of cash and a TI 89 Titanium. *mutters*\r\n\r\nO well, I should be grateful I did as well as I did. :yup:", - "Solution_2": "wow i did bad. only ranked 44.\r\n\r\ni still think that last question was worded ambiguously." -} -{ - "Problem": "If $x$ is a real number such that:\r\n\r\n$1 + x +\\frac {x^2}{2! } + \\frac {x^3}{3!}+\\frac {x^4}{4!} ......+\\frac {x^{2004}}{2004!} = 0 $\r\n\r\nProve that $[x^{2004}] \\equiv 1 \\mod 2005$\r\n\r\nWhere $[x^{2004}] $ , of course means an integer less than or equal to $x^{2004} $\r\n\r\nEdited later: Thanks JBL for editing \r\nP.S. . Please put answers in spoilers ...", - "Solution_1": "That should be, $[x^{2004}] \\equiv 1 \\mod 2005$.", - "Solution_2": "Hmmmm, not my kinda question, but yet intriguing... any hints that dont give away too much?", - "Solution_3": "Gyan, maybe i misunderstood something but why would that polynomial have any real roots at all?\r\n\r\n[hide]\nFirst its clear $x<0$. Also By taylor's theorem, there exists $c\\in\\ (x,0)$ such that $e^x=\\frac{x^{2005}}{2005!}e^c+\\sum_{i=0}^{2004}\\frac{x^i}{i!}$ Now if the summation is 0 then\n$e^x=\\frac{x^{2005}}{2005!}e^c$ .... but the right side is negative and the left is positive...so whats wrong here??\n[/hide]", - "Solution_4": "[b]radagast[/b] is right.Look if $n$ is even $e^{x}>1+x+\\frac{x^{2}}{2!}+\\ldots+\\frac{x^{n}}{n!}$ for $x>0$ and $e^{x}<1+x+\\frac{x^{2}}{2!}+\\ldots+\\frac{x^{n}}{n!}$ for $x<0$.So $1+x+\\frac{x^{2}}{2!}+\\ldots+\\frac{x^{n}}{n!}>0$ for all $x$.That means that the equation\r\n$1+x+\\frac{x^{2}}{2!}+\\ldots+\\frac{x^{n}}{n!}=0$ does not have real solutions for even $n$-s.For odd $n$-s the equation above has one real root because it is a polynomial of odd degree and its derivative is the polynomial $1+x+\\frac{x^{2}}{2!}+\\ldots+\\frac{x^{n-1}}{(n-1)!}>0$.", - "Solution_5": "[quote=\"radagast\"]Gyan, maybe i misunderstood something but why would that polynomial have any real roots at all?\n\n[hide]\nFirst its clear $x<0$. Also By taylor's theorem, there exists $c\\in\\ (x,0)$ such that $e^x=\\frac{x^{2005}}{2005!}e^c+\\sum_{i=0}^{2004}\\frac{x^i}{i!}$ Now if the summation is 0 then\n$e^x=\\frac{x^{2005}}{2005!}e^c$ .... but the right side is negative and the left is positive...so whats wrong here??\n[/hide][/quote]\n\n[hide]No reason at all! But don't you see the beauty of that? :D Since there are no real roots :D the \"theorem\" is proved ~ :lol: ( for all x the second part is true since there are/is no such x ..) (Yes I could have put anything there, but was just looking for some funny looking stuff)\n\n:lol: Okay, ducking for cover now! :lol: \nBTW there was \":)\" in the subject line.\nPS Random House, if you don't know is a MIT dorm where all the nerdy math students live ..\n[/hide]", - "Solution_6": "Boo\r\n :lol: :lol: :lol:", - "Solution_7": "Heh yes i was kinda suspecting something was afoot when seeing the source... :)", - "Solution_8": "well that was rather anti-climactic" -} -{ - "Problem": "I need help....\r\n\r\nFind the number of order triples (a,b,c) where a, b, and c are positive integers. A is a factor of B, A is a factor of C, and A + B + C = 100.", - "Solution_1": ":huh: \r\nIs this AIME?\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=798316#798316\r\n\r\nYou can also do it by counting the possible values of b and c when a=1, 2, 4, 5, 10, and 20", - "Solution_2": "no... I dun think so =) obviously not xD\r\n\r\nLet $ b \\equal{} ka,c \\equal{} na$\r\n\r\nThen... $ a(1 \\plus{} k \\plus{} n) \\equal{} 100\\rightarrow a|100$\r\n\r\nTest... lol.\r\n\r\n$ a \\equal{} 1, b \\equal{} 1\\hdots98, c \\equal{} 1\\hdots98$ $ b$ determines $ c$ obviously =) xD. So $ 98$ possible since the order actually matters (o.O).\r\n\r\n$ a \\equal{} 2,b \\equal{} 2, 4, \\hdots 96, c \\equal{} 2, 4, \\hdots 96$ $ 48$ possible i think.... or something of the sort\r\n\r\n$ a \\equal{} 4, b \\equal{} 4,8,\\hdots 92, c \\equal{} 4, 8, \\hdots 92$ ya blah blah etc etc. $ 23$ possible... something like this lol. might've messed up xD\r\nAND SO ON\r\nIf you really wanted to, you can use balls and urns to find combinations... like... $ k \\plus{} n \\equal{} 99,49,$ etc.", - "Solution_3": "ohh its from aime\r\n\r\nthx" -} -{ - "Problem": "Prove that $\\sqrt{2}+\\sqrt{3}+\\sqrt{5}+\\sqrt{7}$ is not rational.", - "Solution_1": "The solution depends on if you want to do ugly computations or not ;)\r\nSee http://www.mathlinks.ro/Forum/viewtopic.php?t=80213 , especially the links there." -} -{ - "Problem": "What would you do if you would have one free year, without any duty at all? I mean if you would have enough money to live from.. What would you do?", - "Solution_1": "aaaaaaaaaaaaaaa", - "Solution_2": "Well if nothing happened to me, like no weight gaining or heart attacks or strokes, I would just play games all day and eat stuff.", - "Solution_3": "Well I'll buy everything I want because I have infinite money...", - "Solution_4": "Behold the philosophical nexus between prison and retirement.", - "Solution_5": "-Watch every movie that comes up\r\n-Spend time in University bookshop (and buy books)\r\n-Go on Youtube a lot\r\n-Very little sleep\r\n-Eat food I normally won't it (because I normally won't have enough money)\r\n-Facebook a lot\r\n\r\nI don't know. Probably just all the things I love doing.", - "Solution_6": "Visit mathematicians at various universities. :)", - "Solution_7": "[quote=\"t0rajir0u\"]Visit mathematicians at various universities. :)[/quote]\r\n\r\nThat sounds neat too actually :D . I would probably be visiting Chemists though (like Professor Schrock at MIT). Of course assuming that they won't mind :P .", - "Solution_8": "obtain all my technical needs regarding education n then do some research in math :D", - "Solution_9": "8760 hours of sleep", - "Solution_10": "1 free year. Assuming my age remained constant it would be training for the IMO. I would love to go soo much!\r\n\r\nIf my age was affect then probably doing more and more advanced maths", - "Solution_11": "maybe vacation or stay home playing games and watching movies :D", - "Solution_12": "Personal math research, eat, sleep, basketball, gaming, youtube. Put it in any order you want, and you can't go wrong :lol:", - "Solution_13": "You make it sound like taking a free year is some kind of imaginary thing. A lot of people go abroad and do nothing useful for half a year.", - "Solution_14": "Given unlimited funds, I'd travel, practice languages, practice instruments, and walk a lot. I'd probably hire a guide or find very learned friends to teach about wherever I am.", - "Solution_15": "I would buy a tank and drive it on the street as my car(loaded of course). Also I would travel around the world, buy a Farrari and buy the biggest house in the world. After I did all this I would spend my time partying! WOOT", - "Solution_16": "I'd have the ability to go where I'd want, and gain ahold of every possibility. For example, I would buy the books that are new and I'm interested in and read them in the airplane on a ride to wherever I want (probably England). Listening to music (while obtaining it from iTunes), using the laptop, on business class so things can run smoothly, etc. However, I wouldn't be pompous and rent a [i]mansion[/i] for a two-night stay. Something good enough that can withstand a kid, adult, whatever age I'd be by the time I get this free year.", - "Solution_17": "study real hard and apply for MIT, study harder there so I can graduate in a month, buy a huge piece of land, hire a hundred workers to make 20 Earthships there, and 10 container homes. Then get all my relatives to those homes, and we can be a clan!", - "Solution_18": "Assuming that I get to keep the infinite money, buy my way out of school after I'm done. Read a textbook once or twice during the year so I don't make myself feel dumb, and go form my own new nuclear power nation with the money. Hopefully it won't become the next North Korea...hopefully.", - "Solution_19": "I'd probably be really happy, compared to now.\r\nWell, (video games, computer, TV) ^10000000000000\r\nI'd also buy an Xbox360 and play Halo all day after beating Brawl\r\nProbably wouldn't have to buy out of school, just ditch...every day\r\nBuy my way into college sounds fun\r\nBuy a car\r\nBuy my way into a driver's license\r\nBecome the next President\r\nBlow up Iraq and North Korea....just kidding, lol\r\nTHen, I'd flame out after 1 year\r\noh well", - "Solution_20": "An awful lot of computers. A lot of AoPS, games, and programming. No, really. Enter some singing contests, maybe... Junk food. I can't remember the last time I ate a french fry... oh yeah, the day before yesterday I got two.\r\n\r\nMaybe when I get bored of all that, I'll try to accomplish my goal for the year.\r\n\r\nOh yeah, I forgot... buy myself a web domain and begin learning CGI. Learn how to make Firefox addons and themes. Buy a laptop and go back to Paint.NET. Buy Deluxe version of every Popcap game available, write some interesting software, read a lot of books, write a lot of books, and so on. Wonder how that might fit into one year..." -} -{ - "Problem": "Prove \r\n\r\n$ \\int_u^v \\frac{\\ln x}{(x\\plus{}u)(x\\plus{}v)}\\, dx\\equal{}\\frac{\\ln(uv)}{2(v\\minus{}u)}\\, \\ln\\left(\\frac{(u\\plus{}v)^2}{4uv}\\right)$", - "Solution_1": "Using the substitution $ x\\equal{}\\frac{uv}{t}$ we get that\r\n\r\n$ I\\equal{}\\ln(uv)\\int\\limits_{u}^{v}\\frac{dt}{(v\\plus{}t)(t\\plus{}u)}\\minus{}I$,\r\n\r\nand the result follows by a simple computation.", - "Solution_2": "Yes, it works. You can also show $ \\int_0^\\infty \\frac{\\ln x}{(x\\plus{}u)(x\\plus{}v)}\\, dx\\equal{}\\frac{\\ln^2(u)\\minus{}\\ln^2(v)}{2\\, (u\\minus{}v)}$ with this substitution.\r\n\r\nNext task: Show $ \\int_0^1 \\left(\\frac{1}{a\\plus{}1\\minus{}x}\\minus{}\\frac{1}{a\\plus{}x}\\right)\\, \\ln x\\, dx\\equal{}\\frac12 \\ln^2\\left(\\frac{a\\plus{}1}{a}\\right)$", - "Solution_3": "I have found a solution today. See this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=185149]link[/url]." -} -{ - "Problem": "where is the logout button?", - "Solution_1": "logout is at the top \r\n\r\ngo to where it says \r\nforum...latex help...classes\r\n\r\nat the very end of that row is an icon that says resources\r\n\r\nlogout is right under there", - "Solution_2": "dont see it\r\n\r\nmaybe there is a problem", - "Solution_3": "MathLinks skin: Right at the top, under \"Usergroups\" and \"Resources\".\r\nBut it would be good if you send a picture how it looks for you.", - "Solution_4": "in Aops skin, the logout button is at the top. But as zetaX said, it would help if you take a screen shot and post it. :)" -} -{ - "Problem": "The numbers 1 through 36 are arranged in a 6 x 6 square so that the sums of the numbers in each row, column, and diagonal are the same. What is that special sum?", - "Solution_1": "[hide] The total sum of the numbers is $ \\frac{(36)(37)}{2}$. Now, each of the six rows has the same sum, which is just $ \\frac{1}{6}\\cdot\\frac{(36)(37)}{2}$. We can evaluate that expression to be 111. Notice that we might as well have used the six columns instead.[/hide]", - "Solution_2": "These type of square are known as magic square.\r\nif n is no.of rows then sum of each row or column or diagonal=n^2(n^2+1)/2n.", - "Solution_3": "[quote=\"kabi\"]These type of square are known as magic square.\nif n is no.of rows then sum of each row or column or diagonal=n^2(n^2+1)/6.[/quote]\r\n\r\nHm...if we use that formula ($ \\frac{n^2(n^2\\plus{}1)}{6}$), then we get $ \\frac{36(37)}{6}\\equal{}222$, but if we add the highest 6 numbers: $ 36\\plus{}35\\plus{}34\\plus{}33\\plus{}32\\plus{}31\\equal{}201$, so it is impossible to get $ 222$. I don't think your formula works.", - "Solution_4": "[quote=\"007math\"][quote=\"kabi\"]These type of square are known as magic square.\nif n is no.of rows then sum of each row or column or diagonal=n^2(n^2+1)/6.[/quote]\n\nHm...if we use that formula ($ \\frac {n^2(n^2 \\plus{} 1)}{6}$), then we get $ \\frac {36(37)}{6} \\equal{} 222$, but if we add the highest 6 numbers: $ 36 \\plus{} 35 \\plus{} 34 \\plus{} 33 \\plus{} 32 \\plus{} 31 \\equal{} 201$, so it is impossible to get $ 222$. I don't think your formula works.[/quote]\r\nDude who said u 36 35 34 33 32 31 are going to come in a row or column or a diagonal.\r\nGo through magic square.\r\nHopefully u'll get ur quires clear.\r\n\r\nAnyway their is the derivation of that formula.\r\nIn a n*n square ,there are no. from 1,2,3,.....,n^2.\r\nsum of all no. coming in this magic sq.=n^2(n^2+1)/2\r\nsum of any row or column or diagonal=n^2(n^2+1)/2n=n(n^2+1)/2.", - "Solution_5": "Actually, kabi, you're wrong. (You forgot to divide by 2).\r\nSuppose there are $ n$ rows and columns, and the numbers $ 1,2,\\cdots,n^2$ are placed in them to form a magic square.\r\nThen the sum of all the numbers is $ \\sum_{k\\equal{}0}^{n^2}k\\equal{}\\frac{n^2(n^2\\plus{}1)}{2}$.\r\nDivide that sum by the number of rows $ n$ to get $ \\frac{n^2(n^2\\plus{}1)}{2n}\\equal{}\\frac{n(n^2\\plus{}1)}{2}$.\r\n\r\nPlugging in $ n\\equal{}6$ this gives $ \\frac{6(37)}{2}\\equal{}111$ as expected.\r\n\r\n007math's argument makes sense, actually, because the maximum sum of a row of a 6x6 magic square containing $ 1,2,\\cdots,36$ must be $ 36\\plus{}35\\plus{}\\cdots\\plus{}31$.", - "Solution_6": "I apologize for the mistake in writing that formula.I just forgot to write 2.\r\nI have edited them correctly.\r\n007math and grn_trtle sorry about the formula.\r\nFrom now I will check my reply before posting it." -} -{ - "Problem": "Ok, ridiculous subject.. but hopefully it made you all read it.\r\n\r\nI was just wondering the other day.. say you have this point A that starts at (0,0) and moves at a constant speed, say 1 unit per second, along the x axis to positive infinity. Then we have point B that starts at (m,n) and moves at a constant speed k always towards A. What is the graph of B?", - "Solution_1": "[color=cyan]Well, that depends somewhat on k. For example, if k > 1, B will eventually catch up with A. Otherwise, you should be able to design a fairly simple differential equation for the position of B, since you always know its velocity. I'll try the specifics later, but I'm finishing packing for school tomorrow so I should get some sleep to give me ample time and energy to freak out before I go.[/color]", - "Solution_2": "[color=cyan]Actually, scratch the \"easy\" part of previous post. Check me on this, but I think the differential equations you want are\n\ny' / x' = y/(x-t)\n\n(x')^2 + (y')^2 = k^2\n\nx(0) = m\ny(0) = n\n\n\nThe first one says that the direction is towards (t, 0). The second one says that the speed is k. The third and fourth say it starts at (m, n). But that's some ugly equations to solve. Have fun.[/color]", - "Solution_3": "Hm. Nonlinear differential equations...", - "Solution_4": "I believe this is actually a somewhat famous problem, I think it's called the Lion and the Prey problem or something like that. It comes from imagining the path a lion would take while chasing down a prey that moves in a particular path. There was a very good article written about it in Mathematics Monthly a few years ago - I think when I was in high school, which would have been in the early 90s. If I can find some more info about it I will pass it along.\r\n\r\nA good place to ask question like this is on the sci.math newsgroup. Like all newsgroups it can be cluttered with junk, but there are some professional mathematicians that post there who can give good answers to questions like this.", - "Solution_5": "If anyone gets an answer, please post it here! Thanks.", - "Solution_6": "http://mathworld.wolfram.com/PursuitCurve.html" -} -{ - "Problem": "Let triangle $ ABC$ with incircle $ I$ tangent $ BC,CA,AB$ at $ D,E,F$ respectively. Denote $ ID\\cap EF\\equal{}K,AK\\cap BC\\equal{}M$. Prove that: $ M$ be midpoint of $ BC$", - "Solution_1": "Already posted. See [url]http://www.mathlinks.ro/viewtopic.php?t=152959[/url]." -} -{ - "Problem": "Compute the least positive prime $p$ greater than $2$ such that $p^{3}+7p^{2}$ is a perfect square.", - "Solution_1": "[hide=\"Solution\"] If $p^{3}+7p^{2}$ is a perfect square, then $p+7$ must also be a perfect square. Checking, we find that the smallest perfect square for which this is true is $p=29$.[/hide]" -} -{ - "Problem": "$ p(n)$ is the least prime $ >$ n\r\n$ q(n)$ is the biggest prime$ \\leq n$\r\n$ x_n \\equal{} \\sum _{k \\equal{} 2}^n \\frac {1}{p(k).q(k)}$\r\nProve that $ x_n$ converge and find it limit.", - "Solution_1": "More $ h(n)$ is the least prime $ \\geq$ n\r\nLet\r\n$ x_n \\equal{} \\sum_{k \\equal{} 2}^n \\frac {1}{h(k)q(k)}$\r\nProve that $ x_n$converge.", - "Solution_2": "Call $ p_i$ the $ i$-th prime.\r\nNow\r\n\\[ x_{p_m\\minus{}1} \\equal{} \\sum _{k \\equal{} 2}^{p_m\\minus{}1} \\frac {1}{p(k) \\cdot q(k)} \\equal{}\\\\\r\n\\equal{} \\sum_{i\\equal{}1}^{m\\minus{}1} \\sum_{p_i \\leq k < p_{i\\plus{}1}} \\frac {1}{p(k) \\cdot q(k)} \\equal{} \\\\\r\n\\equal{} \\sum_{i\\equal{}1}^{m\\minus{}1} \\sum_{p_i \\leq k < p_{i\\plus{}1}} \\frac {1}{p_i \\cdot p_{i\\plus{}1}} \\equal{} \\\\\r\n\\equal{} \\sum_{i\\equal{}1}^{m\\minus{}1} \\sum_{p_i \\leq k < p_{i\\plus{}1}} \\frac 1{p_{i\\plus{}1}\\minus{}p_i} \\cdot \\left( \\frac 1{p_i} \\minus{} \\frac 1{p_{i\\plus{}1}} \\right) \\equal{}\\\\\r\n\\equal{} \\sum_{i\\equal{}1}^{m\\minus{}1} \\left( \\frac 1{p_i} \\minus{} \\frac 1{p_{i\\plus{}1}} \\right) \\equal{} \\frac 1{p_1} \\minus{} \\frac 1{p_m}.\\]\r\nThus $ x_n$ converges to $ \\frac 1{p_1} \\equal{} \\frac 12$.", - "Solution_3": "Good solution. :wink: \r\nBut I have prove that $ x_n$(in the 2 problem ) converge but now I haven't find out it.\r\nIf we can find out it?", - "Solution_4": "If $ x_n \\equal{} \\sum _{k \\equal{} 2}^n \\frac {1}{p(k) \\cdot q(k)}$ and $ y_n \\equal{} \\sum_{k \\equal{} 2}^n \\frac {1}{h(k) \\cdot q(k)}$, then by $ \\frac {1}{h(k\\plus{}1) \\cdot q(k\\plus{}1)} \\equal{} \\frac {1}{p(k) \\cdot q(k\\plus{}1)} \\leq \\frac {1}{p(k) \\cdot q(k)}$ we get that $ y_k \\leq x_{k\\plus{}1}$, thus $ y_k$ converges as $ x_k$ does.", - "Solution_5": "Yes it easy but which is the ponit converge.?" -} -{ - "Problem": "Let [$a,b,c \\in R^+, ab + bc +ac = 1$. Prove:\r\n$\\sqrt[3]{\\frac{1}{a}+6b} + \\sqrt[3]{\\frac{1}{b}+6c} + \\sqrt[3]{\\frac{1}{c}+6a} \\leq \\frac{1}{abc}$", - "Solution_1": "Not again this one!!! :bomb: I have seldomly seen a more meaningless and idiotic inequality proposed in a competition, and that's not only because I haven't solved it. See http://www.mathlinks.ro/Forum/viewtopic.php?t=30865 for some solutions and you will understand why.\r\n\r\n Darij" -} -{ - "Problem": "it is 5000A.D. and Mars is colonized. The Amulet of Hope has kept Mars thriving and colonized. It has been stolen and shattered into 100 peices. the cities of the planet are falling under seige by the evil Quantomites. You are to vanquish the Quantomite King and restore the Amulet of Hope to Mars. you set of from a village with an armory, a training ground, and a medicine shop.\r\n\r\nWhat weapon do you choose?\r\n\r\nspear+shield(2)\r\nsword+shield(4)\r\nclub+shield(4)\r\nshield(2)", - "Solution_1": "spear and sheild", - "Solution_2": "sword and shield", - "Solution_3": "sword and shield", - "Solution_4": "sword and shield\r\nEDIT: 600th post", - "Solution_5": "spear+Shield", - "Solution_6": "grrr, you took the spear!!!! sword+shield", - "Solution_7": "club+shield(4)\r\nshield bash(2)", - "Solution_8": "sword+shield.", - "Solution_9": "um :maybe: , there are no more swords left wrightboy. choose another weapon", - "Solution_10": "club+shield\r\n\r\nSorry, I didn't realize what the (4)s and (2)s meant. My bad.", - "Solution_11": "club and sheild.", - "Solution_12": "club+shield(2)\r\nrazor shield( spins on your arm and cuts open your opponent, had to make people choose this option and you also have an open spot for a sweet weapon.)(2) oh, and ?(as in an unknown key)+A=secret message\r\n\r\n[color=white][hide]oh, about the message. you get a free choice of spear, sword, club, or pm me and spend 1000 special coins(which only james4l has) to get a lightsaber. but, you have to trade in your primary weapon: spear, club, or nothing to get one of these.[/hide][/color]\r\n\r\n\r\nand your stats are 20HP, (5,5,5) in (strength,speed,defense skill) defense skill is one thing, not two. spear is 5 strength(the amount of damage you do), sword is 5 strength, club is 5 strength, and razor shield is 5 strength.", - "Solution_13": "/is bored\r\n/quits", - "Solution_14": "Hey I pm'ed you too :mad: :mad: :mad: :mad: :mad: :mad: :mad: :mad: :mad: :mad: \r\n\r\n[hide]\nToo many smilies in your post: 10.\nYou can have at most 3 smilies in this message. Please remove some smilies before submitting this message.\nToo many smilies in your post: 10.\nYou can have at most 3 smilies in this message. Please remove some smilies before submitting this message.\nToo many smilies in your post: 10.\nYou can have at most 3 smilies in this message. Please remove some smilies before submitting this message.\nToo many smilies in your post: 10.\nYou can have at most 3 smilies in this message. Please remove some smilies before submitting this message.\nToo many smilies in your post: 10.\nYou can have at most 3 smilies in this message. Please remove some smilies before submitting this message.\n[/hide]", - "Solution_15": "Richard_Min: you play Mozart and the guard falls asleep. you get the other half, but realize you are ina teleport zone :o . you teleport back to town, thankfully.\r\n\r\nButler: you slash at its legs. -5 damage to monkey. you try to restore electricity, but the walls are rubber, you realize :rotfl: .\r\n\r\nMister Geek: you hold it up. You lose concentration and it falls on spikes. -5 damage to monkey. It retaliates and knocksyou flying. -10 damage to you.\r\n\r\n@everyone: If someone could prod eqjj, that would be great because I need him later for the story.", - "Solution_16": "a QWERTY keyboard? to ESCape?", - "Solution_17": "Open the treasure Chest, NOW.", - "Solution_18": "am i dead? if so get butler to electrocute me to restart my hart like those defibrillators \r\n\r\nif not move in with the lightsaber and stab it", - "Solution_19": "Mister Geek, you have 10 health. Also, RTFP!!!!!! NO ELECTRICITY!\r\n\r\nI throw my razor shield at it like a discus. Then I stab it in the eyes to blind it.", - "Solution_20": "eqjj: you answer the riddle. the gaurd looks at you blankly, then flees into a spike pit. you open the door. inside you see jjfun1 holding a dagger. He thinks you are the guard.\r\n\r\n1.persuade him\r\n2.fight him(HARD)\r\n3.Run like heck.\r\n\r\nSpecial OPTION: sit down and cry.\r\n\r\nButler: you move forward. suddenly bars come down in front of you. You ask Mister geek, who was right next to you, \"are we in a cage?\". He replies, \"No, but the monkey is.\"\r\n\r\nMister Geek: you spot an opening in the wall. go through?\r\n\r\nRichard_Min: You have not connected the two sides yet! You find nothing.", - "Solution_21": "I connect the two sides.", - "Solution_22": "I tell it, \"If you give us safe passage and helpp us through this labryinth, then we will heal you and set you free.\"", - "Solution_23": "agree with butler if the answer is not hen leave through the hole", - "Solution_24": "persuade him", - "Solution_25": "Butler: It can't help you. it is trapped.\r\n\r\nMister Geek: You escape and find your self in a room with k00lperson. you guys leave. You have to ask for your reward.\r\n\r\nRichard_Min: it suddenly feels extremely heavy...\r\n\r\neqjj: you persuade him. he agrees to leave, but he must show you something. Do you follow him?", - "Solution_26": "ask for reward\r\n\r\nanyway then go get another quest", - "Solution_27": "Push them together", - "Solution_28": "I start cutting the bars with my lightsaber... BTW, lightsaber isn't electric. It runs on a crystal inside its handle.", - "Solution_29": "ask him what he wants to show me" -} -{ - "Problem": "$ n>2$ is an integer such that $ n^2$ can be represented as a difference of cubes of 2 consecutive positive integers. Prove that $ n$ is a sum of 2 squares of positive integers, and that such $ n$ does exist.", - "Solution_1": "Let $ n^2\\equal{}(x\\plus{}1)^3\\minus{}x^3\\equal{}3x^2\\plus{}3x\\plus{}1$. Multiplying by $ 4$, one gets $ 4n^2\\equal{}12x^2\\plus{}12x\\plus{}4\\iff 4n^2\\minus{}1\\equal{}3(4x^2\\plus{}4x\\plus{}1)$, what is factored as $ (2n\\minus{}1)(2n\\plus{}1)\\equal{}3(2x\\plus{}1)^2$. Since $ 2n\\minus{}1$ and $ 2n\\plus{}1$ are coprime, it must hold either $ 2n\\minus{}1\\equal{}3y^2$ and $ 2n\\plus{}1\\equal{}z^2$, or vice versa. The first case is ruled out because from it would follow $ 2\\equal{}z^2\\minus{}3y^2$, but no square is $ \\equiv 2\\pmod 3$. Therefore, $ 2n\\plus{}1\\equal{}3y^2$ and $ 2n\\minus{}1\\equal{}z^2$. It now follows $ z\\equal{}2a\\plus{}1$, and $ n\\equal{}\\frac{z^2\\plus{}1}2\\equal{}\\frac{4a^2\\plus{}4a\\plus{}2}2\\equal{}a^2\\plus{}(a\\plus{}1)^2$, what was to be proven.", - "Solution_2": "There are infinitely many such $n$. Fix $k \\in \\mathbb{N}_0$. Let $y_k$ and $z_k$ be integers such that $z_k+y_k\\sqrt{3}=(1+\\sqrt{3})(2+\\sqrt{3})^k$. Easy to see that $y_k$ and $z_k$ are odd and that $z_k^2-3y_k^2=-2$. Thus, if $n_k = \\frac{3y_k^2-1}{2} =\\frac{z_k^2+1}{2}$, then $n_k^2=\\left(x_k+1\\right)^3-x_k^3$, where $x_k:=\\frac{y_kz_k-1}{2}$. Examples of $\\left(n_k,x_k\\right)$ are $(1,0)$, $(13,7)$, $(181,104)$, etc." -} -{ - "Problem": "Find all positive integers $ m$ satisfying the property $ m^2 \\minus{} 1$ is a divisor of $ 3^m \\plus{} 5^m$", - "Solution_1": "[quote=\"Allnames\"]Find all positive integers $ m$ satisfying the property $ m^2 \\minus{} 1$ is a divisor of $ 3^m \\plus{} 5^m$[/quote]\r\nNo ideas! :maybe: . It is really nice and not easy!. My result is $ m\\equal{}3$.I am waiting", - "Solution_2": "See here ;) \r\nhttp://www.mathlinks.ro/viewtopic.php?t=275139", - "Solution_3": "[quote=\"TTsphn\"]See here ;) \nhttp://www.mathlinks.ro/viewtopic.php?t=275139[/quote]\r\nIn the fact, This problem should be due to Gabriel Dospinescu. he proposed \" Find all pairs of positive integers $ (m;n)$ such that : $ m^2\\minus{}1| 3^,\\plus{}(n!\\minus{}1)^m$\". After some arguments I got $ n\\equal{}3$ and the equation above. In your link, maybe there was no right solution :maybe: .", - "Solution_4": "It's some mistakes but I can't press \" Edit\".\r\nWe can easy see that $ 3|m; m$is odd. Setting $ m\\equal{}3k. ( k\\in N*).$\r\nSuppose$ k>1.$ \r\nIf $ 5|m$ then $ m^2\\minus{}1$ will be the form of $ 15r\\minus{}1$. Because $ \\left( {\\frac{{ \\minus{} 15}}{{15r \\minus{} 1}}} \\right) \\equal{} \\left( {\\frac{{ \\minus{} 1}}{{15r \\minus{} 1}}} \\right)^2 .\\left( {\\frac{{ \\minus{} 1}}{3}} \\right).\\left( {\\frac{{ \\minus{} 1}}{5}} \\right) \\equal{} \\minus{} 1$, there exist \r\n$ p\\in P; p|m^2\\minus{}1 st (\\minus{}1/p)\\equal{}\\minus{}1$. That gives contradiction.\r\nIf $ m \\equiv 2[5] \\Rightarrow m \\equal{} 15r \\minus{} 3\\equal{}>m\\minus{}1\\equal{}15r\\minus{}4$. From $ \\left( {\\frac{{ \\minus{} 15}}{{15r \\minus{} 4}}} \\right) \\equal{} \\left( {\\frac{{ \\minus{} 1}}{{15r \\minus{} 4}}} \\right)^2 .\\left( {\\frac{{ \\minus{} 4}}{3}} \\right).\\left( {\\frac{{ \\minus{} 4}}{5}} \\right) \\equal{} \\minus{} 1$. That gives contradiction.\r\nIf$ m \\equiv 3[5] \\Rightarrow m \\equal{} 15r \\plus{} 3\\equal{}>m\\minus{}1\\equal{}15r\\plus{}2.$ But $ \\left( {\\frac{{ \\minus{} 15}}{{15r \\plus{} 2}}} \\right) \\equal{} \\left( {\\frac{{ \\minus{} 1}}{{15r \\plus{} 2}}} \\right)^2 .\\left( {\\frac{2}{3}} \\right).\\left( {\\frac{2}{5}} \\right) \\equal{} \\minus{} 1.$\r\nThus $ k\\equal{}1.$" -} -{ - "Problem": "What's a good book for learning amc/aime algebra?", - "Solution_1": "Any AOPS book. This question is in the wrong forum by the way." -} -{ - "Problem": "I failed to make USAPhO even though I got 1a, 2,3,4 right(no joke) on the semifinalist. However, my solutions got there late probably due to the delay of a snowstorm. My coach went to post office on Tuesday and mailed the solution to the AAPT. The mail should arrive by Thursday, but the solution is delivered on Saturday. According to what my coach told me about the phone conversation between Annette Coleman and him, AAPT checked the mailbox on Friday and Monday, and therefore AAPT got my solution packet on Monday instead. Because all the physicists left on Sunday, nobody was available to grade my paper, along with the papers from four other schools. :mad: Now .. great.. I can't even take the semifinal despite doing considerably better than my friends who made the semis.\r\n\r\nI think AAPT should use the system MAA use during USAMO. AAPT should let coaches to fax the solutions in so that there will be NO delay in the delivery...\r\n\r\nHelp.. What should I do? I don't think I can take the semifinal, at least the chance is very very slim. But any advices are appreciated.. :maybe:", - "Solution_1": "I think you are making too big of a deal about this. So you missed USAPhO, it's not the end of the world. There isn't really anything you can do except hope that they read your solutions and accept you. But yes, perhaps they should reform the solution submission system (say by fax as you suggested, or possibly an online test).", - "Solution_2": "JRav, keep in mind that for some students, physics is their one big thing that they've prepared for all year, and it's a really big deal. Think about how a former MOPer would feel if he or she got disqualified on the AIME because of lost mail. But, yes, you're right; there's probably nothing that can be done at this point (the semis have already begun and will end on the 19th).\r\n\r\nAllan Z, I would recommend that your teacher use at least one-day mail or even next-day FedEx. I don't know how that would be affected by a snowstorm, but at least there are [i]some[/i] guarantees. In the case of a snowstorm, I doubt FedEx would delay more than one or two days past the \"one-day\" deadline.", - "Solution_3": "This seems pretty unfair. Reminds me of a time when my mom got a late charge on a bill due on Sunday (why??!) because her check arrived on Sat and they didn't open it until Monday. \r\n\r\nIt would probably be worth making calls to see if they will grade your test (and you should make it then) and you might be able to take the semi before the window closes. You have nothing to lose by calling, I think.", - "Solution_4": "Thank you all for your suggestions. I guess I am going to call AAPT and ask them whether I can take the test provisionally and let them grade my quarterfinal later. If I don't qualify for the semifinal, AAPT can just discard my results.", - "Solution_5": "[quote=\"Allan Z\"]Thank you all for your suggestions. I guess I am going to call AAPT and ask them whether I can take the test provisionally and let them grade my quarterfinal later. If I don't qualify for the semifinal, AAPT can just discard my results.[/quote]\r\n\r\nAllan, are they letting you do this? Because my physics teacher told me yesterday that AAPT sent him semifinal tests for my school to take, but he didnt really tell me anything else. Neither me nor the other kid at my school who took the quarterfinals test are on the AAPT website's list of semifinalists, and i was confused as to what is going on." -} -{ - "Problem": "Calculate the following indefinite integrals.\r\n\r\n[1] $\\int (\\sin x+\\cos x)^4 dx$\r\n\r\n[2] $\\int \\frac{e^{2x}}{e^x+1}dx$\r\n\r\n[3] $\\int \\sin ^ 4 xdx$\r\n\r\n[4] $\\int \\sin 6x\\cos 2xdx$\r\n\r\n[5] $\\int \\frac{x^2}{\\sqrt{(x+1)^3}}dx$", - "Solution_1": "just glancing through, #3 seems very easy, just use a half angle trign stuff..\r\nans:\r\n$\\frac{3x}{8}-\\frac{\\sin{2x}}{4}+\\frac{\\sin{4x}}{32}$", - "Solution_2": "2\r\n$e^x+1-\\ln{(e^x+1)}$", - "Solution_3": "[quote=\"amirhtlusa\"]just glancing through, #3 seems very easy, just use a half angle trign stuff..\nans:\n$\\frac{3x}{8}-\\frac{\\sin{2x}}{4}+\\frac{\\sin{4x}}{32}$[/quote]\r\n\r\n :(", - "Solution_4": "[quote=\"amirhtlusa\"]2\n$e^x+1-\\ln{(e^x+1)}$[/quote]\r\n\r\n :)", - "Solution_5": "[quote=\"kunny\"][quote=\"amirhtlusa\"]just glancing through, #3 seems very easy, just use a half angle trign stuff..\nans:\n$\\frac{3x}{8}-\\frac{\\sin{2x}}{4}+\\frac{\\sin{4x}}{32}$[/quote]\n\n :([/quote]\r\n\r\nIt is correct kunny! (Minus the arbitary constant)", - "Solution_6": "yes i did it again and got the same thing. ;)", - "Solution_7": "FOR 4\r\nlet takeI= $\\int\\sin2x\\cos6xdx$ and J the integral you want.\r\ncalculate I+J and I-J it's too easy..\r\nthan you will got J.\r\n\r\n\r\n\r\n\r\nIT'S TOO EASY. ;)", - "Solution_8": "1)\r\n\r\n$(\\sin(x)+\\cos(x))^4 = (\\sin^2(x)+2\\sin(x)\\cos(x)+\\cos^2(x))^2=(1+\\sin(2x))^2 = 1+2\\sin(2x)+\\sin^2(2x)$\r\n\r\n$\\int 1+2\\sin(2x)+\\sin^2(2x) \\ dx = \\frac{3x}{2}-\\cos(2x)-\\frac{1}{8}\\sin(4x)$", - "Solution_9": "[quote=\"amirhtlusa\"]yes i did it again and got the same thing. ;)[/quote]\r\n\r\nI'm very sorry, you were RIGHT! :blush: \r\n\r\nkunny", - "Solution_10": "[quote=\"staicovitch\"]FOR 4\r\nlet takeI= $\\int\\sin2x\\cos6xdx$ and J the integral you want.\r\ncalculate I+J and I-J it's too easy..\r\nthan you will got J.\r\n\r\nYes. I also have had your solution.", - "Solution_11": "[quote=\"Jimmy\"]1)\n\n$(\\sin(x)+\\cos(x))^4 = (\\sin^2(x)+2\\sin(x)\\cos(x)+\\cos^2(x))^2=(1+\\sin(2x))^2 = 1+2\\sin(2x)+\\sin^2(2x)$\n\n$\\int 1+2\\sin(2x)+\\sin^2(2x) \\ dx = \\frac{3x}{2}-\\cos(2x)-\\frac{1}{8}\\sin(4x)$[/quote]\r\n\r\nThat's right, Jimmy :)", - "Solution_12": "5)\r\n\r\n$\\frac{x^2}{\\sqrt{(x+1)^3}} = \\frac{(x+1)^2}{\\sqrt{(x+1)^3}} - \\frac{2x+1}{\\sqrt{(x+1)^3}}$\r\n\r\n$=\\frac{(x+1)^2}{\\sqrt{(x+1)^3}} - \\frac{(x+1)}{\\sqrt{(x+1)^3}} - \\frac{x}{\\sqrt{(x+1)^3}}$\r\n\r\n$=\\frac{(x+1)^2}{\\sqrt{(x+1)^3}} - \\frac{(x+1)}{\\sqrt{(x+1)^3}} -\\frac{(x+1)}{\\sqrt{(x+1)^3}} +\\frac{1}{\\sqrt{(x+1)^3}}$\r\n\r\n$=\\sqrt{x+1} - \\frac{1}{\\sqrt{x+1}} - \\frac{1}{\\sqrt{x+1}} + \\frac{1}{\\sqrt{(x+1)^3}}$\r\n\r\n$\\int \\frac{x^2}{\\sqrt{(x+1)^3}} \\ dx = \\int \\sqrt{x+1} \\ dx - 2\\int \\frac{dx}{\\sqrt{x+1}} + \\int \\frac{dx}{\\sqrt{(x+1)^3}}$\r\n\r\n$=\\frac{2}{3}\\sqrt{(x+1)^3} - 4\\sqrt{x+1} - \\frac{2}{\\sqrt{x+1}}$\r\n\r\n$=\\frac{2(x+1)^2}{3\\sqrt{x+1}} - \\frac{12(x+1)}{3\\sqrt{x+1}} - \\frac{6}{3\\sqrt{x+1}}$\r\n\r\n$=\\frac{2(x^2-4x-8)}{3\\sqrt{x+1}}$", - "Solution_13": "That's right, Jimmy. :)" -} -{ - "Problem": ":oops: Out of [i]n[/i] straight lines of length 1,2,3.....n units, find the no. of ways in which four may be chosen such that it willl form a quadrilateral in which a circle can be inscribed???\r\n\r\n [hide]Pls give ur ans. with detailed explanation.\nTy[/hide]", - "Solution_1": "for a quad. in which circle can be inscribed, sum of each pair of opp. sides is equal.\r\n(can be easily proved using 'equal tangents' property)\r\nso we just have to find the no. of ways in which each of the nos. from 10(=1+2+3+4) to \r\n4n-6[=n+(n-1)+(n-2)+(n-3)] can be split into 4 distinct nos.(each <= n). \r\nThis should just involve some simple combinatorics (i hope).\r\nActually, I'm just too lazy to work it out, or think about it. :oops:", - "Solution_2": "I forgot that we also have to include the condition that sum of opp. side-pairs is equal. :oops: \r\nthat will further complicate the combinatorics.\r\n\r\nseems the problem is not as simple as I thought it was. \r\ncould someone please work it out? (my mind doesnt seem to be functioning properly today)", - "Solution_3": "[hide]\nIs it \n$ \\sum_{i\\equal{}5}^{2n\\minus{}3} \\binom{[\\frac{(i\\minus{}1)}{2}]}{2}$,[x] is gif\n\nAs keyree10 told,If the sum of opposite sides of a quadrilateral are equal ,then it is a tangential quadrilateral.I was not able to prove it :blush: ,but saw the result in Wikipedia.Someone please post the proof.\nNow,we have to choose 4 numbers,such that sum of two is equal to other two.\nMinimum sum is 5.\n5=1+4,2+3 ---2 ways\n6=1+5,2+4 ---2 ways\n7=1+6,2+5,3+4 ---3 ways\nand so on.\nfor i,it can be reprsented a sum of \n$ [\\frac{(i\\minus{}1)}{2}]$ pairs.\nwhere,[x] is gif.\nwe can choose any two of the $ [\\frac{(i\\minus{}1)}{2}]$ pairs and form a tangential quad with those four sides.\n\nSo,we just add up all from 5 to (2n-3).\n2n-3=(n)+(n-3),(n-1)+(n-2)\n\nI am not sure about the answer.\n[/hide]", - "Solution_4": "@ Arjun\r\n\r\nDraw the inscribed circle\r\n\r\nThen u will get 4 pairs of equal tangents from each of the 4 vertices.\r\n\r\nThen u observe that the sum of opposite sides is indeed equal.", - "Solution_5": "@befuddlers\r\nNo,I meant the proof for the converse of the theorem\r\n(ie)If the sum of opposite sides of a quadrilateral are equal,then it is a tangential quadrilateral (or a circle can be inscribed in it)" -} -{ - "Problem": "Proof that \r\n$\r\n\\sum\\limits_{n = 0}^\\infty {\\frac{{\\left( {4n} \\right)!}}{{\\left( {2n + 1} \\right)!\\left( {n!} \\right)^2 2^{6n + 1} }} = \\frac{{2\\ln \\left( {1 + \\sqrt 2 } \\right)}}{\\pi }} \r\n$", - "Solution_1": "ok\r\nyou are the best!!!!!\r\nWe can see now the solution?", - "Solution_2": "An hint , starting with\r\n$\\sqrt {{\\frac {1-\\sqrt {1-x}}{x}}}=\\frac{1}{\\sqrt {2}}\\sum _{n=0}^{\\infty }\r\n{\\frac {{4\\,n\\choose 2\\,n}{x}^{n}}{{16}^{n} \\left( 2\\,n+1 \\right) }}\r\n$\r\nput $x=sin(t)^2$, integrate from $0$ to $\\pi/2$ and use Wallis", - "Solution_3": "Please remark that\r\n$\\frac{2\\ln(1+\\sqrt(2))}{\\pi}=\\frac{arcsinh(1)}{\\arcsin(1)}$" -} -{ - "Problem": "$\\mbox{Whats the inverse of } \\left(\\begin{array}{cc}12&7\\\\-7&-4\\end{array}\\right) \\mbox{ ?}$", - "Solution_1": "for this problem:\r\n$\\left(\\begin{array}{cc}12&7\\\\-7&-4\\end{array}\\right) . \\left(\\begin{array}{cc}a&b\\\\c&d\\end{array}\\right)= \\left(\\begin{array}{cc}1&0\\\\0&1\\end{array}\\right)$\r\nand you solve the system... There some better algorithms for large matrix...", - "Solution_2": "There's a very curious strategy to find matrix inverse [it is essencially solving the system proposed by igor]. Let me see if I have the domain of TeX so that I can do this by making no mistake\r\n\r\n$\\left(\\begin{array}{cccc}12&7&1&0\\\\-7&-4&0&1\\end{array}\\right)$\r\n\r\nNow you perform some matrix operations to transform the initial matrix into the identity one. The matrix that remains on the right, after the end of the steps is the inverse matrix [why?]\r\n\r\nExample (beggining of the step)\r\n \r\n$\\left(\\begin{array}{cccc}1&\\frac{7}{12}&\\frac{1}{12}&0\\\\-7&-4&0&1\\end{array}\\right)$\r\n$\\left(\\begin{array}{cccc}1&\\frac{7}{12}&\\frac{1}{12}&0\\\\0&\\frac{1}{12}&\\frac{7}{12}&1\\end{array}\\right)$\r\n\r\nand so on...;-)", - "Solution_3": "Because is a 2x2 matrix you can also try \r\nif $A = \\left(\\begin{array}{cc}a&b\\\\c&d\\end{array}\\right)$\r\n\r\nThen $A^{-1}$ be found with formula \r\n\r\n$A^{-1} = \\frac 1{ac-bd} \\left(\\begin{array}{cc}d&-b\\\\-c&a\\end{array}\\right)$" -} -{ - "Problem": "In an obscure card game, each player is dealth six cards from a standard deck of 52 cards. If a player receives exactly two fives, she wins. What's the probability of being dealth a winning hand?", - "Solution_1": "[quote=\"4everwise\"]In an obscure card game, each player is dealth six cards from a standard deck of 52 cards. If a player receives exactly two fives, she wins. What's the probability of being dealth a winning hand?[/quote]\r\n[hide]\nthere are 13 different cards, so.. the chance of getting a specific card is 1/13 or 4/52.\nsince you are getting 6 cards, that multiplies the chance of getting a specific card by 6.\nso the chance of getting one five with 6 cards is 24/52.\n\nBut since you need 2 fives, you mulitply 24/52 by itself = 6/13\n\nSo the probability of winning is 36/169\n[/hide]", - "Solution_2": "[hide=\"maybe\"]\n\n$P = \\frac{4}{52} \\cdot \\frac{3}{51} \\cdot 6 = \\frac{6}{221}$[/hide]", - "Solution_3": "[hide]\n\nSo basically, each card drawn affects the odds of whether the next card will be a five or not. In order for there to be exactly 2 fives in a hand, two of the draws must be fives, and two must not. Remember that the odds of 6 events happenning is equal to the product of the odds of the 6 events. The odds that the first card is a five is 4/52, since there are 4 fives in a deck of 52 cards. Given that the first card is a five, the odds that the second card is a five are 3/51, since there are 3 fives left in a deck of 51 cards. Therefore the odds of drawing a hand beginning with two fives and followed by four non-fives is equal to:\n 4 * 3 *48*47*46*45\n---------------------------\n52*51*50*49*48*47\n\nBut the cards don't necessarily have to be drawn in that order. There are 6 choose 2 possible orders, which is 6*5/2*1=15. So the probability listed above must be multiplied by 15 to get the answer, so:\n\n 4 * 3 *48*47*46*45\nP=15*--------------------------- The 48's and 47's cancel out, leaving:\n 52*51*50*49*48*47\n\n 4 * 3 *46*45\nP=15*------------------ Which leaves:\n 52*51*50*49 \n\nP=621/10829\n[/hide]", - "Solution_4": "This problem is interesting depending on interpretation.\r\n\r\nI think it means:\r\n\r\n[hide]$\\frac{\\binom{4}{2} \\cdot \\binom{48}{4}}{\\binom{52}{6}}$\n\nIn a hand, of 52 choose 6 cards, 2 must be 5's, and 4 others non-fives at random. [/hide]", - "Solution_5": "[hide]I think its...\n$\\frac{\\displaystyle\\binom{48}{4}\\binom{4}{2}}{\\displaystyle\\binom{52}{6}}$. The number of ways the two fives can be dealt is $\\displaystyle\\binom{4}{2}$, and of the 48 non-fives, we must choose 4. So $\\displaystyle\\binom{48}{4}$ over the total number of ways 6 cards can be drawn from 52, $\\displaystyle\\binom{52}{6}$[/hide]" -} -{ - "Problem": "Between 2007 and 2008, the AMC Committee removed calculators from the AMC10 and AMC12. Did the test difficulty change to compensate or not?\r\nThis is for my information in practice scores. Thanks :)", - "Solution_1": "didn't this happen in the amc 8 too??Well..not really, because for me, typing into the calculator takes me longer than me just writing it out by hand :)", - "Solution_2": "The AMC8 was considerably easier, but I'm not sure about 10/12.", - "Solution_3": "i dont believe it was any easier. All the problems on the AMC 10/12 can be solved(and are intended to be solved) without a calculator, so i dont think taking away calculators would change the difficulty at all.", - "Solution_4": "Well, the loss of calculator opens up the possibility of careless errors in computation, but we're careful, right? :D" -} -{ - "Problem": "if we talk about the roots of a polynomial in the sense of when the polinomial =0 mod p, it's obvious that it might have at most n roots where n is the degree, for example $(x-1)(x-2)$ =0 mod 3 for 1 and 2...but if the polinomial cannot be factor, for example $x^{2}+1$ does the theorem is still valid? thanks.", - "Solution_1": "The theorem's still valid though the polynomial cannot be factored.\r\nHere is Lagrange's theorem (according to Sierpinki):\r\n[quote]If $n$ is a natural number and f(x) is a polynomial of degree n with integral coefficient, the coefficient of $x^{n}$ is not divisible by p then the congruence $f(x) \\equiv 0$ (mod $p$) has at most $n$ roots[/quote]", - "Solution_2": "what's the proof?", - "Solution_3": "Use induction and Bezout theorem (if $W(a)=0$ then $W(x)=(x-a)P(x)$)", - "Solution_4": "[quote=\"dondigo\"]Use induction and Bezout theorem (if $W(a)=0$ then $W(x)=(x-a)P(x)$)[/quote]\r\n\r\n\r\n I dont get this stuff, know the Bezout i get bu the other one, there is no proof or work\r\n\r\n\r\n P.S. I NEED WORD TO UNDERSTAND!!!!! :furious: :wallbash_red: :spam:", - "Solution_5": "[hide=\"Discussion\"] Lagrange's Theorem holds in prime moduli because they are integral domains; in other words, if $ab \\equiv 0 \\bmod p$ then either $a \\equiv 0 \\bmod p$ or $b \\equiv 0 \\bmod p$.\n\nThat is why Lagrange's Theorem fails $\\bmod$ composite values. For example, the quadratic $(x-a)(x-b) \\bmod pq$ where $p, q$ are distinct odd primes and $a \\neq b$ has four solutions (obvious by Chinese Remainder Theorem). [/hide]", - "Solution_6": "[quote=\"dondigo\"]Use induction and Bezout theorem (if $W(a)=0$ then $W(x)=(x-a)P(x)$)[/quote]what do you mean? if the polynomial cannot be factored...", - "Solution_7": "[hide=\"Complete proof\"]\n[b]Theorem[/b]\n\n[color=blue]Polynomial with integer coefficients $W(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\\ldots+a_{1}x+a_{0}$ satisfying $a_{n}\\not\\equiv{0}\\pmod{p}$ has utmost $n$ roots modulo $p$\n[/color]\n\nWe will perform an induction on $n$\n\nFor $n=1$ we have $W(x)=ax+b$. Assume, that it has more that one root.\nLet say $r,t$ are roots of $W(x)$ thus:\\[ar+b\\equiv{at+b}\\equiv{0}\\pmod{0}\\Rightarrow a(r-t)\\equiv{0}\\pmod{p}\\Rightarrow r\\equiv{t}\\pmod{p}\\]\nContradiction.\n\nNow assume, that the theorem holds for $n-1$. I'll show that it holds for $n$\nIf $W(x)$ has no roots, we are at home, in other case it has at least one root. \nLet it be $b$\n\n\\[W(x)\\equiv{W(x)-W(b)}\\equiv{a_{n}x^{n}+a_{n-1}x^{n-1}+\\ldots+a_{1}x+a_{0}-(a_{n}b^{n}+a_{n-1}b^{n-1}+\\ldots+a_{1}b+a_{0})}\\pmod{p}\\]\n\n\\[W(x)\\equiv{a_{n}(x^{n}-b^{n})+a_{n-1}(x^{n-1}-b^{n-1})+\\ldots+a_{1}(x-b)}\\pmod{p}\\]\n\\[W(x)\\equiv{(x-b)g(x)}\\pmod{p}\\]\n\nPolynomial $g(x)$ has integer coefficients and it's leading coefficient if equal to $a_{n}$(one can compute it using the formula $a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+\\cdots+ab^{n-2}+b^{n-1}$), so it satisfies the conditions of out theorem.\nAssume by contradiciton, that $W(x)$ has at least $n+1$ roots. It means that it has has at least $n$ roots different than $b$. All of them are roots of $g(x)$. But from the assumption follows, that it has utmost $n-1$ roots. Contradiciton.\n\n\n\n\n\n\n\n\n[/hide]", - "Solution_8": "thanks! :)", - "Solution_9": "In general, this result holds in any field, but not in an arbitrary integral domain", - "Solution_10": "No, it holds in integral domains, too (assumed integral domains to be commutative and having $1$).\r\n\r\n\r\nAnd to srulikbd: You say \"...if the polinomial cannot be factor...\". Note that factoring should be seen as some property of whatever you work in. Especially, $x^{2}+1$ factors in $\\mathbb Z / 5 \\mathbb Z$ as $(x-2)(x+2)$.\r\nThis makes life easier ;)", - "Solution_11": "Ah yes... of course you are right.\r\nToo much noncommutative ring theory can destroy basic knowledge :oops:" -} -{ - "Problem": "Something is wrong. No matter what I do my Karma stays at 1000!", - "Solution_1": "Well it finally started working right. It really wouldnt change for a while, though.", - "Solution_2": "You were probably in diagnostic mode; you hadn't done enough questions. Are you a new user for Alcumus? Look in the instructions.", - "Solution_3": "Yeah I just started going on Alcumus. Its working now.", - "Solution_4": "Please read the [url=http://www.artofproblemsolving.com/Alcumus/Instructions.php]instructions[/url] as it contains information about such things." -} -{ - "Problem": "For $a;b;c \\in R$ and satisfying $x \\leq y \\leq z$.Let $a;b;c$ satisfying $x \\leq a;b;c \\leq y$ and $a+b+c=x+y+z$.Prove that:\r\n$a^{2}+b^{2}+c^{2}\\leq x^{2}+y^{2}+z^{2}$", - "Solution_1": "same idea http://www.mathlinks.ro/Forum/viewtopic.php?t=130332" -} -{ - "Problem": "If $ a,b,c,x,y,z>0$ show that :\r\n\r\n$ xyz\\left(\\frac{1}{ax\\plus{}by\\plus{}cz}\\plus{}\\frac{1}{bx\\plus{}cy\\plus{}az}\\plus{}\\frac{1}{cx\\plus{}ay\\plus{}bz}\\right)\\le \\frac{xy\\plus{}yz\\plus{}zx}{a\\plus{}b\\plus{}c}$", - "Solution_1": "[quote=\"alex2008\"]If $ a,b,c,x,y,z > 0$ show that :\n\n$ xyz\\left(\\frac {1}{ax \\plus{} by \\plus{} cz} \\plus{} \\frac {1}{bx \\plus{} cy \\plus{} az} \\plus{} \\frac {1}{cx \\plus{} ay \\plus{} bz}\\right)\\le \\frac {xy \\plus{} yz \\plus{} zx}{a \\plus{} b \\plus{} c}$[/quote]\r\nIt's obvious because $ f(x) \\equal{} \\frac {1}{x}$ is convex function when $ x>0.$ :wink:", - "Solution_2": "Maybe this requires an explanation. Jensen's inequality gives\r\n\r\n$ \\frac{a}{a \\plus{} b \\plus{} c}\\frac{1}{x} \\plus{} \\frac{b}{a \\plus{} b \\plus{} c}\\frac{1}{y} \\plus{} \\frac{c}{a \\plus{} b \\plus{} c}\\frac{1}{z} \\ge \\frac{a \\plus{} b \\plus{} c}{ax \\plus{} by \\plus{} cz}$\r\n\r\nAdding the inequalities obtained when we change $ a, b$ and $ c$ cyclically, this gives\r\n\r\n$ \\frac{xy \\plus{} yz \\plus{} zx}{xyz} \\equal{} \\frac{1}{x} \\plus{} \\frac{1}{y} \\plus{} \\frac{1}{z} \\ge (a \\plus{} b \\plus{} c)*$ what you want." -} -{ - "Problem": "prove that the sum of the reciprocals of the lengths of the altitudes of a triangle equals the reciprocal of the length of the inradius.", - "Solution_1": "I don't know if you want complete geometric reasoning, but this is my proof in any case:\r\n\r\n[hide]$ \\frac{1}{h_a}\\plus{}\\frac{1}{h_b}\\plus{}\\frac{1}{h_c}\\equal{}\\frac{1}{2S}\\left(AB\\plus{}BC\\plus{}CA\\right)$\n\n$ \\equal{}\\frac{2s}{2S}\\equal{}\\frac{1}{r}$[/hide]" -} -{ - "Problem": "let $0 2$, define $q(k)$ to be the product of all primes less than $p(k)$, otherwise let $q(k) = 1$. Consider the sequence\r\n\\[ \\\\x_0 = 1,\\hspace{.3in}x_{n+1} = \\frac{x_np(x_n)}{q(x_n)},\\hspace{.3in} n\\in\\mathbb{Z^+}\\cup\\{0\\} \\]\r\n\r\nDetermine all natural numbers $n$ such that $x_n = 111111$.\r\n\r\nMasoud Zargar" -} -{ - "Problem": "show that there is no biggest number a such that $ a^2$ $ <$ $ 2$", - "Solution_1": "we get $ a < 1.414$.now it does not matter if we increase the decimal places.so $ (1.414367890.......)^2 <2$.these decimal places can keep stretching on and on so we cant find the largest no.$ <2$ .", - "Solution_2": "@aadil your proof is not that rigorous ,\r\nThere is a very simple following proof given by a 9th standard guy and was asked in a Indian exam conducted by association of mathematics teachers of India.\r\nLet $ a$ be the maximum such number satisfying the condition .\r\nBut we know that $ a < \\frac{a \\plus{} \\sqrt{2}}{2} < \\sqrt{2}$ \r\n$ a^{2} < (\\frac{ a \\plus{} \\sqrt{2}}{2})^{2} < 2$....so we cannot have a maximum number.. :wink:", - "Solution_3": "One can give a stronger result, though: there is no largest [b]rational[/b] number with the desired property.", - "Solution_4": "[quote]@aadil your proof is not that rigorous [/quote]\r\nit was only an idea not a proof", - "Solution_5": "i dont get it man..of course aadil's argument is sufficient to prove it\r\nit actually goes to hyperreal arguments, which provide a quality satisfactory answer.\r\n[url]http://en.wikipedia.org/wiki/Hyperreal_number[/url]", - "Solution_6": "No, it doesn't. aadil's argument only works if he knows that the limit of the sequence he's trying to define is $ \\sqrt {2}$, and that brings in a stronger claim (both the irrationality of $ \\sqrt {2}$ and the existence of its decimal expansion) than is necessary for this problem. This problem is much simpler than that.\r\n\r\n[b]Edit:[/b] I guess I should put my money where my mouth is. If $ a^2 < 2$ then I claim that $ a^2 < \\left( \\frac{2a \\plus{} 2}{a \\plus{} 2} \\right)^2 < 2$." -} -{ - "Problem": "Several sets of prime numbers, such as $ \\{ 7, 83, 421, 659\\}$ use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?\r\n\r\n$ \\textbf{(A)}\\ 193\\qquad\\textbf{(B)}\\ 207\\qquad\\textbf{(C)}\\ 225\\qquad\\textbf{(D)}\\ 252\\qquad\\textbf{(E)}\\ 447$", - "Solution_1": "[quote=\"chess64\"]Several sets of prime numbers, such as $\\{ 7, 83, 421, 659\\}$ use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?\n\n\\[ \\textbf {(A) } 193\\qquad\\textbf {(B) } 207\\qquad\\textbf {(C) } 225\\qquad\\textbf {(D) } 252\\qquad\\textbf {(E) } 447 \\][/quote]\r\n[hide]\ni got C)225\n\nso obviously 2,4,5,6,8 cannot be the ones digits, so the smallest sum (i think) is if they are tens digits\n\n2_, 4_, 5_, 6_, 8_\n\nsince we need one one digit number that is prime, 2 or 5 works\n\nso the ones digit shouldnt matter because they add up to the same anyways and since ten 2s is smaller than ten 5s the single digit prime is the 5 \n\nso randomly placing i got 29+47+61+83+5=225[/hide]", - "Solution_2": "[hide]\n207:\n\n2\n3\n5\n47\n61\n89\n[/hide]", - "Solution_3": "[hide=\"Answer\"]Obviously, the even digits must be in the tens place except for $2$. Therefore, we need three two-digit numbers, and we can place the remaining numbers in the units places so that we have only prime numbers. Therefore, our answer is $40+60+80+1+2+3+5+7+9=207\\Rightarrow \\boxed{B}$. (The primes can be 2, 3, 5, 41, 67, 89.)[/hide]", - "Solution_4": "For reference, this was 2002 AMC 12A Problem 17.\n[hide=my sol]\nwe start by using 2,3,5, and 7 as one digit primes\nthis leaves us with 1,4,6,8, and 9 as possible options\nwe can pair up 4 and 1 to get 41, 8 and 9 to 89, and combine the 6 with 7 (which we already had) to get 67\nthis gives us a sum of 41+89+67+2+3+5=207\nsince 4,6,8 have to be in tens place or higher, it's not hard to see that this is the minimum\n[/hide]" -} -{ - "Problem": "When will the WOOT message boards and classroom be open, and how would one navigate to those pages? \r\n\r\nThanks. :lol:\r\n\r\n--200th post!--", - "Solution_1": "We'll likely activate 2008-09 WOOT Message Boards and classroom towards the end of the month. There won't be any classes until September, but we'll start some message board discussion when we open the board, by posting occasional problems to challenge enrolled students.", - "Solution_2": "i have a question that is somewhat related to this topic and i figured it would be useless to make another topic, but just for conformation, students can type latex in the WOOT classroom? because i remember attending some math jams and only the instructor could use latex", - "Solution_3": "[quote=\"cognos599\"]i have a question that is somewhat related to this topic and i figured it would be useless to make another topic, but just for conformation, students can type latex in the WOOT classroom? because i remember attending some math jams and only the instructor could use latex[/quote]\r\n\r\nStudents can use LaTeX in the classroom, even in the Math Jams (we sometimes don't mention that at the start of Math Jams, because, particularly in large Math Jams, it leads to some confusion, as some students think we mean that they have to know LaTeX, etc).", - "Solution_4": "I (my parents) purchased the WOOT class for me today, but I still cannot find the corresponding message board. Is it under the Online School forum? I looked there, but I can't find it.\r\nedit: Yay, thanks!", - "Solution_5": "You should see it now." -} -{ - "Problem": "Define $ f: \\mathbb{R}\\to\\mathbb{R}$ by\r\n\\[ f(x)\\equal{}\\begin{cases}x&\\text{if }x\\le e\\\\ xf(\\ln x)&\\text{if }x>e\\end{cases}\\]\r\nDoes $ \\displaystyle\\sum_{n\\equal{}1}^{\\infty}\\frac1{f(n)}$ converge?", - "Solution_1": "We prove this by the integral test. Once we know that the integral test applies (more on that below,) we have that $ \\sum_{n\\equal{}1}^{\\infty}\\frac1{f(n)}$ converges if and only if $ \\int_1^{\\infty}\\frac1{f(x)}\\,dx$ converges. Then,\r\n\\[ \\int_1^{\\infty}\\frac1{f(x)}\\,dx\\equal{} \\int_1^e\\frac1x\\,dx\\plus{}\\int_e^{\\infty}\\frac1{xf(\\ln x)}\\,dx\\]\r\nMake the substitution $ u\\equal{}\\ln x$ to turn the second integral into $ \\int_1^{\\infty}\\frac1{f(u)}\\,du.$ Hence:\r\n\\[ \\int_1^{\\infty}\\frac1{f(x)}\\,dx\\equal{}1\\plus{}\\int_1^{\\infty}\\frac1{f(x)}\\,dx\\]\r\nwhich implies that that integral is infinite and hence the sum diverges.\r\n\r\nWe have to work a little harder than that; for one thing, we do have to show that $ f(x)$ is increasing so that we can have $ \\frac1{f(x)}$ decreasing in order to justify the integral test. Define $ I_1\\equal{}(1,e],I_2\\equal{}(e,e^e],I_3\\equal{}(e^e,e^{e^e}],$ and so on, with $ x\\in I_{k\\plus{}1}$ if and only if $ \\ln x\\in I_k.$\r\n\r\n$ f$ is differentiable on the interior of each $ I_k.$ We will show that $ f'(x)>0$ for all such points, so that $ f$ is increasing within each $ I_k.$ Note that $ f'(x)\\equal{}1>0$ in the interior of $ I_1.$ Now suppose that $ f'(x)>0$ for $ x$ in the interior of $ I_k.$ Consider $ x$ in the interior of $ I_{k\\plus{}1}.$ Then $ f'(x)\\equal{}f(\\ln x)\\plus{}f'(\\ln x).$ We have $ f(\\ln x)>0$ since the function is constructed to be positive and and $ f'(\\ln x)>0$ since $ \\ln x\\in I_k.$ Hence $ f'(x)>0$ for $ x$ in the interior of $ I_{k\\plus{}1}.$ Next, we show that if $ x\\in I_k$ and $ y\\in I_{k\\plus{}1},$ then $ f(x)y\\ln e\\equal{}y>e.$ Hence $ f(x) a - b - 1$\r\n\r\nCollecting like terms, $ \\sum_{n = 1}^\\infty\\frac {1}{f(n)} > \\sum_{k = 1}^\\infty(^{k + 1}e - ^ke - 1)(\\prod_{i = 0}{}^k{}^ie)^{ - 1}$\r\n$ = \\sum_{k = 1}^\\infty^{k + 1}e(\\prod_{i = 0}{}^k{}^ie)^{ - 1} - \\sum_{k = 1}^\\infty^ke(\\prod_{i = 0}{}^k{}^ie)^{ - 1} - \\sum_{k = 1}^\\infty(\\prod_{i = 0}{}^k{}^ie)^{ - 1}$\r\n\r\nThe last term goes down as an inverse of a tetration, and can be easily shown to converge. The second last term cancels one of the tetration products but the others remain, and similarly converges. We're left with\r\n\r\n$ = \\sum_{k = 1}^\\infty^{k + 1}e(\\prod_{i = 0}{}^k{}^ie)^{ - 1} - const$\r\n\r\nand invoke the ratio test: $ \\left|\\frac {a_{k + 1}}{a_k}\\right|=\\frac {^{k + 2}e}{^{k + 1}e}\\frac{\\prod_{i = 0}{}^k{}^ie}{\\prod_{i = 0}^{k + 1}^ie}=\\frac {(^{k + 1}e)^e}{(^{k + 1}e)^2}=(^{k + 1}e)^{e - 2} > 1$\r\n\r\nAnd we have bounded the series from below with a rather quickly diverging series (collected from $ ^ke$ terms though)", - "Solution_3": "I correctly got that each I_k integral was equal to 1 by induction, but I forgot to actually justify using the integral test (just used it without justification)...how many points do you think that's worth?", - "Solution_4": "Kent\r\nIs it necessary to verify that f(x) is continuous", - "Solution_5": "It's necessary to verify that it's monotone increasing- which means that showing that the limits from below and from above at the key points are the same, in this case. It's actually differentiable at all of those points except $ 1$ (we could define $ f(x) \\equal{} 1$ on $ [0,1]$ and use the functional equation everywhere above that).\r\n\r\nSee also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244073]Putnam 2002 A6[/url]. I was instantly reminded when I saw this problem.", - "Solution_6": "I found that it diverged using the integral test...but i'm not sure my proof was wordy enough. I think I said something along the lines of the Lim t->infinity of ln(f(t)) approached infinity because the terms would always be the log of t*ln...of a lot of stuff. I really wanted to get the calc related question as I have no experience in anything but calc/diff eqs/a little discrete math.", - "Solution_7": "[quote=\"jmerry\"]It's necessary to verify that it's monotone increasing- which means that showing that the limits from below and from above at the key points are the same, in this case. [/quote]\r\nOne can do a little worse; the limit from the left just needs to be at most the limit from the right.", - "Solution_8": "I ridiculusly did the problem wrong.....\r\nI used integral test and came to the conclution that it converges... :blush:", - "Solution_9": "[quote=\"jmerry\"]It's necessary to verify that it's monotone increasing[/quote]\r\nHm, this seemed obvious to me so I didn't explain it. I mean, just note that $ f$ is increasing on $ [1, e]$ and that $ f$ is greater than 1 when $ x > e$ so if $ x > e$ then $ xf(x)$ is a product of increasing functions and thus still increasing, and also doesn't drop at $ x\\equal{}e$. This works for the first interval ($ [e, e^e]$) and you can just bootstrap up with the previous result to get the next one and cover everything. I hope they don't kill me for just claiming that it's increasing...", - "Solution_10": "Could someone tell me what I did wrong?\r\n\r\n$ \\sum_{n\\equal{}1}^{\\infty} \\frac{1}{f(n)} \\leq \\int_1^{\\infty} \\frac{1}{f(x)} dx \\equal{} \\int_1^e \\frac{1}{x} dx \\plus{} \\int_e^{e^e} \\frac{1}{f(x)} dx \\plus{} \\int_{e^e}^{e^{e^e}} \\frac{1}{f(x)} dx \\plus{} \\cdots$\r\n\r\n$ \\int_1^e \\frac{1}{x} dx \\equal{} 1$\r\n\r\nLet $ x$ be from $ 1$ to $ e$.\r\n\r\nThen define $ f_n(x) \\equal{} f(e^{e^{e^{\\cdots^x}}})$ where there are $ n$ $ e$'s and $ f_0(x) \\equal{} f(x)$. Thus $ f_n(x) \\equal{} e^{e^{\\cdots^x}} f_{n\\minus{}1}(x)$.\r\n\r\n$ f_n(x) \\geq e^{e^{\\cdots^x}}f_0(x) \\geq e^{e^{\\cdots^1}}x$.\r\n\r\n$ \\int_{e^{\\cdots^e}}^{e^{e^{\\cdots^e}}} \\frac{1}{f(x)} dx \\equal{} \\int_{1}^{e} \\frac{1}{f_n(x)} dx$ where the first tower has $ n$ $ e$'s and the second has $ n \\plus{} 1$ $ e$'s.\r\n\r\n$ \\frac{1}{f_n(x)} \\leq \\frac{1}{e^{\\cdots^e}x}$ and so $ \\int_{1}^{e} \\frac{1}{f_n(x)} dx \\leq \\int_1^e \\frac{1}{e^{\\cdots^e}x} dx \\equal{} \\frac{1}{e^{\\cdots^e}}$.\r\n\r\nThus $ \\sum_{n\\equal{}1}^{\\infty} \\frac{1}{f(n)} \\leq 1 \\plus{} \\frac{1}{e} \\plus{} \\frac{1}{e^e} \\plus{} \\frac{1}{e^{e^e}} \\plus{} \\cdots$ which converges.", - "Solution_11": "[quote=\"peter_\"]$ \\int_{e^{\\cdots^e}}^{e^{e^{\\cdots^e}}} \\frac {1}{f(x)} dx \\equal{} \\int_{1}^{e} \\frac {1}{f_n(x)} dx$[/quote]\r\nThis step is not correct; you're changing variables, so you need to multiply by the appropriate conversion factor. I'm not actually sure if your recursion is set up properly; you should be getting that this integral is equal to $ \\int_1^x \\frac{1}{f(x)} \\, dx$.", - "Solution_12": "Claim: $f$ is increasing over $\\mathbb{R}_{\\ge 1}$.\n\nProof: Fix $x>y$. Let $y_0=y, y_i=\\log(y_{i-1})$ if $y_{i-1}\\ge e$. Define $x_i$ similarly. Let $k$ be the maximum $k$ such that $y_k$ is defined. Since $x_j>y_j$ for all $0\\le j\\le k$ it follows that $$f(y) = \\prod\\limits_{j=0}^k y_j < \\prod\\limits_{j=0}^k x_j \\le f(x)$$\n\nThis means we can apply the integral test, and only need to worry about whether $$\\int_1^{\\infty} \\frac{1}{f(x)} dx$$ diverges or not.\n\nTo make $f(x)$ easier to describe, we find the intervals when $f(x)=x, x\\log x, x\\log x \\log \\log x$ and so on, so on.\n\nLet $L$ be the iterated logarithm (i.e. $L_k(x) = L_{k-1}(\\log x)$)\n\nLet $x_0=1, x_j=e^{x_{j-1}}$ for all $j\\in \\mathbb{N}$. Then $f(x) = \\prod\\limits_{j=0}^k L_j(x)$ if and only if $x\\in [x_k, x_{k+1})$.\n\nThus,\n\n$$\\int_1^{\\infty} \\frac{1}{f(x)} dx = \\sum\\limits_{n\\ge 0} \\int_{x_n}^{x_{n+1}} \\frac{1}{x L_1(x) L_2(x) \\cdots L_n(x)} dx $$ \n\nTo compute $$\\int_{x_n}^{x_{n+1}} \\frac{1}{x L_1(x) L_2(x) \\cdots L_n(x)} dx $$ \n\nLet $x=e^y$ then $dx=e^y dy$. Thus,\n\n$$\\int_{x_n}^{x_{n+1}} \\frac{1}{x L_1(x) L_2(x) \\cdots L_n(x)} dx = \\int_{y=x^{n-1}}^{x_n} \\frac{1}{e^y y L_1(y) \\cdots L_{n-1}(y)} e^y dy = \\int_{x_{n-1}}^{x_n} \\frac{1}{x L_1(x) L_2(x) \\cdots L_n(x)} dx$$ \n\nThus, all terms in the sum are equal, so this integral diverges." -} -{ - "Problem": "Hi, i need help with this question\r\nim kinda of stuck on it as there is not enuf info, but here folos...\r\nA triangle has two altitudes h & k. h can't equal k. Determine the upper and lower bounds of the length of the third altitude in terms of h and k.\r\nDo i use triangular inequalties, pythagoras or orthocentre? i dunno where to start. \r\n :blush: :blush: \r\nThnx :D", - "Solution_1": "i think this has been posted before...perhaps you can try to use the search function.", - "Solution_2": "Let third altitude be \"b\", here we have:\r\n\r\n1/h + 1/k > 1/b\r\n(h+k)/hk>1/b\r\nb(h+k)/hk>1\r\nb>hk/(h+k)" -} -{ - "Problem": "Did you make it? ^^\"", - "Solution_1": "you should make an option that says \"i didin't make USAMO\"", - "Solution_2": "I did, and I was pleasantly surprised about it. My index was 204 (144 on AMC 10A, 6 on AIME 1).", - "Solution_3": "And a option for AMC10. :wink:", - "Solution_4": "do they count it as your index if you took the AMC 10 instead of the 12?", - "Solution_5": "lol... a fourth grader from canada made it... that's just scary!! and then, all those seventh graders (like 5, but still!! I HATE THEM ALL !! SHOULD'VE PASSED!!)... ne ways... lol :D", - "Solution_6": "[quote=\"mathcrazed\"]lol... a fourth grader from canada made it... that's just scary!! and then, all those seventh graders (like 5, but still!! I HATE THEM ALL !! SHOULD'VE PASSED!!)... ne ways... lol :D[/quote]\r\n\r\nGo shut your computer off forever.", - "Solution_7": "AMC 12 \r\n150 + 10(9) = 240. I still think this index is too low if it's possible to make it with a 5 on the AIME.", - "Solution_8": "[quote=\"mathcrazed\"]lol... a fourth grader from canada made it... that's just scary!! and then, all those seventh graders (like 5, but still!! I HATE THEM ALL !! SHOULD'VE PASSED!!)... ne ways... lol :D[/quote]\r\n\r\nhey, one of those \"seventh graders\" u hate is me!!! im offended jkjk", - "Solution_9": "didn't you get a 4, mathcrazed? no offense but that was not too close to passing. that's what i got too, but at least i'm not complaining *cough**cough*... my index sucked. it was 175.", - "Solution_10": "[quote=\"mwpl11\"]didn't you get a 4, mathcrazed? no offense but that was not too close to passing. that's what i got too, but at least i'm not complaining *cough**cough*... my index sucked. it was 175.[/quote]\r\n\r\n\r\nuh..well, i dont want to speak for mathcrazed, but i think he made like an addition mistake, a multiplication mistake, and misread a question..and those mistakes are the easiest to catch (and the hardest to forget)..so he probalby wouldve passed if he wasnt so careless..but AIME tests for carelessness with all of its computation.", - "Solution_11": "yeah if it was like usamo then people would get partial credit, but if you screw up by a tiny bit, you get no credit on aime...\r\nthats why i lost 2 potential points, 9 --> 7... :( \r\nbut i guess thats better than some people", - "Solution_12": "[quote=\"Art of Owna\"][quote=\"mwpl11\"]didn't you get a 4, mathcrazed? no offense but that was not too close to passing. that's what i got too, but at least i'm not complaining *cough**cough*... my index sucked. it was 175.[/quote]\n\n\nuh..well, i dont want to speak for mathcrazed, but i think he made like an addition mistake, a multiplication mistake, and misread a question..and those mistakes are the easiest to catch (and the hardest to forget)..so he probalby wouldve passed if he wasnt so careless..but AIME tests for carelessness with all of its computation.[/quote]\r\n\r\nAnd the floor would've been like 9 if everyone didn't do those mistakes so no he was not close to passing.", - "Solution_13": "[quote=\"xxazurewrathxx\"]AMC 12 \n150 + 10(9) = 240. I still think this index is too low if it's possible to make it with a 5 on the AIME.[/quote]\r\n\r\nThat's one of the consequences for having twice as many USAMO tests available for students.\r\n\r\nI guess with that kind of score required when taking the AIME to get into the USAMO, the AIME or USAMO could be tougher next year.", - "Solution_14": "Well, 5 was only 1 point below the floor. Last year, one could have made the floor with a 7 AIME, one point below the floor. It's the score distribution, not just the numbers, that matter.", - "Solution_15": "So a person with a 150 AMC10 and 5 AIME has a better index than a 140 AMC12 and 5 AIME?\r\n\r\nIf you take both tests, what happens? I'm confused by how this works...", - "Solution_16": "[quote=\"LawOfSigns\"]So a person with a 150 AMC10 and 5 AIME has a better index than a 140 AMC12 and 5 AIME?\n\nIf you take both tests, what happens? I'm confused by how this works...[/quote]\r\n\r\nYes. If you take both, you get the higher one.", - "Solution_17": "[quote=\"paladin8\"][quote=\"LawOfSigns\"]So a person with a 150 AMC10 and 5 AIME has a better index than a 140 AMC12 and 5 AIME?\n\nIf you take both tests, what happens? I'm confused by how this works...[/quote]\n\nYes. If you take both, you get the higher one.[/quote]\r\n\r\nAre you sure? I thought index qualifiers were selected only by the AMC12...", - "Solution_18": "[quote=\"K81o7\"][quote=\"paladin8\"][quote=\"LawOfSigns\"]So a person with a 150 AMC10 and 5 AIME has a better index than a 140 AMC12 and 5 AIME?\n\nIf you take both tests, what happens? I'm confused by how this works...[/quote]\n\nYes. If you take both, you get the higher one.[/quote]\n\nAre you sure? I thought index qualifiers were selected only by the AMC12...[/quote]\r\n\r\nThat's true, but if you're in 10th grade or under, for the tiebreaker, your AMC10 index would be higher.", - "Solution_19": "If I had gotten two more AIME questions right, I would have made it. (careless mistakes on 5 and 11)\r\n\r\nAlso, we just learned how to do Logarithms in my Alg2/Trig class, so I went back and did #7 in like 3 minutes, a problem I had no idea how to approach a month ago.", - "Solution_20": "[quote=\"K81o7\"]Are you sure? I thought index qualifiers were selected only by the AMC12...[/quote]\r\n\r\nYeah, that's right, sorry. I was referring to the tiebreaker stuff.", - "Solution_21": "I barely made it. I got a 6 on the AIME." -} -{ - "Problem": "Starting at the origin, a beam of light hits a mirror (in the form of a line) at $ A (4, 8)$ and is reflected to the point $ B (8, 12)$. Compute the exact slope of the mirror.", - "Solution_1": "[hide=\"Solution\"]\nExtend $ AB$ to some point $ B'$ on $ \\vec{AB}$ such that $ OA \\equal{} AB'$, forming an isosceles triangle. The slope of $ OB'$ is the slope of the mirror.\n\n$ \\vec{AB} \\equal{} (4, \\; 4)$ and $ \\frac{\\|OA\\|}{\\|AB\\|} \\equal{} \\frac{\\sqrt{10}}{2}$\n\nHence, $ \\vec{AB'} \\equal{} \\frac{\\sqrt{10}}{2} \\vec{AB} \\equal{} (2 \\sqrt{10}, \\; 2 \\sqrt{10})$\n\nThen, $ OB' \\equal{} (2 \\sqrt{10}\\plus{}4, \\; 2 \\sqrt{10}\\plus{}8)$, and it's slope is $ \\frac{2 \\sqrt{10}\\plus{}8}{2 \\sqrt{10}\\plus{}4} \\equal{} \\frac{ \\sqrt{10}\\plus{}4}{ \\sqrt{10}\\plus{}2}$\n\nWe can go ahead and rationalize this to get $ \\boxed{\\frac{1}{3}(1 \\plus{} \\sqrt{10})}$\n[/hide]", - "Solution_2": "magnitude of the acute angle between the beam and the mirror and the image and the mirror is same.\r\n$ \\left| \\frac {2\\minus{}x}{1\\plus{}2x} \\right|\\equal{}\\left| \\frac {1\\minus{}x}{1\\plus{}x} \\right|$\r\nYou get 2 cases of which one yields nonsense and the other $ \\frac 13(1\\pm \\sqrt {10})$ simple observation tells us that $ \\frac 13(1\\plus{} \\sqrt {10})$ is the slope of the mirror while $ \\frac 13(1\\minus{} \\sqrt {10})$ is the slope of the normal to the mirror." -} -{ - "Problem": "I may just be overly anxious, but has anyone received their mail notification? I have received the email, but not the one in the mail.", - "Solution_1": "I got my mail notification today.", - "Solution_2": "[quote=\"watermaximillion\"]I got my mail notification today.[/quote]\r\n\r\nMe too.", - "Solution_3": "I will follow the trend and announce that, in fact, I did receive my notification today.", - "Solution_4": "Yep got it today as well.", - "Solution_5": "Got it today.", - "Solution_6": "got it\r\n\r\n(what exactly is the point of this thread, other than to make people who haven't gotten it yet freak out unnecessarily?)", - "Solution_7": "Yeah...got it today too. Yeah, there's not too much of point.", - "Solution_8": "[quote=\"tenniskidperson3\"]I may just be overly anxious, but has anyone received their mail notification? I have received the email, but not the one in the mail.[/quote]\r\n\r\nIf you haven't gotten the mail invitation yet, you should probably start PANICKING.", - "Solution_9": "[quote=\"nr1337\"][quote=\"tenniskidperson3\"]I may just be overly anxious, but has anyone received their mail notification? I have received the email, but not the one in the mail.[/quote]\n\nIf you haven't gotten the mail invitation yet, you should probably start PANICKING.[/quote]\r\n\r\nThe only thing in the mail was the forms. You can get them online.", - "Solution_10": "OK, that's all I really needed to know. A mod can lock this now." -} -{ - "Problem": "You are to design a \u201cvehicle\u201d that will safely transport an egg from the third floor of a building to the ground. Your vehicle should be as light and as fast as possible. Any ideas?", - "Solution_1": "A parachute of the right size may do it.\r\nOr you have to use the stairs in the building?\r\nA hot air balloon will do it too. As the air cools down it will go down. \r\nAnd it's apparent weight will be close to zero...." -} -{ - "Problem": "Is there a special section in the college resume that asks what you did over the summer? (listing names of camps, activities, etc each year) :?", - "Solution_1": "yes, I believe so.", - "Solution_2": "I always just listed mine under \"seminars, etc\"... that's where I listed AoPS class, my equine reproduction seminar, and my transportation engineering internship (which was horrible, fyi). When in doubt, put it under whatever's most general.", - "Solution_3": "The [url=http://app.commonapp.org/index.cfm?APP=AppOnline&ACT=Display&DSP=Forms]Common Application[/url] has a space for asking about summer activities.", - "Solution_4": "Having looked at the common application...I'm concluding that we don't list all the camps we've attended during summer?\r\nAlso, what does the resume cover? \r\nThank you!", - "Solution_5": "[quote=\"thisismysn\"]Having looked at the common application...I'm concluding that we don't list all the camps we've attended during summer?[/quote]\r\n\r\nWell, you could. The first general principle to consider is that you list activities since eighth grade. Some middle school accomplishments are impressive, but generally colleges want to know what you did at high school age. So start with the summer after eighth grade, usually. If you run out of space, you can write \"see attachment\" and put on an extra page to list all the summer programs you did. That is an issue of setting priorities: list the summer programs that were most impressive (that is, the ones that you had to take a test to get into) and maybe don't mention unknown local programs except in such terms as \"local chess champ, 2003-2006\" or something like that. But you DEFINITELY want to mention something like MOP, MathCamp, Ross, or PROMYS, so make space for it if it doesn't fit in the standard blanks on the form." -} -{ - "Problem": "[b]An eight-bit binary word is a sequence of eight digits each of whcih is either 0 or 1. Find the number of eight-bit binary words.[/b]\r\n\r\n[hide=\"i got\"]128[/hide]\n\n[hide=\"they got\"]256[/hide]", - "Solution_1": ":?: \r\n[hide]Each spot has 2 possibilities. There are 8 spots, so the total combinations is $ 2^{8}\\equal{}256$.[/hide]", - "Solution_2": "yeah, i understand that. the two possibilities are 1 and 0. however the first spot can't be 0, otherwise it would really be a 7-bit number. :wink: so there is only 1 possibility for the number in first digits place. \r\n\r\nthat's where my problem lies.", - "Solution_3": "[quote=\"mihail911\"]yeah, i understand that. the two possibilities are 1 and 0. however the first spot can't be 0, otherwise it would really be a 7-bit number. :wink: so there is only 1 possibility for the number in first digits place. \n\nthat's where my problem lies.[/quote]\r\nSorry...\r\n :oops: Still, I don't think you're supposed to assume that. Otherwise, they would have said 8-digit binary number instead of \"eight-bit binary word.\"", - "Solution_4": "$ 00000000$ is considered a word.\r\n\r\nI must commend you for being able to think ahead, and assuming a 1 must be in the first place.", - "Solution_5": "hmm..i guess that is the case. thanks fraenkel and archimedes!" -} -{ - "Problem": "A dietician in a hospital must prepare a meal for a patient as was prescribed by a doctor to have 800 calories, 55 g of protein, and 220 mg of vitamin C. How many servings of broccoli ($ b$), potatoes ($ p$), and steak ($ s$) are needed in order to exactly satisfy the physician's requirements? Express your answer as the ordered triple ($ b$, $ p$, $ s$).\n\n\\begin{tabular}{cccc}\n & Calories & Protein (g) & Vitamin C (mg) &\nSteak & 300 & 20 & 0 &\nPotato & 100 & 5 & 20 &\nBroccoli & 50 & 5 & 100\n\\end{tabular}", - "Solution_1": "From the fact that the dietician needs 220 mg of Vitamin C, we can deduce that thedietician needs one serving of potato. Then, directly from that last fact, we deduce that he needs $ 2$ servings of broccoli. Finally, from both of these facts, we can deduce that he needs $ 2$ servings of steak. Thus our ordered triple is $ \\boxed{(2,1,2)}$." -} -{ - "Problem": "I want to know is it that any wave described by a functin of the form y(x,t)=f(x-vt) moves in the +x direction with speed v. How do I show that y(x,t)=f(x-vt) satisfies the wave equation, no matter what the functional form of f.", - "Solution_1": "Just plug your sample functiction into each side of the wave equation and make sure that the result is equal.", - "Solution_2": "so you should note that you'll need some mild differentiability properties on $ f$ assuming you wish to show it's a classical solution. If you're at the level of talking about weak/mild/other generalized solutions, then you've got other things to think about, but I'll assume this isn't where you're headed.", - "Solution_3": "[quote=\"bigman\"]I want to know is it that any wave described by a functin of the form y(x,t)=f(x-vt) moves in the +x direction with speed v. How do I show that y(x,t)=f(x-vt) satisfies the wave equation, no matter what the functional form of f.[/quote]\r\nFirst write down teh wave eqn\r\nhow do u think the wave equation was \"derived'?", - "Solution_4": "Hehe, \r\n\r\nSum up all the posted replies then you will get detailed analysis of the problem! :rotfl:" -} -{ - "Problem": "We wish to construct a circular sequence $ a_{0},...,a_{7}$ (indices $ mod8$) in such a way that a sliding window $ a_{i},a_{i\\plus{}1},a_{i\\plus{}3}$ $ i\\equal{}1,2,...,7$ will contain every possible three-tuple once. Show(not just by trial or error) that this is impossible.", - "Solution_1": "could you elaborate where does the sliding window come from ? and what does this mean \"will contain every possible three-tuple once\"?", - "Solution_2": "Wish, i knew it too. That's why i posted this problem here. Because i cant think anything to solve it and of course because i don't understand what does the problem ask :( :(", - "Solution_3": "I guess either u have got d question wrong or maybe I am missing something (dont take it otherwise..i normally do)\r\n\r\nFor infact, the thing that u ask is possible(regarding debrujin sequences) if i think i understand the problem. Anyway for clarity sake I restate the problem.\r\n\r\n(The ideas below are taken from d wonderful book [b]Mathematics: The man-made universe by Sherman K stein[/b])\r\n\r\nThe problem:\r\n\r\nYou are supposed to make a 2^n-letterd wheel using numbers 0's and 1's only. Can u make this wheel in such a fashion that if i start off rreading every n consecutive symbols in the wheel (mod-2^n) i end up reading evey possible 3-bit combination of 0's and 1's. Let us call them magic wheels.\r\n\r\nUr case sets n=3. Here we will see 8 strings (and in general 2^n-strings) like a0.a1.a2...a1.a2.a3...a2.a3.a4...a3.a4.a5...a4.a5.a6....a5.a6.a7....a6.a7.a0....a7.a0.a1...\r\n\r\nU get d picture...\r\n\r\nNow u may want to know that the answer to the above question is [b]YES.[/b].\r\n\r\n\r\n\r\n[b]Here's the solution[/b]..(It can be genaralised readily)\r\n\r\n\r\n\r\nlet us represent the every 3-digit combination of 0's and 1's as cities on map. U get 8 cities. Now there are roads (directed edges) between 2 cities if u find the last two bits of a city common with first 2 bits of another. clearly there are 2 roads that go out of a city and that go into a city. (There are roads leading back to the city itself or self loops)\r\n\r\nAnd now the fun begins......(Please try to think for Stein's power-packed punch is about to come.)\r\n\r\n\r\nThis graph has an Eulerian circuit as every edge has indegree and outdegree both even and infact same. Thus, if I start at a particular city, I can come back to it following a certain Eulerian path visting every edge exactly once...and if I start out with city1 writing its first 3-letters down and attaching to it the third letter of the city i vist, i will come up with an 8-leterd magic wheel.\r\n\r\nI hope that the above outline suffices (if i recollect it properl), please point out if i made any mistakes..\r\n\r\nThanks and have a nice day", - "Solution_4": "Hi all,\r\n\r\nI thought this could be a nice treat for u all. In spirit with the above problems (well, in continuation of his paper on de-bruijn sequences by de-bruijn) the following problem can be tried.\r\n\r\nYou need to prove that the number of de-bruijn sequences (or magic wheels) when u desire the magic wheel to read through all n-bit combinations of 1's and 0's is 2^n-1 - n.\r\n\r\nI do not know the solution myself.", - "Solution_5": "Hello...anybody...any ideas...(Please go thru d previous few posts)", - "Solution_6": "I think the problem statement is inaccurate. Let me restate the problem:\r\n\r\nWe wish to construct a circular sequence $ a_0,a_1,a_2,...,a_7$(indices $ \\pmod 8$) in such a way that a sliding window $ a_ia_{i\\plus{}1}a_{i\\plus{}3}$ for $ i\\equal{}0,1,2,...,7$ will contain every possible three-tuple once. Show that this is impossible.\r\n\r\nThe edit I made was the index $ i$ must run from 0 to 7 and not 1 to 7 as previously stated. This is because there are 8 possible 3-tuples. Let the explain what the term \"sliding window\" means. The problem asks if there are binary variables $ a_0,a_1,a_2,...,a_7$ such that $ a_0a_1a_3$,$ a_1a_2a_4$,$ a_2a_3a_5$,$ a_3a_4a_6$,$ a_4a_5a_7$,$ a_5a_6a_0$,$ a_6a_7a_1$ and $ a_7a_0a_1$ are 000,001,010,011,100,101,110 and 111 in some order. We wish to show that this is not possible. Suppose it were there must be some $ a_ia_{i\\plus{}1}a_{i\\plus{}3}$ which is 000. Without loss of generality let $ a_0a_1a_3$ be 000.(because if i were not 0 we could have cyclically shifted the sequence to make i=0).\r\n$ a_4a_5a_7$ is the only term that doesn't contain any of $ a_0$,$ a_1$ or $ a_3$. In other words $ a_4a_5a_7$ is the only term that doesn't contain a 0. This implies $ a_4a_5a_7$=111. $ a_1a_2a_4\\neq a_2a_3a_5$ implies $ 0a_21\\neq a_201$ which implies $ a_2\\neq 0$. Also $ a_2a_3a_5\\neq a_7a_0a_2$ implies $ a_201\\neq 10a_2$ which implies $ a_2\\neq 1$. Since $ a_2$ can neither have values 0 and 1 we get the desired contradiction. Hence its impossible to have such $ a_i$'s." -} -{ - "Problem": "Compute: $ S\\equal{}\\frac{1}{1 \\cdot 2}\\plus{}\\frac{1}{2 \\cdot 3}\\plus{}...\\plus{}\\frac{1}{2011 \\cdot 2012}.$", - "Solution_1": "[hide]The general term is $ \\dfrac{1}{k(k\\plus{}1)}$. This is also just $ \\dfrac{1}{k}\\minus{}\\dfrac{1}{k\\plus{}1}$. If we use this in our expression for $ S$, we get\n\n$ \\dfrac{1}{1}\\minus{}\\dfrac{1}{2}\\plus{}\\dfrac{1}{2}\\minus{}\\dfrac{1}{3}\\plus{}\\dfrac{1}{3}\\minus{}\\dfrac{1}{4}\\plus{}...\\plus{}\\dfrac{1}{2011}\\minus{}\\dfrac{1}{2012}$\n\nThis series telescopes, and all the terms between 1 and 1/2012 cancel. We are left with\n\n$ S\\equal{}1\\minus{}\\dfrac{1}{2012}\\equal{}\\boxed{\\dfrac{2011}{2012}}$.[/hide]" -} -{ - "Problem": "The length of the circumradius of regular hexagon ABCDEK equals [i]R[/i]. Find the length of inradius of triangle ABD", - "Solution_1": "Note that in a regular hexagon the lenght of the circumradius is the same as the side of the hexagon.\r\n\r\n :)", - "Solution_2": "[hide=\"hint\"]The sides of the trianglse in question are R, 2R and $ \\sqrt{3} R$.\n$ \\Delta \\equal{} rs \\equal{} \\frac {abc}{4R}$[/hide]" -} -{ - "Problem": "1,1,2,3,5,8,13,21,34,55,89,144,233,....\r\n\r\nIs there a formula for this so as called the \"Fibonnaci Sequence\"\r\n\r\n :rotfl: :blush:", - "Solution_1": "Yes...Binet's Formula: $F_{n}=\\frac{1}{\\sqrt{5}}[(\\frac{1+\\sqrt{5}}{2})^{n+1}-(\\frac{1-\\sqrt{5}}{2})^{n+1}]$... :|", - "Solution_2": "The question's been discussed (and locked, too...)\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=130146[/url]", - "Solution_3": "There's a few formulas.\r\n\r\nBinet's is the well known one.\r\nAnother is the general recursive relationship:\r\nLet $F_{n}$ be defined as the Nth Fibonacci number.\r\nDefine $F_{0}= 1$ and $F_{1}= 1$\r\n\r\nThen the general recursive relation is $F_{n}= F_{n-2}+F_{n-2}$\r\nOr, $F_{n+2}=F_{n+1}+F_{n}$\r\n\r\nHere's how we'd go about solving this for Binet's Formula (beware that this is well above MC Level, but I'll put it in for anyone who is interested:\r\n\r\nFor all linear recursions, we can generate a characteristic polynomial that defines the recursion. For example, the polynomial for the Fibonacci numbers is \r\n$x^{2}= x+1$\r\n(This comes from the fact that the Nth term becomes $x^{0}$, and the N+1th term becomes $x^{1}$)\r\n\r\nRearranging this equation, we get\r\n$x^{2}-x-1 = 0$\r\n\r\nNow we can set up a general recurrence formula, in the form \r\n$F_{n}= A\\cdot(R_{1})^{n}+B\\cdot(R_{2})^{n}$, where A and B are constants, and $R_{1}$ and $R_{2}$ are the roots of the characteristic polynomial.\r\n\r\nThe roots of the above eqn, by the Quadratic Formula, are\r\n$\\frac{1 \\pm \\sqrt{5}}{2}$ Call this $\\phi$\r\n\r\nThus, we can sub these in, and set up a system of equations:\r\n$F_{0}= A(\\phi)^{0}+B(\\phi)^{0}$\r\n$F_{1}= A(\\phi)^{1}+B(\\phi)^{1}$\r\n\r\n$1 = A+B$\r\n$1 = A\\phi+B\\phi$\r\n\r\nSolving this yields values for A and B, which leads us to the explicit formula for the Fibonacci Series (Binet's Formula).\r\n\r\nHere's a practice problem to try.\r\n\r\nThere is a recurrence relation defined as $F_{n}= 3F_{n-1}-2F_{n}$ with initial values $F_{0}= 2$, $F_{1}= 3$\r\n\r\n[hide=\"Step One: Define the characteristic polynomial\"]\n$F_{n+2}= 3F_{n+1}-2F_{n}$\n$x^{2}= 3x-2$\n$x^{2}-3x+2 = 0$[/hide]\n\n[hide=\"Step Two: Find the Roots\"]\n$(x-2)(x-1) = 0$\n$x = 2; 1$[/hide]\n\n[hide=\"Step Three: Set up the recurrence relation\"]\n$F_{n}= A(r_{1})^{n}+B(r_{2})^{n}$\n$F_{n}= A(2)^{n}+B(1)^{n}$\n$F_{n}= A(2)^{n}+B$[/hide]\n\n[hide=\"Step Four: Set up system of equation to find A and B\"]\n$2 = A(2)^{0}+B$\n$3 = A(2)^{1}+B$[/hide]\n\n[hide=\"Step Five: Solve system, getting values of constants A and B\"]\n$2 = A+B$\n$3 = 2A+B$\n$A = 1; B = 1$[/hide]\n\n[hide=\"Step Six: Finish the explicit formula\"]\n$F_{N}= 2^{n}+1$[/hide]", - "Solution_4": "okay guys i was using the above formulas to make a program to find the nth term in the fibonnacci sequence. they didn't work and they were off in terms of the signs. so i went to wolframs math world and i got binets formula and it really does work\r\n\r\nhere is the binets formula that i obtained from math world\r\n\r\n{(1+sqrt50^N)-(1-sqrt5)^N} divided by(2^N times sqrt 5)\r\n\r\nN represents the term in the sequence, e.g. first term, third term", - "Solution_5": "[quote=\"learner1729\"]okay guys i was using the above formulas to make a program to find the nth term in the fibonnacci sequence. they didn't work and they were off in terms of the signs. so i went to wolframs math world and i got binets formula and it really does work\n\nhere is the binets formula that i obtained from math world\n\n{(1+sqrt50^N)-(1-sqrt5)^N} divided by(2^N times sqrt 5)\n\nN represents the term in the sequence, e.g. first term, third term[/quote]\r\n\r\nSorry to dissappoint you but that was [u]just[/u] posted. Please do not post things that have already been posted.", - "Solution_6": "all right but mine is a whole lot easier to understand, just one variable to plug in", - "Solution_7": "and how many are in the first 1?...", - "Solution_8": "i don't understand what u mean", - "Solution_9": "who r u????????venky??????", - "Solution_10": "[quote=\"DiscreetFourierTransform\"]There's a few formulas.\n\nBinet's is the well known one.\nAnother is the general recursive relationship:\nLet $F_{n}$ be defined as the Nth Fibonacci number.\nDefine $F_{0}= 1$ and $F_{1}= 1$\n\nThen the general recursive relation is $F_{n}= F_{n-2}+F_{n-2}$\nOr, $F_{n+2}=F_{n+1}+F_{n}$\n\nHere's how we'd go about solving this for Binet's Formula (beware that this is well above MC Level, but I'll put it in for anyone who is interested:\n\nFor all linear recursions, we can generate a characteristic polynomial that defines the recursion. For example, the polynomial for the Fibonacci numbers is \n$x^{2}= x+1$\n(This comes from the fact that the Nth term becomes $x^{0}$, and the N+1th term becomes $x^{1}$)\n\nRearranging this equation, we get\n$x^{2}-x-1 = 0$\n\nNow we can set up a general recurrence formula, in the form \n$F_{n}= A\\cdot(R_{1})^{n}+B\\cdot(R_{2})^{n}$, where A and B are constants, and $R_{1}$ and $R_{2}$ are the roots of the characteristic polynomial.\n\nThe roots of the above eqn, by the Quadratic Formula, are\n$\\frac{1 \\pm \\sqrt{5}}{2}$ Call this $\\phi$\n\nThus, we can sub these in, and set up a system of equations:\n$F_{0}= A(\\phi)^{0}+B(\\phi)^{0}$\n$F_{1}= A(\\phi)^{1}+B(\\phi)^{1}$\n\n$1 = A+B$\n$1 = A\\phi+B\\phi$\n\nSolving this yields values for A and B, which leads us to the explicit formula for the Fibonacci Series (Binet's Formula).\n\nHere's a practice problem to try.\n\nThere is a recurrence relation defined as $F_{n}= 3F_{n-1}-2F_{n}$ with initial values $F_{0}= 2$, $F_{1}= 3$\n\n[hide=\"Step One: Define the characteristic polynomial\"]\n$F_{n+2}= 3F_{n+1}-2F_{n}$\n$x^{2}= 3x-2$\n$x^{2}-3x+2 = 0$[/hide]\n\n[hide=\"Step Two: Find the Roots\"]\n$(x-2)(x-1) = 0$\n$x = 2; 1$[/hide]\n\n[hide=\"Step Three: Set up the recurrence relation\"]\n$F_{n}= A(r_{1})^{n}+B(r_{2})^{n}$\n$F_{n}= A(2)^{n}+B(1)^{n}$\n$F_{n}= A(2)^{n}+B$[/hide]\n\n[hide=\"Step Four: Set up system of equation to find A and B\"]\n$2 = A(2)^{0}+B$\n$3 = A(2)^{1}+B$[/hide]\n\n[hide=\"Step Five: Solve system, getting values of constants A and B\"]\n$2 = A+B$\n$3 = 2A+B$\n$A = 1; B = 1$[/hide]\n\n[hide=\"Step Six: Finish the explicit formula\"]\n$F_{N}= 2^{n}+1$[/hide][/quote]\r\n\r\ni'm confused\r\n\r\nwhat is the $F_{N}= 2^{n}+1$ for?\r\n\r\nnot for fibonacci i think. i didn't work.\r\n\r\njorian", - "Solution_11": "it's for the recursion he defined.", - "Solution_12": "it was too complex so i had a hard time trying to understand it\r\n\r\n-jorian" -} -{ - "Problem": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14770[/img]\r\n\r\nthis is my 93\u00ba post (5 were double?)\r\n\r\n :(", - "Solution_1": "The length of the hypotenuse is $ AB \\equal{} \\frac {2r^2}{h \\minus{} 2r}$ and is done.", - "Solution_2": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14789[/img]\r\n\\[ \\frac {{AB}}{r} \\equal{} \\frac {{2r}}{{h \\minus{} 2r}}\r\n\\]\r\n[b]AB r h datum[/b]\r\n\r\n[b]and arc capable 90\u00ba of AB[/b]\r\n\r\n :thumbup:" -} -{ - "Problem": "Let $ a,b,c \\in R^{ \\plus{} }$,Proof that $ \\sum_{cyc}\\frac {a}{a^2 \\plus{}2bc} \\leq \\frac {a \\plus{} b \\plus{} c}{ab \\plus{} bc \\plus{} ca}$", - "Solution_1": "[quote=\"apollo\"]Let $ a,b,c \\in R^{ \\plus{} }$,Proof that $ \\sum_{cyc}\\frac {a}{a^2 \\plus{} bc} \\leq \\frac {a \\plus{} b \\plus{} c}{ab \\plus{} bc \\plus{} ca}$[/quote]\r\n $ a\\equal{}b\\equal{}c\\equal{}1$ not true :) \r\nI think it's better is\r\nLet $ a,b,c \\in R^{ \\plus{} }$,Proof that $ \\sum_{cyc}\\frac {2a}{a^2 \\plus{} bc} \\leq \\frac {3(a \\plus{} b \\plus{} c)}{ab \\plus{} bc \\plus{} ca}$", - "Solution_2": "Sorry,It is $ \\sum_{cyc}\\frac{a}{a^2\\plus{}2bc} \\leq \\frac{a\\plus{}b\\plus{}c}{ab\\plus{}bc\\plus{}ca}$", - "Solution_3": "see my posts at http://www.mathlinks.ro/viewtopic.php?t=173359.... for a proof using the Vornicu Schur inequality" -} -{ - "Problem": "Find the maximum value of the expression $13 \\sqrt{x^2-x^4} +9 \\sqrt{x^2+x^4}$ for $0 \\leqslant x \\leqslant 1$.", - "Solution_1": "it's similar with this problem \r\n Let $D=\\sqrt{x+x^3}+\\sqrt{x-x^3}$. For $x\\in[0,1]$, $D\\ge0$, so to maximise $D$ it is sufficient to maximise $D^2=2x(1+\\sqrt{1-x^4})$.\r\n\r\nDifferentiating this, and letting $u=\\sqrt{1-x^4}$, we get $\\frac{d}{dx}D^2=2+2u-\\frac{2x^4}{u}$. \r\n\r\n$\\frac{d}{dx}D^2=0\\Leftrightarrow u+u^2-2x^4=3u^2+u-2=0\\Leftrightarrow u=-1$ or $\\frac{2}{3}$.\r\n\r\nObviously $x\\in[0,1]\\Rightarrow u\\in[0,1]\\Rightarrow u=\\frac{2}{3}\\Rightarrow x=\\sqrt[4]{1-\\frac{2}{3}^2}=\\frac{\\sqrt[4]{5}}{\\sqrt{3}}$. It is easy to confirm that this turning point is in fact the maximum", - "Solution_2": "What's the Mathscope ??\r\nA magazine on math??\r\n :D" -} -{ - "Problem": "As I am Indian myself, I wanted to start a discussion thread on the attacks a couple of days ago.", - "Solution_1": "I heard that the USA FBI from LA is traveling there with a bomb squad and forensics people. But it was pretty bad, except no crazy terrorists crashing into big towers.", - "Solution_2": "Even though the Prime Minister didn't point any fingers and merely said \"the attackers are foreigners\", I am scared about how this event will affect the fragile unity between Hindus and Muslims. On the other hand, I doubt that the terrorists will accomplish their goal of permanently disabling India's international business. :)", - "Solution_3": "Just now I got the news five more hostages are killed.\r\nThings are getting worse.\r\nI don't have words to describe how cruelly they have killed these innocent people.\r\nAll the three hotels had turned into house of death bodies.\r\nI don't care whether the NSG kill those terrorists or arrest them.\r\nBut what I want at this moment is no more killing of innocent people.\r\nMay Jesus give strength to those family who have lost there loving ones in this attack.\r\n\r\nI remembered one of the favorite poem of Gandhi Ji which we used to sing in school\r\nIsswar Allah tere jahaa mein nafrat kuon hain \r\njang hain kuon\r\nTera dil to itna bada hain\r\nInshaan ka dil tang hain kuon hain", - "Solution_4": "Although more intricately planned and with more innocent lives at stake, this event seems similar to the 1980 Iranian Embassy Siege. I hope this will end as soon as possible so officials can figure out exactly why the attacks were carried out and what group is responsible for this.", - "Solution_5": "wait u're Indian ernie? :o \r\n\r\nand yeah its pretty bad over there...\r\npoor them", - "Solution_6": "It's pretty ironic, because the security was loosened a couple of days ago.\r\n\r\nAnd India has verified that the terrorists were Pakistani and some of them were from Europe (possibly those who participated in the London bombings and the LET).\r\n\r\nRemember that this all started up again on July 25th (ironically, my birthday), with the bombings in Bangalore. The next day, there were bombings in my hometown city, Ahmadabad. Since then, there were 2 more bombings in Delhi and Assam. But India was able to pick itself up and move on. However, this attack was much more terrible, killing over 160 people and injuring more than 350 more.\r\n\r\nSince then, we have taken back the Oberoi and the Nariman house. But we have yet to recover the once-beautiful Taj hotel.", - "Solution_7": "You are right. I lived very close to Mumbai and know about the Taj. Seeing it burning is just heartbreaking. I just wonder, why did the terrorists attack a hospital? Do you know they had taken patients hostage there, but released them?", - "Solution_8": "[size=200]I know just the thing to do to those terrorists: Cut off their body parts one by one before killing them. Also remove the organs. That should ensure maximum pain. And this execution should be on camera, so any other terrorists with bad plans know their fate. It serves them right for what they did to India. Maybe we will not kill them, we will let those [/size][size=9]jackasses[/size][size=200] live in pain and suffering. That should put an end to terrorism.[/size]", - "Solution_9": "[quote=\"jain_harshi\"]You are right. I lived very close to Mumbai and know about the Taj. Seeing it burning is just heartbreaking. I just wonder, why did the terrorists attack a hospital? Do you know they had taken patients hostage there, but released them?[/quote]\n\nThis tactic was used in the Ahmadabad bombings. All the injured were admitted to a hospital, which was later bombed.\n\n[quote=\"mathelete\"]I know just the thing to do to those terrorists: Cut off their body parts one by one before killing them. Also remove the organs. That should ensure maximum pain. And this execution should be on camera, so any other terrorists with bad plans know their fate. It serves them right for what they did to India. Maybe we will not kill them, we will let those jackasses live in pain and suffering. That should put an end to terrorism.[/quote]\r\n\r\nNo it won't. But we need to take all the information out of them too.", - "Solution_10": "[quote=\"mathelete\"]I know just the thing to do to those terrorists: Cut off their body parts one by one before killing them. Also remove the organs. That should ensure maximum pain. And this execution should be on camera, so any other terrorists with bad plans know their fate. It serves them right for what they did to India. Maybe we will not kill them, we will let those [/size][size=9]jackasses[/size] live in pain and suffering. That should put an end to terrorism.[/size][/quote]\r\nHoly crap. That's scary.\r\nAnd... no. That would only inspire more violence. As much as some Americans (reminded of Palin here) see the terrorists as evil, they are only fighting for their beliefs. If you think about it, our Founding Fathers were terrorists too...\r\nAnd taking all the information out of them also sounds scary. \"Enhanced interrogation techniques,\" anyone?", - "Solution_11": "[quote=\"Brut3Forc3\"]And taking all the information out of them also sounds scary. \"Enhanced interrogation techniques,\" anyone?[/quote]\r\nDo you have any idea what [b]their[/b] techniques did to us and more specifically to how many of us and to how much things of ours?\r\nThey knew they would die when they started all these. So we need to do something more with them so that no one dares it again.\r\n\r\nAnd yes the Pakistan refused to send ISI chief to India: \r\nhttp://timesofindia.indiatimes.com/World/Pak_to_send_official_not_ISI_chief/articleshow/3771372.cms", - "Solution_12": "The started violence and cruelity, so you are allowed to use it too\u00bf That's a rather dangerous and abandoned view, it just causes hate and an infinite circle of even more hate.\r\nAnd after your \"very reasonable interrogation techniques\" were used against them, the next step will be to use them against people you aren't even sure about their \"terroristic activities\". But after enough \"interrogation\", they surely will confess it.. You know, european witch hunting was very successfull, most of them confessed to be witches... (and the those who didn't confess were obviously guilty for collecting herbs or having birthmarks)\r\nYou didn't understand some implications of human rights; and mathlete's postis one of the stupidiest posts I've ever seen on this forum. He may be angry at them, but this is no reason to talk about cruel murders of persons (even if terrorists). In fact, mathlete is even worse than some terrorists, he puts judge on them and shows some execution video to the world; his post sounds more like some execution video of the taliban than words of an intelligent being that has considered a bit more than his useless revenge. Maybe you haven't thought of this, but you would act no better than them; so why shouldn't they in reverse do the same to you\u00bf What's the exact point in acting exactly like them\u00bf Showing them even more, that this actions are accepted on all sides\u00bf Showing them, that you murder their friends and relatives in a very cruel way, so that they hate you and your nation more than everything else, causing more attacks\u00bf Thinking further, you acting that way may even be their plan: it gives them more supporters (in two ways) and it attacks your society (like you) by alternating them into wishfull murderers.\r\nAnd writing that post in bold and huge letters is just annoying, don't do this ever again.", - "Solution_13": "[quote=\"mathelete\"]I know just the thing to do to those terrorists: Cut off their body parts one by one before killing them. Also remove the organs. That should ensure maximum pain. And this execution should be on camera, so any other terrorists with bad plans know their fate. It serves them right for what they did to India. Maybe we will not kill them, we will let those [/size][size=9]jackasses[/size][size=200] live in pain and suffering. That should put an end to terrorism.[/quote]\r\n\r\nmathelete\r\nWe all are sad.I personally know some people who have lost their loved ones.\r\nThat's not mean that we should behave like them .\r\nI have seen the images of those terrorists .They would be less than 25 years.Can't u think why they are doing this when they know that they would surely be killed by soldiers and police.Remember in Delhi's blast the terrorist were students of Jamiya university .\r\nThe reason is they don't know what humanity is?They just became a destructive machine.\r\nIf we want to get rid of terror,we have to find the root of terror.\r\nOnce I was watching a video in which it was shown that how children nearly 10-15 years old are trained to become a terrorist.\r\nIf u really want to do something ,do something which can save these kids to become a terrorist.\r\n\r\nAt last I just wanna say that \"In real life , two negative never make a positive\".\r\nHopefully u would understand.", - "Solution_14": "[quote=\"ZetaX\"]The started violence and cruelity, so you are allowed to use it too\u00bf That's a rather dangerous and abandoned view, it just causes hate and an infinite circle of even more hate.[/quote]\nhmm, true. But I don't really understand their point of view. I mean they knew they would die, didn't they? Yes there is no point in cruelty. But we can't let them loose. They should be imprisoned, right? the few who are alive?\n\n[quote=\"ZetaX\"]And after your \"very reasonable interrogation techniques\" were used against them, the next step will be to use them against people you aren't even sure about their \"terroristic activities\". But after enough \"interrogation\", they surely will confess it.. You know, european witch hunting was very successfull, most of them confessed to be witches... (and the those who didn't confess were obviously guilty for collecting herbs or having birthmarks)[/quote]\nI never agree to \"interrogating\" ppl on the basis of suspecting. But things like who sent them and similar things should be asked. That's all I wanted to say.\n\n[quote=\"ZetaX\"]You didn't understand some implications of human rights; and mathlete's post is one of the stupidiest posts ... this ever again.[/quote]\r\nI never agreed to mathlete's post. I only replied to the quote of the previous post.\r\n\r\nP.S. Why are all your \"?\"s inverted?", - "Solution_15": "[quote=\"Akashnil\"]I mean they knew they would die, didn't they?[/quote]\r\n\r\nThese people were willing to die.\r\n\r\nLet's see what happened:\r\n\r\nAbout 15 terrorists hijacked 10 different places, including two luxury hotels (Taj and Oberoi), a Jewish center, and a hospital. They just ran in with AK-47's, hand grenades, and started randomly opening fire. They kept hostages, and they killed around 170 people, injuring around 350 more.\r\n\r\nWow, they even hijacked a police car and started shooting from there.\r\n\r\nBut I only got this from the media.\r\n\r\nAnd see this: HT said that there were 15 terrorists. Another news line (24-7?) said that there were 40. You can't clearly trust the media very much.", - "Solution_16": "[quote=\"ernie\"]These people were willing to die.[/quote]\r\nI know that. But my question is WHY?", - "Solution_17": "[quote=\"Akashnil\"][quote=\"ernie\"]These people were willing to die.[/quote]\nI know that. But my question is WHY?[/quote]\r\n\r\nThey are ready to give up their life, they hate the world, and want to destroy it. And how would I know?", - "Solution_18": "Probably they are brainwashed, similar to what happened in Nazi Germany, where children were brought up with an overwhelming deluge of pro-Nazi and anti-Jew propaganda. Terrorists can become fanatic in their beliefs from years of careful conditioning by their leaders to believe they are the saviors of the world. Even ordinary people can fall into this too.\r\nCase in point: the Crusades. The Europeans and the Muslims, on their \"war for religion\", probably never stopped to look at their opponents. If they did, they would have realized they were fighting a people who believed in basically the exact same religion.\r\nThe media is especially good at doing this. People have a tendency to trust everything they hear, as long as it's not contradicted. When the terrorists were in their camps in the Pakistani mountains or wherever, they only heard pro-Muslim and anti-Indian messages, thus leading them to believe they were fighting or a just cause. I am not trying to defend their acts, just explain them. If we can understand why these things occur then it will be easier to stop future occurrences.", - "Solution_19": "[quote=\"kabi\"]\nAt last I just wanna say that \"In real life, two negatives never make a positive\".\nHopefully u would understand.[/quote]\r\n\r\nThen what do we do, let them live so they can attack again?", - "Solution_20": "[quote=\"mathelete\"]\n\nThen what do we do, let them live so they can attack again?[/quote]\r\nDear why don't u understand ?\r\nThere are in millions of people who are in these type of ridiculous works.\r\nO.K. fine if we all start doing what u have said then will they stop all these.They will some other.\r\nThey all have lost their soul.\r\nAll of them who are involved in this should be punished but not in the way in which u want.\r\nIf u really want to help India destroy the roots of terrorism .\r\nThese terrorists group take children from their houses and give them training .Why can't u understand that why they are doing this?\r\nDon't waste ur energy in finding useless works .\r\nUr country need u .Do something which can really prevent this world from terrorism.\r\nI meant to a place where terrorists had made their home but I don't want ur to become the same.I too have lost my near ones in an attack.So my responsibility to stop these all has increased.\r\nI m really feeling sorry for u.", - "Solution_21": "[quote=\"Akashnil\"][quote=\"ernie\"]These people were willing to die.[/quote]\nI know that. But my question is WHY?[/quote]\r\n\r\nMost terrorist groups are started by religious extremists who interpret their religious texts in a rather twisted/irrational way (the most common idea being that violence is justified to eradicate other religions, to put it crudely). However, these leaders obviously need a large force behind them, so they turn to regions of extreme poverty. Usually, they recruit young boys with the lure of money/food/etc for their families. Of course, these young men see no other choice (they'll be taken in by force if they don't agree anyway), so they join the training camp, where they gradually become brainwashed (as someone said). They are literally void of emotions; they're highly trained to kill, even if it means giving up their own life.", - "Solution_22": "I guess I might be violating some of the \"The Round Table\" rules (reviving threads,etc), but I would like to pay homage to all the victims of this ghastly act of terrorism-- perpetrated by Pakistani terrorists. Today is the first anniversary of 26/11 Mumbai terror attacks-- one of the most horrific instances of terrorism the world has ever witnessed. \r\n\r\nOn a discursive note, I would like to note that one can't compare this attack to \"9/11\" (as the [b] OP [/b] had done) primarily because the extent of damage, the number of casualties etc pales in comparison with the \"9/11 attacks\". \r\n\r\n[b] May the souls of all victims rest in peace.[/b]", - "Solution_23": "Don't worry!Such things will continue till Pakistan is there!And we Indians have got used to it!They have been pushing terrorists into India for the last 60 years!\r\nIt took years for people in the west to wake up to the reality of Pak.Nobody cares till one is affected.And Obama talks of nuclear disarmament.", - "Solution_24": "I guess by this time (exactly an year ago!) the two terrorists had finished their massacre at the CST and were heading for Cama hospital. By this time, I guess several Americans and other foreigners were also taken hostage at the Taj. I once again pay homage to all the bereaved families of foreigners who were killed in this savage act of terrorism.\r\n\r\n[quote=\"nobelium\"]They have been pushing terrorists into India for the last 60 years! [/quote]\n\nYou're mistaken. They have been using terrorism as a state policy-- by starting insurgency in Kashmir from almost 1988-- the time by which they had or had developed the N-bomb. As soon as they had obtained the N-bomb-- the levels of insurgency in Kashmir started and increased dramatically over the years. Fomenting unrest in India was always one of their most important state policies. So, by having weapons of Mass Destruction in their arsenal-- they can always resort to nuclear blackmail-- after committing grave and savage outrages like 26/11.\n\n[quote=\"nobelium\"] It took years for people in the west to wake up to the reality of Pak.Nobody cares till one is affected.And Obama talks of nuclear disarmament.[/quote]\r\n\r\nQuite unfortunately, they're still sleeping! They're fighting the Al-Qaeda, but they're not realizing the grave threat posed by terrorist organizations like the Lashkar-e-Taiba (LeT) and others. They have to secure Pakistan's nuclear weapons (as quickly as possible)-- which pose an imminent threat to world peace and stability. Imagine a scenario in which a terrorist organization seizes these weapons. \r\n\r\nThey might not launch these weapons against the West, but the West should understand the fallout of a nuclear war and its effects on global climate. They should also understand the threat posed by nuclear proliferation by various rogues states or scientists. But, I opine that the U.S. should secure or wrest control of these weapons as soon as possible.\r\n\r\n[b] P.S.: [/b] All of my views/opinions are known facts. I don't intend to offend anyone.", - "Solution_25": "[quote=\"Gen8\"]\nYou're mistaken. They have been using terrorism as a state policy-- by starting insurgency in Kashmir from almost 1988-- the time by which they had or had developed the N-bomb. As soon as they had obtained the N-bomb-- the levels of insurgency in Kashmir started and increased dramatically over the years. Fomenting unrest in India was always one of their most important state policies. So, by having weapons of Mass Destruction in their arsenal-- they can always resort to nuclear blackmail-- after committing grave and savage outrages like 26/11.\n[/quote]\r\n\r\nThey pushed the villainous pathans into India immediately after partition.They massacred thousands.It too was a form of terrorism.Pathans were not part of regular army of Pak.\r\nBy the way,which country do you live in?", - "Solution_26": "[quote=\"nobelium\"] They pushed the villainous pathans into India immediately after partition.They massacred thousands.It too was a form of terrorism. [/quote]\r\n\r\nOkay, But we forgot that they instigated Sikhs too. (In the 80s) They fomented Sikh terrorism too. Also, I would concur with your assessment that these terror attacks will continue as long as Pakistan pursues its policy of sponsoring terrorism against its neighbours.", - "Solution_27": "Anyway,let's stop it here.No use being so harsh on them.", - "Solution_28": "[quote=\"nobelium\"]Anyway,let's stop it here.No use being so harsh on them.[/quote]\r\n\r\nWell, I was only stating the facts. (They're all facts!)", - "Solution_29": "I apologize for posting-- which contravenes [b] JBL's [/b] policies regarding the The Round Table. I had just read 'this' sometime ago and it's making me shiver and shudder.\n\nI will just post the gist of the 'news report' over here. A noted U.S. counter-terrorism official had said that if another attack akin to 26/11 were to occur in India, it could lead India and Pakistan to engage themselves in a nuclear war-- a hair-raising prospect indeed. (The newspaper is the link below is extremely well reputed in India; it ain't like the NY Post)\n\n[url]http://timesofindia.indiatimes.com/india/India-may-strike-terror-training-camps-post-Mumbai-II-US-counter-terrorism-expert/articleshow/6570351.cms[/url]\n\nSecondly, the fallout of a regional nuclear war would be devastating to the climate too.", - "Solution_30": "[quote=\"evilhamster\"]\nCase in point: the Crusades. The Europeans and the Muslims, on their \"war for religion\", probably never stopped to look at their opponents. If they did, they would have realized they were fighting a people who believed in basically the exact same religion.[/quote]\n\nL2crusades please...they were mainly to unite Europe politically and maintain the power of the Church. Europe was extremely unstable at the time and a shared goal helped to bring order to the numerous kingdoms." -} -{ - "Problem": "A runner hopes to complete the 10,000 meter run in less than 30.0 minutes. After exactly 27 minutes, there are still 1100 meters to go. The runner must then accelerate at 0.20 $ m/s^2$ for how many seconds in order to achieve the desired time?", - "Solution_1": "So far the runner has run 8900 metres in 27 \u00d7 60 = 1620 seconds; that\u2019s an average speed of $ \\frac {8900}{1620}\\approx5.49\\ \\textrm{m}\\,\\textrm{s}^{ \\minus{} 1}$. However, if he is to run the last 1100 metres in 180 seconds, he will need to do it at an average speed of no less than $ \\frac {1100}{180}\\approx6.11\\ \\textrm{m}\\,\\textrm{s}^{ \\minus{} 1}$. So if he accelerates for $ t$ seconds at $ 0.20\\ \\textrm{m}\\,\\textrm{s}^{ \\minus{} 2}$, we have\r\n\r\n$ 5.49 \\plus{} 0.20\\times t\\ \\geq \\ 6.11$\r\n\r\nNow solve for $ t$.", - "Solution_2": "Jane_Bennet, the runner can run the last 1100 meters in 179.999 seconds if he wishes...\r\n\r\nAlso, the minimum value of $ t$ in your inequality doesn't yield an average speed of 6.11", - "Solution_3": "[hide]\ncontinuing at the rate the runner had been going, \nhe/she would have finished 8900*10/9 meters, leaving\n1000/9 meters\n\nso now, you need to find what's the minimum t that\n$ 0.1t^2\\plus{}(180\\minus{}t)(0.2t) \\ge 1000/9$\n\nby quadratic, you get\n$ t \\ge 360 \\minus{} 80\\sqrt{11}$\n\nhence, the runner needs to accelerate for at least \n$ 360 \\minus{} 80\\sqrt{11} \\equal{} 94.67001 sec$ at a rate of 0.2m/s^2\n\n[/hide]", - "Solution_4": "I see what I did wrong. I ignored the fact that after the runner has accelerated for a time, the distance left to cover and the time left to do it in both change and so the average speed needed also changes. I\u2019ll try again.\r\n\r\nSuppose he accelerates for a time $ t$. His speed goes up from $ 5.49$ to $ 5.49 \\plus{} 0.20t$. The distance he runs in this time is $ 5.49t \\plus{} 0.10t^2$. Thus the distance left to go is $ 1100 \\minus{} 5.49t \\minus{} 0.10t^2$. If he stops accelerating and runs this remaining distance at constant speed, his total time for the last 1100 metres will be $ t \\plus{} \\frac {1100 \\minus{} 5.49t \\minus{} 0.10t^2}{5.49 \\plus{} 0.20t}$. Hence\r\n\r\n$ t \\plus{} \\frac {1100 \\minus{} 5.49t \\minus{} 0.10t^2}{5.49 \\plus{} 0.20t}\\ \\leq\\ 180$\r\n\r\nThat\u2019s what needs to be solved to find $ t$. Am I correct now? :oops:\r\n\r\nNB: $ 5.49$ is an approximation of $ \\frac{8900}{1620}$. The latter should be used to get a more accurate calculation.", - "Solution_5": "The answer is 3.41 seconds, so I'm not sure. :maybe:", - "Solution_6": "All right, let\u2019s solve the quadratic inequality I got:\r\n\r\n$ t \\plus{} \\frac {1100 \\minus{} 5.49t \\minus{} 0.10t^2}{5.49 \\plus{} 0.20t}\\ \\leq\\ 180$\r\n\r\nTo be more accurate, let\u2019s replace the 5.49 by $ \\frac{8900}{1620}\\equal{}\\frac{445}{81}$:\r\n\r\n$ t \\plus{} \\frac {1100 \\minus{} \\frac{445}{81}t \\minus{} 0.10t^2}{\\frac{445}{81} \\plus{} 0.20t}\\ \\leq\\ 180$\r\n\r\nThus\r\n\r\n$ \\frac{445}{81}t \\plus{} 0.20t^2 \\plus{} 1100 \\minus{} \\frac{445}{81}t \\minus{} 0.10t^2\\ \\leq\\ 180\\left(\\frac{445}{81}\\plus{}0.20t\\right)$\r\n\r\nIt looks horrible, but it can actually be simplified to\r\n\r\n$ 9t^2\\minus{}3240t\\plus{}10000\\ \\leq\\ 0$\r\n\r\nHence\r\n\r\n$ \\frac{3240\\minus{}\\sqrt{3240^2\\minus{}360000}}{18}\\ \\leq\\ t\\ \\leq\\ \\frac{3240\\plus{}\\sqrt{3240^2\\minus{}360000}}{18}$\r\n\r\nthat is to say\r\n\r\n$ t\\ \\geq\\ 3.11334$ (we can ignore the upper limit)\r\n\r\nSo my answer (to 2 d.p.) is 3.11 seconds (or 3.13 seconds if using 5.49 for the initial speed). I would say it\u2019s pretty close. :wink:", - "Solution_7": "Yes, Jane Bennet has the right answer." -} -{ - "Problem": "Hi..Who can guids me to a site or a n eBook that contains\r\nsolutions of excersices'Rudin book (real and complex analysis)\r\nthanks.", - "Solution_1": "I'm pretty sure that a complete set of solutions don't exist. Do them yourself.", - "Solution_2": "But I need it .. :o\r\nWhat dose I do then?\r\nThanks." -} -{ - "Problem": "Hi, all. I'm kind of a weirdo in that I really love contest math, but I also really love English and creative writing, specifically poetry. Of the following schools, which have strong undergrad pure math programs compared to say, U. Berkeley or Chicago?\r\n\r\n\r\nBrown\r\nDartmouth\r\nWilliams\r\nJohn Hopkins\r\nU. Chicago\r\nU. Michigan\r\nUPenn\r\nSwarthmore\r\n\r\nSince five of my school choices are already taken up, I'd like to narrow this list to 3 or fewer. Thanks in advance-TBAO", - "Solution_1": "Brown is known more for its applied math and computer science than pure math. The majority of my math professors were wonderful teachers, though. The writing courses are very popular and well supported, though. You can browse through some of the classes offered at mocha.cs.brown.edu. Sorry I can't say anything about the other schools.", - "Solution_2": "I'm a first year at U. Chicago, so I'll put in what I can.\r\n\r\nWe have a good program in pure math, with a lot of emphasis on theory. I'm very happy with my math class. I don't know much about our creative writing program, but you can check out http://creativewriting.uchicago.edu/.\r\n\r\nIIRC, Brown has an open curriculum. I think it's important to factor in the difference between Chicago's Core and Brown's open curriculum. I personally like the Core, but of course it's a matter of preference. I mainly like the Core because I think that (1) the analytic skills developed through the Core complement any field of study and (2) in the words of a friend, the Core forces me to read all the texts I wanted to read but may have set aside in favor of other passions like math.\r\n\r\nLike the last poster, sorry I can't say anything about the other schools.", - "Solution_3": "[quote=\"oab729\"]Hi, all. I'm kind of a weirdo in that I really love contest math, but I also really love English and creative writing, specifically poetry. Of the following schools, which have strong undergrad pure math programs compared to say, U. Berkeley or Chicago?\n\n\nBrown\nDartmouth\nWilliams\nJohn Hopkins\nU. Chicago\nU. Michigan\nUPenn\nSwarthmore\n\nSince five of my school choices are already taken up, I'd like to narrow this list to 3 or fewer. Thanks in advance-TBAO[/quote]\r\n\r\nWow I thought I was the only one. I love both math and poetry as well. :)" -} -{ - "Problem": "Team #3\r\n\r\nSolve for $x$: $\\log_2 (x+2) + 5 = 8 + \\log_2 x$.", - "Solution_1": "[hide]$\\log_2 (x+2) = \\log_2 8 + \\log_2 x = \\log _2 8x$\n\n$8x = x+2$\n\n$x = \\frac 27$[/hide]", - "Solution_2": "[hide]$\\log_2(x+2)+5=8+\\log_2x$\n$\\log_2(x+2)-\\log_2x=3$\n$\\log_2\\frac{x+2}{x}=3$\n$\\frac{x+2}{x}=8$\n\nSolving for $x$, $x=\\frac{2}{7}$.[/hide]", - "Solution_3": "I remember this problem, because I actually participated in NCIML this year. DOes anyone else here participate in it?", - "Solution_4": "I'm going to give you a detailed explanation even if you can't read it :diablo: \r\n\r\nokay:\r\n\r\nlog2 of (x+2) + 5= 8 + log2 of (x)\r\n\r\nlog2 of (x+2)-log2 of (x)=8-5\r\n\r\nlog2 of (x+2)/x=3\r\n\r\nThen exponentiate to get (x+2)/x=2^3\r\n\r\nFrom there, just cross multiply and get [b]x=2/7[/b]", - "Solution_5": "Everyone, You all forgot the condition for anti-logarithm. :D \r\n\r\nkunny", - "Solution_6": "[quote=\"kunny\"]Everyone, You all forgot the condition for anti-logarithm. :D \n\nkunny[/quote]\r\n\r\nHm? Can you expand on that statement? Never heard of that before.", - "Solution_7": "[quote=\"kunny\"]Everyone, You all forgot the condition for anti-logarithm. :D \n\nkunny[/quote]\r\n\r\nUm.. Not sure what you mean because that's what I got and it's what the site had too. ;)", - "Solution_8": "Here's what I found out what AntiLogarithm is:\r\n\r\nIt's the inverse function of a log defined as:\r\n\r\nlog b of (antilog b of z) = z = antilog b of (log b of z)\r\n\r\nThe antilog of base b of z is therefore $b^z$\r\n\r\nHowever, I don't see how we needed this, I think the solutions posted above should be allright :sleep2: :sleeping:", - "Solution_9": "Can anyone solve the following logarithmic equation and inequality?\r\n\r\n[1] $\\log_2(x-2)+\\log_2(7-x)=2$\r\n\r\n[2] $\\log_2(x-2)+\\log_2(7-x)\\leq2$", - "Solution_10": "[hide]\n$(x-2)(7-x) = 4$\n\n$(x-2)(7-x) \\le 4$\n\nThen solve.[/hide]", - "Solution_11": "Then what's your answer,tarquin.", - "Solution_12": "I don't get it...for the first one isn't the answer just $6$ or $3$... :?:", - "Solution_13": "Yep, at least that's what I got.", - "Solution_14": "second one is:\r\n$x<3 \\cup x>6$", - "Solution_15": "[quote=\"amirhtlusa\"]second one is:\n$x<3 \\cup x>6$ :( [/quote]", - "Solution_16": "Second one: $-x^2+9x-18\\leq 4$\r\n\r\nSo $x\\in(-\\infty,3]\\cup[6,\\infty)$", - "Solution_17": "Everyone,you forgot [color=blue]the condition for anti-logarithm[/color], that is to say the condition for which the function of logarithm can be defined. :)", - "Solution_18": "$x>2$ :?", - "Solution_19": "[quote=\"Slizzel\"]$x>2$ :?[/quote]\r\n\r\n :(", - "Solution_20": "[quote=\"kunny\"]Everyone,you forgot [color=blue]the condition for anti=logarithm[/color], that is to say the condition for which the function of logarithm can be defined. :)[/quote]\r\nBut it still doesn't affect Silverfalcon's problem, does it?", - "Solution_21": "Yes, you are right, but in case of logarithmic inequality, you can't skimp. ;)", - "Solution_22": "So the condition is $2< x< 7$, right?", - "Solution_23": "Yes. :)", - "Solution_24": "errr...\r\nyup kunny is right i forgot the defenition of anit logaritm here is the answer:\r\n$2\\le x\\le3 \\cup 6\\le x\\le 7$", - "Solution_25": "[quote=\"amirhtlusa\"]errr...\nyup kunny is right i forgot the defenition of anit logaritm here is the answer:\n$2\\le x\\le3 \\cup 6\\le x\\le 7$ :? :( [/quote]", - "Solution_26": "It doesn't work for $x=2$ and $7$", - "Solution_27": "yes it's right its open interval\r\n$x\\in (2,3]\\cup [6,7)$", - "Solution_28": "That's right. :)", - "Solution_29": "Yeah.. thanks kunny for reminding us about anti-logarithm", - "Solution_30": "Don't mention it. :)", - "Solution_31": "So everytime we deal with logarithms, we have to think of the anti-logarithms too? \r\n\r\nIs that the same with the restriction? \r\n\r\nHow come in the school, my teacher has never mentioned anything like anti-logarithms before? :( :(", - "Solution_32": "[quote=\"YooSam\"]So everytime we deal with logarithms, we have to think of the anti-logarithms too? \n\nIs that the same with the restriction? \n\nHow come in the school, my teacher has never mentioned anything like anti-logarithms before? :( :([/quote]\r\n\r\nYes. We should take into consideration the condition for which the logarithm can be defined, that is to say, \r\n\r\n[color=blue]anti-logarithm is always positive real number[/color] and moreover [color=red]the base is positive real number and not equal 1[/color].\r\n\r\nkunny", - "Solution_33": "[quote=\"kunny\"][quote=\"YooSam\"]So everytime we deal with logarithms, we have to think of the anti-logarithms too? \n\nIs that the same with the restriction? \n\nHow come in the school, my teacher has never mentioned anything like anti-logarithms before? :( :([/quote]\n\nYes. We should take into consideration the condition for which the logarithm can be defined, that is to say, \n\n[color=blue]anti-logarithm is always positive real number[/color] and moreover [color=red]the base is positive real number and not equal 1[/color].\n\nkunny[/quote]\r\n\r\nOh, that is what we call restriction :D . I never knew that people call it anti-logarithms :( . Thanks kunny." -} -{ - "Problem": "Suppose $ x,y,z$ are positive integers satisfying the equation \r\n\\[ \\frac{1}{x} \\minus{} \\frac{1}{y} \\equal{} \\frac{1}{z},\r\n\\] and let $ h$ be the highest common factor of $ x,y,z$.\r\n\r\nProve that $ hxyz$ is a perfect square. \r\n\r\nProve also that $ h(y\\minus{}x)$ is a perfect square.", - "Solution_1": "[hide=\"An Idea?\"]\nLet $ x,y,z \\equal{} hx',hy',hz'$. \n\nThen $ hxyz\\equal{}h^4x'y'z'$, which is a perfect square when $ x'y'z'$ is a perfect square. \n\nSimilarly, $ h(y\\minus{}x) \\equal{} h^2(y'\\minus{}x')$ which is a perfect square when $ y'\\minus{}x'$ is a perfect square.\n\n\nThen $ 1/x \\minus{} 1/y \\equal{} 1/z \\iff \\frac{y'\\minus{}x'}{x'y'z'} \\equal{} \\left( \\frac{1}{z'} \\right) ^2$. \n\nNow what?[/hide]", - "Solution_2": "[hide=\"Solution\"]Rewrite the equation as $ (z \\minus{} x)(z \\plus{} y) \\equal{} z^2$. Thus we conclude that $ z \\minus{} x \\equal{} m^2k, z \\plus{} y \\equal{} n^2k,z \\equal{} mnk$ for some integers $ m,n,k$ such that $ (m,n) \\equal{} 1$ ([url=http://www.mathlinks.ro/weblog_entry.php?t=201834]proof[/url]). That means $ (x,y,z) \\equal{} (mk(n \\minus{} m),nk(n \\minus{} m),mnk)$. Since $ (m,n) \\equal{} (m,n \\minus{} m) \\equal{} (n,n \\minus{} m) \\equal{} 1$, we conclude that $ k \\equal{} h$, and that $ hxyz \\equal{} (mnk^2(n \\minus{} m))^2$, $ h(y \\minus{} x) \\equal{} (k(n \\minus{} m))^2$, which concludes the proof.[/hide]" -} -{ - "Problem": "Tim has three times as many coins as Mike. If Tim gives one coin to Mike, Mike will\nhave a total number of coins equal to half the number of coins Tim started with. How\nmany total coins do Tim and Mike have together?", - "Solution_1": "let's say that Tim has 3x coins (so Mike has x coins)\r\n\r\n$ x\\plus{}1\\equal{}\\frac{3x}{2}$\r\n\r\nand $ 1\\equal{}\\frac{x}{2}$ so x is 2.\r\n\r\nso there are total 6+2=8 coins.\r\n\r\nanswer : 8" -} -{ - "Problem": "Let $R$ be an integral domain and $I\\subset R$ is a nonzero ideal. Show that $M=\\frac{R}{I}$ is not a flat module.\r\n\r\nI know that $M\\otimes_R \\frac{R}{I}\\cong\\frac{M}{IM}$. I just cannot think of an example.", - "Solution_1": "Generally, if $R/I$ was flat, then we would have an exact sequence $0\\to I\\otimes_RR/I\\toR\\otimes_RR/I\\to R/I\\otimes_RR/I\\to 0$.\r\nUp to isomorphisms, this is $0\\to I/I^2\\to R/I\\to R/I\\to 0$. It is not that hard to check that the map $R/I\\to R/I$ in this induced sequence is the identity map, so it is an isomorphism. But then $I/I^2=0$. This does not generally happen, but it is possible. For example if $p$ is a prime ideal in the integral closure of $\\mathbb Z$ in $\\mathbb C$.\r\n\r\nSo what I proved is that if $I^2\\neq I$ then $R/I$ is not a flat $R$-module.", - "Solution_2": "Assume that $R/I$ is flat.\r\nLet $J$ be an arbitrary ideal of $R$. Then we have a short exact sequence $0\\to J\\to R\\to R/J\\to 0$.\r\nUpon tensoring with $R/I$, according to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76587[/url], we have a short exact sequence \r\n$0\\to J/IJ\\to R/I\\to R/(I+J)\\to 0$. The maps are $J/IJ\\to R/I: j+IJ\\to j+I$ and $R/I\\to R/(I+J): r+I\\to r+(I+J)$. Since the first map must be injective, its kernel $\\frac{I\\cap J}{IJ}$ must be $0$. This happens iff $I\\cap J=IJ$.\r\nSo if $R/I$ is flat, then $I\\cap J=IJ$ for every ideal $J$ in $R$.\r\n\r\nLet $J=xR$ with $x\\in I$, $x\\neq 0$. Then $x\\in I\\Rightarrow J\\subset I$. Hence we must have $xR=xI$ for every $x\\in I$, $x\\neq 0$. $xR=xI\\Rightarrow x\\in xI$. Since $R$ is a domain, this implies $1\\in I$ which contradicts something which should have been in the hypothesis i.e. that $I$ is a proper (besides from being nonzero) ideal of $R$. After all what can be more flat than $0$ :D? except maybe when you exclude the $0$ case by definition." -} -{ - "Problem": "If $ n$ is an integer greater than 1 then prove that $ n$ doesn't divide $ 3^{n}\\minus{}2^{n}$", - "Solution_1": "Posted many time : \r\nIf n is a natural satisfy $ n|(a\\plus{}1)^n\\minus{}a^n$ then $ n\\equal{}1$\r\nSolution : Suppose $ n>1$ \r\nCall p is least prime of $ n$ then we have :\r\n$ p|\\gcd((a\\plus{}1)^n\\minus{}a^n,(a\\plus{}1)^{p\\minus{}1}\\minus{}a^{p\\minus{}1})\\equal{}(a\\plus{}1)^{\\gcd(n,p\\minus{}1)}\\minus{}a^{\\gcd(n,p\\minus{}1)}\\equal{}1$ \r\nSo $ p|1$ \r\nTradition!\r\nImply that $ n\\equal{}1$" -} -{ - "Problem": "Suppose I heat up a piece of Iron past the Curie temperature and hold a powerful magnet next to it. When it cools down, will it align to the guage field lines of the powerful magnet and form a magnet?\r\n\r\nThis problem has been plauging me for a while.", - "Solution_1": "Try it! :D" -} -{ - "Problem": "How many three-digit numbers satisfy the property that the middle\ndigit is the average of the first and the last digits?", - "Solution_1": "Let the digits be $ a$, $ b$, and $ c$, where $ b$ is the middle digit and $ a$ and $ c$ are the first and last digit respectively. \r\n\r\n$ a$ and $ c$ must have the same parity, since $ b$ needs to be an integer. There are 9 possibilities for $ a$, 5 for $ c$, and $ b$ depends on what $ a$ and $ c$ are, so just one possibility for $ b$. Thus, there are $ 9 \\cdot 5 \\cdot 1 \\equal{} \\boxed{45}$.", - "Solution_2": "[quote=\"Alcumus\"]...There are $5\\cdot 5=25$ odd-odd combinations for the first and last digits. There are $4\\cdot 5=20$ even-even combinations that do not use zero as the first integer... [/quote]\n\nI don't understand how these equations 5 times 5 and 5 times 4 were created and why they work. Could I please have some help?\n\nThank you.", - "Solution_3": "[quote=\"Python54\"][quote=\"Alcumus\"]...There are $5\\cdot 5=25$ odd-odd combinations for the first and last digits. There are $4\\cdot 5=20$ even-even combinations that do not use zero as the first integer... [/quote]\n\nI don't understand how these equations 5 times 5 and 5 times 4 were created and why they work. Could I please have some help?\n\nThank you.[/quote]\n[hide=\"Explanation\"]For the odd-odd combination, the first number can be $1$, $3$, $5$, $7$, or $9$, and same for the second number. That gives $5\\times5$ ways. For the even-even combination, the first number can be $2$, $4$, $6$, or $8$, but not $0$. The second number can be $0$, $2$, $4$, $6$, or $8$, a total of $4\\times5$ ways for the even-even combination.[/hide]", - "Solution_4": "Solution:\n[hide]Each hundred can only have 5 different combinations: take 100-199 for example. There are 5 different numbers: 111, 123, 135, 147, 159. If you do this for 200-299 you can see that there are also 5 numbers. So, since 100-999 has 9 different hundreds, 5*9 = 45.[/hide]", - "Solution_5": "This is the question that was answered in the Counting&Probability 2.2 Video." -} -{ - "Problem": "Let $ p\\geq3$ is a prime number and $ a_{1}, a_{2},..., a_{p \\minus{} 2}$ are positive integers such that : $ a_{k}$ and $ a_{k}^{k} \\minus{} 1$ don't divide by $ p$, for all $ k \\equal{} 1,2,...,p \\minus{} 2$. Prove that : \r\n$ \\forall m\\in N\\bigcap [1,p \\minus{} 1], \\exists S\\subset N\\bigcap [1,p \\minus{} 2], \\prod_{i\\in S} a_i\\equiv m (mod p)$", - "Solution_1": "[quote=\"conan6032\"]Let $ p\\geq3$ is a prime number and $ a_{1}, a_{2},..., a_{p \\minus{} 2}$ are positive integers such that : $ a_{k}$ and $ a_{k}^{k} \\minus{} 1$ don't divide by $ p$, for all $ k \\equal{} 1,2,...,p \\minus{} 2$. Prove that : \n$ \\forall m\\in N\\bigcap [1,p \\minus{} 1], \\exists S\\subset N\\bigcap [1,p \\minus{} 2], \\prod_{i\\in S} a_i\\equiv m (mod p)$[/quote]\r\nNobody can solve my problem??? :( . Pleases, help me :(" -} -{ - "Problem": "What is your predicted score/\r\n\r\nHow hard do you think its gonna be", - "Solution_1": "extremely hard LOL", - "Solution_2": "is it getting a lot harder each year?...cuz i just did the 2001 one and it was pretty easy...i mean...it wasnt so hard...lol", - "Solution_3": "Some people (including me) believe that odd years are harder than even years. So, it might be hard...", - "Solution_4": "LOL bryant what's ur answer for #2?\r\ni got 65 but i am not sure if it's correct \r\nsomeone in the discussion post got 56 :huh:", - "Solution_5": "65...lol...if ur not sure just put it into the equation [N/3]=[N/5]+[N/7]-[N/35]... :P", - "Solution_6": "It'll probably be same difficulty as last year's. It's made by same country...", - "Solution_7": "coments on your score???", - "Solution_8": "[quote=\"Tomaths\"]coments on your score???[/quote]\r\ndis is me first year...dun kno...probubly around 20...", - "Solution_9": "woah maybe a bit over confident (no offence)", - "Solution_10": "lol...i just skimmed through the 2002 competition...the last problem seems to be the easiest...and the first one is soooo hard(at least to me...)i dont get it...dont they put the problems in the order of increasing difficulty?cuz if the first problem is really hard then i might get really nervous and wont be able to solve any of the problems...i think generally i can solve 2-4 problems...", - "Solution_11": "So, the contest is tomorrow.\r\nGood luck to everybody writing APMO! :) \r\n\r\n[b]Please do not discuss the contest till Wednesday because the people in Asia are writing the contest on Tuesday. NOTE: Last yr, it said on the contest paper that you can't discuss the problems via internet till the problems are posted on the APMO website.[/b]\r\n\r\nMan, I am gonna bomb this contest. I was sick for the past week with a fever and I just got a little better now and probably my parents won't even allow me to write the olympiad. Life sucks. :(", - "Solution_12": "Sports psychology... As corny as it sounds, you're more likely to bomb it if you say you're going to bomb it. A gold medalist never has a doubt in his/her mind that they were going to win. If you end up writing it, you're talented, you'll destroy those questions. \r\n\r\nGood luck to everyone!", - "Solution_13": "[quote=\"bryant124\"]lol...i just skimmed through the 2002 competition...the last problem seems to be the easiest...and the first one is soooo hard(at least to me...)i dont get it...dont they put the problems in the order of increasing difficulty?cuz if the first problem is really hard then i might get really nervous and wont be able to solve any of the problems...i think generally i can solve 2-4 problems...[/quote]\r\n2002 is much easier relative to 2006, don't know why~maybe i'm just more used to the problems on 2002?\r\nand i agree, 1st question on 2002 is the hardest(4me) and 4th question is the easiest tho solved within a few steps.", - "Solution_14": "ya Good Luck! to all APMO writers tomorrow...\r\n\r\nP.S. rem u'll probably pwn!!!!", - "Solution_15": "Not yet. The problems must first appear on the official website:\r\nhttp://www.kms.or.kr/competitions/apmo/", - "Solution_16": "btw...when r the results gonna come out?", - "Solution_17": "They usually E-mail the results( your score and rank) around the beginning of May.", - "Solution_18": "....end of May?? wow they reli spend a loooong time on marking....\r\ni think i got 1, 3 and a part of 5.....but 3 seems too easy (!), so maybe i did wrong....and 5...if i was not wrong...it was a very surprising answer", - "Solution_19": "daaaaaaaaaaaaaaaamn!!!!!!! i cudnt sleep last night and got up and proved #4 in 10 minutes...and i was just 1 step away from solving it in the actual test, but then my head was messed up and i just thought it was impossible and gave up.................................. :ewpu: :wallbash: :wallbash_red: now im soooooo mad i can cut myself....oh nvm...i just did...", - "Solution_20": "Bah, I got #1 (with a really crappy solution that was about 2 pages long >.<), #2 (with a proof, but not a really good one >.<), and #5 without proof. I'm going to get at most 16, but even that's wishful thinking.\r\n\r\nIt beats what I did last year, though, I only got 12 last year (though the 2006 #5 was a joke...)", - "Solution_21": "the australian results are out, and it was rare for people to have scored 7s on question 1, in fact noone did, for q2, a mark was lost for not using the condition that AC>AB or something, which meant that there was only one possibly diagram, so most ppl got 6 for q2, lol everyone got 0 on q4, several ppl got 2 marks on q3 and 1 person solved q5.", - "Solution_22": "Wow...Australia is fast! \r\n\r\nYeah, for #1, I'm told you lost 1 point for not mentioning the integers could be negative. I missed this, and it seems alot of other people did too.", - "Solution_23": "lol...i mentioned the integers could be negative in #1, and i also used the condition that AB>AC(or maybe its the other way, i dont remember) in #2...damn...i should have believed in myself and not give up so fast...#4 was totally easy...", - "Solution_24": "$\\longrightarrow$ bryant: Well you still have two more years so..dont worry i guess...\r\n\r\nhow does australia have their results so fast?", - "Solution_25": "lol...yea...2 years is a reeeeeeally long time...but i dont know...my parents want to move to US... :( ...but theyr really slow in making decisions :P ...\r\n\r\nerrrrrrr...they still havent posted the problems...i wanna see their solution to #4...mine was just geometrically innovative... :rotfl:\r\n\r\nedit: the australians only got 0 or 2 on #3...so i guess there is a lot more to it than we thought...", - "Solution_26": "I'd hope I got that god dang number 4, and well, maybe I'm missing a few steps in the middle, but I'd believe I got parts of it. For number 2, grrr, only after carpo explained to me for 10 min i sorta got it.", - "Solution_27": "hey bryant, are you into kobe?\r\nman i love him.", - "Solution_28": "[quote=\"sen\"]hey bryant, are you into kobe?\nman i love him.[/quote]\r\nlol...yea...huge fan...btw...we can discuss apmo now...i feel my solution to #4 is better than the official solution(which was just tooooo long and boring to read...)...btw i didnt prove it until recently...", - "Solution_29": "more easy than the other years.. i solved the first 4 problems in one hour.\r\n\r\n(i'm not contestant, Brazil aren't participant in APMO :( )" -} -{ - "Problem": "Find the largest real number $k$ such that:\r\n$a^3+b^3+c^3+kabc\\ge \\frac{1}{9}+\\frac{k}{27}$ $\\forall a,b,c\\ge 0$ and $a+b+c=1$", - "Solution_1": "$k=\\frac{15}{4}$.", - "Solution_2": "a generalization result:[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=67915]www.mathlinks.ro/Forum/viewtopic.php?t=67915[/url] ;)", - "Solution_3": "Have you solved the n-variables case?", - "Solution_4": "This seems t be a stupid question: what's the basic of choosing $a=0,b=c=\\frac12 \\implies k\\le \\frac{15}{4}$ (this variable is true), while $a=b=0,c=1 \\implies k \\le 54$, - false?", - "Solution_5": "This is old inequality and easy problem\r\n$k=\\frac{15}{4}$", - "Solution_6": "[quote=\"yptsoi\"]Have you solved the n-variables case?[/quote]\r\nHave you asked me ? ;)" -} -{ - "Problem": "Jeff decides to play with a Magic 8 Ball. Each time he asks it a question, it has a 2/5 chance of giving him a positive answer. If he asks it 5 questions, what is the probability that it gives him exactly 2 positive answers?", - "Solution_1": "This is $ \\binom{5}{2} \\times \\left (\\frac{2}{5} \\right )^2 \\times \\left (\\frac{3}{5} \\right )^3\\equal{}\\frac{10 \\times 4 \\times 27}{3125}\\equal{}\\frac{2 \\times 4 \\times 27}{625}\\equal{}\\boxed{\\frac{216}{625}}$." -} -{ - "Problem": "10 couples to be seated on a row of 20 seats.\r\nfind the number of arrangement such that no spouses sit next each other.\r\n\r\nthis sounds easy. but it got quite irritating when i got stuck with it....\r\nso far the only solution i could think of is to write a program and let the computer solve it.\r\nanyone got the solution?\r\n\r\nthxns and cheers.", - "Solution_1": "[hide=\"probably wrong...\"]\n$10! (10!-9!) * 2^{10}= 12135803387904000$ \nLol, am I just stupid or is this problem really not as easy as it looks like on the first sight?? :roll: \n[/hide]", - "Solution_2": "So, I'm not at all sure how the question ended up following the (wrong) answer, and I really don't want to think too hard about it. This sequence exists in the [url=http://www.research.att.com/~njas/sequences/]Online Encyclopaedia of Integer Sequences[/url], a wonderful resource. The best formula you're likely to find for it is the one that you get if you try to apply the principle of inclusion-exclusion, which gives a sum that probably is expressible in closed form via Bessel functions or hypergeometric functions, but probably not via elementary functions.", - "Solution_3": "well, would someone mind posting a complete solution? :)" -} -{ - "Problem": "I hope this topic is not posted here yet.\r\n\r\n\r\n\r\nIdentify $ x!$ as $ \\Gamma(x\\plus{}1)$. Then, does the equality\r\n\r\n$ \\int_{\\minus{}\\infty}^{\\infty} \\binom{n}{x} \\, dx \\ \\equal{} \\ 2^n$\r\n\r\nhold for a nonnegative real number $ n$? (where $ \\binom{n}{x} \\equal{} \\frac{n!}{x!(n\\minus{}x)!}$ is the binomial coefficient)\r\n\r\n\r\nI proved that it holds for $ n$ nonnegative integer, but the methid I used seems unable to be extended to the general case.", - "Solution_1": "According to numerical tests the formula seems correct for $ n\\in\\mathbb{R^\\plus{}}$.\r\nA partial result: if $ \\phi(n)\\equal{}\\int_{\\minus{}\\infty}^{\\infty}\\binom{n}{x}dx$ for $ n>0$, then $ \\phi(n\\plus{}1)\\equal{}2\\phi(n)$.\r\nProof: $ \\forall{y}\\in\\mathbb{R},\\phi(n)\\equal{}\\int_{\\minus{}\\infty}^{\\infty}\\binom{n}{x\\plus{}y}dx$; moreover $ \\forall{x,y}\\in\\mathbb{R},\\binom{n}{x}\\plus{}\\binom{n}{x\\plus{}1}\\equal{}\\binom{n\\plus{}1}{x\\plus{}1}$. Thus $ \\phi(n)\\plus{}\\phi(n)\\equal{}\\phi(n\\plus{}1)$. QED", - "Solution_2": "It is actually much better than \"a partial result\". Note that the identity is true for complex $ n$ with positive real part as well, so, when we multiply the integral by $ 2^{ \\minus{} z}$, we get an analytic function that is $ 1$-periodic in the right half-plane and, therefore, can be extended to the entire plane. Now all we need to do is to show that this function grows not too fast when we go up the imaginary axis (any estimate short of $ e^{\\pi|z|}$ will suffice) to conclude that it is actually constant. This reduces the problem to the [b]estimation[/b] of the integral instead of its [b]exact computation[/b]. I have no time to do it now but I hope it is not too difficult :)." -} -{ - "Problem": "consider the following scenario: let's say the AMC 12B has more qualifiers for AIME than AMC 12A. However, only like 20-50 people got more than 130 on the 12B. Ignoring any differences there might be between AIME and alternate AIME, is this possible to produce two different USAMO cut off based on the AMC 12 test??\r\n\r\nOr even, let's say AIME and alternate AIME has significant difficulty gap too, will there be 4 different USAMO cut off??\r\n\r\nBecause I (actually the So. Cal. geniuses and me) thought it is a lot harder to get high scores on the 12B, but the first 11-14 questions are easier to answer.", - "Solution_1": "Yes." -} -{ - "Problem": "$ \\sum \\frac{\\cos \\sqrt{n}}{n}$\r\nconverge or .. ?", - "Solution_1": "i have a feeling that it converges, but showing it is an all different matter, obviously. :( \r\n\r\npresume dirichlet should be of help here, i.e. given $ a_n,\\;b_n\\,\\in\\mathbb R,\\;\\;\\;\\text{where}\\;\\;\\;n\\in\\mathbb N$, \r\nif $ a_n\\to 0$ as $ n\\to\\infty$ and if $ S_{partial}\\equal{}\\sum_{n\\equal{}1}^m\\;b_k$ is bounded, then $ \\sum_{n\\equal{}1}^\\infty\\;(a_n\\cdot b_n)$ converges.\r\n\r\n\r\nlet $ a_n\\equal{}\\frac{1}{n}$. clearly, $ a_n\\to 0$ as $ n\\to\\infty$.\r\nnow we look at $ S_{partial}\\equal{}\\sum_{n\\equal{}1}^m\\;\\cos\\,\\sqrt n$.\r\n\r\nshowing that the sequence $ S_{partial}$ has bounded partial sums will now require some thinking.... hmmm.... not easy! :|", - "Solution_2": "this is main moment . I don't agree that this sum $ \\sum \\cos \\sqrt {n}$ is bounded. it's enough to show that $ \\sum \\cos \\sqrt {n} \\equal{} O(n^{\\delta})$ for $ \\delta < 1$ \r\nor $ |\\sum e^{2\\pi i \\sqrt {n} x }|$ is $ O(n^{\\delta})$ but here doesn't work some trick which used when we considered $ \\sum \\frac {\\sin n^{2}}{n}$ there was $ k^{2} \\minus{} m^{2} \\equal{} (k \\minus{} m)(k \\plus{} m) ...k \\minus{} m \\equal{} h...$ but here I can't do nothing $ \\sum e^{...\\sqrt {k} \\minus{} \\sqrt {m}...}$\r\nor generalization for like this sum will be next :\r\nhow to find asymptotic of this \r\n$ |\\sum e^{i f(n) x}|$ where $ x$ is liovillian. :huh:\r\nhere is another problem :\r\n$ \\sum \\frac{\\cos 2^{n}}{n}$ I suppose that here we can use something if we consider $ \\sum e ^{...(2^{k}\\minus{}2^{m})...}$", - "Solution_3": "compare it with the integral of $ \\frac{\\cos( \\sqrt{x})}{x}$. This fonction doesn't oscillate too much, so it will work :)", - "Solution_4": "$ f(x)\\equal{}\\frac{\\cos(\\sqrt{x})}{x}$\r\n\r\nwe have \r\n\r\n(1) $ \\int_{1}^{N\\plus{}1}f\\equal{}\\sum_{n\\equal{}1}^{N}f(n) \\plus{} \\sum_{n\\equal{}1}^{N}a(n)$\r\n\r\n$ a(n)\\equal{}\\int_{n}^{n\\plus{}1}(n\\plus{}1\\minus{}x)f'(x)dx$ \r\n\r\nbut \r\n$ |a(n)|\\leq \\int_{n}^{n\\plus{}1} |f'| \\leq \\frac{1}{2n^{1.5}}\\plus{}\\frac{1}{n^2}$\r\n\r\nthen \r\n\r\n(2) $ \\sum a(n)$ is converge absolutly\r\n\r\n(1) and ( 2) gives $ \\sum f(n)$ is the same nature of $ \\int_{1}^{\\plus{}\\infty}f$ \r\n\r\nthe last integral converge (change variable u^2=x )", - "Solution_5": "yep, if a fonction does not oscillate too much i.e. $ \\int |f'| < \\infty$, then $ \\sum f(n)$ and $ \\int f(x)dx$ have the same behavior.." -} -{ - "Problem": "Let $ n \\geq 3$ be an odd integer. We denote by $ [\\minus{}n,n]$ the set of all integers greater or equal than $ \\minus{}n$ and less or equal than $ n$.\r\nPlayer $ A$ chooses an arbitrary positive integer $ k$, then player $ B$ picks a subset of $ k$ (distinct) elements from $ [\\minus{}n,n]$. Let this subset be $ S$.\r\nIf all numbers in $ [\\minus{}n,n]$ can be written as the sum of exactly $ n$ distinct elements of $ S$, then player $ A$ wins the game. If not, $ B$ wins.\r\nFind the least value of $ k$ such that player $ A$ can always win the game.", - "Solution_1": "k>n+1, since if k{\\equal{}}\\int_0^1{f(x)g(x)dx}$. The seeked polynomial $ q$ is the orthogonal projection of $ g$ over $ \\mathbb{R}_n[x]$. If you want the true equations then use the Lagrange's polynomials $ L_n$ (orthogonal system for this scalar product but over [-1,1]). cf. wikipedia.\r\nSketch of the result:\r\n$ \\forall{i}\\equal{}\\{0,\\cdots,n\\}: {\\equal{}0}$, with $ P_i(x)\\equal{}L_i(2x\\minus{}1)$.", - "Solution_2": "[quote=\"loup blanc\"]If you want the true equations then use the Lagrange's polynomials $ L_n$ (orthogonal system for this scalar product but over [-1,1]).[/quote]\r\nFor a finite interval and a constant weight, the orthogonal system would be an appropriately scaled version of the Legendre polynomials - at least that's the name I know them by." -} -{ - "Problem": "Let $ u_n$ satisfying $ u_0 > 0,u_{n + 1} = u_n - e^{\\frac {- 1}{u_n^2}}$\r\nProve that $ \\limits_{n\\to \\infty} (u_n^2 ln n) = 1$", - "Solution_1": "[quote=\"Allnames\"]Let $ u_n$ satisfying $ u_0 > 0,u_{n + 1} = u_n - e^{\\frac { - 1}{u_n^2}}$\nProve that $ \\limits_{n\\to \\infty} (u_n^2 ln n) = 1$[/quote]\r\nNo help for me :o :o :o", - "Solution_2": "I think It has some problems!\r\n$ (U_n)$ is increase,don't exeist limit -> $ (U_{n}^2)\\minus{}>\\infty$ and $ ln_n \\minus{}>\\infty$\r\n->don't exist limit of your sequence! :wink:", - "Solution_3": "[quote=\"Allnames\"]No help for me :o :o :o[/quote]\r\nIt would have had a better chance had you posted it where it belongs. You could try asking a moderator to move it to the calculus-analysis section.", - "Solution_4": "[quote=\"Math pro\"]I think It has some problems!\n$ (U_n)$ is increase,don't exeist limit -> $ (U_{n}^2) \\minus{} > \\infty$ and $ ln_n \\minus{} > \\infty$\n->don't exist limit of your sequence! :wink:[/quote]\nthank for your comment but my proof is true,and your idea is clearly wrong try it with the sequence $ x_n \\equal{} \\frac {1}{n}$\n[quote=\"Kent Merryfield\"]It would have had a better chance had you posted it where it belongs. You could try asking a moderator to move it to the calculus-analysis section.\n\n[/quote]\r\nthanks,but here is only a problem in a contest,and we must kill it by only all thing that we learned\r\nNow .I proved it but my proof is really and used much hard knowledge ,,who can help me", - "Solution_5": "It is clearly wrong?I don't think so. with $ U_n\\equal{}\\frac{1}{n} \\minus{}> (U_n)$ have limit -> not satisfy your problem!\r\nI can't find any mistake inmy ideal!I explained you in my class but why you don't understand?\r\nI know that you r problem from a book!Can you detail the solution from the book? :maybe:", - "Solution_6": "Uhm,ok,I have just check carefully your idea,and I knew where your mistake is\r\nCan you detail why $ U_n$ doesn't have limit,I know it is increase,and you told me if exist $ c$ satisfies\r\n$ U_n\\ge c$ for all $ n$,we had the wrong thing\r\nBut you should note $ lim_{n\\to \\infty } u_n\\equal{}0$ and it is a small part in my ugly proof(it is not obivious,and to proof it is quite hard)", - "Solution_7": "but if $ (U_n)$ exist limit $ L \\minus{} > L \\equal{} L \\minus{} e^{ \\minus{} \\frac {1}{L^2}} \\minus{} > e^{ \\minus{} \\frac {1}{L^2}} \\equal{} 0$ Is it true? :wink:", - "Solution_8": "[quote=\"Math pro\"]but if $ (U_n)$ exist limit $ L \\minus{} > L \\equal{} L \\minus{} e^{ \\minus{} \\frac {1}{L^2}} \\minus{} > e^{ \\minus{} \\frac {1}{L^2}} \\equal{} 0$ Is it true? :wink:[/quote]\r\nwrong,note $ L\\equal{}L\\minus{}e^{lim \\minus{}\\frac{1}{u_n^2}}$ and if $ lim u_n\\equal{}0$ it is until true.To prove it,we only need Lagrangre theorem,only a few lines" -} -{ - "Problem": "Solve for all real x: (2x^2 x +4)^2 6(2x^2 x) = 19.\r\n\r\nMY bad...", - "Solution_1": "Isn't it just straightforward algebra? \n\n\n\nthe answer I'm getting...\n\n[hide]unless there's a trick the answer I'm getting is -3[/hide]", - "Solution_2": "ya, i dink so", - "Solution_3": "To solve this one:\n\n\n\n[hide]let 2x^2-x=y.\n\nThen, you get a quadratic for y. \n\nAll downhill from there[/hide]", - "Solution_4": "[hide]\n\nLet a = 2x:^2:-x\n\n\n\n(a+4):^2:-6a = a:^2:+2a+16 = 19\n\na:^2:+2a-3 = 0\n\n(a+3)(a-1)\n\n\n\na= -3, 1\n\n\n\n2x:^2:-x = -3\n\n2x:^2:-x-3 = 0\n\n(2x-3)(x+1) = 0\n\n\n\n2x:^2:-x=1\n\n2x:^2:-x-1 = 0\n\n(2x+1)(x-1) = 0\n\n\n\nWe should check for extraneous roots, but I'm too lazy, and at the moment too tired to care if I'm wrong.\n\n\n\nx = 3/2, -1, 1, -1/2\n\n\n\nThat was how I did it..\n\n\n\n[/hide]" -} -{ - "Problem": "In a sequence, each term is obtained by calculating the sum of the preceding two terms. the eighth term is 81, and the sixth term is 31. What is the fourth term?", - "Solution_1": "It's a Fibonacci-like sequence. To calculate the seventh term, we have $ 31 \\plus{} x \\equal{} 81 \\implies x \\equal{} 50$. Working backwards, we find that the fifth term is $ 19$, and the fourth term is $ 31 \\minus{} 19 \\equal{} \\boxed{12}$." -} -{ - "Problem": "What is the value of the sum $ 2^{\\minus{}1}\\plus{}2^{\\minus{}2}\\plus{}2^{\\minus{}3}\\plus{} \\ldots \\plus{}2^{\\minus{}9} \\plus{} 2^{\\minus{}10}$? Express your answer as a common fraction.", - "Solution_1": "hello, after a short computation we get\r\n$ \\sum_{i\\equal{}1}^{10}2^{\\minus{}i}\\equal{}\\frac{1023}{1024}$\r\nSonnhard.", - "Solution_2": "In these types of problems, the numerator will be 1 less than the denominator. Thus, the denominator for this solution will be 1024, and the numerator is 1023. Therefore, $ {%Error. \"division\" is a bad command.\n{1023}/{1024}} $", - "Solution_3": "[hide=\"Geometric sequence solution\"]\n\nThe sum of geometric sequences is as follows:\n\n$S_n=a\\left({\\frac{1-r^n}{1-r}}\\right)$, where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.\n\nPlugging things in:\n\n$S_{10}={\\frac{1}{2}}\\left({\\frac{1-\\left(\\frac{1}{2}\\right)^{10}}{1-{\\frac{1}{2}}}}\\right)$\n\nSimplifying a lot:\n\n$S_{10}=1-{\\frac{1}{1024}} \\implies \\boxed{{\\frac{1023}{1024}}}$[/hide]", - "Solution_4": "[quote=\"HYP135peppers\"]In these types of problems, the numerator will be 1 less than the denominator. Thus, the denominator for this solution will be 1024, and the numerator is 1023. Therefore, $ {%Error. \"division\" is a bad command.\n{1023}/{1024}} $[/quote]\n\n[hide=\"Proof\"]\nWe wish to find the value of $2^{-1}+2^{-2}+\\cdots+2^{-n}$.\n\n[quote=\"newb\"]\nThe sum of geometric sequences is as follows:\n$S_n=a\\left({\\frac{1-r^n}{1-r}}\\right)$, where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.\n[/quote]\n\nPluggin in values:\n\n\\[ S_n={\\frac{1}{2}}\\left({\\frac{1-\\left(\\frac{1}{2}\\right)^{n}}{1-{\\frac{1}{2}}}}\\right) .\\]\n\nThis becomes \\[S_n=\\dfrac{\\frac{2^n-1}{2^n}}{2\\cdot\\frac{1}{2}}.\\] The denominator of the fraction becomes $1$, so we are left with \n\\[\\boxed{S_n=\\dfrac{2^n-1}{2^n}}.\\]\n[/hide]" -} -{ - "Problem": "I've been trying to fax in my USAMTS solutions all evening today and the line was busy. Is it possible that they would allow me to fax them tomorrow? (if I don't succeed within the next hour)", - "Solution_1": "They are due at midnight tonight (Pacific Time)." -} -{ - "Problem": "An icosidodecahedron is a polyhedron with twenty triangular faces and twelve pentagonal faces. Each vertex of an icosidodecahedron is identical with two triangles and two pentagons meeting at each vertex. What is the sum of the number of vertices and the number of edges in an icosidodecahedron?", - "Solution_1": "Say that name fast three time........", - "Solution_2": "[hide]Well.... there's a trick:\n\nThere's a formula (maybe Pascal's theorem???)out there that says\" Verticies + Faces - edges = 2\"\nObviously, there are 20 + 12 = 32 faces.\nOver counting the edges, we see that there is 20*3 + 12*5 = 120 edges (that's a bit too much). Since every edge is shared by two figures, we should divide 120 by 2 (60). plugging this in:\n\nVerticies + 32 - 60 = 2\n There must be 30 verticies\n\n30 + 60 = [b]90[/b][/hide]", - "Solution_3": "[hide]i think its eulers formula , not pascal's where if v = vertices, e = edges, f = faces, V+F=E+2[/hide]\r\n\r\nicosadodecahedron,icosadodecahedron,icosadodecahedron\r\n\r\nThere, I said it three times fast", - "Solution_4": "[quote=\"philB\"][hide]Well.... there's a trick:\n\nThere's a formula (maybe Pascal's theorem???)out there that says\" Verticies + Faces - edges = 2\"\nObviously, there are 20 + 12 = 32 faces.\nOver counting the edges, we see that there is 20*3 + 12*5 = 120 edges (that's a bit too much). Since every edge is shared by two figures, we should divide 120 by 2 (60). plugging this in:\n\nVerticies + 32 - 60 = 2\n There must be 30 verticies\n\n30 + 60 = [b]90[/b][/hide][/quote]\r\nEuler's Formula, not Pascal's (I think)", - "Solution_5": "[hide]first there are 20*3+12*5=120 edges without restrictions, but we overcounted in the polyhedron. so divide by 4 since each edge was counted twice to get 30. there are also 120 edges throughout the polygons, each edge is counted twice, so divide by 2 to get 60. 60+30=90[/hide]", - "Solution_6": "[hide]Similiarly, each vertex is counted four times.... there are 120 verticies UNRESTICED, divided by four equals 30....[/hide]", - "Solution_7": "Extension: how many space diagonals does it have?", - "Solution_8": "are space diags where you connect vertices?\r\n\r\nhow do you figure out how many of the vertices lie on pentagons?\r\n\r\njorian[/hide]", - "Solution_9": "[quote=\"13375P34K43V312\"]Extension: how many space diagonals does it have?[/quote]\r\nIsn't that just $\\binom{n}{2}-5k-e$ where n is the number of vertices, k is the number of pentagons, e is the number of edges?", - "Solution_10": "Yeah, that would be right.\r\n\r\nAs for the solution, we would first select 2 points, which is where the ${n\\choose2}$ comes from, then subtract all of the edges, because those are generated from selecting 2 points, but don't count as space diagonals, which is where the $-e$ comes from. Finally, we have surface diagonals, the diagonals on the pentagons. There are five diagonals per pentagon, so $-5k$.", - "Solution_11": "wow, the solution is pretty easy once you know how to calculate it\r\n\r\njorian" -} -{ - "Problem": "how many kind of dices can be made, ennnumerating the faces of a cube with the numbers 1 2 3 4 5 6 without restrictions? and how many with the restriction that the sum of the numbers in opposite sides is 7?", - "Solution_1": "any solution to this problem?", - "Solution_2": "For the first question there are 30 dices. for the second there are only 2.\r\n\r\nIn this case it is actually very easy to write down an exhaustive solution. In general there are enumerative methods that take advantage of the underlying symmetries (like Polya enumeration theorem)", - "Solution_3": "For the distribution of the numbers we can do it in $ 6!$ ways, and we have to divide it by the number of rotations of the dice, which is $ 6 \\times 4$ because we can take any of the $ 6$ faces appearing up, and for such position we have the four rotations of the lateral sides. So we have $ 6! / 24 \\equal{} 30$, is this correct argument?" -} -{ - "Problem": "\u0394\u03b5\u03af\u03c4\u03b5 \u03c3\u03c4\u03bf \u03c3\u03c5\u03bd\u03b7\u03bc\u03bc\u03ad\u03bd\u03bf \u03b1\u03c1\u03c7\u03b5\u03af\u03bf .", - "Solution_1": "\u039fk,thanks.\u03a4\u03bf \u03ba\u03b1\u03c4\u03b5\u03b2\u03b1\u03c3\u03b1.....\r\n\r\n\u03a1\u0391\u039b\u039b\u0397\u03a3 :D" -} -{ - "Problem": "Prove that if there exists a linear map $T$ on $V$ whose null space and range are both finite-dimensional, then $V$ is finite-dimensional.", - "Solution_1": "[hide=\"My Solution\"]\nI let the nullspace be denoted by $N(T)$ and the range be denoted by $V(T)$. Let $\\dim N(T) = n$, $\\dim V(T) = m$. \n\nAssume that $V$ is infinite-dimensional. Let $e_1,...,e_n$ be a basis for $N(T)$. Since V is infinite-dimensional we can extend this to $e_1,\\cdots,e_n,e_{n+1},\\cdots,e_j$ linearly independent vectors in $V$. \n\nI will show that $T(e_{n+1}), \\cdots , T(e_j)$ are linearly independent. If\n$0 = x_{n+1}T(e_{n+1}) + \\cdots + x_jT(e_j) = T(x_{n+1}e_{n+1} + \\cdots + x_je_j).$\nthen the vector $x_{n+1}e_{n+1} + \\cdots + x_je_j$ belongs to $N(T)$. This is only possible if $x_{n+1} = \\cdots = x_j = 0$, because a vector $x_1e_1 + \\cdots + x_je_j$ belongs to $N(T)$ if and only if $x_{n+1} = \\cdots = x_j = 0$.\n\nWe can choose $j$ so that we get more than $m$ linearly independent vectors $T(e_{n+1}), \\cdots , T(e_j)$. A contradiction.\n[/hide]\r\n\r\nIs this correct? Is there a more constructive proof?", - "Solution_2": "The dimension formula $\\dim(V) = \\dim(ker(T)) + \\dim(Im(f))$ holds for [i]all[/i] vector spaces $V,W$ and all linear maps $V \\to W$. It's just an equation of cardinal numbers, and can be proven like the popular one with the needless restriction to finite-dimensional spaces. From here, the claim is obvious.", - "Solution_3": "Yes, but I don't know about cardinalities or anything. My definition of a infinite-dimensional space is that it cannot be generated by a finite amount of vectors.\r\n\r\nI'd still like to see if my proof holds up or if I forgot something.." -} -{ - "Problem": "The real polynomial p(x) = ax3 + bx2 + cx + d is such that |p(x)| <= 1 for all x such that |x| <= 1. Show that |a| + |b| + |c| + |d| <= 7.", - "Solution_1": "a solution at:\r\nhttp://www.kalva.demon.co.uk/short/soln/sh965.html" -} -{ - "Problem": "Hi,\r\n\r\nI would just like to know when the new and improved Alcumus is coming out. I've been reading that you can choose what type of problems you want.\r\n\r\nThanks,\r\n\r\nrts2007", - "Solution_1": "We will announce any future versions in the sticky post area when we are aware of a timeframe." -} -{ - "Problem": "y is it called that? i mean its not really partial is it?", - "Solution_1": "Because you're taking the derivative with respect to one variable of many.", - "Solution_2": "That method is used in Multivariable Calculus, when you have multiple independent variables and one dependent variable, in which when you do derivatives, you can only do it with one at a time, so that's why it's called partial differentiation.", - "Solution_3": "well, well i didn't know you could do gauss1181 could do calculus!" -} -{ - "Problem": "let $ \\{a_i\\} \\in (R^\\plus{})^n$ s.t. $ \\sum_{i\\equal{}1}^n{a_i^p}\\equal{}n$\r\n\r\nshow that $ \\sum_{i\\equal{}1}^n{a_i^r} \\le \\sum_{i\\equal{}1}^n{a_i^s}, \\forall 0 Inequalities forum.", - "Solution_5": "it is wrong if you take $ p \\equal{} 6,r \\equal{} 2,s \\equal{} 5,n > 81$..", - "Solution_6": "Are you guys sure this is wrong? It appears to me that this is just the power mean inequality: $ 0 < r < s$ implies $ \\bigg(\\sum a^r\\bigg)^{1/r} \\le \\bigg( \\sum a^s\\bigg)^{1/s}$. If we raise both sides of this to the $ s$th power, we have $ \\sum a^r \\le \\bigg(\\sum a^r\\bigg)^{s/r} \\le \\sum a^s$...\r\n\r\nPerhaps the problem arises when $ \\sum a_i^r < 1$ occurs because then removing the $ s/r$ power increases the size of the term instead of decreasing it as usual.\r\n\r\nEDIT: just noticed the pdf I'll read through it later..." -} -{ - "Problem": "Hi,\r\nAny one can point me to good logarithm documention site /book or what ever as long as it going to help me to pass my final exam on LOGARITHM ?\r\n\r\nthanks", - "Solution_1": "Hi,\r\nWhat is I need it some good problms with solution.\r\nthnaks", - "Solution_2": "You should check [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25616]this[/url] out.\r\n\r\nAlso, search the forum. Theres been a lot of different problems with logs.\r\n\r\nFinally, you could google it." -} -{ - "Problem": "I'm absolutely not convinced that this is original or correct or..., but I think it's quite nice and deserves some discussion. I hesitated before posting it in the analysis forum, but knowing what is happening on the linear algebra forum (where a virus, worse than Robertux, is trying to invade all the given space...) and that my solution fully uses analysis, I think I'll post it here: \r\n Let $ K$ be a convex compact set of $ R^3$, with nonempty interior. Suppose that there are infinitely many invertible linear operators leaving $ K$ invariant. Then there is an inner product on $ R^3$ such that $ K$ has a rotation axis for this inner product.", - "Solution_1": "The group $ G$ which fixes $ K$ (as a set, not pointwise) is closed and bounded, hence compact. After conjugation with some invertible linear transformation, we can assume it's orthogonal. After this modification the problem is reduced to showing that some rotation around a line fixes $ K$. We can even show that $ K$ is fixed by the reflection in a line through the origin (i.e. a rotation of $ 180^\\circ$ around a line). \r\n\r\nWe know $ G$ is infinite, so it must be a compact Lie subgroup of $ O(3,\\mathbb R)$ of dimension $ \\ge 1$. As such, it contains a torus of dimension $ \\ge 1$, so, in particular, its identity component contains an element of order $ 2$. This element must be a reflection with respect to a line through the origin, because it belongs to the identity component of $ G$ and hence to $ SO(3,\\mathbb R)$ (the only possibilities are: reflection through a point, a line, or a plane; the first and the last ones have determinant $ \\minus{} 1$).", - "Solution_2": "That pretty much what I did, just that I used the easy fact that the Lie algebra is not zero, so one gets all matrices of the form $ e^{ta}$ for some nontrivial $ a$ and so all rotations having the same axis as $ a$." -} -{ - "Problem": "Calculate the sum in terms of $n$ and $cosx$\r\n$\\dbinom{n}{0}+\\dbinom{n}{4}+\\dbinom{n}{8}+\\dbinom{n}{12}+...$", - "Solution_1": "[hide]We have $2^{\\frac{n}{2}}(\\cos{\\frac{n\\pi}{4}}+\\sin{\\frac{n\\pi}{4}})=(1+i)^{n}=C_{n}^{0}+C_{n}^{1}i+C_{n}^{2}i^{2}+...$ by $i^{4k}=1,i^{4k+1}=i,i^{4k+2}=-1,i^{4k+3}=-i\\forall k=0,1,2,...$. Therefore we have sum=...(note: part real and part imf...)[/hide]", - "Solution_2": "we know\r\n$(1-i)^{n}=^{n}C_{0}-^{n}C_{1}*i+^{n}C_{2}*i^{2}.......$\r\n$Re(1-i)^{n}=^{n}C_{0}-^{n}C_{2}+^{n}C_{4}.......$--------i\r\nwe also know\r\n$(1-1)^{n}=^{n}C_{0}-^{n}C_{1}+^{n}C_{2}.......$ -------ii\r\nadding (i) and (ii)\r\nwe get\r\n$Re(1-i)^{n}+0 = 2*S-(^{n}C_{1}+^{n}C_{3}+^{n}C_{5}.....)$----iii\r\nwhere S we have to find\r\n$(1+1)^{n}=^{n}C_{0}+^{n}C_{1}+^{n}C_{2}.......$\r\n$2*(^{n}C_{1}+^{n}C_{3}+^{n}C_{5}.....) = 2^{n}$\r\n$(^{n}C_{1}+^{n}C_{3}+^{n}C_{5}.....) = 2^{n-1}$\r\nhence substituting the value in the equation iii\r\n$Re(1-i)^{n}= 2*S-2^{n-1}$\r\nhence\r\nby solving we get\r\n$s= \\frac{n}{4}log(2)+2^{n-2}$", - "Solution_3": "$C_{n}^{0}+C_{n}^{4}+...=\\frac{(1+1)^{n}+(1-1)^{n}+(1+i)^{n}+(1-i)^{n}}{4}=2^{n-2}+2^{n/2-1}cos\\frac{\\pi n}{4}$." -} -{ - "Problem": "Assume in a democracy there is a $\\frac{1}{3}$ chance of there being at least one revolution in a given year. What is the probability that over 10 years 3 of those years contained at least one revolution?", - "Solution_1": "[hide]There are $\\binom{10}3$ ways to choose which 10 years contain a revolution. The probability that 7 years do NOT have a revolution is $\\left(\\frac23\\right)^{7}$ and the probability of 3 years having a revolution is $\\left(\\frac13\\right)^{3}$.\n\n$\\therefore\\frac{\\binom{10}32^{7}}{3^{10}}$\n\n$\\frac{40\\times128}{3^{9}}$\n\n$\\frac{5120}{19,683}$[/hide]", - "Solution_2": "[quote=\"b-flat\"][hide]There are $\\binom{10}3$ ways to choose which 10 years contain a revolution. The probability that 7 years do NOT have a revolution is $\\left(\\frac23\\right)^{7}$ and the probability of 3 years having a revolution is $\\left(\\frac13\\right)^{3}$.\n\n$\\therefore\\frac{\\binom{10}32^{7}}{3^{10}}$\n\n$\\frac{40\\times128}{3^{9}}$\n\n$\\frac{5120}{19,683}$[/hide][/quote]\n[hide] wait...wouldn't there be no need for the ten choose 3? because the problem never asks for which 10 years...it simply says what are the chances that we're going to have a revolution in these 10 years...so wouldn't the solution simply be $\\frac{2^{7}}{3^{10}}$? There are $\\frac{1}{3}$ chance of there having a revolution in a year, and $\\frac{2}{3}$ chance of there not being a revolution in a year. so isn't it $\\frac{1^{3}}{3^{3}}$ chance of there being a revolution in 3 years, and then just $\\frac{2^{7}}{3^{7}}$ possibility of there NOT being a revolution in the other 7 years...multiply together to get: \n$\\boxed{\\frac{2^{7}}{3^{10}}}$ \ncorrect me if i'm wrong...I generally am... :oops: \n[/hide]", - "Solution_3": "[quote=\"turboturtle\"][quote=\"b-flat\"][hide]There are $\\binom{10}3$ ways to choose which 10 years contain a revolution. The probability that 7 years do NOT have a revolution is $\\left(\\frac23\\right)^{7}$ and the probability of 3 years having a revolution is $\\left(\\frac13\\right)^{3}$.\n\n$\\therefore\\frac{\\binom{10}32^{7}}{3^{10}}$\n\n$\\frac{40\\times128}{3^{9}}$\n\n$\\frac{5120}{19,683}$[/hide][/quote]\n[hide] wait...wouldn't there be no need for the ten choose 3? because the problem never asks for which 10 years...it simply says what are the chances that we're going to have a revolution in these 10 years...so wouldn't the solution simply be $\\frac{2^{7}}{3^{10}}$? There are $\\frac{1}{3}$ chance of there having a revolution in a year, and $\\frac{2}{3}$ chance of there not being a revolution in a year. so isn't it $\\frac{1^{3}}{3^{3}}$ chance of there being a revolution in 3 years, and then just $\\frac{2^{7}}{3^{7}}$ possibility of there NOT being a revolution in the other 7 years...multiply together to get: \n$\\boxed{\\frac{2^{7}}{3^{10}}}$ \ncorrect me if i'm wrong...I generally am... :oops: \n[/hide][/quote]\n[hide]There is a $2^{7}/3^{10}$ for each arrangement such that 3 yrs have a revolution. We must account for all such arrangements, which there are ${10\\choose3}$ of.[/hide]", - "Solution_4": "eh heh heh...oops. Thanks, Pianoforte... :oops:" -} -{ - "Problem": "Suppose that $ f\\in\\mathbb{Q}[x]$ , that $ f$ is irreducible over $ \\mathbb{Q}$ and that $ \\deg f=3$. Let $ L$ denote the splitting field for $ f$ over $ \\mathbb{Q}$. What are the possible values $ [L: \\mathbb{Q}]$? Give an example of a polynomial $ f$ to illustrate each possibility. Answer the same question if one omits the requirement that $ f$ be irreducible over $ \\mathbb{Q}$.", - "Solution_1": "This isn't too bad. The key here is if $ f$ is irreducible, then one of the roots of $ f$, say $ \\alpha$, has the property so that $ F(\\alpha)$/$ F$ is a degree $ 3$ extension. Thus because the Galois group must be a subgroup of $ S_{3}$, the only two possibilities are $ A_{3}$ and $ S_{3}$, as they both have size divisible by $ 3$. (Note: they also must be transitive, and these are the only two subgroups of $ S_{3}$ that are). To show they are both possible, note that the splitting field of $ x^{3}-2$ of $ \\mathbb{Q}$ is simply $ \\mathbb{Q}(\\omega, 2^{\\frac{1}{3}})$, where $ \\omega$ is a cube root of unity. This extension is of degree $ 6$ and corresponds to the Galois group $ S_{3}$. On the other hand, the splitting field of the polynomial $ x^{3}+x^{2}-2*x-1$ is irreducible and has Galois group $ A_{3}$, though this is less obvious. The key to this being only a degree $ 3$ extension is that the discriminant\r\n$ D=(ab)^{2}-4b^{3}-4a^{3}c-27c^{2}+18abc$\r\n$ =(-2)^{2}-4(-2)^{3}-4(-1)-27(-1)^{2}+18(2)=49$ is a square.\r\n\r\n\r\nIf $ f$ does not need to be irreducible, then in addition to the extensions above, we can also have two more cases. The first is trivial, where $ f$ has $ 3$ rational roots and so the extension is of degree 1. The second is when $ f$ splits into a rational linear factor and an irreducible quadratic. Hence the extension is of degree $ 2$, and the Galois group is simply $ \\mathbb{Z}_{2}$." -} -{ - "Problem": "It is once again that time of year for MATHCOUNTS. Chapter competitions have already taken place across the nation.\r\n\r\nDue to the popularity of the format for our last MATHCOUNTS Math Jam we will again be running a countdown round style Math Jam on Wednesday, Feb. 18. We will keep a tally of points based on the first three students to submit correct answers as well as the best explanations of the solutions offered. To see how this worked last time, check out the transcript:\r\n\r\nhttp://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=20\r\n\r\nThere will be a slight change in the way the game will be run. No split points will be awarded for best solutions. If two or more solutions are equally nice, the student who answered the problem the quickest will be awarded the extra point.\r\n\r\nCome and challenge some of the top MATHCOUNTS students in the nation!", - "Solution_1": "The questions will come from all levels. I think they will be appropriate practice for the upcoming competitions.", - "Solution_2": "darn. the math jam is four days after our chapter competition.", - "Solution_3": "I just checked the Math Jams Page and it doesn't list the MATHCOUNTS Jam on the 11th, but on the 18th. Was that a typo, or have you added one and just haven't listed it yet?\r\n\r\nthanks", - "Solution_4": "[quote=\"frost13\"]I just checked the Math Jams Page and it doesn't list the MATHCOUNTS Jam on the 11th, but on the 18th. Was that a typo, or have you added one and just haven't listed it yet?\n\nthanks[/quote]\r\n\r\nThank you for noticing my mistake. The Math Jam will indeed by on the 18th. I will edit my post." -} -{ - "Problem": "Hello,\r\n\r\nCan someone please state me what is the use of Heron's formula and how can I use in a simple example from you?\r\n\r\nPlease since we are talking about geometry can someone please till me how can I find the area of a triangle by just knowing the sides only?\r\n\r\nThanking you", - "Solution_1": "Heron's:\r\n\r\n$ s \\equal{} \\frac {a \\plus{} b \\plus{} c}{2}$\r\nA-area\r\n$ A \\equal{} \\sqrt {s(s \\minus{} a)(s \\minus{} b)(s \\minus{} c)}$\r\n\r\nExample: Find the area of triangle with sides 13,14,15\r\n\r\nApply Heron's to get 84.\r\n\r\nEdited, thanks to Rofl..", - "Solution_2": "Herons Fromula does give the area of any triangle by just knowing the side lengths. The relation between the two is $ A\\equal{}\\sqrt{s(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)}$ where $ s$ is the semi-perimeter or $ s\\equal{}\\frac{a\\plus{}b\\plus{}c}{2}$. A simple example may be like: what is the area of triangle $ ABC$ if $ AB\\equal{}6$, $ BC\\equal{}7$, and $ AC\\equal{}9$. We first compute $ s$ as $ \\frac{6\\plus{}7\\plus{}9}{2}\\equal{}11$ Then, we have that the area is equal to: \\[ \\sqrt{11(5)(4)(2)}\\equal{}\\boxed{2\\sqrt{110}}.\\]", - "Solution_3": "Try the [url=http://www.artofproblemsolving.com/Wiki/index.php/Heron's_formula]Wiki.[/url]\r\n\r\nI'm sure you can come up with some problems on your own. For [hide=\"example,\"]\nFind the inradius of a triangle with side lengths $ 13,14,15.$[/hide]\r\n\r\nCorrection for Bugi. The area is actually \\begin{align*}\\sqrt{\\frac{39}{2}\\left(\\frac{39}{2}-12\\right)\\left(\\frac{39}{2}-13\\right)\\left(\\frac{39}{2}-14\\right)}&=\\sqrt{\\frac{39}{2}\\cdot\\frac{15}{2}\\cdot\\frac{13}{2}\\cdot\\frac{11}{2}}\\\\&=\\frac{39\\sqrt{55}}{4}.\\end{align*}", - "Solution_4": "Heron's formula states that $ \\sqrt{s(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)}$ gives the area of triangle ABC where $ s$ is the semiperimeter.\r\n\r\nThis formula is often useful when trying to find the altitude/height of the triangle. A simple example would be:\r\n\r\nTriangle ABC has lengths as follows: $ AB\\equal{}6, BC\\equal{}7, CA\\equal{}11$. Find the length of the altitude from $ A$ to $ BC$. You would set up an equation comparing two expressions for the area: Heron's and the basic one.\r\n\r\n$ \\sqrt{12(12\\minus{}6)(12\\minus{}7)(12\\minus{}11)}\\equal{}\\frac{7h}{2}$\r\n\r\n$ \\frac{7h}{2}\\equal{}\\sqrt{360}$\r\n\r\n$ 7h\\equal{}2\\sqrt{360}$\r\n\r\n$ h\\equal{}\\frac{2\\sqrt{360}}{7}$\r\n\r\n$ h\\equal{}\\frac{12\\sqrt{10}}{7}$\r\n\r\nThere are many other applications, but this is a very simple one.", - "Solution_5": "Wonderful :D \r\nThanks a lot :)" -} -{ - "Problem": "For what value of K>0, does S(4kx-5k)dx=x^2 ?\r\nS(4kx-5k)dx is an integral", - "Solution_1": "There is no such number, but there is a function $ k(x)$ with that property:\r\n\r\n$ \\int(4k(x)x\\minus{}5k(x))dx\\equal{}x^2$\r\n\r\n$ \\left(\\int(4k(x)x\\minus{}5k(x))dx\\right)^{'}\\equal{}2x$\r\n\r\n$ 4k(x)x\\minus{}5k(x)\\equal{}2x$\r\n\r\n$ k(x)\\equal{}\\frac{2x}{4x\\minus{}5}$", - "Solution_2": "I'm sorry I typed in the question incorrectly. \r\nIt should be....\r\n\r\nFor what value of K>0, does S(4kx-5k)dx= [b]k^2[/b]?\r\n\r\nBy the way how do you type the mathematics out like the post above this one?", - "Solution_3": "anyone can help me?", - "Solution_4": "As written, that's nonsense. The indefinite integral of a nonzero function can't be a constant.\r\n\r\nYou probably mean to ask about a specific definite integral- so what are the limits?\r\n\r\nThe [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]pretty math[/url] on this forum is $ \\text{\\LaTeX}$." -} -{ - "Problem": "a) Prove that if $\\begin {small}X=\\left[ \\begin{array}{cccccccc} X_1&|& X_2&|& \\dots &|& X_n \\end{array}\\right]\\end{small}$ is a $n\\times n$ matrix and $A\\in \\mathcal M_n(\\mathbb F)$. Prove that :\r\n\\[ \\begin {small}\\det \\left[ \\begin{array}{cccccccc} AX_1&|& \\dots &|&X_n \\end{array}\\right]+\\ \\dots\\ +\\det \\left[\\begin{array}{cccccccc} X_1&|& \\dots &|&AX_n \\end{array}\\right]= \\mbox {Tr}(A)\\det (X)\\end{small} \\]\r\n\r\nb) Suppose that $A,X$ are $n\\times n$ matrices which $X$ is invertible and $\\mbox{Tr}(A)\\neq 0$. Prove one of the matrices $\\begin {small}\\left[ \\begin{array}{ccccccc} AX_1&|& \\dots &|&X_n \\end{array}\\right]\\end{small}$, ... , $\\begin {small}\\left[ \\begin{array}{ccccc} X_1&|& \\dots &|&AX_n \\end{array}\\right]\\end{small}$ are invertible.\r\n\r\nc) Can we make the last statement stronger?", - "Solution_1": "Nobody Interested :?: :?:", - "Solution_2": "Interested.\r\n\r\nAny hints :?:", - "Solution_3": "Well the second statement is obvious with the first one. And for the first one you must do some clever computations.", - "Solution_4": "But why b) follows from a)? Probably you meant that $X$ is invertible... And for a), one doesn't need computations: just observe that the LHS is a n-linear alternated application and thus we know it is proportional with the determinant application. Then one reduces to the case when $X_i$ are the canonical basis, case which is trivial (this is exactly the argument used to find the wronskian in linear differential equations).", - "Solution_5": "Great Proof Harazai !!! :) I didn't even thought that this function is n-linear alternated. For the last one I think we could make a weaker assumption and prove the same thing." -} -{ - "Problem": "Sarah intened to multiply a two digit number and a three digit number, but she left out the multplication sign and simply placed the two digit number to the left of the three digit number, Thereby forming a five digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum of the two digit number and the three digit number?", - "Solution_1": "Does ACU have anything to do with computer programming. Maybe that is why no one answered this. Anyways you did not to write program and this was probably posted in the wrong section if you did not use program. Probably a Pre-Olympiad question because you need to use division formula.\r\n\r\nSo my solution is number theory style and I used division formula to solve for the numbers. Two digit number is 14 and three digit number is 112.", - "Solution_2": "i don't know if this will work or not but give it a shot.\r\nlets just say the two digit number was ab (this is not a times b).\r\nand the three digit number was cde. then you can write ab as:\r\n(10 a +b)and cde as:(100c+10d+e). then you can write ab*cde as: 1000ac+100ad+10ae+100bc+10bd+be then set this equal to: 90000a+9000b+900C+90d+9e. \r\nthen solve for 100c+10(a+d)+b+e" -} -{ - "Problem": "This thing is troubling me since the late February that is when I heard the news that 7 HODs of Bansal left and opened a new institute named Vibrant.\r\nI wanted to go to Bansal because of its best faculty and highly competitive atmosphere. Now, the Bansal faculties have left but still most of the students want to remain in Bansal only. So, the advantages of Bansal (for me) have split.\r\nNow, what should I do? Go for Bansal or Vibrant?\r\nAlso, if you know when and where will Vibrant conduct its entrance exam for students of X (going to XI), then please tell me?", - "Solution_1": "well well this would put many in doubts!!!!", - "Solution_2": "Please answer as soon as possible as seats are limited in Vibrant Academy.\r\n\r\nIf anybody of you has studied under Nitin Jain sir or Narendra Avasthi sir, then please tell your experiences under them.", - "Solution_3": "well yes this is true that top 7 faculty teaching p1-p2-p3 batches has left bansal but still it has tried to almost compensate the loss.........they have bougth many good teachers........\r\n\r\na d gupta of physics..........\r\nbest of carrer point as maths.........\r\nand others........", - "Solution_4": "I don't know much but some of my acquaintances have left bansals and they were in gud enuf batches.\r\n\r\nbansals needs to fix it quickly", - "Solution_5": "from my vantage point now, I can say coaching institutes suck, they only help eat up your precious plus two's, but of course by the time you understand that, it's always too late.\r\nI wish I could convey to you youngsters in some way the utter contempt we hold these 'Bansal sir's in now.", - "Solution_6": "I strongly disagree with going to Kota for coaching. But if you have some truly awesome coaching in your hometown like me, which finihes course by september-october and doesn't screw your boards, I'm more than sattisfied. Besides our coaching teachers are so nice, they hold extra classes at our request to teach us the school topics. :)", - "Solution_7": "And i wanted to screw up my boards somehow!!! :D :)", - "Solution_8": "Vinrant all dway..\r\nteachers r super!!!!", - "Solution_9": "I also want to screw up my boards and I want to study under great teachers................but for Vibrant one thing that is troubling me is that there's always a chance that Vibrant teachers are not able to manage and fight and break away. In that way, what shall we do?\r\n\r\nI again repeat my request, please if anybody has studied under Nitin Jain or Narendra Avasthi sir, please tell your experiences.", - "Solution_10": "1 of my ex-classmates in P2 is leaving for Vibrant with her friends coz those were all their fav teachers. And another good friend is leaving bansal for Resonance in 12th.\r\nI'd strongly suggest neither... Just join a good coaching in your hometown and do more self study.", - "Solution_11": "Thats i wanted to say!!\r\nbeing a nerd wont help ur coz!! :D", - "Solution_12": "@nuclear.............\r\nu mean shivangi with ex classmeate???????", - "Solution_13": "Ermmm firstly, don't mention this to anyone I said about Shivangi. She didn't want to. Thats why I never took her name and I thought you'd not remember lol. Anyways, the girl in P2 was Shivangi and the other guy- Shantanu. Forgot Shantanu's batch lol. He's in Lucknow giving school exams these days, but we ambush his house every other day for multiplayer games :P" -} -{ - "Problem": "Guess the next number in the following series:\r\n\r\n172, 468, 317, 246, ?", - "Solution_1": "Belongs [url=http://www.artofproblemsolving.com/Forum/index.php?f=337]here[/url] (this is not a very mathematical question). I would guess $ 831$." -} -{ - "Problem": "hi,\r\ncalculate:\r\n$\\lim_{n\\to\\infty}n \\cdot (\\frac{\\pi^{2}}{8}-\\sum_{k=1}^{n}\\frac{1}{(2k-1)^{2}})$.", - "Solution_1": "\\[\\lim_{n\\to\\infty}n \\cdot (\\frac{\\pi^{2}}{8}-\\sum_{k=1}^{n}\\frac{1}{(2k-1)^{2}})=\\lim_{n\\to\\infty}n\\sum_{k=n+1}^{\\infty }\\frac{1}{(2k-1)^{2}}=\\frac{1}{4}. \\]" -} -{ - "Problem": "[b](1) Find Max. and Min. value of (a-x)*(x+ (x^2+b^2)^1/2)[/b]", - "Solution_1": "As $ x$ goes to $ \\infty$, $ y$ goes to $ \\minus{}\\infty$. As $ x$ goes to $ \\minus{}\\infty$, $ y$ goes to $ 0$. So there should be a local maximum (I think) somewhere. Perhaps differentiation should find it.", - "Solution_2": "hello, we have\r\n$ f'(x)\\equal{}\\minus{}x\\minus{}\\sqrt{x^2\\plus{}b^2}\\plus{}(a\\minus{}x)\\left(1\\plus{}\\frac{x}{\\sqrt{x^2\\plus{}b^2}}\\right)$\r\nand we get\r\n$ f'(x)\\equal{}0 \\Leftrightarrow \\minus{}b^2(\\minus{}2ax\\plus{}a^2\\minus{}b^2)\\equal{}0$\r\nand from here you can go on.\r\nSonnhard." -} -{ - "Problem": "Find all polynomials $ P(x,y)$ such that for all reals $ x$ and $y$,\n\\[P(x^{2},y^{2}) =P\\left(\\frac {(x + y)^{2}}{2},\\frac {(x - y)^{2}}{2}\\right).\\]", - "Solution_1": "[quote=\"khashi70\"]Find All Polynomials $ P(x,y)$ such that for all real $ x,y$ we have :\n$ P(x^{2},y^{2}) = P(\\frac {(x + y)^{2}}{2},\\frac {(x - y)^{2}}{2})$[/quote]\r\n\r\nThe equation may be written $ P(\\frac {x^2 + y^2}{2} + \\frac {x^2 - y^2}{2},$ $ \\frac {x^2 + y^2}{2} - \\frac {x^2 - y^2}{2})$ $ = P(\\frac {x^2 + y^2}{2} + xy,$ $ \\frac {x^2 + y^2}{2} - xy)$\r\n\r\nAnd so : $ P(x + y,x - y) = P(x + \\sqrt {x^2 - y^2},x - \\sqrt {x^2 - y^2)}$ $ \\forall x,y$ such that $ x\\geq |y|$\r\n\r\nLet then $ Q(x,y) = P(x + y,x - y)$ We have $ Q(x,y) = Q(x,\\sqrt {x^2 - y^2})$ $ \\forall x,y$ such that $ x\\geq |y|$\r\n\r\nSo $ Q(x,y) = Q(x, - y)$ $ \\forall x,y$ such that $ x\\geq |y|$ and so $ \\forall x,y$ ($ Q$ is a polynomial). So $ \\exists$ $ R(x,y)$ such that $ Q(x,y) = R(x,y^2)$ And :\r\n\r\n$ R(x,y^2) = R(x,x^2 - y^2)$ $ \\forall x,y$ such that $ x\\geq |y|$ and so $ \\forall x,y$ ($ R$ is a polynomial). \r\n\r\nAnd a general solution of such an equation is $ R(x,y) = H(x,y) + H(x,x^2 - y)$ With $ H(x,y)$ any polynomial.\r\n\r\nAnd so a general solution of initial equation : $ P(x,y) = H(\\frac {x + y}{2},(\\frac {x - y}{2})^2) + H(\\frac {x + y}{2},(\\frac {x + y}{2})^2 - (\\frac {x - y}{2})^2)$\r\n\r\nWhich may be written in a simplier manner : $ P(x,y) = H(x + y,(x - y)^2) + H(x + y,4xy)$ (for any polynomial $ H(x,y)$)", - "Solution_2": "we have $ P(x^{2},(-y)^{2})=P(x^{2},y^{2})=P(\\frac{(x+y)^{2}}{2},\\frac{(x-y)^{2}}{2})=P(\\frac{(x-y)^{2}}{2},\\frac{(x+y)^{2}}{2}) $. so $ P(x,y) $ is symmetric.\nthen we can write $ P $ in the form $ P(x,y)=Q(x+y)+xyR(x,y) $.\nthen we obtain that $ R(x,y)=\\frac{(x-y)^{2}}{4}S(x,y) $ with $ S $ satisfying the initial equation.\nhow can I finish the problem now?", - "Solution_3": "[quote=\"pco\"][quote=\"khashi70\"]Find All Polynomials $ P(x,y)$ such that for all real $ x,y$ we have :\n$ P(x^{2},y^{2}) = P(\\frac {(x + y)^{2}}{2},\\frac {(x - y)^{2}}{2})$[/quote]\nAnd a general solution of such an equation is $ R(x,y) = H(x,y) + H(x,x^2 - y)$ With $ H(x,y)$ any polynomial.\n\nAnd so a general solution of initial equation : $ P(x,y) = H(\\frac {x + y}{2},(\\frac {x - y}{2})^2) + H(\\frac {x + y}{2},(\\frac {x + y}{2})^2 - (\\frac {x - y}{2})^2)$\n\nWhich may be written in a simplier manner : $ P(x,y) = H(x + y,(x - y)^2) + H(x + y,4xy)$ (for any polynomial $ H(x,y)$)[/quote]\nhow can we get the general solution ?????", - "Solution_4": "[quote=TheBeatlesVN]how can we get the general solution ?????[/quote]\nWe have to solve $R(x,y)=R(x,x^2-y)$ $\\forall x,y$\n\n1) Any $R(x,y)=H(x,y)+H(x,x^2-y)$ is a solution, whatever is $H(x,y)$\nTrivial result : just check\n\n2) any such polynomial may be written as $R(x,y)=H(x,y)+H(x,x^2-y)$ for some $H(x,y)$\nTrivial result : just set $H(x,y)=\\frac 12R(x,y)$\n\nAnd so $R(x,y)=H(x,y)+H(x,x^2-y)$ indeed is a general solution for such an equation.\nQ.E.D.\n", - "Solution_5": "The answer is \n$$ P(x,y) \\equiv Q \\left( x+y, xy(x-y)^2 \\right) $$\nfor any $Q(x,y) \\in \\mathbb R[x,y]$. First we prove all these solutions work. Note that both \n$$ P(x) \\equiv x+y \\qquad , \\qquad P(x) \\equiv xy(x-y)^2 $$\nsatisfy the given assertion. It follows all specified solutions indeed work. \n\nNow we prove the converse direction. Changing $y$ to $-y$, it isn't hard to get $f$ is symmetric, i.e. \n$$ f(x,y) = f(y,x) ~~ \\forall ~ x,y \\in \\mathbb R $$ Note if we subtract from $P$ any polynomial $A(x+y)$, then $P$ would still satisfy the assertion. By doing this, we may WLOG assume $P(x,0) \\equiv 0$. As $P$ is symmetric, hence $xy$ divides $P(x,y)$. Now putting $y=x$ gives\n$$ P(x^2,x^2) = P(2x^2,0) = 0 $$\nIt follows $P$ is divisible by $(x-y)$. As $P$ is symmetric, thus $P$ is divisible by $(x-y)^2$. So $xy(x-y)^2$ divides $P$. Again, we cancel a factor of $xy(x-y)^2$ from $P$. Then $P$ still satisfies the given assertion. We keep proceeding like this, and we are done. $\\blacksquare$\n\n\n[color=#f00][b][color=#f00]Remark: [/color][/b][/color] I am not exactly sure how this solution set is same as the one mentioned by [b]pco[/b] , though I cannot find any flaw in my solution (maybe someone else can comment on this). ", - "Solution_6": "[quote=guptaamitu1]I am not exactly sure how this solution set is same as the one mentioned by [b]pco[/b] , though I cannot find any flaw in my solution (maybe someone else can comment on this).[/quote]\nI did not check your proof/result but it is very common that many totally equivalent general form for a given F.E. exist.\nAs soon as each is a general solution (any function in this form is a solution and any solution may be written in this form), they are quite equivalent.\n\nExample with the \"trivial\" FE : \"find all the functions from $\\mathbb R\\to\\mathbb R$ such that $f(x)=f(-x)\\quad\\forall x$\"\nDifferent general solution :\n1) $g(x)$ whatever is even function $g(x)$ from $\\mathbb R\\to\\mathbb R$ :D \n2) $g(x)+g(-x)$ whatever is function $g(x)$ from $\\mathbb R\\to\\mathbb R$\n3) $g(|x|)$ whatever is function $g(x)$ from $\\mathbb R_{\\ge 0}\\to\\mathbb R$\n4) $\\forall x\\ge 0$ $f(x)=g(x)$ and $\\forall x<0$ $f(x)=g(-x)$ whatever is function $g(x)$ from $\\mathbb R_{\\ge 0}\\to\\mathbb R$\n\nBut for example $g(x)g(-x)$ whatever is $g(x)$ from $\\mathbb R\\to\\mathbb R$ is not a general solution (some solutions can not be written in this form)" -} -{ - "Problem": "im checking the telescopical way of getting the algebraic expresion for this summation.1^n +2^n +3^n + . . . + k^n =?? i get to explaining to n=3 but i dont know how to express a generalisation. I could use as much help posible. Thanks\r\n\r\nBen", - "Solution_1": "Not only have you already posted this question, but you repost after there have been several responses in the first post. Bad nettiquite....\r\nDouble posting and crying urgency is a recipe for a thread that goes ignored...\r\n\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=154838[/url]\r\nAccept the answers you've been given and expect nothing else.\r\nEDIT: Make that TRIPLE POST.... that's enough ignore his questions.. :furious: \r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=154862[/url]\r\n[b]EDIT: Make that a QUADRUPLE POST!!! [/b]\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=870784#870784[/url]\r\n\r\nCan a mod please delete his triple ( now quadruple) posts.\r\n\r\nAs far as I'm concerned I'm ignoring this and any future questions by him..." -} -{ - "Problem": "Let $ L$ be linear isomorphism of space $ R^{k}$ into $ R^{k}$. Prove that $ \\lim_{\\mathbf{x}\\rightarrow \\infty}\\parallel{}L\\mathbf{x}\\parallel{}\\equal{}\\infty$. We define $ \\phi(\\mathbf{x})\\equal{}\\frac{\\mathbf{x}}{\\parallel{}\\mathbf{x}\\parallel{}^{2}}$ for $ \\mathbf{x}\\neq 0$ We define also function f as $ f(\\mathbf{x})\\equal{}\\phi(L\\phi(\\mathbf{x}))$ and $ f(\\mathbf{0})\\equal{}\\mathbf{0}$. For which isomorphisms $ L$ function $ f$ has derivative in $ \\mathbf{0}$", - "Solution_1": "Anybody ? ? ? ? ? ? ? ? ? ? ? ?" -} -{ - "Problem": "Consider the power series:\r\n1. $ \\sum\\frac{n!}{n^n}x^n$ and \r\n2. $ \\sum\\frac{n^n}{n!}x^n$\r\n\r\nShow that sum 1 diverges at $ x\\equal{}\\pm e$\r\n\r\nShow that sum 2 diverges at $ x\\equal{}e^{\\minus{}1}$ and converges at $ x\\equal{}\\minus{}e^{\\minus{}1}$", - "Solution_1": "Since all terms are non negative, we have\r\n\r\n$ \\lim \\frac{a_{n\\plus{}1}}{a_n} \\equal{} \\lim \\frac{(n\\plus{}1)!e^{n\\plus{}1}}{(n\\plus{}1)^{n\\plus{}1}} \\cdot \\frac{n^n}{n!e^n} \\equal{} e \\lim \\left(\\frac{n}{n\\plus{}1}\\right)^n \\equal{} \\lim \\frac{e}{\\left(1 \\plus{} \\frac1n\\right)^n} \\equal{} 1^\\plus{}$, \r\n\r\nsince the sequence on the denominator is monotone increasing to $ e$. Therefore, the first sum diverges (for both $ x \\equal{} e$ and $ x \\equal{} \\minus{}e$).", - "Solution_2": "To decide the second question, we seem to need something close to the full strength of Stirling's approximation:\r\n\r\n$ n!\\approx \\sqrt{2\\pi n}\\,n^ne^{\\minus{}n}.$\r\n\r\nHence if $ x\\equal{}e^{\\minus{}1},$ we have that $ \\sum\\frac{n^n}{n!}x^n$ is limit comparable to $ \\sum\\frac{1}{\\sqrt{n}}$ and hence diverges.\r\n\r\nWhen $ x\\equal{}\\minus{}e^{\\minus{}1},$ the terms are the same size but now the alternating series test applies and the series converges conditionally." -} -{ - "Problem": "Hey guys, plz help me with the last part of question 10E (linear algebra) of paper 3 in this:\r\nhttp://www.maths.cam.ac.uk/teaching/pastpapers//2008/Part_IB/index.html\r\n\r\n(This part should be done in around 5 minutes, as the previous parts take quite long already. So it should be sth obvious but uuurrrgh... :mad: i don't know...)\r\n\r\nThanks a lot", - "Solution_1": "It's not obvious... the Tripos is supposed to be difficult (though mostly it isn't).\r\n\r\nProve that if $ k$ is an algebraically closed field (of characteristic $ \\neq 2$, to be safe), then any quadratic space of dimension at least $ 2$ over $ k$ has a nonzero isotropic vector (i. e. a nonzero vector $ v$ such that $ \\left < v,v\\right > \\equal{} 0$). (Feel free to assume $ k\\equal{}\\mathbb{C}$ here and below.)\r\n\r\nProve that if $ k$ is an algebraically closed field (of characteristic $ \\neq 2$, to be safe), then any quadratic space of dimension $ n$ over $ k$ has an isotropic subspace (i. e. a vector subspace consisting of isotropic vectors only) of dimension $ \\left\\lfloor\\frac {n}{2}\\right\\rfloor$. (Construct such a space recursively, each time adding a new isotropic vector orthogonal to the isotropic subspace already constructed. For how long will this work?)\r\n\r\nConclude by induction.\r\n\r\n darij" -} -{ - "Problem": "which place do you think will host the 2016 olympics. right now the 4 places they are deciding on are Chicago :lol: , Tokyo :P Madrid :mad: and Rio De janeiro( i think thats how you spell it) \r\n and don't just pick chicago because we live in america", - "Solution_1": "Tokyo: ~14% chance of winning = they have hosted it before\r\nChicago: ~50% chance of winning = in 1904 they won but win was transferred to St. Louis\r\nMadrid: ~34% chance of winning = they have bidded twice without success\r\nRio De Janeiro: ~2% chance of winning = Is this a serious bid?", - "Solution_2": "wahh. chicago cuz i live like 40 min away!! umm... yeah i agree with james though on the percentage thing...", - "Solution_3": "I'm going to say Madrid. Although, I hope it's Chicago -- then I can go watch what they do!", - "Solution_4": "I dont give the last place much of a shot. However the final 3 split the chances pretty evenly. Here's my percentages.\r\n\r\nChicago-40%\r\nTokyo-35%\r\nMadrid-25%", - "Solution_5": "Comments:\r\n\r\nTokyo has hosted this before. Since the others haven't had a chance, less chance.\r\nMadrid... actually has quite a bit of chance. 2 unsuccessful bids, 1976 and 2012. An unsuccessful bid in 2012 gives it lower chances. Chicago actually won the bid in 1904, but due to the World Fair it was moved to St. Louis.\r\n\r\n(Slightly) Updated Probability\r\n\r\nTokyo: 20%\r\nChicago: 43%\r\nMadrid: 37%", - "Solution_6": "PEOPLE you're forgetting Rio de janeiro. it still ahs a chance. also fyi in 2010 china gets to hold the WORLD EXPEDITION THING. maybe its the same as teh world fair. w/e.I think its in shanghai", - "Solution_7": "Err, I think we have kinda ruled out Rio De Janeiro. \r\n\r\n1) One of the cities has to be removed in Round 1, likelier to be Rio De Janeiro than any other city.\r\n\r\n2) Besides: its starting score: which determines \"feasibility\" of the Olympics was even lower than Doha, Qatar [Score was 6.1 compared to Doha's 6.7] and is lower than any of the other 3 cities.\r\n\r\nJust a thought.", - "Solution_8": "um this isn't a competition its just a prediction.who do you think will be hosting the 2016 olympics. not a competition.thers no city getting vote out", - "Solution_9": "Actually this is the process used by the IOC to determine Host Cities.", - "Solution_10": "chicago!! because i live in the city. but to say the truth, chicago's chances dont seem to be that high right now" -} -{ - "Problem": "Find all $ n$ such that $ [\\sqrt[n]{100}]$ divides $ 111$", - "Solution_1": "Since $ 111\\equal{}3\\cdot37$ we have to make some distinguishing:\r\n\r\na.) $ [\\sqrt[n]{100}] \\equal{} 1$\r\nSo we have \r\n$ 1\\leq\\sqrt[n]{100}< 2$\r\n$ 1\\leq 100 < 2^{n}$\r\nThis inequality holds if $ n$ is bigger than or equal 7.\r\n\r\nb.) $ [\\sqrt[n]{100}] \\equal{} 3$\r\nSo we have\r\n$ 3\\leq\\sqrt[n]{100}< 4$\r\n$ 3^{n}\\leq 100 < 4^{n}$\r\n\r\nRight inequality holds if $ n$ is bigger that or equal 4, left inequality holds if $ n$ is smaller than or equal 4, so $ n\\equal{}4$.\r\n\r\nc.) $ [\\sqrt[n]{100}] \\equal{} 37$\r\nSo\r\n$ 37\\leq\\sqrt[n]{100}< 38$\r\n$ 37^{n}\\leq 100 < 38^{n}$\r\nClearly, no integer $ n$ satisfies this condition.\r\n\r\nd.) $ [\\sqrt[n]{100}] \\equal{} 111$\r\nAlso here we can't have solutions since $ \\sqrt[n]{100}\\leq 100\\leq 111$\r\n\r\nSo we have solutions for $ n\\equal{}4$ and $ n\\geq 7$" -} -{ - "Problem": "First, a little background. I'm a junior. I took the AMCs this year for the first time since 8th grade, when I took the AMC 10. This year, I just barely got into the AIMEs with a 103.5, and got an abysmal 4 on the AIME itself.\r\n\r\nI hadn't put much effort into preparation for the AMC as I should have, and yet I was the school winner for it by a longshot, mainly because, from what I observe, I'm the only one at my school that really wants to perform well on these contests; I apologize for the globalizing statement, but I'm good friends with a lot of the people who took the AMCs, and it's generally this feeling throughout the junior and senior class. When I was at the lecture hall before the AMC, I overheard people saying that they were only taking the test so they could get out of class. Also, at risk of sounding like I'm pushing the guilt around, none of my teachers are as much into these as I've gathered from other posts in these forums; there are no post-contest review of problems, and some of the teachers I talked to didn't even know that the test was being administered. Some of the posts on this forum allude to math teams, with \"coaches\" that help teach problem-solving techniques and what-not. Some allude to teachers even bringing donuts on the day of the AIME (not that this is important to scoring well, but, hey, it's something).\r\n\r\nI have a friend that attends a private school in the state, and him and a lot of his friends are much more into it, as are their teachers; definitely more than one person qualifying for the AIMEs, and achieving the top score on the AMCs is a competition they have amongst each other -- nothing like my school.\r\n\r\nSo I'm just curious: I'd like to know how other schools treat these competitions. Specifically, how many people usually qualify for the AIMEs at your school per year? What's usually the top-AMC score for your school? Any information would be greatly appreciated.\r\n\r\nAgain, although I am disgruntled, this post is not intended to vent my anger or shift the blame for my poor performance onto my school; I completely accept the fact that my lack of preparation lead to my low scores. I simply want an idea of what kind of experiences people have in preparing for the AMCs.", - "Solution_1": "i think it depends on how much interest there is and how competitive the school is.", - "Solution_2": "Yea, my school requires all students enrolled in the upper lane math class to take AMC10 or 12, and we have a quite hardy math crowd, and a good deal of beastly people, but no one else actually cares. I don't think a good percentage of people ANYWHERE care about math, so don't feel bad or anything.", - "Solution_3": "mmm, my school hadn't done amc's for a few years before i came, and we only do the 12a test, this year and last. most people just don't care. the people who care somewhat don't have the experience of doing these kind of competitions.\r\n\r\nthese two years, we have had less than 20 people take the 12a test. the high scores last year were 132(me), 88.5, and high 70s. this year, 97.5 (me, yeah, deproved a bit), 94.5, 91.5. the same person got second both years, and she is a senior this year.\r\n\r\nthis area (KS) isn't very competitive math wise. more or less, the (relatively) good math people, students and coaches, around here are scattered.", - "Solution_4": "Not my school, but another school in my district requires their honors math class students to take the amc 10 or 12. Of 200+ participants last year in the 12, not a single aime qualifier.", - "Solution_5": "My school, compared to others around my area, isn't all that math-competitive. I've managed to get school winner for the last two years (I'm a sophomore) with a 112.5 on the 12A both years. The total number of AIME qualifiers has been around 2-3 both years as well. And I'm pretty much the only person who cares at my school, but I have quite a few friends at other schools, where the math culture is a lot more competitive.", - "Solution_6": "[quote=\"pianoforte\"]mmm, my school hadn't done amc's for a few years before i came[/quote]\r\n\r\nWhat caused them to finally resume the testing? Did you approach a faculty member yourself? I was also thinking of asking my school to participate in the Mandelbrot Competition...\r\n\r\nThanks for the responses, guys!", - "Solution_7": "My school is one of the biggest in the US \r\nand for the last couple of years, we usually have like 100 people that take the AMC with about 30 qualifying.\r\n\r\nHowever, this year, since no math teachers announced it to their math classes, and it was only in the announcements which no one listens to during school, we only had like 40 takers with like 20 passing.\r\n\r\nBut even though we have alot of people taking the AMC and even AIME, I think we've only had 2 USAMO qualifiers in our history...\r\n\r\nHowever, I think like 2 people from my school made it this year (if the cutoff is 8-) and I think like 5 will make it next year because everyones like getting super good all of a sudden...\r\n\r\nBut just compare us to some nearby schools; we usually have more qualifiers than them but somehow, the other schools manage to produce more top notch USAMO qualifers...\r\n\r\nThis is the same with our MATHCOUNTS (middle school), we never had a guy to to nationals but one of our middle schools always gets top 6 state (and this state's finished top 3 for like the last 5 years at Nationals-guess what it is) and sometimes 2 or 3 finish top 6.", - "Solution_8": "[quote=\"the future\"]{...} we usually have like 100 people that take the AMC with about 30 qualifying. {...}\n\n{...} we only had like 40 takers with like 20 passing. {...}\n\n{...} we've only had 2 USAMO qualifiers in our history... {...}\n[/quote]\r\n\r\n20 AIME qualifiers out of 40 AMC participants -- those seem like some impressive statistics to me! What roles do your teachers play in the AMCs, if any? Do you have afterschool math clubs or things like that? I'm part of a Math League at my school that participates in competitions in the region, but the problems pale in comparison to even the first few AMC 10 problems from past tests I've looked at.", - "Solution_9": "Here's how things work:\r\n\r\nThe AMC is announced in our math classes. It is described as a test that is challenging, but participation looks good on a college transcript. My school offers the AMC 8, AMC 10, and the AMC 12. However, I was only given the opportunity to take the AMC 8 in 8th grade.\r\n\r\nAny student who wants to take the AMC can take it. However, there are not a lot of people in the upper grades who take the AMC 12, and everyone in 10th grade or below takes the AMC 10. That's probably because they know they're going to do terribly on it anyway.\r\n\r\nThis year, the three best math students at my school including myself (based on a local math contest) scored 132, 93, and 88.5 (I was the 132) on the AMC 12. On the AMC 10, the highest scorer got an 84.\r\n\r\nI think I am the only person in my school to ever qualify for USAMO.\r\n\r\nI don't think that some of the kids really care about how well they do or improving. I once offered to hold an AMC solutions seminar one day after school, and it didn't attract many kids. *sigh*", - "Solution_10": "[quote=\"jvenezuela716\"][quote=\"the future\"]{...} we usually have like 100 people that take the AMC with about 30 qualifying. {...}\n\n{...} we only had like 40 takers with like 20 passing. {...}\n\n{...} we've only had 2 USAMO qualifiers in our history... {...}\n[/quote]\n\n20 AIME qualifiers out of 40 AMC participants -- those seem like some impressive statistics to me! What roles do your teachers play in the AMCs, if any? Do you have afterschool math clubs or things like that? I'm part of a Math League at my school that participates in competitions in the region, but the problems pale in comparison to even the first few AMC 10 problems from past tests I've looked at.[/quote]\r\n\r\nthats because only the people good at math take it (like 30/40 in math club)\r\nits not a requirement at our school to take it so the stats are going to be a little skewed.\r\n\r\nthe 100/30 is more realistic but even that, its the people in honors (2 math classes above their grade level that signed up) that take it.\r\nof course like i said, this year, it was only on the announcements at school so only the really enthusiastic (for a lack of a better term take it)\r\n\r\nyea we have math clubs, we do like mandelbrot, and some state math competitions.\r\n\r\nalso many people have a math tutor...(if you live in IN you know which one Im talking about)...\r\n\r\nand our middle schools are very good (like i said, atleast one gets top 6 in IN, one of the best MATHCOUNTS states), and our school is pretty well known in the US (its got 4000+ kids...)", - "Solution_11": "aaaaaaaaaaaaaaaaaa", - "Solution_12": "I think distribution of strong people is somewhat random, despite patterns. For example in Santa Clara County, we have a ridiculous amount of math people. Go up a little north to San Mateo County, and there's almost none. I have no reason why that is so. But then you have some random small states with a good few people who are very good.\r\nAt our school, I'm estimating we had at least 600 people who took the test, but 33 made AIME through it (heh compulsory AMC taking take that! But hey it matches the statistics, 6.6% of students in our school made AIME (the people in non-honors had no chance in the first place, nor did many of the people in honors either) and our school is \"smart\"). Actually it was 34 since junggi qualified through USAMTS, but then only like 7 of those people actually care about math.\r\nVery few people care about math, as you can see. Our school has a \"lot\" of people into math, and we only have 8? (2000 people almost at our school).", - "Solution_13": "You can change how your school views these tests. You, as a student, can get the school to take it more seriously.\r\n\r\nI say this because I was the one who instated AMC at my school. I'm a senior now, but when I was a sophomore, I asked one of the math teachers if he would be interested in proctoring it. I did my research on what the AMC was, how much it cost, how it needed to be administered etc. I actually ended up paying the 60 bucks or whatever it is to register the school that year. I told all the math teachers to tell students about it. Only the AMC12 was offered, but it was something.\r\n\r\nWe've had about 10-15 students each year take the AMC. This year is the 3rd year it's been run at my school. We had like 5 AIME qualifiers 2 years ago, and 3 qualifiers last year and this year. Nobody has made USAMO yet, but I hope to change that this year *crosses fingers on 213 index*\r\n\r\n\r\nIf you want your school to take it more seriously, take some actions. Ask if you can set up a time where students who want to take it can meet to improve their problem solving. Things don't just happen by themselves.", - "Solution_14": "Hey jvenezuela716 I totally feel your pain. I faced the same problems, as you seem to be dealing with right now. My high school (which is in San Mateo County) had had a dead math team/ program since 2000, when our sole USAMO/ IMO qualifier graduated. Since then my school had had 1-2 AIME qualifiers every year with no USAMO qualifications. Thus, last year, as an eighth grader, I spoke to the math department head of my current high school, and created a Math Club/Team, in which I hold weekly meetings to explore math concepts. This year has been relatively successful for us, with an AMC 10 School Score of 291/300, 4 AIME qualifiers and 2-3 USAMO qualifiers. (yes, i do realize that this is pathetic by standards at good schools, but it's an improvement :) )\r\n\r\nTherefore, from my experience I encourage you to voice your concerns to the math department head, principle, and/or other authorities at your school.", - "Solution_15": "[quote=\"jvenezuela716\"][quote=\"mathling235\"]Some random freshman kid offered to pay for 10A, and he didn't even make AIME.[/quote]\n\nThat shows at least an interest, so that's something to build on. A lot of the things that you need to know to do well on the AMCs are things that are not covered, at least not very deeply, in high school classes, so people that have an interest in math but are just not exposed to that type of problem solving may not necessarily perform well either. Just the fact that that kid offered to pay for the test himself shows that he must have some interest in math... why not show him AoPS if he doesn't know about it already? Maybe over the next three years he'll make USAMOs. :)[/quote]\r\n\r\nThat's the attitude we should all have. The math community in general always could have more people, and it's hard to know of the existence of AoPS if all you're exposed to is high school classes. I myself didn't know about this until I was fortunate enough to have AoPS I as my textbook in 8th grade.", - "Solution_16": "I can think of someone at my school who finds math really cool but doesn't do especially well on contests (94.5 on both 12A and 12B - ouch). I often show him cool stuff I find on AoPS and Wikipedia. You don't need to be a genius to like math.\r\n\r\nMy school puts a lot of energy into athletics. While we're lucky to have a friendly math department, they can only do so much with budgetary restrictions, etc. I'm sure the per-student expenditures for athletic stuff is far, far higher than it is for math team.", - "Solution_17": "[quote=\"worthawholebean\"]I'm sure the per-student expenditures for athletic stuff is far, far higher than it is for math team.[/quote]\r\n\r\nOf course... what do you need to buy for the math team that could possibly compare to equipment you would need for most sports? :wink: \r\n\r\nOn a more serious note, though, it's one thing to have an interest in math, and an entirely different thing to have the drive to learn math outside of the curriculum (imho). And about building on the AoPS community, I'm not saying you have to go and tell everyone you know how great AoPS is (which it is, don't get me wrong), but if there's someone that is, as in the previous example, paying for the exams themselves, or at least shows some signs they are taking the AMCs or other competition for something other than getting out of class, why not help them get their time/money's worth?", - "Solution_18": "wow reading all this, i go to a pretty lucky school...\r\nyou can just ask my school and theyll order as many AMCs as you want (and any academic competition)\r\nIt offers every AP class avaliable on collegeboard and it does mostly all the prestigious academic competitions in high school.\r\nand its the same with athletics.", - "Solution_19": "I'm the only who was eligible for AIME at my school. But then again, it's middle school, albeit one of the better ones around. My high school next year will have a bit of competition for AMC's. But most freshmen will be from my current school, so I can still have a good overall position for my grade. My maths teachers know about the AMC's, but other than that, they probaby have no clue about it.", - "Solution_20": "My school's AMC10 score range this year was 19.5 - 112.5.\r\nOur top three AMC12 scores were 132 (me), 112.5, 91.5.\r\n\r\nI really don't know how that 19.5 happened. If he was just messing around and answering all of the questions blindly, his score would have been a multiple of 6. So he was probably taking the test semi-intelligently. :huh: Next year, if I manage to make a 150 on the 12A, I'll go for a 0 on the 12B. :-D", - "Solution_21": "lol....our lowest was like a 50, and that's because he had to leave early...and we're a middle school.\r\n\r\nWe only offer the AMC 10, and top was 132 (me), 120, and 109.5.\r\n\r\nI wonder if our high school offers the 12...", - "Solution_22": "[quote=\"Twiz\"]My school's AMC10 score range this year was 19.5 - 112.5.\nOur top three AMC12 scores were 132 (me), 112.5, 91.5.\n\nI really don't know how that 19.5 happened. If he was just messing around and answering all of the questions blindly, his score would have been a multiple of 6. So he was probably taking the test semi-intelligently. :huh: Next year, if I manage to make a 150 on the 12A, I'll go for a 0 on the 12B. :-D[/quote]\r\n\r\n19.5? Wow...I feel badly for him. If he had just put his name on the paper, twiddled his thumbs, and took a nap, he would have gotten almost twice that score. But then again, maybe he was trying to get a 0 but had some really bad luck by guessing some right. :lol:", - "Solution_23": "[quote=\"Emton\"]But then again, maybe he was trying to get a 0 but had some really bad luck by guessing some right.[/quote]\r\nYes, but since his score was not a multiple of six, he did not answer all of the questions, so if he was trying to get a zero, he didn't use a very good strategy. :|", - "Solution_24": "Then again the global scores go from 6-150. 6 is even harder to get.", - "Solution_25": "Back when I was taking the AHSME myself (1969-71), the scoring system was 0 for a blank some positive number (possibly varying by groups of questions) for a correct answer, and some negative number for a wrong answer. I did hear that there were negative scores in my school.", - "Solution_26": "[quote=\"Kent Merryfield\"]Back when I was taking the AHSME myself (1969-71), the scoring system was 0 for a blank some positive number (possibly varying by groups of questions) for a correct answer, and some negative number for a wrong answer. I did hear that there were negative scores in my school.[/quote]\r\nIt was $ 30\\plus{}4c\\minus{}w$ if I remember correctly, there was an AIME problem about it I think.", - "Solution_27": "my school takes AMC 8 seriously, but to most teachers, 10/12 is a joke. i asked for statistics about AMC 10/12 in past years and they said that only 20 people have taken it in somemthing like 4 years. this and last year got snowed out", - "Solution_28": "[quote=\"worthawholebean\"][quote=\"Kent Merryfield\"]Back when I was taking the AHSME myself (1969-71), the scoring system was 0 for a blank some positive number (possibly varying by groups of questions) for a correct answer, and some negative number for a wrong answer. I did hear that there were negative scores in my school.[/quote]\nIt was $ 30 \\plus{} 4c \\minus{} w$ if I remember correctly, there was an AIME problem about it I think.[/quote]\r\n\r\nUm...that was in the 30-problem era. Lowest then was a 0.", - "Solution_29": "None of you guys replying to me have it right. For one thing, it wasn't the 30 problem era - it was 35 problems. Here's a quote from [i]The Contest Problem Book III,[/i] by Salkind and Earl:\r\n\r\n[quote]On the 1968-1972 examinations, the questions are divided into four parts of 10, 10, 10, and 5 questions, respectively with weights 3, 4, 5, and 6. Examinations are scored by the formula $ \\textstyle R\\minus{}\\frac14W,$ where $ R$ and $ W$ denote weighted counts of correct and incorrect responses, respectively.[/quote]\r\nI can assure you that negative scores were possible and negative scores did in fact occur. And the expected score from completely guessing on everything would be the same as the score for leaving everything blank: zero. (That is, the test was guessing-neutral, as opposed to later AHSME and AMC exams which became guessing-negative.)" -} -{ - "Problem": "Let $ a,b > 0$. Prove that:\r\n\\[ [5a] \\plus{} [5b]\\ge [3a \\plus{} b] \\plus{} [3b \\plus{} a]\\plus{}[a]\\plus{}[b]\r\n\\]\r\nNotice that:\r\n\\[ x \\equal{} [x] \\plus{} \\left\\{ x \\right\\}\r\n\\]", - "Solution_1": "Not hard.\r\nUse $ [5x]\\equal{}[x]\\plus{}[x\\plus{}\\frac{1}{5}]\\plus{}[x\\plus{}\\frac{2}{5}]\\plus{}[x\\plus{}\\frac{3}{5}]\\plus{}[x\\plus{}\\frac{4}{5}]$ and consider some case." -} -{ - "Problem": "Circles of radius 3 and 6 are externally tangent to each other and are internally tangent to a circle of radius 9. The circle of radius 9 has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord.", - "Solution_1": "[hide=\"Solution.\"]Draw a line from the center of the radius 3 circle perpendicular to the radius of the radius 6 circle that connects to the tangent. you have a right triangle with two known sides and a rectangle. Find the unknown side of the right triangle and you get $\\boxed{72}$.!!![/hide]", - "Solution_2": "[quote=\"bigboy82892\"][hide=\"Solution.\"]Draw a line from the center of the radius 3 circle perpendicular to the radius of the radius 6 circle that connects to the tangent. you have a right triangle with two known sides and a rectangle. Find the unknown side of the right triangle and you get $\\boxed{72}$.!!![/hide][/quote] I think you only found the length of the tangent, not the chord.", - "Solution_3": "[quote=\"lotrgreengrapes7926\"][quote=\"bigboy82892\"][hide=\"Solution.\"]Draw a line from the center of the radius 3 circle perpendicular to the radius of the radius 6 circle that connects to the tangent. you have a right triangle with two known sides and a rectangle. Find the unknown side of the right triangle and you get $\\boxed{72}$.!!![/hide][/quote] I think you only found the length of the tangent, not the chord.[/quote]\r\nOops! Sorry! Then I don't know the answer. :blush:", - "Solution_4": "Im sry to bring this back, but since Im studying for the AIME, Im trying to do all the problems I can. For this, the answer is not posted yet and I get 224, I just want to know if this is correct.", - "Solution_5": "Ya, thats right", - "Solution_6": "Could somebody please post a solution?", - "Solution_7": "I've only had a very brief look, but here is a solution. Let the centres of the circles with r=3, 6 be A, B respectively. Therefore |AB|=9. By a bit of scrutiny, one realises that AB is a diameter of the big circle, and so the centre of the big circle lies 6 from A and 3 from B, and therefore 3 from the point of tangency. This is getting a bit confusing because there is no diagram, but by Pythagoras, we know that the square of HALF of the chord = 9^2 - 3^2 = 72. Therefore the square of the whole chord = 288.", - "Solution_8": "Um, well, the answer on the website says its 224", - "Solution_9": "Let the chord have endpoints $X$ and $Y$ (let $Y$ be nearest the circle with radius $6$).\r\nThe big circle is centered at $O$, the circle of radius $6$ at $A$ and the circle of radius $3$ at $B$.\r\nDraw perpendiculars from points $O$ and $A$ to $O'$ and $A'$ on chord $XY$.\r\nDraw the perpendicular from $B$ to $B'$ on $AA'$.\r\n$OO'$ intersects $BB'$ at $O''$.\r\n\r\n[hide]$BO''B \\sim BB'A$\nSo,$OO'' = 2$ and $OO' = 5$. Now Pythagorean theorem tells us $(O'Y)^{2}= (OY)^{2}-(OO')^{2}\\Rightarrow (O'Y)^{2}= 56$, and $(XY)^{2}= (2(O'Y))^{2}= 4(56) = 224$. [/hide]", - "Solution_10": "[quote=\"white_horse_king88\"]Let the chord have endpoints $ X$ and $ Y$ (let $ Y$ be nearest the circle with radius $ 6$).\nThe big circle is centered at $ O$, the circle of radius $ 6$ at $ A$ and the circle of radius $ 3$ at $ B$.\nDraw perpendiculars from points $ O$ and $ A$ to $ O'$ and $ A'$ on chord $ XY$.\nDraw the perpendicular from $ B$ to $ B'$ on $ AA'$.\n$ OO'$ intersects $ BB'$ at $ O''$.\n\n[hide]$ BO''B \\sim BB'A$\nSo,$ OO'' \\equal{} 2$ and $ OO' \\equal{} 5$. Now Pythagorean theorem tells us $ (O'Y)^{2} \\equal{} (OY)^{2} \\minus{} (OO')^{2}\\Rightarrow (O'Y)^{2} \\equal{} 56$, and $ (XY)^{2} \\equal{} (2(O'Y))^{2} \\equal{} 4(56) \\equal{} 224$. [/hide][/quote]\r\n\r\nSorry but i couldn't figure out what did you mean by $ BO''B \\sim BB'A$" -} -{ - "Problem": "A convex quadrilateral has one vertex at $ (2,\\minus{}1)$. The other three vertices are at the reflections of $ (2,\\minus{}1)$ about the $ x$-axis, the $ y$-axis, and the line $ y\\equal{}x$, respectively. What is the area of this quadrilateral?", - "Solution_1": "By the given conditions, the points are $ (2,\\minus{}1),(2,1),(\\minus{}2,\\minus{}1),(\\minus{}1,2)$. Now, divide this into two triangles by drawing the diagonal connecting $ (\\minus{}1,2)$ and $ (2,\\minus{}1)$. The sum of these triangles' areas can easily be computed as (using lines parallel to the axes as bases) $ \\frac12\\cdot(2\\minus{}(\\minus{}2))\\cdot(2\\minus{}(\\minus{}1))\\plus{}\\frac12\\cdot(1\\minus{}(\\minus{}1))\\cdot(2\\minus{}(\\minus{}1))\\equal{}\\boxed9$." -} -{ - "Problem": "Im working on a fairly long PDF in Texnic Center, and today when I entered new text and built the file nothing was changing; I couldn't make changes. Then I copied my code and put it in a new doc and when I tried to build it said 'file cannot be found' or something like that. HELP!!\r\nNote: I may have pressed Ctrl 4 but then I undid it.", - "Solution_1": "Find the tex file you were working on and then open it. You can look for the tex file in Windows Explorer or by using Open in TexnicCenter. It may or may not have the most recent changes in it.", - "Solution_2": "that did not do anything. Basically i press build and i have 5 error. Then I press view and it build a file, but with not all the code included. However, if I delete a particularly dense section of code it builds and works. does it not build if there are too many errors or something? Because in the past it just builds it.", - "Solution_3": "If you have 5 errors then you [i]must[/i] correct them first. That is why parts of the document are not being shown. TexnicCenter tells you where the errors are by pressing F9 repeatedly or the ! button on the toolbar.\r\nThe errors are also in the log file which is kept in the same directory as the tex file.", - "Solution_4": "So now there are three errors, but when I hit build it produces that same old file, when I copy and paste the code into another new doc it cant be \"found\" when i try to view output, and when I check for more errors it says \"File ended while scanning the use of \\frac\"", - "Solution_5": "I just tried this:\r\n\r\n[quote]\\documentclass[11pt]{article}\n\n\\begin{document}\n\nFraction test\n\n$ \\$$\\frac{x}{$ \\$$\n\n\\end{document} [/quote]\n\nAnd got \"Paragraph ended before \\frac was complete\"\n\nOn the other hand, if I try\n\n[quote]\\documentclass[11pt]{article}\n\n\\begin{document}\n\nFraction test\n\n$ \\$$\\frac{x}$ \\$$\n\n\\end{document} [/quote]\r\n\r\nThen I get \"Missing } inserted. }\" and then the name of the line on which the offending command is.\r\n\r\nIn neither case did I get a .pdf file.\r\n\r\nSo I haven't exactly reproduced your error message - but you've got something like that going on. It's still your error, and it halts compilation of your file. You have to track down and fix [b]all[/b] of your syntax mistakes.", - "Solution_6": "Yes I am getting:\r\nFile ended while scanning use of \\frac\r\n\r\n \\par\r\nI'll try and fix all my mistakes \r\n*tedious hope this is worth it for the promys app", - "Solution_7": "This is hindsight, but it's probably best to compile your file after each half page or even each paragraph, and after any particularly hairy formula or figure. Then you'll find your problems immediately and know where they are.\r\n\r\nIf your error message doesn't seem particularly helpful in terms of finding your mistake, you can take a formula or paragraph you're suspicious of and put % in front of it. That turns it into a comment, and the compiler ignores it. If the file now compiles, you've narrowed down the location or locations of the problems.\r\n\r\nThere's probably someone, somewhere, who can type errorless $ \\text{\\LaTeX}$ first time, every time - but I'm not that person. Debugging is a necessary part of writing in $ \\text{\\TeX}$ - get used to it.", - "Solution_8": "If you already have a long document with errors and you don't want to copy and paste bits into another document, then add \\end{document} after the section you want to check. Then LaTeX will only compile the part before \\end{document} and you can look to see if there are any errors there. Once that's error-free you can move \\end{document} to further along the document.\r\nThis is only for an already written document. I wholeheartedly agree with Kent's suggestion of frequent compiling as you write the document.", - "Solution_9": "[quote=\"mathemonster\"]Yes I am getting:\nFile ended while scanning use of \\frac[/quote]Sounds like a missing }. In TexnicCenter if you click to the right of a { or } it will highlight in green the matching bracket. If the { isn't matched it will be highlighted in red. You'll find this very useful for finding orphan brackets particularly nested ones. You can also do this for other brackets such as (),[].", - "Solution_10": "More insidious are errors that aren't fatal to the compilation but still aren't what you want. If I amend my minimal example in #6 to make the key line \\frac{x}{} (with dollar signs around it), then the file does compile and produce output - but that output is kind of funny-looking. There's no substitute for taking a long, hard look at your output to make sure it's what you intended." -} -{ - "Problem": "Let $n \\in \\mathbb N$, $n \\geq 2$.\r\n\r\n(a) Give an example of two matrices $A,B \\in \\mathcal M_n \\left( \\mathbb C \\right)$ such that \\[ \\textrm{rank} \\left( AB \\right) - \\textrm{rank} \\left( BA \\right) = \\left\\lfloor \\frac{n}{2} \\right\\rfloor . \\]\r\n\r\n(b) Prove that for all matrices $X,Y \\in \\mathcal M_n \\left( \\mathbb C \\right)$ we have \\[ \\textrm{rank} \\left( XY \\right) - \\textrm{rank} \\left( YX \\right) \\leq \\left\\lfloor \\frac{n}{2} \\right\\rfloor . \\]\r\n\r\n[i]Ion Savu[/i]", - "Solution_1": "I really don't know how to do this problem... I tried time and time again, but I can't find the right idea.\r\n\r\n :blush:", - "Solution_2": "the second part was solved here http://www.mathlinks.ro/Forum/viewtopic.php?highlight=sylvester+rank&t=6544", - "Solution_3": "For the first part we can take \r\n$\\ A_{2k}=\\left(\\begin{array}{cc} I_k & O_k \\\\ O_k & O_k \\end{array}\\right)$ $\\ B_{2k}=\\left(\\begin{array}{cc} O_k & I_k \\\\ O_k & O_k\\end{array}\\right)$ and $\\ A_{2k+1}=\\left(\\begin{array}{cc} A_{2k} & 0 \\\\ 0 & 0\\end{array}\\right)$ $\\ B_{2k+1}=\\left(\\begin{array}{cc} B_{2k} & 0 \\\\ 0 & 0\\end{array}\\right)$", - "Solution_4": "Very interesting \n\n[url= https://artofproblemsolving.com/community/c7h2576955_matrix_and_rank] see [/url]\nhttps://artofproblemsolving.com/community/c7h2576955_matrix_and_rank", - "Solution_5": "I swear... most of your most recent College Math posts are literally post farming with two word sentences... maybe think about contributing something helpful?", - "Solution_6": "[quote=pezzeepenguinstalker]I swear... most of your most recent College Math posts are literally post farming with two word sentences... maybe think about contributing something helpful?[/quote]\n\nwho are you talking to, the OP?\n", - "Solution_7": "It was referencing #5 who revived a 15 year thread just to say it was interesting", - "Solution_8": "pezzeepenguinsta , honestly I don't see the point of your remark. When someone reports an old result that he considers interesting, I find it very good and, moreover, very common in a forum like ours. If that doesn't interest you then, please, don't disgust others. \nIn conclusion, keep your strength to perform more constructive acts. ", - "Solution_9": "He's saying that simply saying a problem is interesting and no other remark is basically post farming, and I have to agree. At least make some comment regarding how to solve the problem or its origin or something similar. Also, his username isn't 'pezzeepenguinsta'" -} -{ - "Problem": "If $ B(X,Y) \\equal{} \\{f: X\\longrightarrow Y \\mid f$ linear, continuous and bounded$ \\}$ is a Banach (normed and complete) space, then $ Y$ is a Banach space ($ X$ and $ Y$ are normed linear spaces).", - "Solution_1": "wrong (take e.g. X = 0).", - "Solution_2": "But if $ X\\not \\equal{}0$, is it true? :maybe:", - "Solution_3": "Let $ (y_n)_n$ be a Cauchy sequence in $ Y$. Knowing that $ X\\not \\equal{} 0$, for $ x_0\\in X\\setminus \\{0\\}$, we take $ f: X\\longrightarrow \\mathbb{R}$ s.t. $ f(x_0) \\equal{} \\| x_0\\|$ and $ \\|f\\| \\equal{} 1$ (this function exists, using a corollary of the Hahn-Banach theorem - probably for this we must add $ X$ complete in the statement, i'm not sure). \r\n\r\nThen, we consider the sequence $ (u_n)_n \\in \\mathcal B(X,Y)$, $ u_n(x) \\equal{} f(x)y_n$ and it is easy to prove that $ (u_n)_n$ is Cauchy. By the hypothesis, it is convergent. So $ u_n\\longrightarrow u\\in \\mathcal B(X,Y)$ and, finally, $ y_n\\longrightarrow \\frac {u(x)}{\\|x_0\\|}\\in Y$. \r\nThis means $ Y$ is a Banach space. \r\n\r\n q.e.d." -} -{ - "Problem": "Let $ p\\in\\mathbb{Q}[x]$ be an irreducible polynomial of degree $ r$ and $ A\\in\\mathcal{M}_{r\\plus{}1}(\\mathbb{Q})$ be s.t. $ det(p(A))\\equal{}0$.\r\nProve that there exist $ u,v\\in\\mathbb{Q}^{r\\plus{}1}$ s.t. $ p(A)\\equal{}uv^T$ and $ v^Tu\\not\\equal{}0$.", - "Solution_1": "It's false.\r\nCounterexample: $ r\\equal{}1,p(x)\\equal{}x,A\\equal{}0$.\r\n\r\nIf $ r$ is at least 2, this basically says that $ A$ has rank 1, because we can specify its characteristic polynomial as $ pq$ with $ q$ of degree 1, relatively prime to $ p$. $ v^Tu$ is $ p(\\lambda)$ for the sole eigenvalue $ \\lambda$ of $ A$, and $ \\mathbb{Q}$ is irrelevant except for providing the existence of enough irreducibles." -} -{ - "Problem": "For positive real numbers $ a$, $ b$, $ c$ and $ d$ such that $ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} d^2 \\equal{} 1$ prove that\r\n\r\n$ a^2b^2cd \\plus{} \\plus{}ab^2c^2d \\plus{} abc^2d^2 \\plus{} a^2bcd^2 \\plus{} a^2bc^2d \\plus{} ab^2cd^2 \\le 3/32,$\r\n\r\nand determine the cases of equality.", - "Solution_1": "[quote=\"bnick17\"]For positive real numbers $ a$, $ b$, $ c$ and $ d$ such that $ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} d^2 \\equal{} 1$ prove that\n\n$ a^2b^2cd \\plus{} \\plus{} ab^2c^2d \\plus{} abc^2d^2 \\plus{} a^2bcd^2 \\plus{} a^2bc^2d \\plus{} ab^2cd^2 \\le 3/32,$\n\nand determine the cases of equality.[/quote]\r\n$ a^2b^2cd \\plus{} \\plus{} ab^2c^2d \\plus{} abc^2d^2 \\plus{} a^2bcd^2 \\plus{} a^2bc^2d \\plus{} ab^2cd^2\\equal{}$\r\n$ \\equal{}abcd(ab\\plus{}ac\\plus{}ad\\plus{}bc\\plus{}bd\\plus{}cd)\\leq\\frac{1}{16}\\cdot\\frac{3}{2}\\equal{}\\frac{3}{32}.$\r\nThe equality occurs when $ a\\equal{}b\\equal{}c\\equal{}d\\equal{}\\frac{1}{2}.$", - "Solution_2": "[quote=\"arqady\"]$ abcd(ab \\plus{} ac \\plus{} ad \\plus{} bc \\plus{} bd \\plus{} cd)\\leq\\frac {1}{16}\\cdot\\frac {3}{2}$[/quote]\r\n\r\nWhy? Did you get it via AM-GM or something similar?", - "Solution_3": "[quote=\"GioMott\"][quote=\"arqady\"]$ abcd(ab \\plus{} ac \\plus{} ad \\plus{} bc \\plus{} bd \\plus{} cd)\\leq\\frac {1}{16}\\cdot\\frac {3}{2}$[/quote]\n\nWhy? Did you get it via AM-GM or something similar?[/quote]\r\nwe have with AM-GM $ a^2\\plus{}b^2\\plus{}c^2\\plus{}d^2 \\ge 4\\sqrt{abcd}$,and \r\n$ \\frac{a^2\\plus{}b^2\\plus{}a^2\\plus{}c^2\\plus{}a^2\\plus{}d^2\\plus{}b^2\\plus{}c^2\\plus{}b^2\\plus{}d^2\\plus{}c^2\\plus{}d^2}{2} \\ge (ab \\plus{} ac \\plus{} ad \\plus{} bc \\plus{} bd \\plus{} cd)$ ,\r\nso $ abcd \\le \\frac{1}{16}$ and $ ab \\plus{} ac \\plus{} ad \\plus{} bc \\plus{} bd \\plus{} cd \\le \\frac{3}{2}$ ,multiplying we get the result.", - "Solution_4": "This is really weak... (abcd) is weaker than SUM a^2, and SUM ab is weaker than SUM a^2..." -} -{ - "Problem": "Please vote for an ARML shirt if you are going.\r\nThe one I created is MRDraft, and ccy created the other one.\r\nYou should vote for my shirt (please?) :P , but unfortunately I can't vote for my own design (MRDraft) because I can't make it to the competition this year.\r\n\r\nOr vote for Charles. Or neither.\r\nWhatever. They're both good!", - "Solution_1": "I'm excited for the shirt either way.\r\n\r\nWhat happened to combining the two, with ccy's design on the front, and your fractal thingy on the back?", - "Solution_2": "Honestly, I see that as the best solution.\r\n(maybe cuz i'm losing :P )\r\nBut I really, really like his design too.\r\nCan you extend the poll for a (both) option?", - "Solution_3": "I can't, but send an email to Ted. Mention that there was a fair bit of support on AoPS for it.", - "Solution_4": "I'm not really sure if the combining would work. The colors don't really match in my opinion... bright for your design and dark for ccy's.", - "Solution_5": "also, to add an option, ted would need to start over the poll, which he is pretty reluctant to do, i bet" -} -{ - "Problem": "I am trying to write an algorithm which, given two or more strings, searches to find any characters from A to Z which are common to at least two of the strings. Once it finds a match, it needs to output a string with the relation in it, such as \"string_1[2]==string_2[5]\". For example here is a sample input/output:\r\n\r\nInput:\r\n\"*_A436B56\"\r\n\"0231A15534846B\"\r\n\"A4\"\r\n\r\nOutput:\r\n\"string_1[2]==string_2[4]==string_3[0]\"\r\n\"string_1[6]==string_2[13]\"\r\n\r\nOne last thing I forgot to mention: If a letter is contained in a string, then all letters preceding it alphabetically must also be contained in a string. i.e. if 'D' is in a string, then A,B, and C must also be inside some string.\r\n\r\nI mainly need speed for this algorithm, since I am going to try to implement it on a 15MHz device. I was thinking of just searching each string for A and finding matches, then B, and so on. This is pretty slow if there are a lot of characters, though. Any ideas for how to do it faster?", - "Solution_1": "Well why don't u do this:\r\n\r\nfor each char from A to Z keep array flag[ max number of strings]\r\n\r\nthat meens u r to declrae array: flag[26][max # of strings]\r\n\r\nthen flagging :P", - "Solution_2": "I'm not sure if there is a faster way, but as you search for each letter you could cut down the string length, for example:\r\n\r\nCreate an array of LinkedLists or something similar one representing each letter\r\nEach time a letter is found you can add it to its corresponding linked list in the array and shorten the length of the string:\r\n\"*_A436B56\" \r\nWould become\r\n\"(3)436B56\"\r\nAfter the first search\r\nThen\r\n\"(7)56\"\r\nSo you are not searching the same values over and over\r\n\r\nThis is basically just using the array of linked lists and then you will only have to search through the string once. When you find a letter add it to its corresponding linked list.", - "Solution_3": "Basically, you have to use the Dynamic Programming Principle. \r\nUsing this technique, you don't need to search each of the given strings for every member of the alphabet string (which takes a huge amount of energy). Search everything once only and store it for later use.", - "Solution_4": "If any letter after 'A' is in the string, then 'A' is in the string, right? So why don't you just search the string for 'A's? Or do you have to output all matches?" -} -{ - "Problem": "Hey all when this is me in 3 years:\r\n\r\nNational Honor Society, Mu Alpha Theta, Cum Laude Society\r\nNational Merit Scholar\r\n2350 SAT, 35 ACT\r\n4.0 GPA\r\nCaptain of Scholars Bowl Team, Science Bowl team, Science Olympiad, Math Team\r\n#1 Singles player on Varsity Boys Tennis Team\r\nPiano for 6 years\r\nFew hundred hours of volunteering\r\n10 or 11 APs, all 4+ on exam\r\nValedictorian\r\n\r\nWill I be able to get into Harvard or MIT or any of those? If not, why?", - "Solution_1": "Probably, but not necessarily. Why? That's a good start. But you might not get it, and you wouldn't get it if you fail to demonstrate on your essays that your numerical credentials are matched by your ability to think critically and be an interesting person. \r\n\r\nAlso, some of those might be completely worthless, depending on your school. One time when I was in high school I tutored a college student in physics. He was top of his class in high school, but he needed help in the intro physics class at a mediocre state school. Why? His high school just wasn't that great. So don't assume that admissions officers are dumb enough to view \"valedictorian\" or \"captain of XXX\" really highly without bothering to check how good the other people at your school probably were. \r\n\r\nPersonally, I think if you're at a school where all of those actually mean something, there's not a snowball's chance in h3ll that you'll actually manage to obtain every last one of those credentials.\r\n\r\n[hide=\"More details\"]Here's my personal opinion of each of those: \nNational Honor Society: worthless. At most schools you can be in this by doing almost nothing. \n\nMu Alpha Theta: almost worthless. Admissions officers will be able to figure out that you're good at math from more important criteria, probably. (Hopefully you will have some decent competition results to mention, and good AP and SAT math scores.) \n\nCum Laude Society: I'm not sure what this is, but I'm assuming it's similar to NHS, so the same comment applies. \n\nNational Merit: It's not that hard. Way more people get this than get into Ivies+MIT/Caltech/Stanford. It certainly helps, but it's not a buy-in. \n\n2350 SAT / 35 ACT: Come on, you're so optimistic about everything else, why not assume you're going to get a perfect score? Anyway, all you really need to do is not bomb the standardized tests. If they have two students that are pretty much equivalent, I'll wager the [i]last[/i] thing they'd use to choose between the two of them would be a difference of 20 or 30 points in SAT score. By the way, lots of students with perfect standardized test scores get rejected. Why? Because they don't have anything else to show, since they spent too much time studying for standardized tests and not having any kind of life. \n\n4.0 GPA: This really depends a lot on the grading system at your school. Based on my knowledge of typical American high school grading standards, this is probably almost worthless. My school gives a +1.0 weight to honors and AP classes, and sometimes the number of people who graduate summa cum laude (>4.0) is more than the top 10% of the class. And I went to a decent high school. \n\nCaptain of Scholars Bowl Team: Not sure what this is. If it's similar to quiz bowl or science bowl, then being the captain of the team isn't much different from being on the team, really. It's a position of slightly superior responsibility. It won't mean much unless your team also has good results to show for your efforts. \n\n(Captain of?) Science Bowl team: see what I said above. \n\n(Captain of?) Science Olympiad: again, this is a contest that a lot of people do, so it's not really going to mean much at a place like MIT unless you and your team actually do pretty well at it. \n\n(Captain of?) Math Team: as above. \n\n#1 singles player on varsity tennis team: Is your tennis team actually any good? It is very helpful that you play a varsity sport. But again, this isn't a shoe-in even if your team is pretty good, and if your team is a no-name in the tennis world then it's probably almost no different than if you say \"I play tennis a lot.\" \n\nPiano for 6 years: Can't hurt. Not a buy-in. I'd say about 80% of the people I know played an instrument when they were in high school. It's not an unusual thing. (Also, just so you know, piano is the most common instrument. If you want to get admitted to a school because of the instrument you play, play something weird like a bassoon or a bagpipe. And be good at it.) \n\nFew hundred hours of volunteering: Probably good. Volunteering at what? If it's BS volunteering, like tutoring for extra credit after class or playing piano for your church/other religious institution/some social institution, then it's probably not really worth that much. If it's like Habitat for Humanity and stuff, serious volunteering, then that's pretty good. But again, no single thing can buy your way in. \n\n10 or 11 APs, all 4+: Ho hum. Do 20 APs and that'll be pretty spectacular. Or get 5s on all 11 and that might draw a bit of attention. But it's not really that hard to get a 4 (or even a 5, in most cases) on an AP test, so unless you do it for a LOT of them, it's not going to make a big difference. If you took ones that you had to study on your own for (and did well), this might help somewhat because it shows that you're motivated to learn things on your own (and can actually succeed too). \n\nValedictorian: As I said before, depends entirely on your school. At my school that would be pretty good -- it's a class of 500+ and we tend to have 10+ National Merit finalists, so there's decent competition. At the high school across town, though, probably any of the top 20 people at my school could have been valedictorian if they transfered over there. Maybe more than that. And that high school wasn't even that bad. \n\nSorry if I sound overly critical. I just have the impression that at top schools, it's personality and motivation and passion that are the strongest traits, not numbers. If you put in too much effort to craft a perfect resume, admissions officers might realize this and doubt whether you have actual useful skills to back up what you've done. They don't want someone who appears to be a perfect student on paper, but is actually only good at making themselves look good on paper. They want people who actually are good at things, even if they aren't so great at making themselves look better than they really are. \n\nFor the record, I'm at Princeton, and here's how little I had of your list: \n\nNational Merit\nGood SAT/SATII scores\nHigh GPA\nCaptain of Science Bowl team; vice president or president or something of Math/Science Team. (Note: these teams had good results in state and national contests.) \nSwimming (not varsity. no good times or competition results or anything.) \n17 AP tests. One 3, the rest 4 or 5. \n\nThat's not even half of what you \"plan\" to have. I think I got in because I had good essays and recommendations, and because they thought that I was motivated and would take advantage of the opportunities available at Princeton. Not because my list of accomplishments got a high score in their \"applicant rating equation\". [/hide]\r\n\r\nDo what you like, not what you think will get you into a good college. The first is more likely to get you into college than the second, and you'll enjoy it more too.", - "Solution_2": "Xev left out the most important thing that I think people neglect to take into account: college admissions are a probabilistic process. There is no grand deterministic algorithm applied by Harvard (or MIT, or Yale, or (nowadays) the University of Michigan). For any statistic you like, each of these schools gets more applicants above the mean of its student body than it has spots available to freshmen. (My personal favorite is version of this is that Harvard gets enough applicants that it could fill its freshman class entirely with 1600 SAT scores* (probably not 2400 in the new system, but I don't know) or entirely with high school valedictorians, if it so chose.) As a result, \"anything you do\" is likely to enhance you chances of getting in and nothing you is likely to guarantee you a place at any given school. The trick, of course, is that you don't have to apply to just a single school -- people who raise their likelyhood of getting into Harvard are probably also raising their likelyhood of getting into MIT, and Yale, and the University of Michigan.\r\n\r\n\r\n* Apparently this is not (maybe \"is no longer\"?) true -- see Tokenadult's post that follows. Having actual data is a wonderful thing.", - "Solution_3": "Mathematically, based on the size of its entering class, Harvard could definitely NOT fill its class with students who have 2400s on the three-section SAT. \r\n\r\nhttp://www.collegeboard.com/prod_downloads/highered/ra/sat/composite_CR_M_W_percentile_ranks.pdf \r\n\r\nIt couldn't even quite fill its class with students who score 1600 based on two-section consideration of math and critical reading scores. \r\n\r\nhttp://www.collegeboard.com/prod_downloads/highered/ra/sat/SAT_composite_CR_M_percentile_ranks.pdf \r\n\r\nBut even though the pigeonhole principle \"forces\" Harvard to admit some students with less-than-perfect SAT I scores, what I find interesting is that Harvard (and every other college worthy of note in this regard) passes over some perfect scorers to admit some students with other desirable characteristics. The linear extension considered by admission offices of the partially ordered set of college applicants does not just lie along the dimension of test scores. \r\n\r\nJBL has it right: you can do things to increase your probability of admission, but you can't do much of anything to guarantee admission. Apply widely to all of the colleges that are of interest to you. While you are still in high school, enjoy your studies and your activities and do your personal best in them. When you apply, be sure to mention anything unique about you that makes you a good match for each college. Then submit the applications, and enjoy high school while you wait for the response of each admission office.", - "Solution_4": "Lol wow thanks for the truth Xevarion. Heres some extra info to clear up:\r\n\r\nI say I'd have a close to perfect if not perfect standardized test scores because i scored a 30 on the ACT in 7th grade, 2026 on SAT in 8th grade, 31/32 PLAN in 9th, and 221 on PSAT in 9th. Look im not bragging or boasting or anything, im just telling you.\r\n\r\nBy 4.0 GPA I mean unweighted, if it was weighted, it would only be a 4.6 because thats the highest my school gives with APs (kinda stupid)\r\n\r\nI can only take 10 or 11 APs because you can only start taking APs in 11th grade at my school (another stupid thing)\r\n\r\nGoogle Cum Laude Society\r\n\r\nI am in Habitat for Humanity, but im not doing so much right now because im only 14\r\n\r\nI agree national honor societ and MAT are pretty worthless\r\n\r\nCurrently, our Science Olympiad team almost always goes to national, Scholars Bowl team finishes as finalists at state tourney, Science Bowl team almost always goes to nationals, Math Team the best in the state for our school size.\r\n\r\nOur tennis team was runner up at state last year, i was #5 ( but ya ure probably right our team wont be that great when im #1) \r\n\r\nWe have a class size of 72, and we have approximately have 10 naitonal merit semifinalists every year(thats amazing) And my grade has always had the highest test scores and is expected to have almost 20% of the class be national merit semifinalists (with me being the smartest in our grade, and one of the only AZNs XD)\r\n\r\nAnd of course, i perfectly know that statistics and numbers arent the most important, but they're up there.\r\nI just want a heads-up\r\n\r\nSo what else did u do that made Princeton accept you?\r\n\r\nBtw I live in Alabama and attend Randolph School", - "Solution_5": "Sounds like you do go to a pretty good school. Did your school go to NSB in 2004 or 2005? I don't remember playing your school, but there were a lot of matches and it's been a while... \r\n\r\nAs JBL and tokenadult and I already said, chance is a part of it. I had good essays and good recommendations. It probably helped that my dad went to Princeton. I had some leadership kind of stuff (like I created a chess club). That's about it, I guess...", - "Solution_6": "[quote=\"Xevarion\"]at a mediocre state school.[/quote]\r\nThat's a harsh assessment, Mr. Princeton. :cool:", - "Solution_7": "[quote=\"mlok\"]That's a harsh assessment, Mr. Princeton. :cool:[/quote]\r\nSorry. :P You should take it only as a criticism of the intro physics class, which (in my opinion) was a bit horrific.", - "Solution_8": "[quote=\"lifesapain\"]2026 on SAT in 8th grade[/quote]\r\n\r\nThis is the most impressive statistic on your list. :wink:\r\n\r\nAnyway, everything you have is good, but do you see something that if someone (an admissions officer, specifically) read they would be like \"Wow, I really want this guy at my school.\" Once you do, I think you'll have a much better chance at getting into Harvard/MIT. But even if you don't, it's not that big of a deal. The point is that you do something unique that you enjoy and that's what sets you apart.", - "Solution_9": "I'm going to echo what a couple of other people have said and tell you to do what interests you, not what you think will get you into college. I'll add that you don't look like the kind of person who needs to worry about tests and other statistics - just don't slack off too much in your classes and you'll be fine in that regard. :) There are more important things to worry about!", - "Solution_10": "As it's been said before in this thread (and tons more times in this forum) the top schools really like to admit students that have something on their application that stands out in some way. From the people I know at MIT, here are a few things I've noted that probably attributed a lot to why they're here (the rest of their app is also solid, of course):\r\n\r\nResearch - If you can do some research in high school, that'll go a long way. If you can get into RSI, then you're doing [i]really[/i] well (clearly, since they only accept the top of the top). But doing research at a state school near you for a summer or so does a lot as well. Some places have programs that pair up students with mentors. If there's nothing like that around, try emailing a prof at a local university to see if you they'd be willing to mentor you. I admit that it'd be hard to find a prof that'll accept a high school student, but if you were successful, just think about how impressed elite schools would be.\r\n\r\nStarting a Business - Starting a business and getting it of the ground requires a lot of hard work and determination. Colleges definitely want this type of person.\r\n\r\nSo that's just a few ideas. There are other ways to stand out. Be creative and good luck.", - "Solution_11": "joml88 is right about research. There are also lots of research contests you can participate in; some of those are team contests, so you could do them with your friends. I think Toshiba has a contest, there are the famous Siemens and Intel competitions, and probably other corporate-sponsored ones too that I don't know about. You can look for more these kinds of opportunities yourself, of course. Find something that looks fun, and then do a really good job of it. That's the way to go! ;-)", - "Solution_12": "[quote]If there's nothing like that around, try emailing a prof at a local university to see if you they'd be willing to mentor you. I admit that it'd be hard to find a prof that'll accept a high school student, but if you were successful, just think about how impressed elite schools would be.[/quote]\r\n\r\nSeveral professors at my local college (not a slouchy one, either) mentored me while I was still a high school student. It was one of the best things that ever happened to me. The intellectual stimulation was wonderful, the experience I gained helped me immensely in college, and I'm still friends with some of the people I met (in fact, one of the professors still writes letters of recommendation for me). \r\n\r\nI'd say, if you want to try to find a professor to work with, you need to show him or her that you already have an exceptionally good knowledge base and are willing to work hard, that you know what the professor does and are really interested in it (reading some of their papers really helps), and that you can't get any of this from your high school. So, basically, you should show why you'd be rewarding to mentor and why you wouldn't be wasting their time. :D", - "Solution_13": "Oh yea definitely, dont' get me wrong i enjoy scholar's bowl and math team and such. I like academic competitions and competing with other students across the nation. those are my interests. And i keep my grades up because i always want to be #1 anyway, not just so it will look better for college. \" i strive to be the best\" -lol i know sounds corny\r\n\r\nAnyway, im only taking the non-science and no-math APs purely for college ( i h8 english and social studies)\r\n\r\nWhat can I do during the summer to help me out? The researching part?", - "Solution_14": "Q: What's sillier than asking someone to handicapping your chances of getting into Harvard?\r\n\r\nA: Asking someone to handicap the chances of your hypothetical future self.\r\n\r\nSorry, I don't mean to be so obnoxious. I really do sympathize with today's students. It's a ridiculous academic arms race out there, and it's only getting harder and harder to get in to these schools.\r\n\r\nHere's my honest answer: Sounds like a strong application, but it's no shoe-in. Almost nothing is. These top schools do not admit by the numbers. Your raw info is more impressive than that of lots of people who get accepted and less impressive than that of lots of people who get rejected. I've been doing interviews for Harvard for 5 years now, and I scarcely have a clue what Harvard admissions is doing. You would be terrified to know about some of the students I've seen Harvard reject. The fact is that Harvard admits a *class*. It is not exactly a meritocracy, and it's not even really intended to be, as far as I can tell.\r\n\r\nI don't believe in giving advice on how to get into college, because I don't believe I have any useful advice to offer. I think that's what they talk about over at College Confidential.", - "Solution_15": "okay from now on everyone can refer to this person as \"Bob\" and not me. This is random person bob's application and he needs help.", - "Solution_16": "not too shoot down your hopes or anythingk, but you better do something much more impresive than just be cdaptain of your math team. i know randolph school, and its pretty mediocre at best when ti comes to math. kurt was leaps and bounds ahead of everyone else on that math team and he wasn't even that good just in teh spectrum of AL math students.", - "Solution_17": "ummm seva? i didnt say i was \"captain\" of the math team. There is no captain and it wouldnt mean anything either. ya and unlike vestavia hills, which is only good at math (cuz of ure 24/7 practicing), randolph has other strengths too...", - "Solution_18": "Let's not make dumb college threads, mmk?", - "Solution_19": "I'm not usually one to defend Vestavia, but shoot. They aren't even that good in the spectrum of national mathematics. They are good as a product of practice. Just like you are good at tennis as a product of practice. I think most college admissions officers can spot a filler extra-curricular activity. NHS/MAO are pretty useless. Cum Laude Society? really. Can you quantify a few hundred hours of volunteering? (my suggestion for that is to have a set cause and work towards it.)\r\n\r\nAlso, you're banking a lot on things remaining the same. I understand that Randolph is decent at Scholars Bowl, to some extent, behind the beast that is Brindlee Mountain, but is your current class going to be that good in a few years? You could be an amazing quiz bowl player that a place like MIT would want. Eh.\r\n\r\nI think focus is kind of important. There are thousands upon thousands of people who are exactly like you or better based on what you've said. \r\n\r\nWhat makes you special in the eyes of the college admissions officer?\r\n\r\nThats harsh, but you know, there is some truth to it. You can bring out alot of that in your essays but it needs to show on your application as well.", - "Solution_20": "Interesting information, i have similar credentials to \"Bob\" and i found this very helpful. \"Be yourself\" is the new catch phrase of college admisssions advisers everywhere yet no one seems to understand this simple approach.\r\n I personally have been doing things that interest me and have found that i do well in school because i like school, every part from the romantic language of shakespeare to the beautiful understanding that came with physics. \r\nHowever, because this is such a long post i stayed up reading this rather my new john grisham book, The Appeal, therefore hindering me from my pursuit to do what i like, as many have suggested =(. \r\nBy the way I enjoyed the criticism put forth by the princeton guy, xev. It seems that too many high schools lack any real challenges and therefore instill an intellectual bravado in its myriad of 4.0 students.\r\nI look forward to applying to Harvard next year and wish \"bob\" luck as well." -} -{ - "Problem": "Prove that :\r\nx 2 y+y 2 z+z 2 x \\leq x ^3 +y ^3 +z ^3 \\leq 1+1/2(x^4+y^4+z^4).\r\nfor non negative real numbers x,y,z satisfying the condition :x+y+z=2", - "Solution_1": "sum(x^3)>=sum(x^2y) was discussed on the forum .. it's post ineq C51 .. it's done by rearrangement.. the condition x+y+z=2 for this ineq is extra... it's probably for the 2nd one.. I haven't tried it yet! :D", - "Solution_2": "I won't insist on the left one, since it is easy. I say only that it can be proved very nice with Cauchy. The sceond inequality is really nice:\r\n Write it as sum x^3(1-x/2)<=1, or sum x^3(y+z)<=2, or sum xy(x^2+y^2)<=2. Do you recognise now IMO 1999? \r\n We have sum xy(x^2+y^2)<=(xy+zy+zx)(x^2+y^2+z^2)=1/2*(2sum xy)*(4-2sum xy)<=1/2*(4/2)^2=2, by AM-GM. Very nice problem!", - "Solution_3": "Very nice sln to very nice problem. \r\n\r\nI just want to warn A1lqdSchool: Dont suggest in the forum recent problems of journal \"Mathematics and Youth\" (THTT in Vietnamese), because it is rule of our forum. You can put problems from M&Y, but only \r\nold one", - "Solution_4": "Harazi, what do you means saying:\r\n\r\n\"I say only that it can be proved very nice with Cauchy.\" \r\n\r\nPerhaps\r\n\r\na^3+b^3+c^3= 1/3(2a^3 +b^3) +1/3(2b^3+c^3)+1/3(2c^3+a^3) \\geq a^2*b+ b^2*c+c^2*a\r\n\r\n(but this uses AM-GM and not Cauchy)\r\n\r\nor something else using Cauchy.\r\n\r\nThank you very much.", - "Solution_5": "Indeed, the solution with Cauchy is a little bit non-trivial:\r\n (x^2y+y^2z+z^2x)^2<=(x^3+y^3+z^3)(xy^2+yz^2+zx^2) and\r\n (xy^2+yz^2+zx^2)^2<=(x^3+y^3+z^3)(x^2y+y^2z+z^2x). If x^2y+y^2z+z^2x>=xy^2+yz^2+zx^2 everything comes from the first relation. Otherwise, use the second relation.", - "Solution_6": "Very interesting and very nice trick.", - "Solution_7": "When we have equality on the second inequality?\r\nThanks", - "Solution_8": "I agree with Namdung that one should not post the most recent problems of a magazine in order not to harm anybody. And please be more specific about the source: not only THTT, but also the year and volume please. It is the same with the olympiads: e.g. not just Iran but the Iran Olympiad, year, round and author if available. Thank you very much for keeping to the rule.", - "Solution_9": "nevermind of my question....we have equality iff two of x,y,z are 1 and the other one is 0\r\nThanks", - "Solution_10": "Let $x,y,z$ be nonnegative real numbers satisfying the condition $x+y+z=2$, then\r\n\r\n$x^{3}+y^{3}+z^{3}+\\frac{11}{12}\\min\\{x^{2},y^{2},z^{2}\\}\\leq 1+\\frac{1}{2}(x^{4}+y^{4}+z^{4})$,\r\nwith equality if $x=y=z=\\frac{2}{3}$." -} -{ - "Problem": "Whenever I leave the forum (ex. click my classes) I get logged off?", - "Solution_1": "Make sure you have cookies enabled." -} -{ - "Problem": "How can we prove any two basis of a given vector space have the same cardinality??", - "Solution_1": "Look just a few posts below this one. This question was also posted by alekk. Myth and I gave solutions.", - "Solution_2": "Thanks, man!!Problems Solved :D", - "Solution_3": "[quote=\"harazi\"]Myth and I gave solutions.[/quote]\r\nI have already completely forgotten about this. :blush:" -} -{ - "Problem": "For arbitrary $a_1,\\ldots,a_n \\in \\mathbb{C} \\setminus \\{0\\}$, does there necessarily exist a $z \\in \\mathbb{C}$ such that $1+e^{a_1z} + \\cdots + e^{a_nz} = 0$?", - "Solution_1": "Yes, because there is a theorem (due to Hadamard, if I don't mistake) stating that any non-constant entire function $F(z)$ of order $1$ can be represented as $ce^{\\beta z}$ times a canonical product $z^m\\prod_j \\left(1-\\frac z{z_j}\\right)e^{z/z_j}$ over its zeroes. In your situation, the function is, clearly, not a pure exponent, so the canonical product must be non-trivial. Moreover, in this particular case zeroes are infinitely many and one can say a lot about their location.\r\n\r\nThe proof of the Hadamard factorization theorem can be found in any decent textbook on entire functions.", - "Solution_2": "Thank you! But, come to think of it, I just found a completely elementary solution that does not neither Hadamard's result nor Rouche's theorem -- but only the fundamental theorem of algebra. \r\n\r\nFor simplicity, we assume in the sequel that the $a_i$ are reals; the proof easily extends to the general case. Of couse, we may then assume that $a_1 \\leq a_2 \\leq \\cdots \\leq a_s < a_{s+1} \\leq \\cdots \\leq a_n = a$, with $a_n > 0$.\r\n\r\n For every $k$, we consider the polynomial\r\n\\[\r\nf_k(x) := 1 + x^{[ka_1]} + \\cdots + x^{[ka_n]} \\in \\mathbb{C}[x].\r\n\\]\r\nLet $x_k$ be any root of $f_k$, and define $z_k$ by $e^{z_k} = x_k^k$ and $\\text{Im} \\, z_k \\in [0,2\\pi)$ (that is, $z_k = k \\, \\textrm{Log}_0 \\, x_k$). \r\n\r\nWe claim that the sequence $\\{ |z_k| \\}$ is bounded. Assuming the contrary, for every $M>0$ there is a $k=k(M)$ with $| \\Re (z_k)| > M$ (because the imaginary part of $z_k$ is bounded by our choice of the principal branch of the logarithm). For this $k$, we either have $|e^{z_k}|>e^M$ or $|e^{z_k}| < e^{-M}$, hence either $|x_k|^k>e^{M}$ or $|x_k|^k < e^{-M}$. We see that both cases are impossible when $M$ is large enough. Indeed, for the first case (the second one is treated analogously), $f_k(x_k)=0$ entails\r\n\\[\r\n(n-s+1)x_k^{[ak]} = - x_k^{[a_sk]} - x_k^{[a_{s-1}k]} \\cdots - 1\r\n\\]\r\nBy the triangle inequality, $|x_k|^{[ak]} \\leq 1 + |x_k|^{[a_{s-1}k]} + \\cdots + |x_k|^{[a_1k]}$, so\r\n\\[\r\n|x_k|^{-b_s} + |x_k|^{-b_{s-1}} + \\cdots + |x_k|^{-b_1} \\geq 1,\r\n\\]\r\nwhere $b_s = [ka]; b_{s-1} = [ka] - [ka_{s-1}], \\ldots, b_1 = [ka] - [ka_1]$ are all $> ck$ for some $c>0$ and large enough $k$. But then, $|x_k|^{-b_i} < |x_k|^{-ck}< e^{-cM}$ for all $i$, giving $se^{-cM} > 1$. Clearly, this cannot hold when $M$ is large, a contradiction. Hence, we have the desired boundedness of $\\{ |z_k| \\}$.\r\n\r\nIn this manner, we proved that the all points $\\{z_k\\}_{k \\geq 1}$ are contained in a bounded, and hence in a compact subset of the plane (notice that our choices of $x_k$ as roots of $f_k$ were [i]arbitrary[/i]). By the compactness, the set $\\{z_k\\}_{k \\geq 1}$ has at least one limit point, and let $z$ be such point. In other words, there is a subsequence $z_{k_1},z_{k_2},\\ldots$ of $\\{z_k\\}$ converging to $z$. For this $z$, we have\r\n\\[\r\n0 = \\lim_{i \\to \\infty} f_{k_i} (e^{\\frac{z}{k_i}}) = \\lim_{i \\to \\infty} \\big( 1 + e^{z\\frac{[k_ia_1]}{k_i}} + \\cdots + e^{z \\frac{[k_ia_n]}{k_i}} \\big) = 1 + e^{a_1z} + \\cdots + a^{a_nz},\r\n\\]\r\nas desired.\r\n\r\nThe general case when $a_i \\in \\mathbb{C}$ can be treated in exactly the same way.\r\n\r\n--Vesselin", - "Solution_3": "Furthermore, in the case when the $a_i$ are reals that we described above, and when $a_n>a_{n-1}$ strictly (that is, the maximal $a_i$ occurs just once, or equivalently, $s=n$), each $f_k$ for $k$-large enough has a root of absolute value $\\geq 1$ (because the roots multiply to $\\pm 1$); choosing $x_k$ as this root ($|x_k| \\geq 1$), we get that $\\Re (z_k) \\geq 0$, i.e. we bound the $z_k$ in the right half-plane, so we have $\\Re (z) \\geq 0$. In this manner, we obtain from the above proof the following:\r\n\r\n[b]Result. [/b] Let $0 < a_1 < a_2 < \\cdots < a_n$ be distinct positive reals, and let $b_0,b_1,\\ldots,b_{n-1} \\in \\mathbb{C}$ be arbitrary complex numbers with $|b_0| \\geq 1$. Then, the exponential sum\r\n\\[\r\nf(z) := e^{a_nz} + b_{n-1}e^{a_{n-1}z} + \\cdots + b_1e^{a_1z} + b_0\r\n\\] \r\nhas a zero $z$ with $\\Re (z) \\geq 0$. Moreover, there is a $B>0$ such that $|\\Re (z)| < B$ for any zero $z$ of $f$. \r\n\r\nIn particular, there is a $z$ with $\\Re (z) > 0$ satisfying $1 + e^{z \\ln{2}} + e^{z \\ln{3}} = 0$, providing another solution to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19406]this[/url] problem. This result also implies the following\r\n\r\n[b]Corollary. [/b] Let $a_1,\\ldots,a_n, n \\geq 2,$ be arbitrary distinct positive reals. Then, there exists a non-trivial continuous function $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying\r\n\\[\r\n f(a_1x) + f(a_2x) + \\cdots + f(a_nx) = 0\r\n\\]\r\nfor all $x$.\r\n\r\n--Vesselin", - "Solution_4": "What do you think, does this simple polynomial approximation stand a chance of being correct? :o It's 3a.m. local time here and I'm very sleepy, so I may have overlooked something ... but it seems clear to me, anyway.", - "Solution_5": "Actually, it is clear that if $f$ is an entire function with no zeros, then $f(z)=e^{g(z)}$, where $g$ is entire. In the case when $f$ has order $1$, $g$ must be a linear function, and it is clear that our exponential sum does not have the form $e^{az+b}$. So, that every non-trivial exponential polynomial (not of the form $ce^{\\beta z}$) has a zero over the complex numbers (infinitely many, in fact) can be shown without Hadamard's theorem (or Rouche's theorem)." -} -{ - "Problem": "[u][b]Problem (zaizai-hoang):[/b][/u]\r\nLet a,b,c are positive number such that: $ab+bc+ca+abc=4$. Prove that:\r\n$3(a^{2}+b^{2}+c^{2})+abc\\ge 10$", - "Solution_1": "Here's my proof. \r\n\r\n[b]Lemma.[/b] for $a,b,c \\in R^{+}$, $a^{2}+b^{2}+c^{2}+2abc+1 \\geq 2(ab+bc+ca)$\r\n\r\nLemma-pf) Let $a=x^{3}, b=y^{3}, c=z^{3}$. then ineq is changed to \r\n\r\n$x^{6}+y^{6}+z^{6}+2x^{3}y^{3}z^{3}+1 \\geq 2\\sum_{cyclic}x^{3}y^{3}$\r\n\r\nusing AM-GM, we get $x^{3}y^{3}z^{3}+x^{3}y^{3}z^{3}+1 \\geq 3x^{2}y^{2}z^{2}$\r\n\r\nso ETS $x^{6}+y^{6}+z^{6}+3x^{2}y^{2}z^{2}\\geq 2\\sum_{cyclic}x^{3}y^{3}$\r\n\r\nHere, from Schur ineq case $n=1$ and AM-GM, we get $x^{6}+y^{6}+z^{6}+3x^{2}y^{2}z^{2}\\geq \\sum_{sym}x^{4}y^{2}\\geq 2\\sum_{cyclic}x^{3}y^{3}$\r\n\r\nso the lemma is proved. \r\n\r\n[b][color=blue]Let's start to prove the problem ! [/color][/b]\r\n\r\nIf $abc>1$ then $ab+bc+ca\\geq 3(abc)^{\\frac{2}{3}}> 3$\r\n\r\nso the condition $ab+bc+ca+abc=4$ can't be satisfied. Therefore, $abc \\leq 1$\r\n\r\n$3(a^{2}+b^{2}+c^{2})+abc \\geq 10$\r\n\r\n$\\Leftrightarrow$ $3(a^{2}+b^{2}+c^{2}+2abc+1 ) \\geq 13+5abc$\r\n\r\nUsing lemma, ETS $6(ab+bc+ca) \\geq 13+5abc$\r\n\r\n$\\Leftrightarrow$ $24-6abc \\geq 13+5abc$\r\n\r\n$\\Leftrightarrow$ $1 \\geq abc$\r\n\r\n$Q.E.D.$\r\n\r\n\r\n\r\nI was fool... :blush: \r\n\r\n[b]Second proof.[/b]\r\n\r\n$abc \\leq 1$ as I showed above. \r\n\r\ncuz $a^{2}+b^{2}+c^{2}\\geq ab+bc+ca$\r\n\r\nETS $3(ab+bc+ca)+abc \\geq 10$.\r\n\r\n$\\Leftrightarrow$ $12-2abc \\geq 10$\r\n\r\n$\\Leftrightarrow$ $1 \\geq abc$\r\n\r\n$Q.E.D.$ \r\n\r\n :blush: :blush:", - "Solution_2": "A simpler way ;) (it helps sometimes to know nothing complicated)\r\n\r\nChange a 4 on the RHS to $ab+bc+ac+abc$. We get\r\n$2(a^{2}+b^{2}+c^{2}) \\ge 6$\r\n\r\nBut we need $ab+bc+ac+abc = 4$. If $ab+bc+ac < 3$, then we have $3 > ab+ac+bc \\ge 3(abc)^{2/3}$, whence $abc < 1$ and our condition is contradicted. $ab+bc+ac \\ge 3$.\r\n\r\nWe conclude by $2(a^{2}+b^{2}+c^{2}) \\ge 2(ab+ac+bc) \\ge 6$.\r\n\r\nedit: I see [b]Chang Woo-JIn[/b] has a better solution also, faster than mine! :P", - "Solution_3": "[quote=\"zaizai-hoang\"]Let a,b,c are positive number such that: $ab+bc+ca+abc=4$. Prove that:\n$3(a^{2}+b^{2}+c^{2})+abc\\ge 10$[/quote]\r\n\r\nFrom $ab+bc+ca+abc=4$, there's positive $p,q,r$ such that $a=\\frac{2p}{q+r}, b=\\frac{2q}{r+p}, c=\\frac{2r}{p+q}$. Applying this, the original one becomes \\[\\sum \\frac{6p^{2}}{(q+r)^{2}}+\\frac{4pqr}{(p+q)(q+r)(r+p)}\\geq 5\\] or \\[6\\sum_{cyc}p^{6}+12\\sum_{sym}p^{5}q+\\sum_{sym}p^{4}q^{2}+14\\sum_{cyc}p^{4}qr \\geq 10 \\sum_{cyc}p^{3}q^{3}+14 \\sum_{sym}p^{3}q^{2}r+24p^{2}q^{2}r^{2}\\] which can be obtained by summation of the following: \\[6\\sum_{cyc}p^{6}\\geq 6\\sum_{cyc}p^{3}q^{3}\\] \\[2\\sum_{sym}p^{5}q \\geq 4\\sum_{cyc}p^{3}q^{3}\\] \\[10\\sum_{sym}p^{5}q \\geq 10 \\sum_{sym}p^{3}q^{2}r\\] \\[\\sum_{sym}p^{4}q^{2}\\geq \\sum_{sym}p^{3}q^{2}r\\] \\[6\\sum_{cyc}p^{4}qr \\geq 3\\sum_{sym}p^{3}q^{2}r\\] \\[8\\sum_{cyc}p^{4}qr \\geq 24p^{2}q^{2}r^{2}.\\]", - "Solution_4": "[quote=\"Sung-yoon Kim\"][quote=\"zaizai-hoang\"]Let a,b,c are positive number such that: $ab+bc+ca+abc=4$. Prove that:\n$3(a^{2}+b^{2}+c^{2})+abc\\ge 10$[/quote]\n\nFrom $ab+bc+ca+abc=4$, there's positive $p,q,r$ such that $a=\\frac{2p}{q+r}, b=\\frac{2q}{r+p}, c=\\frac{2r}{p+q}$. Applying this, the original one becomes \\[\\sum \\frac{6p^{2}}{(q+r)^{2}}+\\frac{4pqr}{(p+q)(q+r)(r+p)}\\geq 5\\] [/quote]\nThen just use: $8pqr\\ge (p+q)(q+r)(r+p)$ and $\\sum\\frac{p^{2}}{(q+r)^{2}}\\ge \\frac{3}{4}$[/quote]", - "Solution_5": "[quote]\nThen just use: $8pqr\\ge (p+q)(q+r)(r+p)$ and $\\sum\\frac{p^{2}}{(q+r)^{2}}\\ge \\frac{3}{4}$[/quote]\r\n\r\nI think you have a mistake. $(p+q)(q+r)(r+p)\\ge 8pqr$ by AM-GM", - "Solution_6": "[quote=\"Chang Woo-JIn\"]Here's my proof. \n\n[b]Lemma.[/b] for $a,b,c \\in R^{+}$, $a^{2}+b^{2}+c^{2}+2abc+1 \\geq 2(ab+bc+ca)$\n\nLemma-pf) Let $a=x^{3}, b=y^{3}, c=z^{3}$. then ineq is changed to \n\n$x^{6}+y^{6}+z^{6}+2x^{3}y^{3}z^{3}+1 \\geq 2\\sum_{cyclic}x^{3}y^{3}$\n\nusing AM-GM, we get $x^{3}y^{3}z^{3}+x^{3}y^{3}z^{3}+1 \\geq 3x^{2}y^{2}z^{2}$\n\nso ETS $x^{6}+y^{6}+z^{6}+3x^{2}y^{2}z^{2}\\geq 2\\sum_{cyclic}x^{3}y^{3}$\n\nHere, from Schur ineq case $n=1$ and AM-GM, we get $x^{6}+y^{6}+z^{6}+3x^{2}y^{2}z^{2}\\geq \\sum_{sym}x^{4}y^{2}\\geq 2\\sum_{cyclic}x^{3}y^{3}$\n\nso the lemma is proved. \n\n :blush: :blush:[/quote]\r\nNice solution. I have a solution to prove that your [b]Lemma[/b]\r\nWLOG we assume\r\n $(a-1)(b-1)\\ge 0 \\Leftrightarrow ab+1-a-b\\ge 0 \\\\ \\Leftrightarrow abc+c-ac-bc\\ge 0$ (because $c\\ge 0$)\r\nNow the inequality equavalent:\r\n$(a^{2}-2ab+b^{2})+2(abc+c-ac-bc)+(c^{2}-2c+1)\\ge 0 \\\\ \\Leftrightarrow (a-b)^{2}+2c(a-1)(b-1)+(c-1)^{2}\\ge$\r\nIt true. :P I think this lemma as same as VMO 2006, group B.\r\nThis problem my solution use mixing variable :D", - "Solution_7": "[quote=\"zaizai-hoang\"][quote=\"Chang Woo-JIn\"]Here's my proof. \n\n[b]Lemma.[/b] for $a,b,c \\in R^{+}$, $a^{2}+b^{2}+c^{2}+2abc+1 \\geq 2(ab+bc+ca)$\n\nLemma-pf) Let $a=x^{3}, b=y^{3}, c=z^{3}$. then ineq is changed to \n\n$x^{6}+y^{6}+z^{6}+2x^{3}y^{3}z^{3}+1 \\geq 2\\sum_{cyclic}x^{3}y^{3}$\n\nusing AM-GM, we get $x^{3}y^{3}z^{3}+x^{3}y^{3}z^{3}+1 \\geq 3x^{2}y^{2}z^{2}$\n\nso ETS $x^{6}+y^{6}+z^{6}+3x^{2}y^{2}z^{2}\\geq 2\\sum_{cyclic}x^{3}y^{3}$\n\nHere, from Schur ineq case $n=1$ and AM-GM, we get $x^{6}+y^{6}+z^{6}+3x^{2}y^{2}z^{2}\\geq \\sum_{sym}x^{4}y^{2}\\geq 2\\sum_{cyclic}x^{3}y^{3}$\n\nso the lemma is proved. \n\n :blush: :blush:[/quote]\nNice solution. I have a solution to prove that your [b]Lemma[/b]\nWLOG we assume\n $(a-1)(b-1)\\ge 0 \\Leftrightarrow ab+1-a-b\\ge 0 \\\\ \\Leftrightarrow abc+c-ac-bc\\ge 0$ (because $c\\ge 0$)\nNow the inequality equavalent:\n$(a^{2}-2ab+b^{2})+2(abc+c-ac-bc)+(c^{2}-2c+1)\\ge 0 \\\\ \\Leftrightarrow (a-b)^{2}+2c(a-1)(b-1)+(c-1)^{2}\\ge$\nIt true. :P I think this lemma as same as VMO 2006, group B.\nThis problem my solution use mixing variable :D[/quote]\r\n\r\nThe lemma was posted by Darij:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=19666", - "Solution_8": "If $ a,b,c$ are [color=red]real[/color] numbers such that $ abc \\plus{} bc \\plus{} ca \\plus{} ab \\equal{} 4,$ then\r\n\r\n$ abc \\plus{} 3\\left(a^2 \\plus{} b^2 \\plus{} c^2\\right) \\geq 10.$\r\n\r\n[b]Proof[/b] From $ abc \\plus{} bc \\plus{} ca \\plus{} ab \\equal{} 4,$ there's [color=blue]real[/color] numbers $ p,q,r$ such that \r\n\r\n$ a \\equal{} \\frac {2p}{q \\plus{} r}, b \\equal{} \\frac {2q}{r \\plus{} p}, c \\equal{} \\frac {2r}{p \\plus{} q}.$ \r\n\r\nApplying this, the original one becomes\r\n\r\n$ \\frac {4pqr}{(q \\plus{} r)(r \\plus{} p)(p \\plus{} q)} \\plus{} \\sum{\\frac {6p^2}{(q \\plus{} r)^2}}\\geq 5,$\r\n\r\nwhich can be obtained by the following :\r\n\r\n$ \\frac {4pqr}{(q \\plus{} r)(r \\plus{} p)(p \\plus{} q)} \\plus{} \\sum \\frac {6p^{2}}{(q \\plus{} r)^{2}} \\minus{} 5$\r\n\r\n$ \\equal{} \\frac {(p \\plus{} q \\plus{} r)^2\\{5(q \\minus{} r)^2(r \\minus{} p)^2(p \\minus{} q)^2 \\plus{} 3[(q \\plus{} r)(q \\minus{} r)^2 \\plus{} (r \\plus{} p)(r \\minus{} p)^2 \\plus{} (p \\plus{} q)(p \\minus{} q)^2]^2\\}}{(q \\plus{} r)^2(r \\plus{} p)^2(p \\plus{} q)^2[(q \\minus{} r)^2 \\plus{} (r \\minus{} p)^2 \\plus{} (p \\minus{} q)^2]}$\r\n\r\n$ \\plus{} \\sum{\\frac {(q \\minus{} r)^2(q \\plus{} r \\minus{} 2p)^2}{(q \\plus{} r)^2[(q \\minus{} r)^2 \\plus{} (r \\minus{} p)^2 \\plus{} (p \\minus{} q)^2]}}\\geq 0.$\r\n\r\nBy the way, if $ a,b,c$ are [color=green]positive[/color] numbers, then\r\n\r\n$ (a\\plus{}b\\plus{}c)\\left(abc \\plus{} 3a^2 \\plus{}3 b^2 \\plus{} 3c^2\\minus{}10\\right)\\geq 7(abc \\plus{} bc \\plus{} ca \\plus{} ab \\minus{}4).$", - "Solution_9": "$ 2(a^2\\plus{}b^2\\plus{}c^2) \\ge 10\\minus{}abc\\equal{}6\\plus{}ab\\plus{}bc\\plus{}ca$\r\n$ (a\\minus{}b)^2\\plus{}(b\\minus{}c)^2\\plus{}(c\\minus{}a)^2\\plus{}2(a^2\\plus{}b^2\\plus{}c^2) \\ge 12$\r\nit's enough to prove ineq:\r\n$ a^2\\plus{}b^2\\plus{}c^2 \\ge 6$\r\nwe have $ ab\\plus{}bc\\plus{}ca\\plus{}abc\\equal{}4$ so \r\n$ \\frac{ \\sqrt{ab} }{2}\\equal{}cos \\alpha$ \r\n$ \\frac{ \\sqrt{bc} }{2}\\equal{}cos \\beta$ \r\n$ \\frac{ \\sqrt{ca} }{2}\\equal{}cos \\gamma$ \r\nwith $ \\alpha \\plus{} \\beta \\plus{} \\gamma \\equal{} \\pi$\r\nnow we have inequality:\r\n$ \\sum_{cyc}^{} \\frac{(cos \\alpha)^2 (cos \\beta)^2}{(cos \\gamma^2)} \\ge \\frac{3}{4}$\r\nit is well known that $ cos^2 \\alpha\\plus{}cos^2 \\beta\\plus{}cos^2 \\gamma \\ge \\frac{3}{2}$\r\nso\r\n$ \\sum_{cyc}^{} \\frac{(cos \\alpha)^2 (cos \\beta)^2}{(cos \\gamma^2)} \\ge cos^2 \\alpha\\plus{}cos^2 \\beta\\plus{}cos^2 \\gamma \\ge \\frac{3}{2}$\r\nDone!" -} -{ - "Problem": "\u0395\u03bb\u03c0\u03af\u03b6\u03c9 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c4\u03b7\u03bd \u03be\u03ad\u03c1\u03b5\u03c4\u03b5 \u03ce\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03ad\u03c7\u03b5\u03b9 \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd.\r\n\r\n\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03b1\u03bd $a,b,c,d>0$ \u03bc\u03b5 $abcd=1$ \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03c4\u03b5 \u03cc\u03c4\u03b9\r\n\r\n$\\frac{1}{(1+a)^2}+\\frac{1}{(1+b)^2}+\\frac{1}{(1+c)^2}+\\frac{1}{(1+d)^2}\\geq 1$", - "Solution_1": "\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03ad \u03c0\u03bf\u03bb\u03c5 \u03ba\u03b1\u03bb\u03ae \u03b7 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b1\u03c5\u03c4\u03ae \u03ba\u03b1\u03b9 \u03bc\u03c0\u03bf\u03c1\u03ce \u03bd\u03b1 \u03c3\u03bf\u03c5 \u03c0\u03c9 \u03bf\u03c4\u03b9 \u03b5\u03bd\u03ce \u03c3\u03c4\u03b7\u03bd \u03b1\u03c1\u03c7\u03ae \u03bc\u03bf\u03c5 \u03c6\u03ac\u03bd\u03b7\u03ba\u03b5 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b7 \u03c4\u03b5\u03bb\u03b9\u03ba\u03ac \u03c7\u03c1\u03b5\u03b9\u03ac\u03c3\u03b8\u03b7\u03ba\u03b5 \u03b1\u03c1\u03ba\u03b5\u03c4\u03cc \u03c7\u03c1\u03cc\u03bd\u03bf \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c4\u03b7\u03bd \u03bb\u03cd\u03c3\u03c9.\r\n\r\n\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03b1\u03c0\u03cc \u03b3\u03bd\u03c9\u03c3\u03c4\u03ae \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 $\\frac{1}{(1+a)^{2}} + \\frac{1}{(1+b)^{2}} \\geq \\frac{1}{1+ab}$ (1) \u03b5\u03bd\u03ce \u03bf\u03bc\u03bf\u03af\u03c9\u03c2 $\\frac{1}{(1+c)^{2}} + \\frac{1}{(1+d)^{2}} \\geq \\frac{1}{1+cd}$ (2) \u03ad\u03c4\u03c3\u03b9 \r\n\u03c0\u03c1\u03bf\u03c3\u03b8\u03ad\u03c4\u03bf\u03bd\u03c4\u03b1\u03c2 \u03ba\u03b1\u03c4\u03b1 \u03bc\u03ad\u03bb\u03b7 \u03c4\u03b9\u03c2 (1) \u03ba\u03b1\u03b9 (2) \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03bf\u03c4\u03b9 \r\n\r\n$\\frac{1}{(1+a)^{2}} + \\frac{1}{(1+b)^{2}} + \\frac{1}{(1+c)^{2}} + \\frac{1}{(1+d)^{2}} \\geq\\frac{1}{1 +ab} + \\frac{1}{1+cd}$ \u03cc\u03bc\u03c9\u03c2 \r\n$\\frac{1}{1 +ab} + \\frac{1}{1+cd} = {\\frac{2 + ab + cd }{1 + ab + cd +abcd} = \\frac{ 2 + ab + cd}{ 2 + ab + cd } = 1}$ \u03ac\u03c1\u03b1 $\\frac{1}{(1+a)^{2}} + \\frac{1}{(1+b)^{2}} + \\frac{1}{(1+c)^{2}} + \\frac{1}{(1+d)^{2}} \\geq 1$", - "Solution_2": "\u0391\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2. \u0391\u03c5\u03c4\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03b7 \u03bb\u03cd\u03c3\u03b7 \u03c0\u03bf\u03c5 \u03b5\u03af\u03c7\u03b1 \u03ba\u03b1\u03b9 \u03b5\u03b3\u03ce. \u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03b8\u03b1 \u03c4\u03b1 \u03c0\u03bf\u03cd\u03bc\u03b5." -} -{ - "Problem": "In a quadrilateral ABCD ,it is given that AB is parallel to CD and the diagonals AC and BD are perpendicular to each other.Show that \r\n1. AD.BC>=AB.CD\r\n2. AD+BC>=AB+CD", - "Solution_1": "First of all i would like to say that your problem misses a part. In your problem you don't specificate whether $ AB \\geq CD$ or otherwise. Still, the solution to the problem is :\r\n\r\n Using pithagora in AOB, COB, COD and AOD triangles you express AB, BC, CD or AD. so\r\n\r\n\r\n\r\n$ \\sqrt{BO^2\\plus{}CO^2}\\plus{}\\sqrt{AO^2\\plus{}DO^2}\\geq \\sqrt{BO^2\\plus{}CO^2}\\plus{} \\sqrt{AO^2\\plus{}DO^2}$\r\n\r\n\r\n\r\n $ 2\\sqrt{(BO^2\\plus{}CO^)(AO^2\\plus{}DO^2)}\\geq 2\\sqrt{(BO^2\\plus{}CO^2)(AO^2\\plus{}DO^2)}$\r\n\r\n\r\n\r\n $ (BO^2\\plus{}CO^2)(AO^2\\plus{}DO^2)\\geq (BO^2\\plus{}CO^2)(AO^2\\plus{}DO^2)$ \r\n\r\n\r\n\r\n $ BO^2AO^2\\plus{}BO^2OD^2\\plus{}OC^2AO^2\\plus{}OC^2OD^2\\geq BO^2CO^2\\plus{}BO^2DO^2\\plus{}AO^2CO^2\\plus{}AO^2OD^2$\r\n\r\n\r\n\r\n Then, you'll get:\r\n\r\n\r\n$ AO^2(BO^2\\minus{}OD^2)\\geq CO^2(BO^2\\minus{}OD^2)$ \r\n\r\n\r\n $ AO^2\\geq CO^2$ \r\n\r\n $ AO\\geq CO$ \r\nThis isn't always true. Now you see why you need to know whether $ AB \\geq CD$. If your problem would specify that $ AB \\geq CD$ , you're problem will be true. And, regarding the second part of the problem you just need to look at the first part, because it has already been done. I hope i didn't mess up the expresions to bad.:)", - "Solution_2": "It is [url=http://www.isical.ac.in/~rmo/rmo97.pdf]Regional Mathematical Olympiad 1997; India; problem 4[/url]. :) \r\nI posted it in Mathlinks a long time ago, and this solution is by ocha:\r\n[b](RMO-1997, India) In quadr. ABCD; $ AB\\parallel{}CD$ & $ AC\\perp BD$. Show that :\n(a)$ AD\\cdot BC\\ge AB\\cdot CD$\n(b)$ AD \\plus{} BC\\ge AB \\plus{} CD$[/b]\r\n[hide=\"Solution:(by OCHA)\"]\n\n[asy]draw((0,50)--(0,-100));\ndraw((-50,0)--(100,0));\ndraw((-50,0)--(0,50));\ndraw((0,-100)--(100,0));\ndraw((0,50)--(100,0));\ndraw((-50,0)--(0,-100));\nlabel(\"$a$\", (0,25), E);\nlabel(\"$b$\", (-25,0), N);\nlabel(\"$ka$\", (0,-50), E);\nlabel(\"$kb$\", (50,0), N);\nlabel(\"$A$\", (0,50), N);\nlabel(\"$B$\", (-50,0), W);\nlabel(\"$C$\", (0,-100), S);\nlabel(\"$D$\", (100,0), E);\nlabel(\"$O$\", (0,0), SW);[/asy]\nLet ABCD be the quadr.\n$ \\triangle AOD\\sim \\triangle DOC$\nSo that $ \\frac {OC}{AO} \\equal{} \\frac {OD}{BO} \\equal{} k$ (say; where k is the scaling factor of these two triangles) \nLetting $ AO \\equal{} a, BO \\equal{} b; \\implies OC \\equal{} ka, OD \\equal{} kb$\nWe have some relations:\n$ AB^2 \\equal{} a^2 \\plus{} b^2$\n$ AD^2 \\equal{} a^2 \\plus{} k^2b^2$\n$ CD^2 \\equal{} k^2(a^2 \\plus{} b^2)$\n$ BC^2 \\equal{} b^2 \\plus{} k^2a^2$\nHence\n$ AD^2\\cdot BC^2 \\equal{} a^2b^2 \\plus{} k^2a^4 \\plus{} k^2b^2 \\plus{} k^4a^2b^2$\n$ \\equal{} a^2b^2(1 \\plus{} k^4) \\plus{} (ka^2)^2 \\plus{} (kb^2)^2\\ge 2a^2b^2k^2 \\plus{} (ka^2)^2 \\plus{} (kb^2)^2$\n$ \\equal{} k^2(a^2 \\plus{} b^2)^2 \\equal{} AB^2\\cdot CD^2$\n$ \\therefore \\boxed{AD\\cdot BC\\ge AB\\cdot CD}$\nAlso; $ AD^2 \\plus{} BC^2 \\equal{} AB^2 \\plus{} CD^2 \\equal{} (a^2 \\plus{} b^2)(1 \\plus{} k^2)$\n$ \\implies AD^2 \\plus{} BC^2 \\plus{} 2AD\\cdot BC\\geq AB^2 \\plus{} CD^2 \\plus{} 2AB\\cdot CD$\n$ \\implies (AD \\plus{} BC)^2\\ge (AB \\plus{} CD)^2$\n$ \\boxed{\\implies AD \\plus{} BC\\geq AB \\plus{} CD}$\nQED[/hide]\n[hide]For more information on this useful method using similarities, you can visit [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=295946]Here[/url] :blush: . [/hide]" -} -{ - "Problem": "Find all sets $A$ and $B$ that satisfy the following conditions:\r\n\r\na) $A \\cup B= \\mathbb{Z}$;\r\n\r\nb) if $x \\in A$ then $x-1 \\in B$;\r\n\r\nc) if $x,y \\in B$ then $x+y \\in A$.\r\n\r\n[i]Laurentiu Panaitopol[/i]", - "Solution_1": "...................", - "Solution_2": "I believe it has been discussed before on the forum. It was used in the first Romanian TST in 2002, if I am not mistaken.", - "Solution_3": "[quote=\"socrates\"]A is the set of the even and B of the odd numbers![/quote]\r\n$A=B=\\mathbb{Z}$ is also the answer.", - "Solution_4": "If 0 was in B, by c) 0 is also in A, so 0 is in A. By b) -1 is in B, by c) -2 is in A, by c) -3 is in B. By induction if \r\n-2k+1 is in B, by c) -2k is in A and -2k-1 is in B, so all negative odd numbers are in B. By b) , all negative even numbers are in A. \r\nIf 1 iwas not in B, 1 would be in A, by b) 0 would be in B, so by c) if x is in B, x is also in A, so all numbers would be in A. Because of that all numbers would be in B to, contradiction. So 1 is in B. By c) 2 is in b. By induction it's simple to show that all positive odds are in B and all positiv evens in A.\r\nIf there were no other numbers in the sets, the solution wolud be set of evens and set of odds.\r\nIf there was an odd number in A, there would be by b) an even number in A. So in any other solution then set of evens and set of odds there must be an even number in B. But then 1 is in A, 0 in B so all the numbers from B are in A too and by b) all are in A as well.\r\nSo the only one solutions are (evens,odds), ($\\mathbb Z$,$\\mathbb Z$).", - "Solution_5": "See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=30568" -} -{ - "Problem": "let $x,y,z >0$ and $x+y+z=1$ then\r\n$ x^2y+y^2z+z^2x \\leq \\frac{1}{9} $ or $ \\geq \\frac{1}{9}$", - "Solution_1": "I think it's uncertain. Take x,y ->0 we get l.h.s. -> 0, take x->0 y=z->1/2 then l.h.s. -> 1/8", - "Solution_2": "It will be 1/27 if my remember is not wrong .", - "Solution_3": "However, if we have $x\\geq y\\geq z\\geq 0$ with $x+y+z=1$, then the inequality\r\n\r\n$xy^2+yz^2+zx^2\\leq \\frac {1}{8}$ holds", - "Solution_4": "[quote=\"blahblahblah\"]However, if we have $x\\geq y\\geq z\\geq 0$ with $x+y+z=1$, then the inequality\n\n$xy^2+yz^2+zx^2\\leq \\frac {1}{8}$ holds[/quote]\r\nit is not hard ;)", - "Solution_5": "the original questin is this(i think it was canadian math olympiad)i solved it by lagragne inequality,but there shoud be another solution :) \r\nifx,y,z>0 and x+y+z=1then prove that\r\n$xy^2+yz^2+zx^2\\leq \\frac {4}{27}", - "Solution_6": "the original questin is this(i think it was canadian math olympiad)i solved it by lagragne inequality,but there shoud be another solution :) \r\nifx,y,z>0 and x+y+z=1then prove that\r\n$xy^2+yz^2+zx^2\\leq \\frac {4}{27}", - "Solution_7": "discussed before...\r\nAnyways, the easiest way to prove that maximum is 4/27 is this:\r\nlet x>=y>=z, then show that if you change y to y+d and z to z-d, the expression increases....\r\nSo the maximum is when z=0." -} -{ - "Problem": "pf)\r\nKnowledge is Power. Time is Money. From physics, we know that $Power = \\frac{Work}{Time}$ \r\nSubstitution gives us, $Knowledge = \\frac{Work}{Money}$.\r\nSolving for Money, we get $Money = \\frac{Work}{Knowledge}$.\r\nSo, as long as work is positive(even though it's tiny), money reaches infinity as knowledge reaches 0.\r\nTherefore, less you know, more money you make.\r\n :rotfl:", - "Solution_1": "Well it could also be seen as Money = Work over Knowledge :)", - "Solution_2": "proof that women are, in fact, evil\r\n\r\nwomen require both time and money to keep them happy so we can define women as a product of time and money\r\nW=M*T\r\nnow we know that time is money so\r\nW=M^2\r\nthen we can see that money is the too of all evil so\r\nmoney=evil^.5\r\nmoney^2=evil\r\ntherefore women=evil", - "Solution_3": "[quote=\"Chigr\"]proof that women are, in fact, evil\n\nwomen require both time and money to keep them happy so we can define women as a product of time and money\nW=M*T\nnow we know that time is money so\nW=M^2\nthen we can see that money is the too of all evil so\nmoney=evil^.5\nmoney^2=evil\ntherefore women=evil[/quote]\r\n :rotfl: Good one!", - "Solution_4": "http://www.albinoblacksheep.com/flash/einsteinjokes/\r\n\r\nThese flash movies are also worth a laff.", - "Solution_5": "[quote=\"Chigr\"]proof that women are, in fact, evil\n\nwomen require both time and money to keep them happy so we can define women as a product of time and money\nW=M*T\nnow we know that time is money so\nW=M^2\nthen we can see that money is the too of all evil so\nmoney=evil^.5\nmoney^2=evil\ntherefore women=evil[/quote]\r\n\r\nit's money is the root of all evil, not \"too of all evil.\"\r\nI was thinking of that one when I saw lightrhee's first proof, but I figured most people had heard of it..", - "Solution_6": "[quote=\"ambierona\"]\n...when I saw lightrhee's first proof...[/quote]\r\nIt's not my proof. I've seen it in some other place. :D", - "Solution_7": "This topic is a deja-vu :P", - "Solution_8": "[quote=\"Valentin Vornicu\"]This topic is a deja-vu :P[/quote]\r\n\r\nit's deja-vu all over again! :P", - "Solution_9": "Another joke: A man thinks that the probability of a bomb in an airplane is low but not low enough. So whenever he travels in a plane, he carries a bomb because he reasons that the probability of two bombs in an airplane is infentessimly small.", - "Solution_10": "[quote=\"jimli\"]Another joke: A man thinks that the probability of a bomb in an airplane is low but not low enough. So whenever he travels in a plane, he carries a bomb because he reasons that the probability of two bombs in an airplane is infentessimly small.[/quote]\r\n\r\nHaha. Wow. That's a really good one. :rotfl:", - "Solution_11": "Lol\r\n\r\nThe first proof and bomb thing are hilarious", - "Solution_12": "[quote=\"jimli\"]Another joke: A man thinks that the probability of a bomb in an airplane is low but not low enough. So whenever he travels in a plane, he carries a bomb because he reasons that the probability of two bombs in an airplane is infentessimly small.[/quote]\r\n\r\nI've never heard of that one before, but thats just Hilarious! :rotfl: :rotfl: :winner_first:", - "Solution_13": ":rotfl: :rotfl: :rotfl: :rotfl:", - "Solution_14": "[quote=\"jimli\"]Another joke: A man thinks that the probability of a bomb in an airplane is low but not low enough. So whenever he travels in a plane, he carries a bomb because he reasons that the probability of two bombs in an airplane is infentessimly small.[/quote]\r\n\r\n :lol:", - "Solution_15": "Time to un-Dejavu. :D \r\n\r\nA physicist, a biologist, and a mathematician see two men enter a building. A little later the scientists see three men walking out the building.\r\n\r\n\"The initial measurement was incorrect\" said the physicist.\r\n\"They have multiplied\" said the biologist.\r\n\"If one more person enters the building there will be exactly no people in the building\" said the mathematician.", - "Solution_16": "[quote=\"A+MATH\"]Time to un-Dejavu. :D \n\nA physicist, a biologist, and a mathematician see two men enter a building. A little later the scientists see three men walking out the building.\n\n\"The initial measurement was incorrect\" said the physicist.\n\"They have multiplied\" said the biologist.\n\"If one more person enters the building there will be exactly no people in the building\" said the mathematician.[/quote]\r\n\r\nAck!!! This is deja-vu too. :ninja:", - "Solution_17": "Mathematician: 3 is a prime, 5 is a prime, 7 is a prime, and by induction all odd numbers higher than 2 are prime.\r\n\r\nPhysicist: 3 is a prime, 5 is a prime, 7 is a prime, 9 is an experimental error, 11 is a prime. Just to be sure, let's try some random numbers: 23 is a prime, 17 is a prime, 41 is a prime.\r\n\r\nEngineer: 3 is a prime, 5 is a prime, 7 is a prime, 9 is approximately a prime...\r\n\r\nPsychologist: 3 is a prime, 5 is a prime, 7 is a prime, 9 is a prime but tries to suppress it...\r\n\r\nAdvertiser: 3 is a prime, 5 is a prime, 7 is a prime, 11 is a prime, 13 is a prime...\r\n\r\nBiologist: 3 is a prime, 5 is a prime, 7 is a prime, 9... results have not arrived yet...\r\n\r\nComputer Salesman: 3 is a prime, 5 is a prime, 7 is a prime, 9 will be a prime in the next release...\r\n\r\nAny others?", - "Solution_18": "[quote=\"jimli\"]Another joke: A man thinks that the probability of a bomb in an airplane is low but not low enough. So whenever he travels in a plane, he carries a bomb because he reasons that the probability of two bombs in an airplane is infentessimly small.[/quote]\r\n\r\nThat's a good one! :) :) :) \r\n\r\nDidn't someone just do the girls=evil proof a little while ago?(BTW it is NOT true :) )", - "Solution_19": "Engineer: 3 is a prime, 5 is a prime, 7 is a prime, 9 is a prime, 11 is a prime, ...\r\n\r\nHumorist: 3 is an odd prime, 5 is an odd prime, 7 is an odd prime, 9 is a VERY odd prime, ...\r\n\r\nComputer programmer (reading output): 3 is a prime, 3 is a prime, 3 is a prime, ...\r\n\r\nComputer programmer (reading output from a pentium): 2.992837 is a prime, 4.973937 is a prime, ...\r\n\r\nPolitician: (closes shades, locks door) do you WANT 9 to be a prime?", - "Solution_20": "Politician: Some numbers are prime... but the goal is to create a kinder, softer society where ALL numbers are prime.", - "Solution_21": "What is purple and commutes? An Abelian grape. :rotfl: \r\n\r\n(From Mathworld)", - "Solution_22": "What is yellow and equivalent to the axiom of choice?", - "Solution_23": "Is there a reason all the Mathworld jokes are designated as bad? :rotfl:", - "Solution_24": "[quote=\"MysticTerminator\"]What is yellow and equivalent to the axiom of choice?[/quote]\r\n\r\nZorn's Lemon!", - "Solution_25": "Mathworld has JOKES?! Whoa... where have I been?\r\n\r\nTheorem: If you are given an open-book exam you will forget your book.\r\nCorollary: If you are given a take-home exam you will forget where you live.\r\n\r\nThe Pass-Fail Theorem:\r\nSince AM is the opposite of both PM and FM, PM must equal FM, by the opposite of opposites property. Therefore P=F, and passing equals failing. QED", - "Solution_26": "Just search Mathworld for joke...", - "Solution_27": "knowledge=power\r\npower=corrupt\r\ncorruption=evil\r\nknowledge=evil :lol:", - "Solution_28": "To err is human, but humans are intelligent beings. Therefore to err is intelligent. QED", - "Solution_29": "[quote=\"mathnerd314\"]Humans are intelligent beings.[/quote]\r\nYour conclusion is based on false premises.", - "Solution_30": "Well, then, some humans are intelligent, and therefore some errors are intelligent.\r\n\r\nCalculus Instructor: Well, obviously, we see that the number of primes approaches infinity as the upper bound does. Therefore the number of primes is arbitrary. We integrate random functions and get that either all numbers are prime, no numbers are prime, or 42 numbers are prime. Etc.\r\n\r\nWhy did the chicken cross the Mobius strip? To get to the other... uh...", - "Solution_31": "[quote=\"mathnerd314\"]Why did the chicken cross the Mobius strip? To get to the other... uh...[/quote]\r\n\r\nTo get to the same side!", - "Solution_32": "[quote=\"Chigr\"]proof that women are, in fact, evil\n\nwomen require both time and money to keep them happy so we can define women as a product of time and money\nW=M*T\nnow we know that time is money so\nW=M^2\nthen we can see that money is the too of all evil so\nmoney=evil^.5\nmoney^2=evil\ntherefore women=evil[/quote]\r\n\r\nHey! I did this joke in a past thread... :mad: \r\n\r\nKeep going back and check it out. :( \r\n :P", - "Solution_33": "There's a REALLY dirty one involving intergration but I don't think i'm allowed to post it. PM me if you wanna hear it (don't bother unless you know intergration well).", - "Solution_34": "OK, xdl615x... I'd rather not. :police: \r\n\r\nProof: Math blows things up.\r\n\r\nThoughts are energy passing through neurons (at least we think they are). Energy can be converted into mass. Some people think math is so boring that when they do it, it results in a complete stop of thought. Therefore, math destroys mass and is a weapon of mass destruction. By definition, a weapon of mass destruction is a nuclear warhead, which blows things up. Therefore, math blows things up. Q.E.D.\r\n\r\nSource: MathPath2005 yearbook", - "Solution_35": "[quote=\"jimli\"]Another joke: A man thinks that the probability of a bomb in an airplane is low but not low enough. So whenever he travels in a plane, he carries a bomb because he reasons that the probability of two bombs in an airplane is infentessimly small.[/quote]\r\nThat's my favorite math joke! :lol: I heard a longer version of it a while ago too. :P", - "Solution_36": "[quote=\"mathnerd314\"]OK, xdl615x... I'd rather not. :police: \n\nProof: Math blows things up.\n\nThoughts are energy passing through neurons (at least we think they are). Energy can be converted into mass. Some people think math is so boring that when they do it, it results in a complete stop of thought. Therefore, math destroys mass and is a weapon of mass destruction. By definition, a weapon of mass destruction is a nuclear warhead, which blows things up. Therefore, math blows things up. Q.E.D.\n\nSource: MathPath2005 yearbook[/quote]\r\nrofl math is a weapon of mass destruction :rotfl: \r\nWhy spend millions on nuclear warheads when you could dish out a few thousand for a math degree? :P" -} -{ - "Problem": "January 1, 2007 was a Monday. In 2007, how many Fridays are\nthere?", - "Solution_1": "Since only the first (and in leap years, second as well) go into the $ 53^\\text{rd}$ week of an year, the are as many Fridays as there are full weeks: $ \\fbox{52}$." -} -{ - "Problem": "The circumference of a circle is $ 50\\pi$ units. In terms of $ \\pi$, how many square units are in its area?", - "Solution_1": "50pi = 2pi*r, making r 25. $ 25^2 \\cdot \\pi \\equal{} \\boxed{625}$", - "Solution_2": "[quote=\"Ebob\"]50pi = 2pi*r, making r 25. $ 25^2 \\cdot \\pi \\equal{} \\boxed{625}$[/quote]\r\n\r\nactually, it's $ 625\\pi$" -} -{ - "Problem": "Wait... the polynomial is of the third degree, so it has 3 roots. The problem says there are 2 real roots. This means there is 1 imaginary root. But don't imaginary roots have to come in pairs... complex conjugates?\r\n\r\nEdit: oh, ok, I now see that it just says that they are real roots. The third root is real as well.", - "Solution_1": "use vieta\u00b4s relations.", - "Solution_2": "[hide=\"Simplification\"]Note that by Vieta's formula's, $ a\\plus{}b\\plus{}c \\equal{} \\minus{}1$ or $ a\\plus{}b \\equal{} \\minus{}(c\\plus{}1)$. Plugging into the desired expression, it simplifies to \n\n$ \\minus{}c^3 \\minus{} c^2 \\plus{} 2008c \\plus{} 2008$\n\nNote that since $ c$ is a root of $ x^3 \\plus{} x^2 \\minus{} 2009x \\equal{} 232$, we can simplify it as\n\n$ 1776 \\minus{} c$[/hide]", - "Solution_3": "It's not like we specified which of the roots a and b were (all 3 roots are real), so c could have 3 different values, and that would give 1776 - c three different values." -} -{ - "Problem": "this is for the countdown cup results.\r\n\r\ni will post the matchups either 2day or 2morrow.\r\n\r\ngood luck.", - "Solution_1": "[b]Round 1[/b]\r\n\r\n14 skaldjfhrulez DEFEATS 19 mathblitz, 3-1\r\n4 FantasyLover VS 29 vamathletes\r\n1 superllamel VS 32 blue1421\r\n10 RomanianGenius VS 23 isabella2296\r\n13 Brut3Forc3 VS 20 bob einstein\r\n6 yongyi781 VS 27 PWNER Express\r\n17 hurdler DEFEATS 16 budi713, 3-1\r\n5 alexhhmun VS 28 mathmagician1729\r\n3 ra5249 VS 30 Meta Knight\r\n12 lotsofmath VS 21 ash-Pokemon\r\n9 ernie VS 24 AIME15\r\n7 vallon22 DEFEATS 26 nikeballa96, 3-0\r\n2 mewto55555 DEFEATS 31 Mind Storm, 3-0\r\n15 kl2836 VS 18 violin321\r\n11 BOGTRO VS 22 thdanh90\r\n8 Jongao VS 25 kookamunga\r\n\r\n[b]Round 2[/b]\r\n\r\nskaldjfhrulez VS 4/29\r\n1/32 VS 10/23\r\n13/20 VS 6/27\r\n(16)hurdler VS 5/28\r\n3/30 VS 12/21\r\n9/24 VS 7 vallon22\r\n2 mewto55555 VS 15/18\r\n11/22 VS 8/25", - "Solution_2": "oops, ignore my question on the other thread about the pairings....", - "Solution_3": "nooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooI will get pwned by mathblitz *sobs*", - "Solution_4": "DO NOT SPAM ON THIS THREAD! :bomb: :bomb: :bomb:", - "Solution_5": "Since i lost hurdler is now 16th seed?\r\n :( :(", - "Solution_6": "kl2836 won 3-2 (overall)", - "Solution_7": "MEWTO55555(2) DEFEATS KL2836(15) 3-0\r\n\r\nM(4-0)\r\nM(4-1)\r\nM(4-2)", - "Solution_8": "SUPERLLAMEL (1) DEFEATS BLUE1421 (32)\r\n\r\nS(4-0)\r\nS(4-1)\r\nS(4-0)\r\n\r\nBlue can confirm this.", - "Solution_9": "ROMANIANGENIUS(10) DEFEATS ISABELLA2296(23) 3-0\r\n\r\nR(4-1)\r\nR(4-0)\r\nR(4-1)", - "Solution_10": "[b]Round 1[/b] \r\n\r\n14 skaldjfhrulez DEFEATS 19 mathblitz, 3-1 \r\n4 FantasyLover VS 29 vamathletes \r\n1 superllamel DEFEATS 32 blue1421, 3-0\r\n10 RomanianGenius DEFEATS 23 isabella2296, 3-0\r\n13 Brut3Forc3 VS 20 bob einstein \r\n6 yongyi781 DEFEATS 27 PWNER Express, 3-0 \r\n17 hurdler DEFEATS 16 budi713, 3-1 \r\n5 alexhhmun VS 28 mathmagician1729 \r\n3 ra5249 VS 30 Meta Knight \r\n12 lotsofmath DEFEATS 21 ash-Pokemon, 3-1\r\n9 ernie VS 24 AIME15 \r\n7 vallon22 DEFEATS 26 nikeballa96, 3-0 \r\n2 mewto55555 DEFEATS 31 Mind Storm, 3-0 \r\n15 kl2836 DEFEATS 18 violin321, 3-2 \r\n11 BOGTRO VS 22 thdanh90 \r\n8 Jongao VS 25 kookamunga \r\n\r\n[b]Round 2[/b]\r\n\r\n14 skaldjfhrulez VS 4 FantasyLover/29 vamathletes\r\n1 superllamel VS 10 RomanianGenius \r\n13 Brut3Forc3/20 bob einstein VS yongyi781\r\n(16)hurdler VS 5 alexhhmun/28 mathemagician1729\r\n3 ra5249/30 Meta Knight VS l12 otsofmath\r\n9 ernie /24 AIME15 VS 7 vallon22 \r\n2 mewto55555 DEFEATS 15 kl2836, 3-0\r\n11 BOGTRO/22 thdanh90 VS 8 Jongao/25 kookamunga\r\n\r\n[b]Quarter-Finals[/b]\r\n\r\n????? VS 1 superllamel/10 RomanianGenius\r\n????? VS ?????\r\n????? VS ?????\r\n2 mewto55555 VS ?????\r\n\r\n[b]Semi-Finals[/b]\r\n\r\n????? VS ?????\r\n????? VS ?????\r\n\r\n[b]Finals[/b]\r\n\r\n????? VS ?????\r\n\r\n@budi713- no, the parenthesis indicate that he upset that seed[code][hide=Index][/hide][/code]", - "Solution_11": "12 LOTSOFMATH DEFEATS 21 ASH-POKEMON\r\n\r\n1: L(4-3)\r\n2: A(4-3)\r\n3: L(4-3)\r\n4:L(4-2)", - "Solution_12": "UGH MEWTO WENT TO QUARTER FINALS BUT I DIDNT EVEN FINISH ROUND 1\r\n\r\nWHERE IS VAMATHLETES???\r\n\r\nok im gonna pm him", - "Solution_13": "FANTASYLOVER PWNS VAMATHLETES\r\n4-3 FL WINS\r\n4-0 FL PWNS\r\n4-1 FL PWNS", - "Solution_14": "Round 1 \r\n\r\n14 skaldjfhrulez DEFEATS 19 mathblitz, 3-1 \r\n4 FantasyLover DEFEATS 29 vamathletes, 3-0\r\n1 superllamel DEFEATS 32 blue1421, 3-0 \r\n10 RomanianGenius DEFEATS 23 isabella2296, 3-0 \r\n13 Brut3Forc3 VS 20 bob einstein \r\n6 yongyi781 DEFEATS 27 PWNER Express, 3-0 \r\n17 hurdler DEFEATS 16 budi713, 3-1 \r\n5 alexhhmun VS 28 mathmagician1729 \r\n3 ra5249 VS 30 Meta Knight \r\n12 lotsofmath DEFEATS 21 ash-Pokemon, 3-1 \r\n9 ernie VS 24 AIME15 \r\n7 vallon22 DEFEATS 26 nikeballa96, 3-0 \r\n2 mewto55555 DEFEATS 31 Mind Storm, 3-0 \r\n15 kl2836 DEFEATS 18 violin321, 3-2 \r\n11 BOGTRO VS 22 thdanh90 \r\n8 Jongao VS 25 kookamunga \r\n\r\nRound 2 \r\n\r\n14 skaldjfhrulez VS 4 FantasyLover\r\n1 superllamel VS 10 RomanianGenius \r\n13 Brut3Forc3/20 bob einstein VS yongyi781 \r\n(16)hurdler VS 5 alexhhmun/28 mathemagician1729 \r\n3 ra5249/30 Meta Knight VS 12 lotsofmath \r\n9 ernie /24 AIME15 VS 7 vallon22 \r\n2 mewto55555 DEFEATS 15 kl2836, 3-0 \r\n11 BOGTRO/22 thdanh90 VS 8 Jongao/25 kookamunga \r\n\r\nQuarter-Finals \r\n\r\n14 skaldjfhrulez/4 FantasyLover VS 1 superllamel/10 RomanianGenius \r\n????? VS ????? \r\n????? VS ????? \r\n2 mewto55555 VS ????? \r\n\r\nSemi-Finals \r\n\r\n????? VS ????? \r\n????? VS ????? \r\n\r\nFinals \r\n\r\n????? VS ?????", - "Solution_15": "Why am I against two people? :?: :huh:", - "Solution_16": "You against winner of 13v20 game yongyi.", - "Solution_17": "[quote=\"ernie\"]Round 1 \r\n\r\n14 skaldjfhrulez DEFEATS 19 mathblitz, 3-1 \r\n4 FantasyLover DEFEATS 29 vamathletes, 3-0\r\n1 superllamel DEFEATS 32 blue1421, 3-0 \r\n10 RomanianGenius DEFEATS 23 isabella2296, 3-0 \r\n13 Brut3Forc3 VS 20 bob einstein \r\n6 yongyi781 DEFEATS 27 PWNER Express, 3-0 \r\n17 hurdler DEFEATS 16 budi713, 3-1 \r\n5 alexhhmun VS 28 mathmagician1729 \r\n3 ra5249 VS 30 Meta Knight \r\n12 lotsofmath DEFEATS 21 ash-Pokemon, 3-1 \r\n9 ernie VS 24 AIME15 \r\n7 vallon22 DEFEATS 26 nikeballa96, 3-0 \r\n2 mewto55555 DEFEATS 31 Mind Storm, 3-0 \r\n15 kl2836 DEFEATS 18 violin321, 3-2 \r\n11 BOGTRO VS 22 thdanh90 \r\n8 Jongao VS 25 kookamunga \r\n\r\nRound 2 \r\n\r\n4 FantasyLover DEFEATS 14 skaldjfhrulez, 3-0 \r\n1 superllamel VS 10 RomanianGenius \r\n13 Brut3Forc3/20 bob einstein VS yongyi781 \r\n(16)hurdler VS 5 alexhhmun/28 mathemagician1729 \r\n3 ra5249/30 Meta Knight VS 12 lotsofmath \r\n9 ernie /24 AIME15 VS 7 vallon22 \r\n2 mewto55555 DEFEATS 15 kl2836, 3-0 \r\n11 BOGTRO/22 thdanh90 VS 8 Jongao/25 kookamunga \r\n\r\nQuarter-Finals\r\n\r\n4 FantasyLover VS 1 superllamel/10 RomanianGenius \r\n????? VS ????? \r\n????? VS ????? \r\n2 mewto55555 VS ????? \r\n\r\nSemi-Finals \r\n\r\n????? VS ????? \r\n????? VS ????? \r\n\r\nFinals \r\n\r\n????? VS ?????", - "Solution_18": "...now that i found vamathletes and played with him...\r\n\r\nI NEED TO FIND SUPER!!!!\r\n\r\nWHERE IS SUPER??? I HAVE 2 TOURNEYS WITH HIM!!!", - "Solution_19": "WHAT AN UPSET!!!!!!!!!!!!!!!!! OMGGGGGGGGGGG!!!!!!\r\n\r\nROMANIANGENIUS(10) DEFEATS SUPERLLAMEL(1)\r\n\r\nR(4-2)\r\nS(4-2)\r\nR(4-1)\r\nR(4-0)\r\n\r\n\r\nI HAVE SUCCEEDED WHERE MY BROTHER HAS FAILED! BRING IT ON FANTASY!", - "Solution_20": "AHHH! I've been inactive! Sorry!\r\nWhen can I play bob einstein?", - "Solution_21": "14 skaldjfhrulez DEFEATS 19 mathblitz, 3-1 \r\n4 FantasyLover DEFEATS 29 vamathletes, 3-0 \r\n1 superllamel DEFEATS 32 blue1421, 3-0 \r\n10 RomanianGenius DEFEATS 23 isabella2296, 3-0 \r\n13 Brut3Forc3 VS 20 bob einstein \r\n6 yongyi781 DEFEATS 27 PWNER Express, 3-0 \r\n17 hurdler DEFEATS 16 budi713, 3-1 \r\n5 alexhhmun VS 28 mathmagician1729 \r\n3 ra5249 VS 30 Meta Knight \r\n12 lotsofmath DEFEATS 21 ash-Pokemon, 3-1 \r\n9 ernie DEFEATS 24 AIME15, 3-0\r\n7 vallon22 DEFEATS 26 nikeballa96, 3-0 \r\n2 mewto55555 DEFEATS 31 Mind Storm, 3-0 \r\n15 kl2836 DEFEATS 18 violin321, 3-2 \r\n11 BOGTRO VS 22 thdanh90 \r\n8 Jongao VS 25 kookamunga \r\n\r\nRound 2 \r\n\r\n4 FantasyLover DEFEATS 14 skaldjfhrulez, 3-0 \r\n10 RomanianGenius DEFEATS 1 superllamel, 3-1\r\n13 Brut3Forc3/20 bob einstein VS yongyi781 \r\n(16)hurdler VS 5 alexhhmun/28 mathemagician1729 \r\n3 ra5249/30 Meta Knight VS 12 lotsofmath \r\n9 ernie VS 7 vallon22 \r\n2 mewto55555 DEFEATS 15 kl2836, 3-0 \r\n11 BOGTRO/22 thdanh90 VS 8 Jongao/25 kookamunga \r\n\r\nQuarter-Finals \r\n\r\n4 FantasyLover VS (1)RomanianGenius \r\n????? VS ????? \r\n????? VS 9 ernie/7 vallon22 \r\n2 mewto55555 VS ????? \r\n\r\nSemi-Finals \r\n\r\n????? VS ????? \r\n????? VS ????? \r\n\r\nFinals \r\n\r\n????? VS ?????", - "Solution_22": "FANTASYLOVER(3) DEFEATS ROMANIANGENIUS(10)\r\n\r\nR(4-1)\r\nF(4-2)\r\nF(4-2)\r\nF(4-2)", - "Solution_23": "...then im going to semi-finals?\r\n\r\neh that was fast...", - "Solution_24": "[quote=\"RomanianGenius\"]FANTASYLOVER(3) DEFEATS ROMANIANGENIUS(10)\n\nR(4-1)\nF(4-2)\nF(4-2)\nF(4-2)[/quote]\r\n\r\nonly fantasy is seeded #4 :P\r\n\r\n____________________________________________________________________________________________\r\n\r\n14 skaldjfhrulez DEFEATS 19 mathblitz, 3-1 \r\n4 FantasyLover DEFEATS 29 vamathletes, 3-0 \r\n1 superllamel DEFEATS 32 blue1421, 3-0 \r\n10 RomanianGenius DEFEATS 23 isabella2296, 3-0 \r\n13 Brut3Forc3 VS 20 bob einstein \r\n6 yongyi781 DEFEATS 27 PWNER Express, 3-0 \r\n17 hurdler DEFEATS 16 budi713, 3-1 \r\n5 alexhhmun VS 28 mathmagician1729 \r\n3 ra5249 VS 30 Meta Knight \r\n12 lotsofmath DEFEATS 21 ash-Pokemon, 3-1 \r\n9 ernie DEFEATS 24 AIME15, 3-0 \r\n7 vallon22 DEFEATS 26 nikeballa96, 3-0 \r\n2 mewto55555 DEFEATS 31 Mind Storm, 3-0 \r\n15 kl2836 DEFEATS 18 violin321, 3-2 \r\n11 BOGTRO VS 22 thdanh90 \r\n8 Jongao VS 25 kookamunga \r\n\r\nRound 2 \r\n\r\n4 FantasyLover DEFEATS 14 skaldjfhrulez, 3-0 \r\n10 RomanianGenius DEFEATS 1 superllamel, 3-1 \r\n13 Brut3Forc3/20 bob einstein VS yongyi781 \r\n(16)hurdler VS 5 alexhhmun/28 mathemagician1729 \r\n3 ra5249/30 Meta Knight VS 12 lotsofmath \r\n9 ernie VS 7 vallon22 \r\n2 mewto55555 DEFEATS 15 kl2836, 3-0 \r\n11 BOGTRO/22 thdanh90 VS 8 Jongao/25 kookamunga \r\n\r\nQuarter-Finals \r\n\r\n4 FantasyLover DEFEATS (1)RomanianGenius, 3-1\r\n????? VS ????? \r\n????? VS 9 ernie/7 vallon22 \r\n2 mewto55555 VS ????? \r\n\r\nSemi-Finals \r\n\r\n((1)) FantasyLover VS ????? \r\n????? VS ????? \r\n\r\nFinals \r\n\r\n????? VS ?????\r\n____________________________________________________________________________________________\r\n\r\nGames that can be played currently:\r\n\r\n13 Brut3Forc3 VS 20 bob einstein\r\n5 alexhhmun VS 28 mathemagician1729\r\n3 ra5249 VS 30 Meta Knight\r\n11 BOGTRO VS 22 thdanh90 \r\n8 Jongao VS 25 kookamunga\r\n7 vallon22 VS 9 ernie", - "Solution_25": "Vallon22 beats Ernie\r\n\r\n4-2 E\r\n4-2 V\r\n4-1 V\r\n4-2 V", - "Solution_26": "is it too late for me to join?", - "Solution_27": "[quote=\"AIME_is_hard\"]is it too late for me to join?[/quote]\r\n\r\nyeah its too late, AIME\r\n\r\nvallon, lemme tell u that i typoed 6 TIMES\r\n\r\nHOLY COW! :lol:", - "Solution_28": "I will be out of town until july 13th, starting tomorrow, so i forfeit to alexhmun or mathemagician, (whoever should face me next, unless they wish to wait for me.......)", - "Solution_29": "[quote=\"ernie\"][quote=\"AIME_is_hard\"]is it too late for me to join?[/quote]\n\nyeah its too late, AIME\n\nvallon, lemme tell u that i typoed 6 TIMES\n\nHOLY COW! :lol:[/quote]\r\n\r\nu wish! :)", - "Solution_30": "bump...\r\n\r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting. \r\n \r\nThe message is too small. Please make the message longer before submitting.", - "Solution_31": "Is the message big enough yet?", - "Solution_32": "OMG!!!\r\nit's spam!!! :spam: :spam: :spam:", - "Solution_33": "ITS MORE SPAM!\r\n\r\n:spam: :spam: :spam: \r\n \r\nToo many smilies in your post: 2359.\r\nYou can have at most 3 smilies in this message. Please remove some smilies before submitting this message.", - "Solution_34": "it's DDR!!! :ddr: \r\nand yoda!!! :yoda: \r\nand a sick guy being carried off on a stretcher!!! :stretcher:", - "Solution_35": "I believe you are spamming. :spam: :spam: :spam:", - "Solution_36": "o well\r\n :clap2: :harhar:", - "Solution_37": "The second one (:harhar:) is one of my favorites.", - "Solution_38": "DId this just die? Have the new tournaments made everyone forget this one? Ernie you should pm those who need to play.", - "Solution_39": "No, I have cancelled this tourney, as I have to Ernie's Tourney (the new one).", - "Solution_40": "Yay and my tournament is alomst over. I'll keep PPT up though, as it won't start for probably a month.", - "Solution_41": "idk y i'm still posting in this one...", - "Solution_42": "Please lock this thread.", - "Solution_43": "[i]or[/i] we could just stop posting in this thread", - "Solution_44": "True, but an instinct tells me we'll never stop posting on this thread. :D" -} -{ - "Problem": "prove there is not x,y that:\r\nx^2 + (x+1)^2 = y^4 + (y+1)^4", - "Solution_1": "Why? There are infinetely many real solutions.\r\nInteger solutions $ (x,y)\\equal{}(\\minus{}1,\\minus{}1),(\\minus{}1,0),(0,\\minus{}1),(0,0)$. :D", - "Solution_2": "First observe that the pairs $ (0,0),(0,\\minus{}1),(\\minus{}1,0),(\\minus{}1,\\minus{}1)$ are solutions and that $ xy\\equal{}0$ will yield one of these solutions. We claim that these are all solutions. Observe that for arbitrary $ n\\in\\mathbb{N}$ the function $ g(t)\\equal{}t^{2n}\\plus{}(t\\plus{}1)^{2n}$ fulfills $ g(\\minus{}1\\minus{}t)\\equal{}g(t)$. Therefore, we can assume from now on $ x,y> 0$ and we will perform a case distinction: \r\n\r\nI) The condition $ x\\leq y^2\\plus{}y$ will yield the contradiction $ x^2\\plus{}(x\\plus{}1)^2\\equal{}2y^4\\plus{}4y^3\\plus{}4y^2\\plus{}2y\\plus{}1y^4\\plus{}(y\\plus{}1)^4$. \r\n\r\nSince the case distinction is exhaustive, the proof is complete.", - "Solution_3": "I'm sorry, I should say x,y are from natural set.", - "Solution_4": "More directly:\r\n\r\n$ 2x^2 \\plus{} 2x \\plus{} 1 \\equal{} 2y^4 \\plus{} 4y^3 \\plus{} 6y^2 \\plus{} 4y \\plus{} 1 \\Leftrightarrow$\r\n$ x^2 \\plus{} x \\plus{} 1 \\equal{} y^2 \\plus{} 2y^3 \\plus{} 3y^2 \\plus{} 2y \\plus{} 1 \\equal{} (y^2 \\plus{} y \\plus{} 1)^2$\r\n\r\nBut $ x^2 < x^2 \\plus{} x \\plus{} 1 < (x \\plus{} 1)^2$, contradiction." -} -{ - "Problem": "The sides of a triangle form an arithmetic progression, prove that also the cotangents of the half angles opposite to the sides form a arithmetic progression.", - "Solution_1": "See [b][url=http://www.mathlinks.ro/viewtopic.php?p=1602760#1602760][color=red]here[/url][/b][/color] ...", - "Solution_2": "thank you very much", - "Solution_3": "this is an easy one.\r\n\r\na,b,c are in A.P means that s-a,s-b,s-c are also A.P.\r\n\r\nr /s-a=tan(A/2)\r\nr/s-b=tan(B/2)\r\nr/s-c =tan(C/2)\r\nso,tan(A/2),tan(B/2),tan(C/2)are in H.P\r\nso,cot(A/2),cot(B/2),cot(C/2)are in A.P \r\n\r\nQ.E.D", - "Solution_4": "[quote=\"vaibhav2903\"]this is an easy one.\n\na,b,c are in A.P means that s-a,s-b,s-c are also A.P.\n\nr cs-a=tan(A/2)\nr/s-b=tan(B/2)\nr/s-c =tan(C/2)\nso,tan(A/2),tan(B/2),tan(C/2)are in H.P\nso,cot(A/2),cot(B/2),cot(C/2)are in A.P \n\nQ.E.D[/quote]\r\n\r\nVery very very nice solution.\r\nWe can also write it like this:\r\nBecause a, b, c are in A.P. then also s-a/r, s-b/r, s-c/r are in A.P. so cot(A/2),cot(B/2),cot(C/2) are in A.P" -} -{ - "Problem": "What rank are you in?\r\n\r\nPointcare Conjecture is where I am in. I am almost in Riemann hypothesis.", - "Solution_1": "i'm Riemann Hypothesis. i think 11 people are at the top rank.", - "Solution_2": "I'm posting this to see what my rank is.\r\n\r\nEdit: I'm in Navier Stokes equations", - "Solution_3": "I'm in Birch and Swinnerton Dyer (just today!). \r\nI noticed, I'm the only one in this rank that joined in 2005 :?", - "Solution_4": "how long do you think it will take you to be number 1 in posts? :)", - "Solution_5": "[quote=\"h78reg\"]how long do you think it will take you to be number 1 in posts? :)[/quote]\r\n\r\nforever, grobber has too much of an advantage. :(", - "Solution_6": "Im a riemman hypothosis :) they have weird rank names", - "Solution_7": "[quote=\"jhollenbeck\"]Im a riemman hypothosis :) they have weird rank names[/quote]\r\n\r\nthey are millenium prizes names. There are 6 or 7 of them", - "Solution_8": "[quote=\"236factorial\"][quote=\"h78reg\"]how long do you think it will take you to be number 1 in posts? :)[/quote]\n\nforever, grobber has too much of an advantage. :([/quote]\r\n\r\nIt shouldn't be too hard, since you're packing in a whopping 15 posts a day. Yeesh, so many relatively new members with astronomical posts per day...", - "Solution_9": "lately, 80-120 posts per day :o", - "Solution_10": "Im a Pointcare Conjecture as of rightnow.", - "Solution_11": "im hodge. :mad:", - "Solution_12": "[quote=\"236factorial\"]lately, 80-120 posts per day :o[/quote]\r\nWhoa, that's [i]a lot[/i]. :D \r\n\r\nI'm a Poincare Conjecture right now.", - "Solution_13": "I'm Poincare now but I'm gonna get to the next pretty soon. I average around 3 posts total per day, though on active days I get about 20. I try not to post if it really wouldnt contribute, else I'd have a higher post count.", - "Solution_14": "[quote=\"frt\"][quote=\"236factorial\"]lately, 80-120 posts per day :o[/quote]\nWhoa, that's [i]a lot[/i]. :D \n[/quote]\r\n\r\nIt helps when you're on several hours a day *sigh*", - "Solution_15": "[quote=\"Spencer\"]I thought that a lot of people would be in New Member.\n'cause I think that this site is becoming very popular.[/quote]\r\n\r\nIt's just a lot of the New Members aren't active, and therefore wouldn't be voting in this poll.", - "Solution_16": "[quote=\"h_s_potter2002\"][quote=\"Spencer\"]I thought that a lot of people would be in New Member.\n'cause I think that this site is becoming very popular.[/quote]\n\nIt's just a lot of the New Members aren't active, and therefore wouldn't be voting in this poll.[/quote]\r\n\r\n[hide=\"click here for an embarrasing fact\"]There are 6769 people with no posts :( [/hide]\r\n\r\nAfter all, there isn't too many users that are actually [i]active[/i]", - "Solution_17": "just out of curiosity, how many of those 6500 users have become users over 3 weeks ago? ;)", - "Solution_18": "Out of the 9000+ new members, only about 600 have joined in the last 4 weeks. :( That means most of our new members aren't really flourishing. Maybe math too boring for them? It sure is for 90% of the people.", - "Solution_19": "the new members probably havn't discovered this forum and if they did would be too embarressed to vote on being a new person.", - "Solution_20": "I know of one person who doesn't have any posts, but is still active (or at least was) active. Like he/she did some of the Mock AMC's and AIME's and stuff, and probably just lurks, and participates in the classes.", - "Solution_21": "P and something...\r\n\r\nwhy didn't u put that in your poll?", - "Solution_22": "[quote=\"h78reg\"][quote]anybody know the post count to pass hodge conjecture? \n\n[edit] isnt there also a P vs NP rank?[/quote]\n\nsee the second post ;)[/quote]\r\n\r\nerrrr...there's nothing on the second post that tells us this?", - "Solution_23": "Pointcare Conjecture sometimes post counts dont matter its their contribution to the site and a total spammer could get a high post count and who is Tare", - "Solution_24": "It takes time for new people to get used to the forums. I wasn't very active for the first few months of my registration either.", - "Solution_25": "[quote=\"white_horse_king88\"]It takes time for new people to get used to the forums. I wasn't very active for the first few months of my registration either.[/quote]\r\n\r\nI think I mentioned this too, but the only forum i thought existed at first was Mathcounts :blush: Therefore, I wasn't postinig a lot in the beginning.", - "Solution_26": "i like this forum because its all fun and games and polls :)", - "Solution_27": "[quote=\"jhollenbeck\"]i like this forum because its all fun and games and polls :)[/quote]\r\n\r\nyeah, but it's nothing about math, so don't get hooked only on this forum. :) \r\nBut i do see you posting elsewhere, that's good.", - "Solution_28": "that's the problem. What i do, is taht i aswer about 10 questions, and then i come here for a break. :)", - "Solution_29": "last time i checked this poll, there were no posters below Rimmean Hypothosis. Now there is a majority below Rimmean Hypothosis :lol: :o" -} -{ - "Problem": "sin(x-10)=sin-40", - "Solution_1": "you mean\r\n\r\n$\\sin{(x-10)}=\\sin{(-40)}$ right..", - "Solution_2": "[quote=\"begimix\"]sin(x-10)=sin-40[/quote]\r\n\r\n$\\sin(x-10)=\\sin(-40)$\r\n\r\nI haven't taken trig that long but can't you have?\r\n\r\n$\\sin x-\\sin 10 = \\sin(-40)$\r\n\r\nand this should not be in HSB...", - "Solution_3": "NO!\r\n\r\n$\\sin(a+b)\\neq\\sin(a)+\\sin(b)$", - "Solution_4": "[quote=\"SM4RT\"]NO!\n\n$\\sin(a+b)\\neq\\sin(a)+\\sin(b)$[/quote]\r\n\r\nthat's what i thought i'm getting confused with logs and multiplication :blush:...stupid three letter math terms lol\r\n\r\nedit: isn't the answer just $x=-30$? when you plug it in it makes sense :lol:", - "Solution_5": "sorry \r\nit is sin40", - "Solution_6": "Its 50+2(180)x and 150+2(180)x for any integer x\r\n\r\nThink of it this way: siny=sin40, y can be 40 or 140 if x-10 is 40 or 140, then x is 50 or 150", - "Solution_7": "[quote=\"Art of Owna\"]Its 50+2(180)x and 150+2(180)x for any integer x\n\nThink of it this way: siny=sin40, y can be 40 or 140 if x-10 is 40 or 140, then x is 50 or 150[/quote]\r\n\r\nsince begimix changed his problem...i agree..the solution is either 50 or 150", - "Solution_8": "it seems you didn't understand me\r\nthere are many solutions : x=kp,x=p/4+2kp,x=3p/4+kp", - "Solution_9": "No the solutions are what I stated, plus 2(180)x because sin repeats every 360 degrees.." -} -{ - "Problem": "(A.Zaslavsky, 9--10) Given four points $ A$, $ B$, $ C$, $ D$. Any two circles such that one of them contains $ A$ and $ B$, and the other one contains $ C$ and $ D$, meet. Prove that common chords of all these pairs of circles pass through a fixed point.", - "Solution_1": "It's the reverse of this problem\n\nhttp://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO1995Problem3" -} -{ - "Problem": "[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=86162[/url] is a topic I started about the recent nat geo results among young people in the US and abroad. \r\n\r\nYou can test yourself here : \r\n\r\n[url]http://geosurvey.nationalgeographic.com/geosurvey/templates/question_1.html[/url] to take a 20 question test on the nat geo site.\r\n\r\n[url]http://www.sheppardsoftware.com/geography.htm[/url] is a cool site where you can quiz yourself\r\n\r\n\r\nNow do you know more sites? I didn't wanna go off topic in Round Table so I started a topic here.\r\n\r\n\r\nI am looking for geo games in general, but especially games that test you in knowing which countries border which and provinces of China or provinces in general.", - "Solution_1": "There was a game like that in Encarta 2004.\r\n(but surely, you [i]are[/i] serious after all!)", - "Solution_2": "[quote=\"bubka\"]There was a game like that in Encarta 2004.\n(but surely, you [i]are[/i] serious after all!)[/quote]\r\n\r\n :| Your messages start getting weirder and weirder.\r\nWhat do you mean?", - "Solution_3": "why nothing could be simpler. Why else would geography of all things bother you? :D \r\n\r\nLet us have a vote. Who else thinks fred is serious?", - "Solution_4": ":|\r\nWow\r\n :?", - "Solution_5": "i mean others", - "Solution_6": "wow..... thats kinda funny lol", - "Solution_7": "I'm 10, and I could locate ALL of the countries on the first quiz!\r\nP.S. I got 70% of the test correct." -} -{ - "Problem": "Find $\\lfloor-{17}\\rfloor$.\r\n\r\nshouldn't $\\lfloor{-17}\\rfloor$ be -4, 5.\r\n$\\sqrt{17}= \\pm{4....}$ \r\nso floor of $-4...=-5$\r\nand floor of $4....=4$,\r\nbut the answer is only -5. \r\n\r\n\r\nthanks :)", - "Solution_1": "[hide]\nFor this problem, we want to find an integer that is less than or equal to $-17$ but also has the least difference when it is subtracted from $-17$. Since 0 is the least difference between two numbers, and $-17$ is an integer, the answer would be $-17-0 = \\boxed{-17}$.[/hide]", - "Solution_2": "[quote=\"myc\"]Find $\\lfloor-{17}\\rfloor$.\n\nshouldn't $\\lfloor{-17}\\rfloor$ be -4, 5.\n$\\sqrt{17}= \\pm{4....}$ \nso floor of $-4...=-5$\nand floor of $4....=4$,\nbut the answer is only -5. \n\n\nthanks :)[/quote]\r\n\r\nThe question is actually:\r\nFind $\\lfloor-\\sqrt{17}\\rfloor$\r\n\r\nWhat you're doing is taking both the positive and negative root of 17. However, (as stated on page 6 of AOPS book 1), \"A good rule of thumb is: if the radical sign or the fractional power was in the problem to start with, then we are only looking for the positive root; if we force the problem to have a fractional exponent, then we must find all answers.\"\r\n\r\nSo we would only take the floor of 4....., which would be 4.", - "Solution_3": "[quote=\"Klebian\"]\n\nSo we would only take the floor of 4....., which would be 4.[/quote]\r\nactually, we would take the [b]opposite[/b] of the square root of 17, (notice negative sign outside of radical but inside floor function) or [b]-[/b]4......, the floor of which would be -5.", - "Solution_4": "[quote=\"myc\"]Find $\\lfloor-{17}\\rfloor$.\n\nshouldn't $\\lfloor{-17}\\rfloor$ be -4, 5.\n$\\sqrt{17}= \\pm{4....}$ \nso floor of $-4...=-5$\nand floor of $4....=4$,\nbut the answer is only -5. \n\n\nthanks :)[/quote]\r\n\r\nthe reason is that the floor function does round but it doesnt round like the way you are looking at it. It rounds down.\r\n\r\n$\\lfloor{sqrt{17}}\\rfloor=4$\r\nBUT\r\n$\\lfloor{-sqrt{17}}\\rfloor=-5$\r\n\r\nbecause look at it this way\r\n$sqrt 17$ is alittle above 4 but once it is negative it is to round to the smaller of the 2 numbers in the negatives the larger the smaller it is so common sense would be that $-1>-2$\r\nso $-4>-sqrt{17}>-5$\r\nflooring goes to $-5$", - "Solution_5": "[quote=\"dodobird\"][quote=\"myc\"]Find $\\lfloor-{17}\\rfloor$.\n\nshouldn't $\\lfloor{-17}\\rfloor$ be -4, 5.\n$\\sqrt{17}= \\pm{4....}$ \nso floor of $-4...=-5$\nand floor of $4....=4$,\nbut the answer is only -5. \n\n\nthanks :)[/quote]\n\nthe reason is that the floor function does round but it doesnt round like the way you are looking at it. It rounds down.\n\n$\\lfloor{sqrt{17}}\\rfloor=4$\nBUT\n$\\lfloor{-sqrt{17}}\\rfloor=-5$\n\nbecause look at it this way\n$sqrt 17$ is alittle above 4 but once it is negative it is to round to the smaller of the 2 numbers in the negatives the larger the smaller it is so common sense would be that $-1>-2$\nso $-4>-sqrt{17}>-5$\nflooring goes to $-5$[/quote]\r\n\r\nHe was rounding correctly, what he was doing incorrectly was he was taking both the postive and negative root of 17.\r\n\r\nBy the way, k8reindeer is correct.", - "Solution_6": "[quote=\"Klebian\"][quote=\"myc\"]Find $\\lfloor-{17}\\rfloor$.\n\nshouldn't $\\lfloor{-17}\\rfloor$ be -4, 5.\n$\\sqrt{17}= \\pm{4....}$ \nso floor of $-4...=-5$\nand floor of $4....=4$,\nbut the answer is only -5. \n\n\nthanks :)[/quote]\n\nThe question is actually:\nFind $\\lfloor-\\sqrt{17}\\rfloor$\n\nWhat you're doing is taking both the positive and negative root of 17. However, (as stated on page 6 of AOPS book 1), \"A good rule of thumb is: if the radical sign or the fractional power was in the problem to start with, then we are only looking for the positive root; if we force the problem to have a fractional exponent, then we must find all answers.\"\n\nSo we would only take the floor of 4....., which would be 4.[/quote]\r\n\r\nI don't think that's a rule of thumb, I think that the radical is defined as positive square root only." -} -{ - "Problem": "can you solve :D ( for $ 0<\\alpha,\\beta<1$ ) ,\r\n\r\n$ \\int_0^\\infty\\int_0^\\infty\\;x^{\\alpha \\minus{} 1}\\;y^{\\beta \\minus{} 1}\\;e^{ \\minus{} (x \\plus{} y)}\\;\\textbf dx\\;\\textbf dy$\r\n\r\nuse it to show $ \\int_{ \\minus{} \\infty}^\\infty\\;e^{ \\minus{} x^2}\\;\\textbf dx\\; \\equal{} \\;\\sqrt \\pi$", - "Solution_1": "I wonder why there was reply... because it really looks easy (or I'm wrong :rotfl: )\r\n\r\n$ \\int_0^\\infty\\int_0^\\infty\\;x^{\\alpha \\minus{} 1}\\;y^{\\beta \\minus{} 1}\\;e^{ \\minus{} (x \\plus{} y)}\\;\\textbf dx\\;\\textbf dy \\equal{}$\r\n $ \\equal{} \\int_0^\\infty x^{\\alpha \\minus{} 1} e^{ \\minus{} x} \\, \\textbf d x \\cdot \\int_0^\\infty y^{\\beta \\minus{} 1} e^{ \\minus{} y} \\, \\textbf d y \\equal{} \\Gamma ( \\alpha ) \\Gamma ( \\beta )$\r\n\r\nNow using $ \\Gamma (a) \\Gamma ( 1 \\minus{} a) \\equal{} \\frac {\\pi}{\\sin a \\pi}$ we have $ \\Gamma \\left( \\frac {1}{2} \\right) \\equal{} \\sqrt {\\pi}$\r\nand if in the integral\r\n$ \\int_0^{ \\plus{} \\infty} x^{\\frac {1}{2} \\minus{} 1}e^{ \\minus{} x} \\, \\textbf d x \\equal{} \\sqrt {\\pi}$\r\nwe substitute $ x \\equal{} y^2$ we obtain\r\n$ \\int_0^\\infty e^{ \\minus{} y^2} \\, \\textbf d y \\equal{} \\frac {\\sqrt {\\pi}}{2}$\r\nfinally bacause $ \\int_0^\\infty e^{ \\minus{} y^2} \\, \\textbf d y \\equal{} \\frac {1}{2} \\int_{ \\minus{} \\infty}^\\infty e^{ \\minus{} y^2} \\, \\textbf d y$ it yields desired formula." -} -{ - "Problem": "let $ D = {z : |z| < 1}$ be the unit disc \r\nand $ f: D\\rightarrow\\mathbb{C}$ be analytic function, $ f$ is non-constant function\r\nsuppose that $ Re(f(z))\\geq 0$ for all $ z\\in D$, \r\nI need to prove that[b] $ Re(f(z)) > 0$ for any $ z\\in D$[/b]\r\n\r\nthe following is what I am doing : \r\nwe know that $ 0\\leq |Im(f(z))|\\leq |f(z)|$ for any $ z$ . \r\nSuppose that there is \r\n$ z_{0}\\in D$ st $ Re(f(z_{0})) = 0$ then $ |Im(f(z_{0})| = |f(z_{0})|$\r\n\r\n from here I have some thing similar to [b]minimal modulus theorem[/b] but I can not figure it out.\r\n\r\nAny idea will be appreciated", - "Solution_1": "A slightly stronger statement of the same principle: analytic functions are open maps. The image of the disk is an open set in the plane; if it includes a point on the axis, it necessarily contains nearby points on the wrong side of the axis.", - "Solution_2": "thanks jmerry. I got it" -} -{ - "Problem": "Prove that:$(a^{2}+a+1)(b^{2}+b+1)(c^{2}+c+1) \\leq 27$\r\nwhere $a;b;c$ are non-negative and satisfying $a+b+c=3$", - "Solution_1": "It enogh to prove that\r\n\\[(\\frac{\\sum a^{2}+\\sum a+3}{3})^{3}\\geq 27\\] $\\leftrightarrow$\r\n\\[a^{2}+b^{2}+c^{2}\\geq 3\\] $\\leftrightarrow$ \r\n\\[3(a^{2}+b^{2}+c^{2})\\geq (a+b+c)^{2}\\] which is Cauchy", - "Solution_2": "[quote=\"Svejk\"]It enogh to prove that\n\\[(\\frac{\\sum a^{2}+\\sum a+3}{3})^{3}\\geq 27 \\]\n[/quote]\r\nI think, enough to prove that\r\n$(\\frac{\\sum a^{2}+\\sum a+3}{3})^{3}\\leq 27 ,$ which wrong. :wink:", - "Solution_3": "Yes you are right.I flipped the the sign.Thank you" -} -{ - "Problem": "If the plane is partitioned into two disjoint subsets, show that one of the subsets contains three points forming the vertices of an equilateral triangle.\r\n\r\nI would like to know what others make of this problem.", - "Solution_1": "[hide=\"Hint\"]\nCall the subsets \"black\" and \"white\". Look at a regular hexagon, along with its center. Assume the center is black. What can you say about the colors of the vertices? Can you add another small triangle to finish this off?\n[/hide]", - "Solution_2": "I had a solution, I just didn't know whether there were any other ways of doing it.\r\n\r\n[hide]Call the subsets S and T. If |S|<2 then it's clear that T has the triangle, otherwise choose two points in S, say A and F. Some nearby points in the plane (which happen to make a regular hexagon) are B,C,D,E,F,G (The midpoint of AF) and center O, as in the image. If C is in S then the triangle ACF is made, otherwise C is in T.\n\nCase: O is in S. In that case, if either one of G or E is in S then the triangle is GOF or OEF. Otherwise G,C,E are all in T and GCE is the triangle.\n\nCase: O is in T. In that case, if either one of B or D is in T then the triangle is BCO or CDO. Otherwise B,D,F are all in S, leaving the triangle BDF.[/hide]", - "Solution_3": "If $ G$ and $ E$ are not in $ \\textbf{S}$, that does not mean that $ C$ is in $ \\textbf{T}$. ... Right? :huh:", - "Solution_4": "[quote=\"CornFlakes\"]If C is in S then the triangle ACF is made, otherwise C is in T.[/quote]\r\n\r\nFrom this point, we are working under the assumption that C is in T." -} -{ - "Problem": "if $||xy||\\leq C||yx||$ for every pair of elements $x,y\\in B$,where $B$ is banach algebra ,with some constant $C$ ,then $B$ is commutative algebra.", - "Solution_1": "Hint: this has practically nothing to do with superior algebra, it's analysis. ;)", - "Solution_2": "solution? :lol:", - "Solution_3": "isn't this standard ? e.g.,\r\n\r\nLet $ x,y$ be arbitrary elements of the banach algebra and pick any linear functional $ \\phi$ of $ B$. Now define an entire function via $ f(z)\\equal{}\\phi\\left(e^{zx}ye^{\\minus{}zx}\\right)$ and observe the following inequality:\r\n\r\n$ \\left|f(z)\\right|\\leq \\parallel \\phi\\parallel \\,\\left|\\left|e^{zx}ye^{\\minus{}zx}\\right|\\right|\\leq C\\,\\parallel \\phi\\parallel \\,\\parallel ye^{\\minus{}zx}e^{zx}\\parallel \\equal{}C\\,\\parallel \\phi\\parallel \\,\\parallel y\\parallel $\r\n\r\nAs a bounded entire function $ f$ must be a constant, i.e. $ 0\\equal{}f'(z)\\equal{}\\phi\\left(e^{zx}(xy\\minus{}yx)e^{\\minus{}zx}\\right)$ and setting $ z\\equal{}0$ we see that the commutator $ xy\\minus{}yx$ must vanish on each linear functional, which is only possible for $ xy\\equal{}yx$." -} -{ - "Problem": "Hello everyone, I am a new member here and am slightly intimidated. For I have heard a lot of good things about this forum and finally decided to sign up. Well onto my question, I hope one day to become a mathematician but I am slightly worried that I am \"behind the game\" so to speak. Currently I am seventeen years of age and will be entering my senior year of high school this upcoming year. I am currently studying Calculus III (partial differentiation, multiple integrals, etc.) and plan to complete it as well as have a precursory knowledge of Vector Calculus (Topics such as the curl, divergence theorem, etc.) by the end of summer and continue from there. Now I know that I am above average, but to become a mathematician you have to be a tad better than just above average I would assume. So am I on par for a career in Mathematics or do I need to buckle down and learn a little faster?\r\n\r\nSincerely,\r\nAlex.", - "Solution_1": "You're worried that you're behind since you'll only know calc III after high school? I had only AP Calculus after high school, and then took 3 classes my first year of college. You're fine.", - "Solution_2": "You will probably be enormously busy this year, but if you have time, self study some \"real\" calculus based completely on proofs. That will probably help you more than trying to chug down knowledge of linear algebra. Also, I don't think being a mathematicians is about \"who does it first,\" so I think you are fine also.", - "Solution_3": "[quote=\"Anaxerzia\"]You will probably be enormously busy this year, but if you have time, self study some \"real\" calculus based completely on proofs. That will probably help you more than trying to chug down knowledge of linear algebra. Also, I don't think being a mathematicians is about \"who does it first,\" so I think you are fine also.[/quote]\r\nThank you both very much for your responses, your reassurances are well received :D ..but just out of curosity Anaxerzia I know I should understand the proofs of my course work (which I do, I usually try to derive them on my own), but it sounds more like what you are talking about is something akin to Real Analysis where one goes back and formalizes all the reasons behind the workings of the calculus you have already learned. If so, shouldn't I learn Linear Algebra etc. first, for sequentially doesn't Real Analysis come after these courses?\r\n\r\nMathstud.", - "Solution_4": "When I said \"real\" calculus I meant in opposition to what you learn in AP Calculus. I've used Apostol (the textbook) for Calculus and now I am for Multivariable, and I think I'm learning a lot more about mathematics than if I just learned some matrix computations (i.e., chugged down linear algebra).\r\n\r\nThat said, you could also learn about something interesting like Number Theory, which requires rather little prerequisites :).\r\n\r\nEDIT: Also, look at the math portion of this site: ocw.mit.edu for resources for college math.", - "Solution_5": "[quote=\"Anaxerzia\"]When I said \"real\" calculus I meant in opposition to what you learn in AP Calculus. I've used Apostol (the textbook) for Calculus and now I am for Multivariable, and I think I'm learning a lot more about mathematics than if I just learned some matrix computations (i.e., chugged down linear algebra).\n\nThat said, you could also learn about something interesting like Number Theory, which requires rather little prerequisites :).\n\nEDIT: Also, look at the math portion of this site: ocw.mit.edu for resources for college math.[/quote]\r\nOnce again thank you for your responses, they have been helpful.", - "Solution_6": "Get a hold of Spivak's \"Calculus\". Great book, almost entirely proofs. Assumes you know almost nothing, and then completely develops calculus in a rigorous, proof-based manner.\r\n\r\nIf you want a proof-based book for multivariable calculus and/or linear algebra, here are the titles I used:\r\n\r\nMVC: Colley's \"Vector Calculus\"-proves most of the theorems used in the book, such as Green, Stokes, Gauss, and many of the homework problems are proofs. But it also gives you a good amount of computational practice. Great, great book.\r\n\r\nLin Alg: Friedberg, Insel, Spence \"Linear Algebra\"-very rigorous treatment of linear algebra. Provides some computational practice but is mostly proofs. If you can complete this book and do a fair amount of the problems, you can be sure that you \"get\" linear algebra.", - "Solution_7": "[quote=\"JRav\"]Get a hold of Spivak's \"Calculus\". Great book, almost entirely proofs. Assumes you know almost nothing, and then completely develops calculus in a rigorous, proof-based manner.\n\nIf you want a proof-based book for multivariable calculus and/or linear algebra, here are the titles I used:\n\nMVC: Colley's \"Vector Calculus\"-proves most of the theorems used in the book, such as Green, Stokes, Gauss, and many of the homework problems are proofs. But it also gives you a good amount of computational practice. Great, great book.\n\nLin Alg: Friedberg, Insel, Spence \"Linear Algebra\"-very rigorous treatment of linear algebra. Provides some computational practice but is mostly proofs. If you can complete this book and do a fair amount of the problems, you can be sure that you \"get\" linear algebra.[/quote]\r\nThank you very much.", - "Solution_8": "if u participate in competitions like u probably have and get pretty high in the ranks then u should be fine.", - "Solution_9": "Lin Alg is not a pre-req for Real Analysis. As with any class, the more tools you have under your belt, the better off you will be (mostly due to exposure of proofs), but there is nothing I can think of that would make RA easier with a course in Lin Alg first. But I certainly agree some coursework in Number THeory, Graph Theory, and Enumeration would give you a new view on what math is all about.", - "Solution_10": "Real Analysis is more calculus-based. At my school, they require Lin Alg as a pre-requisite but that may be because lin alg is a good course to start learning proofs in." -} -{ - "Problem": "Let $ n, k$ be positive integers and suppose that the polynomial $ x^{2k}\\minus{}x^k\\plus{}1$ divides $ x^{2n}\\plus{}x^n\\plus{}1$. Prove that $ x^{2k}\\plus{}x^k\\plus{}1$ divides $ x^{2n}\\plus{}x^n\\plus{}1$.", - "Solution_1": "we can find roots of the first polynom x^2k-x^k+1 and it alows as to find a relation between n and k.\r\nwe check easily that roots of x^2k+x^k+1 are roots of x^2n+x^n+1.", - "Solution_2": "Er... I may have miscalculated, but are you sure you don't mean $ x^{2n} \\minus{} x^n \\plus{} 1$ for the first condition?", - "Solution_3": "[quote=\"t0rajir0u\"]Er... I may have miscalculated, but are you sure you don't mean $ x^{2n} \\minus{} x^n \\plus{} 1$ for the first condition?[/quote]I think the stated problem is correct as well.", - "Solution_4": "We can calculate the roots of $ x^{2k} \\minus{} x^{k} \\plus{} 1$: $ cis (\\frac {\\pm60}{k} \\plus{} \\frac {360i}{k}), 0 \\le i \\le k \\minus{} 1$\r\nThey are also roots of $ x^{2n} \\plus{} x^{n} \\plus{} 1$, hence: $ \\frac {\\pm 60n}{k} \\plus{} \\frac {360in}{k} \\equal{} 120$ or $ \\minus{} 120 \\pmod {360}, 0 \\le i \\le k \\minus{} 1$ (1)\r\n\r\nIt is sufficient and necessary to prove that k divides n, and $ (\\frac {n}{k},3) \\equal{} 1$ (sufficiency can be seen considering the polynomial $ P(x) \\equal{} x^{2n/k} \\plus{} x^{n/k} \\plus{} 1 : x^2 \\plus{} x \\plus{} 1|P(x)$ because $ {2,1,0} \\equal{} {\\frac {2n}{k},\\frac {n}{k},1} \\pmod 3$, and we choose $ x \\equal{} y^k$)\r\n\r\nThe equation (1) is equivalent to:$ \\frac {\\pm n \\plus{} 6in}{k} \\equal{} 2$ or $ \\minus{} 2 \\pmod {6}$ for every $ 0 \\le i \\le k \\minus{} 1$.\r\nfor i=0 it implies: $ \\frac {n}{k} \\equal{} \\pm2 \\pmod 6$, and than k divides n and also the ratio is relatively prime to 3, and we are done.\r\n\r\nEDIT: we only used the fact that $ cis \\frac{60}{k}$ is a root of the polynomial. Can anyone validate the solution\\present a more elegant one?", - "Solution_5": "Try the following\r\n\r\n[b]Lemma:[/b] Given a prime $ p$ and a positive integer $ n$, we have\r\n\r\n- $ p | n \\implies \\Phi_n(x^p) \\equal{} \\Phi_{np}(x)$\r\n- $ p \\not | n \\implies \\Phi_n(x^p) \\equal{} \\Phi_{np}(x) \\Phi_n(x)$.\r\n\r\nThis can be proven by root-counting or by Mobius inversion." -} -{ - "Problem": "Let $ X=\\left\\{1,2,...,n^{2}\\right\\}$. How many permutations $ \\pi$ of $ X$ have the property that there does not exist a strictly monotonically increasing sequence $ \\left(a_{1},a_{2},...,a_{n+1}\\right)$ of $ n+1$ distinct elements of $ X$ satisfying $ \\pi\\left(a_{1}\\right)<\\pi\\left(a_{2}\\right)<...<\\pi\\left(a_{n+1}\\right)$ or $ \\pi\\left(a_{1}\\right)>\\pi\\left(a_{2}\\right)>...>\\pi\\left(a_{n+1}\\right)$ ?", - "Solution_1": "There is an answer in the OEIS, [url]http://www.research.att.com/~njas/sequences/A079402[/url], but no proof (although a proof is claimed). Also, please try to name your posts in a more useful way." -} -{ - "Problem": "If we have two unequal rationals $r E,Y,O collinear. Similarly X,O,E collinear. We are done \r\n Sorry if I'm wrong", - "Solution_2": "i cant understand your solution.", - "Solution_3": "[img] http://img130.imageshack.us/img130/4411/mlro.jpg [\\img]", - "Solution_4": "i cant see the image that you give me :(", - "Solution_5": "MJ GEO can you post your solution! And @livetolovemath030894 where is point X in the figure?", - "Solution_6": "hi.now my nice solution\r\n$ AD,BC$ intersect at $ Z$. $ ZM,ZN$ are perpendicular to$ AC,BD$ such that $ M,N$ lie on $ AC,BD$. now we must prove that $ \\frac{MA}{AC}\\equal{}\\frac{NB}{BD}$. by sin theorem we have $ \\frac{BD}{AC}\\equal{}\\frac{ZB}{ZA}$. and $ \\frac{NB}{AM}\\equal{}\\frac{ZB}{ZA}$ :)", - "Solution_7": "Similarly we'll have E,O,X collinear. You can paint it yourself :) :) :)", - "Solution_8": "[quote=\"MJ GEO\"]$ ABCD$ is cyclic.$ YA,YB$ are perpendicular to $ AC,BD$ and $ XD,XC$ are perpendicular to $ BD,AC$.prove that $ AD,BC,XY$ are concur at one point.(I have simple and nice solution)[/quote]\r\n\r\nHere is mine\r\nDenote $ YA,YB$ meet $ (O)$ at $ P,Q$\r\nUsing Pascal theorem in $ PABQCD$ with : $ PA$ meets $ QB$ at $ Y$, $ DA$ meets $ BC$ at $ M$ and $ DQ$ meets $ CP$ at $ O$ then $ M,X,O$ are collinear\r\n\r\nSimilarly,$ A,X,O$ are collinear :)" -} -{ - "Problem": "let a,b,c be the length of the sides of triangle ABC\r\nprove that:\r\n$ \\frac{1}{(a\\plus{}b\\plus{}c)^2}\\plus{}\\frac{1}{(ab\\plus{}bc\\plus{}ca)^2}\\geq \\frac{1}{3(a^2b\\plus{}b^2c\\plus{}c^2a)}\\plus{}\\frac{1}{3(ab^2\\plus{}bc^2\\plus{}ca^2)}$", - "Solution_1": "any idea????", - "Solution_2": "It's an old and nice problem. Can_hang had a nice proof for it. But I can't find his link. Sorry :wink:", - "Solution_3": "[quote=\"nguoivn\"]It's an old and nice problem. Can_hang had a nice proof for it. But I can't find his link. Sorry :wink:[/quote]\r\ncan you ask him??thanks\r\nI really want to know the solution of this inequality.", - "Solution_4": "[quote=\"tuandokim\"]let a,b,c be the length of the sides of triangle ABC\nprove that:\n$ \\frac {1}{(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {1}{(ab \\plus{} bc \\plus{} ca)^2}\\geq \\frac {1}{3(a^2b \\plus{} b^2c \\plus{} c^2a)} \\plus{} \\frac {1}{3(ab^2 \\plus{} bc^2 \\plus{} ca^2)}$[/quote]\r\n$ \\frac {1}{(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {1}{(ab \\plus{} bc \\plus{} ca)^2}\\geq \\frac{2}{(a\\plus{}b\\plus{}c)(ab\\plus{}ac\\plus{}bc)}$ and easy SOS. :wink:", - "Solution_5": "[quote=\"arqady\"][quote=\"tuandokim\"]let a,b,c be the length of the sides of triangle ABC\nprove that:\n$ \\frac {1}{(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {1}{(ab \\plus{} bc \\plus{} ca)^2}\\geq \\frac {1}{3(a^2b \\plus{} b^2c \\plus{} c^2a)} \\plus{} \\frac {1}{3(ab^2 \\plus{} bc^2 \\plus{} ca^2)}$[/quote]\n$ \\frac {1}{(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {1}{(ab \\plus{} bc \\plus{} ca)^2}\\geq \\frac {2}{(a \\plus{} b \\plus{} c)(ab \\plus{} ac \\plus{} bc)}$ and easy SOS. :wink:[/quote]\r\nCan you tell me more about your solution???thanks :blush:", - "Solution_6": "[quote=\"tuandokim\"][quote=\"arqady\"][quote=\"tuandokim\"]let a,b,c be the length of the sides of triangle ABC\nprove that:\n$ \\frac {1}{(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {1}{(ab \\plus{} bc \\plus{} ca)^2}\\geq \\frac {1}{3(a^2b \\plus{} b^2c \\plus{} c^2a)} \\plus{} \\frac {1}{3(ab^2 \\plus{} bc^2 \\plus{} ca^2)}$[/quote]\n$ \\frac {1}{(a \\plus{} b \\plus{} c)^2} \\plus{} \\frac {1}{(ab \\plus{} bc \\plus{} ca)^2}\\geq \\frac {2}{(a \\plus{} b \\plus{} c)(ab \\plus{} ac \\plus{} bc)}$ and easy SOS. :wink:[/quote]\nCan you tell me more about your solution???[/quote]\r\nOf cause!\r\nWe need to prove that\r\n$ \\frac {2}{(a \\plus{} b \\plus{} c)(ab \\plus{} ac \\plus{} bc)}\\geq\\frac {1}{3(a^2b \\plus{} b^2c \\plus{} c^2a)} \\plus{} \\frac {1}{3(ab^2 \\plus{} bc^2 \\plus{} ca^2)}.$\r\nBut $ \\frac {2}{(a \\plus{} b \\plus{} c)(ab \\plus{} ac \\plus{} bc)}\\geq\\frac {1}{3(a^2b \\plus{} b^2c \\plus{} c^2a)} \\plus{} \\frac {1}{3(ab^2 \\plus{} bc^2 \\plus{} ca^2)}\\Leftrightarrow$\r\n$ \\Leftrightarrow6\\sum_{cyc}(a^4bc \\plus{} a^3b^3 \\plus{} a^2b^2c^2)\\geq\\sum_{cyc}(a^2b \\plus{} a^2c)\\sum_{cyc}(a^2b \\plus{} a^2c \\plus{} abc)\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}( \\minus{} a^4b^2 \\minus{} a^4c^2 \\plus{} 4a^3b^3 \\plus{} 4a^4bc \\minus{} 5a^3b^2c \\minus{} 5a^3c^2b \\plus{} 4a^2b^2c^2)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow \\minus{} \\sum_{cyc}(a^4b^2 \\minus{} 2a^3b^3 \\plus{} a^2b^4) \\plus{} \\sum_{cyc}(c^3a^3 \\minus{} c^3a^2b \\minus{} c^3ab^2 \\plus{} c^3b^3) \\plus{}$\r\n$ \\plus{} 2\\sum_{cyc}(a^4bc \\minus{} a^3b^2c \\minus{} a^2b^3c \\plus{} b^4ac) \\minus{} 2\\sum_{cyc}(a^3bc^2 \\minus{} 2a^2b^2c^2 \\plus{} ab^3c^2)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(a \\minus{} b)^2( \\minus{} a^2b^2 \\plus{} c^3(a \\plus{} b) \\plus{} 2abc(a \\plus{} b) \\minus{} 2abc^2)\\geq0.$\r\nLet $ a\\geq b\\geq c$ and $ S_c \\equal{} \\minus{} a^2b^2 \\plus{} c^3(a \\plus{} b) \\plus{} 2abc(a \\plus{} b) \\minus{} 2abc^2.$\r\nHence, $ S_a\\geq0,$ $ S_b\\geq0$ and $ (a \\minus{} c)^2\\geq(a \\minus{} b)^2.$\r\nId est, $ \\sum_{cyc}(a \\minus{} b)^2S_c\\geq(a \\minus{} c)^2S_b \\plus{} (a \\minus{} b)^2S_c\\geq(a \\minus{} b)^2\\left(S_b \\plus{} S_c\\right).$\r\nThus, we need to prove that $ S_b \\plus{} S_c\\geq0.$\r\nBut $ S_b \\plus{} S_c \\equal{} \\minus{} a^2c^2 \\plus{} b^3(a \\plus{} c) \\plus{} 2abc(a \\plus{} c) \\minus{} 2ab^2c \\minus{}$\r\n$ \\minus{} a^2b^2 \\plus{} c^3(a \\plus{} b) \\plus{} 2abc(a \\plus{} b) \\minus{} 2abc^2 \\equal{}$\r\n$ \\equal{} a(b^3 \\plus{} c^3) \\plus{} b^3c \\plus{} c^3b \\plus{} 4a^2bc \\minus{} a^2b^2 \\minus{} a^2c^2\\geq$\r\n$ \\geq a^2(b^2 \\minus{} bc \\plus{} c^2) \\plus{} b^3c \\plus{} c^3b \\plus{} 4a^2bc \\minus{} a^2b^2 \\minus{} a^2c^2 \\equal{}$\r\n$ \\equal{} b^3c \\plus{} c^3b \\plus{} 3a^2bc\\geq0.$ Done! :)", - "Solution_7": "nice solution arqady.thanks :D" -} -{ - "Problem": "Could someone please explain what the following means?\r\n\r\n$ \\text{rate} \\sim \\frac{\\text{decrease in reactant concentration}}{\\text{time}} \\sim \\frac{\\text{increase in product concentration}}{\\text{time}}$\r\n\r\nThanks in advance!", - "Solution_1": "[quote=\"elcric\"]Could someone please explain what the following means?\n\n$ \\text{rate} \\sim \\frac {\\text{decrease in reactant concentration}}{\\text{time}} \\sim \\frac {\\text{increase in product concentration}}{\\text{time}}$\n\nThanks in advance![/quote]With \"=\" instead of \"~\" it wouold look like a definition of rate of reaction.", - "Solution_2": "Why would\r\n\\[ \\frac{\\text{amt. of decrease in reactant concentration}}{\\text{time}} \\equal{} \\frac{\\text{amt. of increase in product concentration}}{\\text{time}}\r\n\\]\r\n\r\nThanks!", - "Solution_3": "Well why don't we consider a generic reaction:\r\n\r\n$ \\ce{aA} \\plus{} \\ce{bB} \\rightarrow \\ce{cC} \\plus{} \\ce{dD}$\r\n\r\nWhat does it mean to measure the rate of the reaction? It really just means to measure how fast a reaction is going and since we know that during a chemical reaction the reactants are converted to products wouldn't it logically follow that measuring the rate at which the reactants are used up (or the products are produced) will give an indication of the rate of reaction?\r\n\r\nTry and see if you can deduce why the following holds:\r\n\r\n$ \\frac{\\minus{}1}{a} \\frac{d[A]}{dt} \\equal{} \\frac{\\minus{}1}{b} \\frac{d[B]}{dt} \\equal{} \\frac{1}{c} \\frac{d[C}{dt} \\equal{} \\frac{1}{d} \\frac{d[D]}{dt}$." -} -{ - "Problem": "Each vertex of a cube is assigned an 1 or a -1, and each face is assigned the product of the numbers assigned to its vertices. Determine the possible values the sum of these 14 numbers can attain.", - "Solution_1": "See the product of the eight numbers in the faces of the cube. :)", - "Solution_2": "[hide=\"Solution\"]\nSuppose that for a given configuration, the sum of the vertexes and faces is a number $n$. If we change the number in a vertex from $a$ to $-a$, the total sum of the vertexes can vary by $\\pm 2$ and the total sum of the faces by $\\pm 6, 2$. This means that in total, $n$ can vary by $\\pm 8, 4, 0$. Note that the new result is congruent modulo 4 to the old one.\n\nNow, since all results must be congruent modulo 4, and if we place ones in all the vertexes we obtain 14, the only attainable numbers are:\n$\\pm 2, 6, 10, 14$\n\nDoing some casework:\n-All ones, 14.\n-One minus one, 6.\n-Using two minus ones:\n *Opposite corners, -2\n *In the same edge, 6\n *Same face, different edges, 2\n-We have checked the only three cases that could give 10, a higher number of minus ones will always result in a number smaller than 10, therefore 10 is not attainable. \n-Five minus ones, -6\n-Two ones, in opposite corners, -10.\n-Finally, all minus ones, -2. This means that -14 is not attainable. \n\nAll the possible values are: -10, -6, -2, 2, 6, 14.\n\n\n[/hide]" -} -{ - "Problem": "Prove that |sin nx| :le: n|sin x|, for n = 0, 1, 2, ..., and any real number x.", - "Solution_1": "I am not sure if this is right\n\n\n\n[hide]Using induction,when n=0, we have the inequality true.\n\nSuppose \n\n|sin nx| :le: n|sin x| is true,\n\nthen\n\n|sin(nx)cos(x)| :le: n|sin(x)| because |cosx| :le: 1\n\n|sin(x)cos(nx)| :le: |sinx| because of same reason\n\nadd them up we have\n\n|sin(nx+x)| :le: (n+1)|sinx| as desired.[/hide]", - "Solution_2": "Wow, that's really great! It looks good to me!", - "Solution_3": "using some kind of calculus argument, couldn't you show it is true for any real number n (instead of just integers). I mean the result is obvious if you just look at the graph y = sin(x) and y = x. The slope of sin(x) is always less than or equal to 1.", - "Solution_4": "It's not true for any n. Try n = 1/2 and x = :pi:.\r\n\r\nBy the way, well done Beta! :-)", - "Solution_5": "beta wrote:I am not sure if this is right\n\n[hide]Using induction,when n=0, we have the inequality true.\nSuppose \n|sin nx| :le: n|sin x| is true,\nthen\n|sin(nx)cos(x)| :le: n|sin(x)| because |cosx| :le: 1\n|sin(x)cos(nx)| :le: |sinx| because of same reason\nadd them up we have\n|sin(nx+x)| :le: (n+1)|sinx| as desired.[/hide]\n\n\n\nThis proof needs one more step.\n\n\n\nBy the triangle inequality,\n\n|sin(nx)cos(x) + sin(x)cos(nx)| :le: |sin(nx)cos(x)| + |sin(x)cos(nx)|.\n\n\n\nYou couldn't simply add them straight up since the left hand sides are different functions within the absolute values.", - "Solution_6": "Thanks, Mr.Crawford.", - "Solution_7": "fine, 0 < x < pi/2" -} -{ - "Problem": "A target consists of three squares, measured in inches, as shown. A dart hits the target randomly. What is the probability that it hits the blackened region? Express your answer as a common fraction.\n\n[asy]path A = (3,3)--(3,-3)--(-3,-3)--(-3,3)--cycle;\npath B = scale(4/3)*A; path C = scale(5/3)*A;\ndraw(A); draw(B);draw(C);\nfill(B,black);fill(A,white);\nlabel(\"10\",(0,3),S); label(\"12\",(0,3.8),N); label(\"14\",(0,5),N);[/asy]", - "Solution_1": "Since the area of the whole target is $ 14\\times14\\equal{}196$ that will be the denominator of the answer.\r\n\r\nThen, to find the area of the blackened region you can do the area of the most inner square minus the area of the inner square plus the blackened region.\r\n\r\nThe area of the inner square is: $ 10\\times10\\equal{}100$\r\n\r\nThe area of the inner square plus the blackened region is: $ 12\\times12\\equal{}144$\r\n\r\nThen: $ 144\\minus{}100\\equal{}44$\r\n\r\nSo the answer is: $ \\frac{44}{196}\\equal{}\\frac{11}{49}$" -} -{ - "Problem": "Well they are choosing the U.S. team...slowly. Theres at least 2 maybe 3 Texans on the team this year. A Gymnast from Grand Prarie and a runner from Texas Tech I think. Anyone know if they have chosen the equestrian team yet? If so whose on it. For the U.S. that is.\r\n\r\nFor all you peoples from other coutries Do you know whose representing your country in the games?", - "Solution_1": "Well the only Canadian that I know on our team is one gymnast who was in my class in 7th grade (she was in 8, but we had a split level class). However, she moved away before highschool since her hometown got a proper gymnastics center. The rest I only know by name/reputation...", - "Solution_2": "Well winter Olympiad happened half a year ago!", - "Solution_3": "[quote=mathleticguyyy]Well winter Olympiad happened half a year ago![/quote]\n\nPLEASE STOP ", - "Solution_4": "[quote=mathleticguyyy]Well winter Olympiad happened half a year ago![/quote]\n\n[quote=mikimoto12][quote=mathleticguyyy]Well winter Olympiad happened half a year ago![/quote]\n\nPLEASE STOP[/quote]\n\nTHIS THREAD IS 14 YEARS OLD, wait I'm actually older than it wow", - "Solution_5": "It's kinda sad: this thread os older than me....." -} -{ - "Problem": "Does a \"convex, equilateral hexagon\" mean a regular hexagon?", - "Solution_1": "No, try varying the lengths of opposite sides while keeping the length the same.", - "Solution_2": "Equilateral only means the sides are congruent. A regular polygon also has congruent angles.\r\nA rhombus is an equilateral quadrilateral, but only squares are regular quadrilaterals." -} -{ - "Problem": "$ a,b,c > 0$ then prove:\r\n\r\n\r\n$ \\sum_{cyc} \\sqrt {a^2 \\plus{} b^2 \\plus{} ab} \\ge \\sqrt {3}.(a \\plus{} b \\plus{} c)$", - "Solution_1": "$ \\sum\\sqrt{a^2\\plus{}ab\\plus{}b^2}\\ge \\sum\\sqrt{\\frac 34(a\\plus{}b)^2}\\equal{}\\sqrt3(a\\plus{}b\\plus{}c)$", - "Solution_2": "oke I think it is easy!!\r\na\u00b2+b\u00b2+ab\u2265$ \\frac {3$\\{a + b}^2}${ }{4}$ \r\nsimilar ............." -} -{ - "Problem": "Prove that $ \\ \\frac {1}{3} < \\int_0^1 x^{(\\sin x \\plus{} \\cos x)^2}\\ dx < \\frac {1}{2}$.", - "Solution_1": "We have \r\n\r\n$ (\\sin x+\\cos x)^{2}=(\\sqrt{2}\\sin(x+\\pi/4))^{2}\\leq 2$, and hence,\r\n\r\n$ \\int\\limits_{0}^{1}x^{(\\sin(x)+\\cos(x))^{2}}dx\\geq \\int\\limits_{0}^{1}x^{2}dx=1/3$.\r\n\r\nOn the other hand,\r\n\r\n$ (\\sin x+\\cos x)^{2}=1+\\sin(2x)\\geq 1$ since $ 0\\leq 2x\\leq 2<\\pi$.\r\n\r\nIt follows that\r\n\r\n$ \\int\\limits_{0}^{1}x^{(\\sin x+\\cos x)^{2}}dx\\leq \\int\\limits_{0}^{1}xdx=1/2$.", - "Solution_2": "When does the equality hold?", - "Solution_3": "It does not hold. You have strict inequality there. I simply used, as usual, $ \\leq, \\geq$ instead of using $ <, >$." -} -{ - "Problem": "How many ways to prove this inequality ?\r\n$ \\sum_{i\\equal{}1}^n a_i \\geq n \\sqrt[n]{a_1a_2...a_n}$\r\nall $ a_i \\geq 0$ and $ i \\equal{} 1 , 2 , ..., n$\r\n Can you post a way ?", - "Solution_1": "You will search it in \"arithmetic-geometric mean\" or \"AM-GM\".", - "Solution_2": "Can you prove this ?\r\nI know this inequality is \"AM-GM\" but why don't we post prof this inequality on this topic for everyone to know together", - "Solution_3": "Simple proof\r\n$ \\log \\left(\\frac{\\sum x_i}{n}\\right) \\ge \\frac{\\sum (\\log x_i)}{n} \\equal{} \\frac{\\log (\\prod x_i)}{n} \\equal{} \\log \\sqrt[n]{\\prod x_i}$\r\nby Jensen's inequality\r\nSo, $ \\frac{\\sum x_i}{n} \\ge \\sqrt[n]{\\prod x_i}$\r\n\r\nAnother proof\r\n[url=http://en.wikipedia.org/wiki/AM-GM]Inequality of arithmetic and geometric means[/url]", - "Solution_4": "I read that there are more than 50 elementary ways to prove that. I think the shortest is that one:\r\n\r\nFirst take:\r\n$ A\\equal{}\\frac{\\sum_{i\\equal{}1}^{n}a_{i}}{n}$\r\nThen:\r\n$ (\\prod_{i\\equal{}1}^{n}a_{i})^{\\frac{1}{n}}\\equal{}A(\\prod_{i\\equal{}1}^{n}(1\\plus{}\\frac{a_{i}\\minus{}A}{A}))^{\\frac{1}{n}}\r\n\\leq A(\\prod_{i\\equal{}1}^{n}e^{\\frac{a_{i}\\minus{}A}{A}})^{\\frac{1}{n}}\\equal{}A*e^{\\frac{\\sum_{i\\equal{}1}^{n}(\\frac{a_{i}}{A}\\minus{}1)}{n}}\\equal{}A$\r\nProof by Lev Kourliantchik :wink:", - "Solution_5": "Thanks so much \r\nso intersting . Other ways ? Can you ? \r\nOh , do you know the way to prove this inequality by inductive method of Cauchy ?\r\nIf you know , can you post this way ? thanks a lot", - "Solution_6": "What about:\r\n\r\n$ (1,0,0,.......,0) \\succ (\\frac{1}{n}, \\frac{1}{n}, \\frac{1}{n}, ........, \\frac{1}{n})$\r\n\r\n :rotfl:", - "Solution_7": "[quote=\"BanishedTraitor\"]What about:\n\n$ (1,0,0,.......,0) \\succ (\\frac {1}{n}, \\frac {1}{n}, \\frac {1}{n}, ........, \\frac {1}{n})$\n\n :rotfl:[/quote]\r\nI don't understand what you mean ? What do you mean ?", - "Solution_8": "Muirhead Inequality:\r\n\r\nhttp://en.wikipedia.org/wiki/Muirhead_inequality#Deriving_the_arithmetic-geometric_mean_inequality", - "Solution_9": "Jacobstahl's inequality gives direct solution to the proof of A.M.-G.M.\r\n\r\nI remember that someone said Rado's inequality. :lol: \r\n\r\nA example: 1978 Kyoto University entrance exam/Humanities and Science\r\n\r\nLet $ a,\\ b,\\ c$ be positive numbers. Prove that : $ 2\\left(\\frac{a\\plus{}b}{2}\\minus{}\\sqrt{ab}\\right)\\leq 3\\left(\\frac{a\\plus{}b\\plus{}c}{3}\\minus{}\\sqrt[3]{abc}\\right)$.\r\n\r\nWhen does the equality hold?", - "Solution_10": "Who can prove this inequality by inductive method of Cauchy ? I really need to know this way . Can you hepl me ?" -} -{ - "Problem": "$x,y,z>0$ and $xyz=(1-x)(1-y)(1-z)$\r\nfind least value of expression : $A=x^3+y^3+z^3$", - "Solution_1": "it is easy to see should only do when $0=k such that c(n,0) = c(n,n) = 1 for all n>=0 and\r\nc(n+1,k) = (2^k)(c(n,k)) + c(n,k-1) for n>=k>=1. \r\nProve that c(n,k) = c(n,n-k) for n>=k>=0", - "Solution_1": "As I remember this one is ISL1998,and the solution can be found in John Scholes' website http://www.kalva.demon.co.uk/short/soln/sh98a4.html", - "Solution_2": "Besides, it has been posted by orl on the forum. Try the \"Algebra\" section." -} -{ - "Problem": "Discuss the continuity of $f$\r\n$f(x)=\\lim_{n \\to \\infty}\\left\\{ \\begin{array}{ll}\\cos^{2n}{x}&{ \\text{if}\\ \\ x < 0}\\\\ \\sqrt[n]{1+x^{n}}&{ \\text{if}\\ \\ 0 \\leq x \\leq 1}\\\\ \\frac{1}{1+x^{n}}&{ \\text{if}\\ \\ x > 1}\\end{array}\\right.$", - "Solution_1": "The condition $\\sqrt[n]{1+x^{n}}$ produces a limit of 1 for all numbers less than 1, and the condition $\\frac{1}{1+x^{n}}$ produces a value of 0 for all inputs greater than 1. The function cannot be continuous because if it were, the Intermediate Value Theorem would imply that the function has multiple outputs for the input 1, contradicting the definition of a function.", - "Solution_2": "what about the continuity at point $x=0$? and for $x<0$?", - "Solution_3": "For $x>1,\\,\\lim_{n\\to\\infty}\\frac1{1+x^{n}}=0.$\r\n\r\nFor $0\\le x\\le 1,\\ 1\\le\\sqrt[n]{1+x^{n}}\\le\\sqrt[n]{2}.$ Since $\\lim_{n\\to\\infty}\\sqrt[n]{2}=1,$ we then have that $\\lim_{n\\to\\infty}\\sqrt[n]{1+x^{n}}=1.$\r\n\r\nFor $x<0,\\ f(x)=\\lim_{n\\to\\infty}\\cos^{2n}x$ equals $1$ whenver $x=k\\pi$ for some negative integer $k$ and equals zero otherwise.\r\n\r\nSo in the end, $f$ is discontinuous at nonpositive integer multiples of $\\pi$ and at $1$ and continuous everywhere else. $f$ takes on only the values $1$ and $0.$", - "Solution_4": "Thanks, very nice! Now it's clear. I was wondering in indeterminate forms like $0^{\\infty}$ and $\\pm 1^{\\infty}$ but now i understand it, the $2n$ in $\\cos^{2n}x$ is not there for nothing... By the way, there are any other way to compute $\\lim_{n \\to \\infty}\\sqrt[n]{1+x^{n}}\\ \\ \\text{for}\\ 0\\leq x \\leq 1$ without squeeze theorem?", - "Solution_5": "[quote=\"math_music\"]By the way, there are any other way to compute $\\lim_{n \\to \\infty}\\sqrt[n]{1+x^{n}}\\ \\ \\text{for}\\ 0\\leq x \\leq 1$ without squeeze theorem?[/quote]\r\n\r\nFor $0\\le x< 1,\\ \\lim_{n\\to\\infty}x^{n}=0$ so $1+x^{n}\\to 1.$ Take the $n$th root and we have a \"$1^{0}$\" case. Think about that for a moment and you'll realize that it's not indeterminate. Then handle $x=1$ separately." -} -{ - "Problem": "I'm interested in learning precalc on my own this summer so that I can take this really amazing postcalc elective my senior year. Any advice on how to cover an honors precalc course in three months?", - "Solution_1": "I was thinking about doing that last summer too, but I really don't think it's worth it. Trigonometry is one of the most fundamental parts of math and trying to learn it in 3 months really will not give you a good mastery of it. Ofcourse I'm sure it's *possible* to learn everything and learn it well, but you'll be doing like 4-5 hours of precal everyday, so in my opinion it's a waste of time, just wait for your senior year :)", - "Solution_2": "I took a precalc course over the summer...but it ended up being somewhat useless because it missed out several key things (3-d land, vectors, that whole section). Anyways, the stuff which I did learn wasn't really connected and I wasn't given any applications or reasons. Anyways, I'm learning many new things in precalc now.", - "Solution_3": "i skipped pre calc ... if you're going to learn it on your own probably some of the more important things to go over are the trig aspects and polar graphing... a lot of the rest you'll get to see again during calculus.", - "Solution_4": "It's not unheard of to take pre-calc and calculus simultaneously. Ask if you can take two math courses a year.\r\nI don't remember what exactly was covered in pre-calc, but definitely not all of it was required for a decent understanding of calculus. We did things like mathematical induction, fractals, conic sections and some series. Only the series turned up in calculus and we reviewed them in calculus before actually using them in a calculus context.\r\nYou don't need to know about vectors or 3-space for first year calculus...", - "Solution_5": "precalc was actually my favorite math subject in high school. you do alot of little things that may not be necessarily connected but nonetheless interesting. even though they may not be very important for a good understanding of calculus, some things covered in precalc are just plain cool and you don't get to see them in any other context.", - "Solution_6": "nonono Trig is actually pretty easy to learn. If you work hard, you can learn trig in about 1 month and a half. then learn the rest", - "Solution_7": "You don't learn much of anything in precalc. Precalc is basically the class where you go like \"Hey there are a bunch of interesting concepts in math let's name them and draw pictures and then ignore them.\" You shouldn't have trouble learning it in a short period of time. Don't worry about the trig; there's really not much there (although apparently there's this rational trig thing or something which looks really cool but utterly unrelated to precalc). In fact, http://www.mathworld.wolfram.com might be enough to teach you most of it.", - "Solution_8": "[quote=\"probability1.01\"]You don't learn much of anything in precalc. Precalc is basically the class where you go like \"Hey there are a bunch of interesting concepts in math let's name them and draw pictures and then ignore them.\" You shouldn't have trouble learning it in a short period of time. Don't worry about the trig; there's really not much there (although apparently there's this rational trig thing or something which looks really cool but utterly unrelated to precalc). In fact, http://www.mathworld.wolfram.com might be enough to teach you most of it.[/quote]\r\n\r\nMy precalc must be very different...", - "Solution_9": "Get a precalc book, figure out how much of it you need to get through each day to cover the subject, and then do it. \r\n(That's what I'd do anyway)", - "Solution_10": "[quote=\"Feynmaniac\"]I'm interested in learning precalc on my own this summer so that I can take this really amazing postcalc elective my senior year. Any advice on how to cover an honors precalc course in three months?[/quote]\r\n\r\nThe content of precalculus courses varies wildly from school to school. (I didn't take precalc, myself; at my school it was a hybrid of honors algebra II and what is now usually the beginning of calculus.) My advice would be to approach the current honors precalc teacher and ask which sections of the book are covered in the course, and what topics in the course don't appear in the book. Ask also if you can borrow a copy of the text for a while. Then you'll have a much better idea of how massive/difficult a task you have ahead of you.", - "Solution_11": "I think that precalc is a complete waste of time. The only part of it that you actually need is limits, which you can learn in about 5 minutes. My school has all of the students on the honors math track skip it so we can devote more time to calculus.", - "Solution_12": "Can someone tell me what exactly is in pre-calc? How much of calculus, how much trig, how much Alg II? I'm not skipping it cause I don't have any time during the summer but I just want to know what's in it so I can be a little ahead.", - "Solution_13": "[quote=\"SplashD\"]Can someone tell me what exactly is in pre-calc? How much of calculus, how much trig, how much Alg II? I'm not skipping it cause I don't have any time during the summer but I just want to know what's in it so I can be a little ahead.[/quote]\r\n\r\n30% TRIG\r\n60% alg 2\r\n\r\n10% calculus (limit and that h->0 derivative thing which i dont remember the name of)\r\n\r\njust my opinion which is a crude estimation.\r\n\r\nThe \"calculus\" you learn will be useless very soon though.", - "Solution_14": "[quote=\"ShoeFactory\"][quote=\"SplashD\"]Can someone tell me what exactly is in pre-calc? How much of calculus, how much trig, how much Alg II? I'm not skipping it cause I don't have any time during the summer but I just want to know what's in it so I can be a little ahead.[/quote]\n\n30% TRIG\n60% alg 2\n\n10% calculus (limit and that h->0 derivative thing which i dont remember the name of)\n\njust my opinion which is a crude estimation.\n\nThe \"calculus\" you learn will be useless very soon though.[/quote]\r\n\r\nIs there calculus in USAMO though?", - "Solution_15": "No but it might help. It has helped me in a few places for math competitions and tests. Calculus is very powerful :O\r\n\r\nPrecalculus calculus is useless though, except for maybe limits which you may or may not consider calculus.", - "Solution_16": "I just completed \"Trigonometry/Math Analysis\" (fancy name for precalculus) at my high school. in which i just graduated :) . Anyways, the course was lame and pretty pointless. I will show you why:\r\n\r\nUnit 1: Linear and quadratic functions (covered in algebra 1 and algebra 2)\r\n\r\nUnit 2: Inequalities (everything in this unit except polynomial inequalities was covered in previous algebra courses)\r\n\r\nUnit 3: Functions (The only new thing we learned was about period ic functions. Everything else such as reflections, translations, and composition of functions etc.. was covered in algebra 2)\r\n\r\nUnit 4 through Unit 8 was all trigonometry, which was new. (trigonometric functions, inverse trigonometric functions, the unit circle, trigonometric equations and identities, right triangle trigonometry and other triangle trigonometry such as the law of sines and cosines, the graphs of all the trigonometric equations, inclination of lines, polar coordinates and graphs, and de moivre's theorem)\r\n\r\nUnit 9: Polynomial functions (Covered in Algebra 2, except with the addition of Descartes rule of signs)\r\n\r\nUnit 10: Exponential and Logarithmic Functions (LAME. COMPLETE ALGEBRA 2 REVIEW)\r\n\r\nUnit 11: Vectors (This chapter was new, although i took the basic high school physics class last year so alot of the material in this chapter was review to me)\r\n\r\nUnit 12: Sequences and Series (Algebra 2 review with the exception of intervals of convergence, limits of sequences, recursive definitions, and mathematical induction)\r\n\r\nUnit 13: Random bits and pieces from the 3 chapters on combinatorics, probability, and statistics (this was also a complete algebra 2 review except for the discussion of normal distributions in the statistics chapter.However i was also concurrently enrolled in AP statistics so i have had lectures on normal distributions up the ass....so yeah. i slept through this unit...as well as most of the other ones)\r\n\r\nUnit 14: Matrices (This was a new unit)\r\n\r\nUnit 15: Limits, continuity, and rational functions (this was also new)\r\n\r\nSo yeah....this class was basically 50% new material, and the other 50% was Algebra 2 all over again. So basically i think it would be alright for you to take precalc over the summer. I'm assuming you have had algebra 2, so you can skim through alot of precalc very quickly so you can concentrate most of your time on trig. The trig isn't very hard at all. It's just a pain to memorize all the identities, sum+difference formulas, and the double and half angle formulas. In fact, i already forgot most of that stuff except for the stuff i use regularly D:", - "Solution_17": "dude i skipped trig just like your thinking. I think that taking it on my own I have a better foundation in trig/precalc than anyone else in my school that took the class. you have to take it serious though. for like 3 full chpts. of the book i made sure i did almost all the odd problems (so i could check my answer) it went by pretty fast though. maybe like 2 hrs. a day.", - "Solution_18": "I'm planning to skip precalculus so that I can do calculus in sophmore year. To do so, I must do reasonably well (80 or above) on both the (honors) precalculus midyear and final exams at our school.\r\n\r\nIn my case, I'm already familiar with most of the concepts in precalculus - it's only a matter of brushing up on a few random details and getting practice. Thus, I'm doing most of the problems in the book we use, which the course follows closely. I don't particularly recommend skipping it unless you already have a general understanding of the concepts involved.\r\n\r\n---------------------------------------------------\r\nAbout the actual course:\r\n\r\nThe course of our school uses the book \"Advanced Mathematics: Precalculus with Discrete Mathematics and Data Analysis, Brown\" and is called Adv. Algebra/Trigonometry/Analytic Geo Honors (citing certain topics that the normal American Precalculus course apparently doesn't cover, as per our department head). As for what it contains, it's about 50% review of earlier courses, 25% trigonometry, and 25% other stuff (polar coordinates, vectors, limits, intro to calc). \r\n\r\nThe exact breakdown of the chapters is:\r\nSemester 1:\r\nChapter 1 - Linear Equations and Quadratics (basic Algebra 1/2 stuff, including complex numbers)\r\nChapter 2 - Polynomials (various stuff about polynomials - factors, remainder theorem, etc)\r\nChapter 3 - Inequalities - linear inequalities, polynomial inequalities where you have to factor, absolute value inequalities, and graphing 2-variable inequalities\r\nChapter 4 - functions (inverses, symmety, translating graphs, periodic functions)\r\nChapter 7 - Trigonometry 1\r\nChapter 8 - Trigonometry 2\r\nChapter 9 - Trigonometry 3\r\nChapter 10 - Trigonometry 4\r\nSemester 2:\r\nChapter 11 - Polar Coordinates and DeMoivre Theorem (new)\r\nChapter 12 - Vectors and Parametric Equations (new)\r\nChapter 13 - Sequences and Series\r\nChapter 14 - Matrices\r\nChapter 19 - Limits\r\nChapter 20 - Derivatives/Intro to calc\r\nChapter 5 - Logarithms\r\nChapter 6 - conics\r\n\r\nSo Chapter 1-4 is basic Alg 1/Alg 2, Chap 7-10 is trig, Chapters 11, 12, 19, and 20 are other new things, and Chapters 13, 14, 5, and 6 are pretty much review Alg 2.\r\n\r\nIf you'd like to see the course description, go to [url]http://ab.mec.edu/abrhs/math/trigonometry331.html[/url]\r\n--------------------------------------------------", - "Solution_19": "i took honors precalc this year and it was rather boring and we didn't learn much; the only new thing was matrices and...some real easy calculus. the trig was all review from the previous year (everyone has to take trig first, then most people forget everything so we learn it all again in precalc). still, it might be different/harder in other schools depending on the exact definition of precalc.", - "Solution_20": "Hey Calc Rulz, did you guys use the textbook \"Advanced Mathematics: Precalculus with discrete mathematics and data analysis\" by Richard G. Brown? lol because we used that textbook and it had the same exact chapters as what you mentioned and in that order as well. My class wasn't the honors class so we didn't cover everything, but we covered most of it. We skipped a section in chapter 4 (functions of several variables), chapter 6 (conic sections) entirely for some stupid reason, even though we covered it in algebra 2, bits and pieces of chapters 15 through 17 (we did select sections from each of those chapters and crammed it all into one unit), and we skipped the last half of chapter 19 and all of chapter 20 (even though i worked most of chapter 20 on my own).", - "Solution_21": "I haven't taken trig yet. I learned a lot on my own, but I think it would be beneficial to take the actual class, so I will.", - "Solution_22": "Would a month to prepare to skip precalc be enough? For me, you have to get a 90 or above to pass and get into calc honors the assessment they give you is the midterm combined with the final.", - "Solution_23": "You're good at math so yes.", - "Solution_24": "[quote=\"SplashD\"]Would a month to prepare to skip precalc be enough? For me, you have to get a 90 or above to pass and get into calc honors the assessment they give you is the midterm combined with the final.[/quote]\r\n\r\nBased on what I've seen you do, I'd say you already know a bunch of the topics in the precal curriculum already. I think it's feasible for you to pick up the remaining topics and prepare for the exam in a month. You'll probably be really bored in pre calculus so I highly suggest you try and do this. Good luck :)", - "Solution_25": "really precalc is a waste of time.\r\nif you learned algebra 2 well and understood it then basically you dont learn much. the only really new stuff is the trig, depending on what kind of geometry background you have. and vectors are easy to learn.\r\nso yes i personally think it is possible to \"learn\" precalc in 3 months. however if you are planning on using what you learned in math competitions then you might want to take precalc to make sure that you didn't skip over anything important.\r\nalso it will be an easy class, somewhere you can study for whatever else it is you need to.", - "Solution_26": "Pre-calc is basically (if you fully know \"Algebra 2\"):\r\n\r\nAlgebra II Review\r\nUnderstand Trig\r\nMemorize Trig Formulas/Identities\r\nBasic knowledge of Vectors\r\nBasic knowledge of Limits\r\nBasic sum/sequence (if you haven't learned from Algebra II)" -} -{ - "Problem": "Hello everyone, \r\n\r\nI'm curious to know:\r\n\r\n[b]A different set of natural numbers has median $ 20$ and medium $ 17$.\n\nWhat will be the largest number in this series?[/b]\r\n\r\nWhat I have done:\r\n\r\nBecause they are natural numbers then are positive;\r\n\r\nalso for the calculation of the median must be sorted in increasing.\r\n\r\nAssuming that the number of terms is odd, then the formula for calculating the median is: $ \\displaystyle\\frac {n \\plus{} 1 }{2} \\equal{} 20$, where $ n \\equal{} 39.$\r\n\r\nAnd the formula for calculating the average we have:\r\n\r\n$ \\displaystyle\\frac {\\displaystyle\\sum_{i \\equal{} 1}^n{X_i}}{39} \\equal{} 17\\longrightarrow{\\displaystyle\\sum_{i \\equal{} 1}^n{X_i \\equal{} 663}}$\r\n\r\nI guess as I now know how many terms and their sum, I calculate the nth term,\r\n\r\nfor if we assume that the numbers are ordered increasingly, the latter being the largest.\r\n\r\nAny ideas?\r\n\r\nThank you very much.\r\n\r\nGreetings", - "Solution_1": "You seem to be thoroughly confused about what the median is ;).", - "Solution_2": "whats the biggest POSSIBLE number in this set i think is your question..\r\n\r\n\r\n$ Y_n \\equal{} X_n \\minus{} 20$ are integers bounded strictly from below by -20, and median 0, with average -3\r\n\r\n$ \\sum_{n \\equal{} 1}^{N} Y_n \\equal{} \\minus{} 3N$\r\n\r\nif we are allowed to repeat numbers:\r\nso we need half of the numbers to be negative or zero, and the other half to be positive or zero.\r\nthus, if we make the sets to have N zeros, N values of -7, and one value of x=N-3, we get the desired properties..\r\nsince you have no restriction on N, this means we can pick as big of a value as we like..\r\n\r\nif we cant repeat numbers:\r\nwe should never use a zero, since we get less space to raise up the highest value\r\nif we use $ N\\le 19$ negatives we should use $ \\minus{} 19, \\minus{} 18, ..., N \\minus{} 20$\r\nand then we can allow N numbers above zero such that we average to -3 overall,\r\nsince $ \\sum_{k \\equal{} 20 \\minus{} N}^{19} \\minus{} k \\equal{} \\minus{} \\frac {(39 \\minus{} N)N}{2}$, we need to pick N positive numbers which add up to $ \\minus{} 3(2N) \\plus{} \\frac {(39 \\minus{} N)N}{2}$\r\nin order to maximize the highest value, we make the rest as low as possible, giving us $ 1,2,..., N \\minus{} 1$\r\nwhich add up to $ \\frac {N(N \\minus{} 1)}{2}$, and thus the last number must be $ \\minus{} 3(2N) \\plus{} \\frac {(39 \\minus{} N)N}{2} \\minus{} \\frac {N(N \\minus{} 1)}{2}$\r\n$ \\equal{} \\minus{} N^2 \\plus{} 14N$\r\nthis comes to a max at $ N \\equal{} 7$, so the maximum top value we can pick is 49\r\nthats of course.. for the Y values, so shifting back to Xs we can pick a high value of 69" -} -{ - "Problem": "Find an entire function $ f \\ne const$ such that $ f$ equal to $ 0$ in the unit circumference $ S^1$.", - "Solution_1": "Such a function does not exist, by the identity theorem." -} -{ - "Problem": "The positive number $ a$ is chosen such that the terms $ 20, a, \\frac{5}{4}$ are the first, second and third terms,\nrespectively, of a geometric sequence. What is the value of $ a$?", - "Solution_1": "We know that $ \\frac{20}{a}\\equal{}\\frac{a}{\\frac{5}{4}}$. Cross multiplying yields\r\n\\[ a^2\\equal{}25\\\\\r\na\\equal{}\\pm 5\\]\r\nSince $ a$ is positive, our answer is simply $ \\boxed{5}$" -} -{ - "Problem": "Let triangle ABC have incenter I . Let A' the incenter of triangle IBC, B' of IAC and C' of IAB prove that ABC and A'B'C' have the same circumcenter", - "Solution_1": "Maybe your problem is wrong,I think A' should be the circumcenter of the triangle IBC,etc.", - "Solution_2": "[quote=\"marko avila\"]Let triangle ABC have incenter I . Let A' the incenter of triangle IBC, B' of IAC and C' of IAB prove that ABC and A'B'C' have the same circumcenter[/quote]\r\n\r\nIndeed, this is wrong; however, if you actually wanted to assume A', B', C' to be the circumcenters of triangles IBC, IAC and IAB (and not incenters), then it is trivial: According to http://www.mathlinks.ro/Forum/viewtopic.php?t=6095 post #2, the midpoint of the arc BC on the circumcircle of triangle ABC is the circumcenter of triangle BIC. In other words, the circumcenter A' of triangle BIC is the midpoint of the arc BC on the circumcircle of triangle ABC. Hence, the point A' lies on the circumcircle of triangle ABC. Similarly, the points B' and C' lie on this circumcircle as well, and consequently, the triangles ABC and A'B'C' have the same circumcircle, and thus the same circumcenter.\r\n\r\n darij", - "Solution_3": "no wonder i couldnt solve it!!!!! very sorry darij and thanks!!! :blush:" -} -{ - "Problem": "Solve for $ x$: \n\n$ 5\\minus{}[x\\plus{}(3\\minus{}2x)]\\equal{}2[x\\minus{}3(2\\minus{}x)]$.", - "Solution_1": "$ 5\\minus{}x\\minus{}3\\plus{}2x\\equal{}8x\\minus{}12$\r\n$ x\\equal{}2$" -} -{ - "Problem": "Let be $ x,y,z\\in \\mathbb{R}_\\plus{}$ such that $ xyz\\equal{}1$ . Prove that :\r\n\r\n$ \\frac{1\\plus{}xy}{1\\plus{}z}\\plus{}\\frac{1\\plus{}yz}{1\\plus{}x}\\plus{}\\frac{1\\plus{}zx}{1\\plus{}y}\\ge 3$", - "Solution_1": "$ \\sum \\frac {1 \\plus{} yz}{1 \\plus{} x}\\geq \\sum \\frac {1 \\plus{} \\frac {1}{x}}{1 \\plus{} x} \\equal{} \\sum\\frac {1}{x}\\geq \\frac {3}{\\sqrt [3]{xyz}} \\equal{} 3$.\r\n\r\n\r\nAlternative Solution:\r\n\r\n$ \\boxed{1 \\minus{} \\frac {1}{z}\\leq \\ln z\\leq z \\minus{} 1\\ (z > 0)}\\Longrightarrow \\frac {1}{z} \\plus{} \\ln z\\geq 1$, the rest is left for you.", - "Solution_2": "very easy,\r\nby AM-GM,\r\n$ S\\equal{}\\frac{1\\plus{}xy}{1\\plus{}z}\\plus{}\\frac{1\\plus{}yz}{1\\plus{}x}\\plus{}\\frac{1\\plus{}zx}{1\\plus{}y}\\ge 3\\sqrt[3]{\\frac{(1\\plus{}xy)(1\\plus{}yz)(1\\plus{}zx)}{(1\\plus{}x)(1\\plus{}y)(1\\plus{}z)}}$\r\nand we have $ (1\\plus{}xy)(1\\plus{}yz)(1\\plus{}zx)\\equal{}\\left(1\\plus{}\\frac{1}{x}\\right)\\left(1\\plus{}\\frac{1}{y}\\right)\\left(1\\plus{}\\frac{1}{z}\\right)\\equal{}(1\\plus{}x)(1\\plus{}y)(1\\plus{}z)$ then $ S \\geq 3$" -} -{ - "Problem": "The sequence $\\{a_{n}\\}$ is defined as $a_{1}=1$ and $a_{n}=\\frac{n}{a_{n-1}}$.\r\nFind $\\lim_{n\\to\\infty}\\frac{a_{n}}{\\sqrt{n}}$\r\n1) for even $n$;\r\n2) for odd $n$.\r\n\r\nI'm not sure that there is a limit, but they appear to be approaching values that are reciprocals of each other.", - "Solution_1": "Well, by some observation $a_{n}=\\frac{n(n-2)(n-4)\\ldots}{(n-1)(n-3)\\ldots}$. I don't know what that converges to, of it it even converges, but from that we divide by $\\sqrt{n}$ to get $\\frac{\\sqrt{n}}{n-1}\\frac{(n-2)(n-4)\\ldots}{(n-3)(n-5)\\ldots}= \\frac{\\sqrt{n}}{n-1}a_{n-2}$\r\n\r\nSo if the individual $a_{n}$ converge to some real number, the entire thing will approach $0$. But then again, the $a_{n}$ might be diverging, which would change the problem a bit.", - "Solution_2": "[hide]\nTo simplify things, we can set $b_{n}=\\frac{a_{n}}{\\sqrt{n}}$, then\n\\[b_{n}b_{n-1}=\\sqrt{\\frac{n}{n-1}}\\]\nThat explains the part that $b_{2k}$ and $b_{2k+1}$ approach reciprocals of each other as $k$ gets arbitrarily large.\n\nNow we only need to consider the limit for odd $n$. If $n=2m+1$, we have\n\\begin{eqnarray*}a_{2m+1}&=&\\frac{3}{2}\\cdot \\frac{5}{4}\\cdot \\frac{7}{6}\\cdot \\cdots \\cdot \\frac{2m+1}{2m}\\\\ a_{2m+1}^{2}&=&\\frac{3}{2}\\cdot \\frac{3}{2}\\cdot \\frac{5}{4}\\cdot \\frac{5}{4}\\cdot \\cdots \\cdot \\frac{2m+1}{2m}\\cdot \\frac{2m+1}{2m}\\\\ &\\geq &\\frac{3}{2}\\cdot \\frac{4}{3}\\cdot \\frac{5}{4}\\cdot \\frac{7}{6}\\cdot \\cdots \\cdot \\frac{2m+1}{2m}\\cdot \\frac{2m+2}{2m+1}\\\\ &=&m+1\\\\ b_{2m+1}&\\geq &\\sqrt{\\frac{m+1}{2m+1}}\\\\ &>&\\frac{1}{2}\\\\ \\end{eqnarray*}\nBut also, from our formula for $b_{n}b_{n-1}$, we can get\n\\[b_{2m+1}=\\frac{\\sqrt{(2m+1)(2m-1)}}{2m}b_{2m-1}= 6.5x\r\n[b]x >= 6.5/12[/b]\r\n\r\nAnd with the right side.\r\n\r\n8-12x <=10/3 \r\n4.66 >= 12x \r\n[b]4.66/12 >= x [/b]\r\n\r\nLeaving, [b] 6.5/12 <= x <= 4.66/12 [/b] ?\r\n\r\nIm very unsure about the 6.5/12.", - "Solution_7": "You have to work on all three parts of the inequality at once. For example, what you have right now is\r\n$\\frac{5}{2}\\le 8-12x\\le \\frac{10}{3}$\r\nIf you plan to add 12x, you have to do it to all three sides, not just left and center. I don't think that's a very good idea anyways, because then you have a variable term in two of the parts instead of one. It would be better to get the constant out of the center by subtracting 8 from [i]all three parts[/i] and then dividing by -12. (Oh yeah and don't forget to reverse the signs when multiplying/dividing by a negative number. I used to have serious trouble with that.)\r\nDid I explain that clearly enough?", - "Solution_8": "[quote=\"LynnelleYe\"]You have to work on all three parts of the inequality at once. For example, what you have right now is\n$\\frac{5}{2}\\le 8-12x\\le \\frac{10}{3}$\nIf you plan to add 12x, you have to do it to all three sides, not just left and center. I don't think that's a very good idea anyways, because then you have a variable term in two of the parts instead of one. It would be better to get the constant out of the center by subtracting 8 from [i]all three parts[/i] and then dividing by -12. (Oh yeah and don't forget to reverse the signs when multiplying/dividing by a negative number. I used to have serious trouble with that.)\nDid I explain that clearly enough?[/quote]\r\n\r\nThanks for your help so far.\r\n\r\nWould it be something like this.\r\n\r\n$\\frac{-11}{2}\\le -12x\\le \\frac{-14}{3}$\r\n\r\nand then, divide by 12: $\\frac{-45}{100}\\le -x\\le \\frac{-38}{100}$\r\n\r\nreverse the signs.\r\n\r\n[b]0 .45 >= x >= 0.38[/b] ??\r\n\r\nI could leave it as the original fractions though.", - "Solution_9": "Almost...you did something weird when you divided by 12, because -11/2 divided by 12 should be -11/24 and -14/3 divided by 12 should be -14/36=-7/18, but you've got the concept down.", - "Solution_10": "[quote=\"LynnelleYe\"]Almost...you did something weird when you divided by 12, because -11/2 divided by 12 should be -11/24 and -14/3 divided by 12 should be -14/36=-7/18, but you've got the concept down.[/quote]\r\n\r\nYea ive been using my calculator to help me!\r\n\r\nThanks for the help and that, im not usually this rusty at maths. Im very glad of your help :)", - "Solution_11": "Oopps I need to check just one more problem.\r\n\r\n16-9x/2 > 1-6x/3\r\n\r\ntimes by 2: 16-9x > 2-12x/3\r\n\r\ntimes by 3: 48 - 27x > 2-12 x\r\n\r\ngives me: 46 > 15x \r\n[b]\n46/15 > x[/b] ??", - "Solution_12": "[quote=\"yeoviltown_fc\"]Oopps I need to check just one more problem.\n\n16-9x/2 > 1-6x/3\n\ntimes by 2: 16-9x > 2-12x/3\n\ntimes by 3: 48 - 27x > 2-12 x\n\ngives me: 46 > 15x \n[b]\n46/15 > x[/b] ??[/quote]\r\n\r\nThe first time, when you multiplied by 2, you forgot to multiply the 16 by 2 as well. And it would probably be easier to multiply by 6 instead of 3 and 2...it gives you twice as many places to make arithmetic mistakes otherwise.\r\n\r\nSo try that problem over again, remembering to multiply everything." -} -{ - "Problem": "The number $2000=2^{4}\\cdot 5^{3}$ is the product of seven not necessarily distinct prime factors. Let $x$ be the smallest integer greater than 2000 with this property and let $y$ be the largest integer less than 2000 with this property. Find $x-y$", - "Solution_1": "$x=2^{7}$, still thinking about $y$", - "Solution_2": "[quote=\"M4RI0\"]$x=2^{7}$, still thinking about $y$[/quote]\r\n$2^{7}= 128$ (too small).\r\n\r\nOn the other hand, $3^{7}=2187.$", - "Solution_3": "$2^{6}\\cdot 31=1984$\r\n\r\n$2^{5}\\cdot 5 \\cdot 13 = 2080$\r\n \r\nmaybe?", - "Solution_4": "Oops, i didn't notice it had to be greater than 2000 :blush:", - "Solution_5": "The question remains, how do we methodically tackle the problem? Trial and error doesn't look too good. \r\n\r\nThe numbers pointed out by [b]sen[/b] look very good!", - "Solution_6": "Strategy:\r\n\r\n$2000 = 2^{4}.5^{3}$\r\n\r\nTake all $x=p_{1}p_{2}..p_{n}$ ($7 > n \\ge 0$) where $p_{i}= 2, 5$. There's 1 empty subset, 2 of size 1, 3 of size 2, 4 of size 3, 4 of size 4, 3 of size 5, 2 of size 6.\r\n\r\nThen take the smallest integer $y$ larger than $x$ with $n$ prime factors for each $x$. You have 18 numbers $\\frac{2000}{x}.y$ and you find the smallest of these to find the smallest integer greater than 2000 with 7 prime factors. You have to bash 18 but it beats bashing.. 79.. or trying one by one, I suppose.\r\n\r\nSimilar process for smaller." -} -{ - "Problem": "Determinati x,y intregi daca:\r\n\r\n3x - y = 56 \r\n[x;y] - 4(x;y) = 154", - "Solution_1": "$\\ [x,y]\\cdot\\ (x,y)=xy$\r\n$\\ [x,y]=d\\Rightarrow\\ x=dm,y=dn, (m,n)=1$" -} -{ - "Problem": "assume $ P(x) \\in Z[x]$ and we have $ deg(P)\\geq 2$\r\nlet $ A$ = {$ p(i)|i\\in Z$}\r\nprove there exist $ a,b \\in Z$ such that if we define:$ B$ = {$ an \\plus{} b|n\\in Z$}\r\nwe have $ A\\cap B \\equal{}\\emptyset$", - "Solution_1": "Solution...\r\n[hide]Let $ k$ be a natural number such that $ |P(k \\plus{} 1) \\minus{} P(k)| \\equal{} s \\geq 2$.\nYou can easy to prove that this $ k$ is exist (because $ degP > 1$).\nLook at the numbers:\n\n$ P(k), P(k \\plus{} 1), ..., P(k \\plus{} s \\minus{} 1)$.\n\nThere are $ s$ numbers, and $ P(k) \\equal{} P(k \\plus{} 1) [mod s]$, so there are not all rests of division by $ s$ in $ P(k), P(k \\plus{} 1), ..., P(k \\plus{} s \\minus{} 1)$.\n\nLet $ t$ - natural number such that $ P(k), P(k \\plus{} 1), ..., P(k \\plus{} s \\minus{} 1) \\neq t (mod s)$.\n\nSo, let $ a \\equal{} s$, $ b \\equal{} t$.\n\nObviously that $ P(ms \\plus{} r) \\equal{} P(r) \\neq t (mod s)$ for all integer $ m$ and $ r \\equal{} 0,1,2,...,s \\minus{} 1$.\n\nSo, we find $ a \\equal{} s$ and $ b \\equal{} t$.[/hide]" -} -{ - "Problem": "\u03a0\u03b9\u03c3\u03c4\u03b5\u03c5\u03c9 \u03c0\u03c9\u03c2 \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c4\u03b5 \u03bd\u03b1 \u03c4\u03b7 \u03b2\u03b3\u03b1\u03bb\u03b5\u03c4\u03b1\u03b9 \u03c7\u03c9\u03c1\u03b9\u03c2 \u03bd\u03b1 \u03ba\u03b1\u03bd\u03b5\u03c4\u03b5 \u03c7\u03c1\u03b7\u03c3\u03b7 \u03b5\u03be\u03c9\u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03c9\u03bd \u03b8\u03b5\u03c9\u03c1\u03b7\u03bc\u03b1\u03c4\u03c9\u03bd:\r\n\r\n\u0391\u03bd \u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03c5\u03c2 $ x,y,z$ \u03b9\u03c3\u03c7\u03c5\u03bf\u03c5\u03bd:\r\n\r\n$ 1/5^x \\plus{} 1/5^y \\plus{} 1/5^z \\equal{} 3$ \u03ba\u03b1\u03b9 $ x \\plus{} y \\plus{} z \\equal{} 0$\r\n\r\n\u03c4\u03bf\u03c4\u03b5 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b5\u03c4\u03b1\u03b9 \u03bf\u03c4\u03b9 $ x \\equal{} y \\equal{} z \\equal{} 0$", - "Solution_1": "[quote=\"NickNafplio\"]\u03a0\u03b9\u03c3\u03c4\u03b5\u03c5\u03c9 \u03c0\u03c9\u03c2 \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c4\u03b5 \u03bd\u03b1 \u03c4\u03b7 \u03b2\u03b3\u03b1\u03bb\u03b5\u03c4\u03b1\u03b9 \u03c7\u03c9\u03c1\u03b9\u03c2 \u03bd\u03b1 \u03ba\u03b1\u03bd\u03b5\u03c4\u03b5 \u03c7\u03c1\u03b7\u03c3\u03b7 \u03b5\u03be\u03c9\u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03c9\u03bd \u03b8\u03b5\u03c9\u03c1\u03b7\u03bc\u03b1\u03c4\u03c9\u03bd:\n\n\u0391\u03bd \u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03c5\u03c2 $ x,y,z$ \u03b9\u03c3\u03c7\u03c5\u03bf\u03c5\u03bd:\n\n$ 1/5^x \\plus{} 1/5^y \\plus{} 1/5^z \\equal{} 3$ \u03ba\u03b1\u03b9 $ x \\plus{} y \\plus{} z \\equal{} 0$\n\n\u03c4\u03bf\u03c4\u03b5 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b5\u03c4\u03b1\u03b9 \u03bf\u03c4\u03b9 $ x \\equal{} y \\equal{} z \\equal{} 0$[/quote]\r\nM\u03b5 Jensen \u03b3\u03b9\u03b1 \u03c4\u03c1\u03b5\u03b9\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd\u03c2? (\u03bf\u03c4\u03b1\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b7 \u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1)", - "Solution_2": "[quote=\"r_boris\"][quote=\"NickNafplio\"]\u03a0\u03b9\u03c3\u03c4\u03b5\u03c5\u03c9 \u03c0\u03c9\u03c2 \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c4\u03b5 \u03bd\u03b1 \u03c4\u03b7 \u03b2\u03b3\u03b1\u03bb\u03b5\u03c4\u03b1\u03b9 \u03c7\u03c9\u03c1\u03b9\u03c2 \u03bd\u03b1 \u03ba\u03b1\u03bd\u03b5\u03c4\u03b5 \u03c7\u03c1\u03b7\u03c3\u03b7 \u03b5\u03be\u03c9\u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03c9\u03bd \u03b8\u03b5\u03c9\u03c1\u03b7\u03bc\u03b1\u03c4\u03c9\u03bd:\n\n\u0391\u03bd \u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03c5\u03c2 $ x,y,z$ \u03b9\u03c3\u03c7\u03c5\u03bf\u03c5\u03bd:\n\n$ 1/5^x \\plus{} 1/5^y \\plus{} 1/5^z \\equal{} 3$ \u03ba\u03b1\u03b9 $ x \\plus{} y \\plus{} z \\equal{} 0$\n\n\u03c4\u03bf\u03c4\u03b5 \u03bd\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03be\u03b5\u03c4\u03b1\u03b9 \u03bf\u03c4\u03b9 $ x \\equal{} y \\equal{} z \\equal{} 0$[/quote]\nM\u03b5 Jensen \u03b3\u03b9\u03b1 \u03c4\u03c1\u03b5\u03b9\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03cd\u03c2? (\u03bf\u03c4\u03b1\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b7 \u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1)[/quote]\r\n\r\nmallon me am-gm kalytera kai periptosh isotitas.\r\n\r\nps. k.bori ksero oti einai to idio pragma, alla kalytera na mi fovizoume ton kosmo me th jensen.", - "Solution_3": "\u03bc\u03c0\u03bf\u03c1\u03b5\u03b9 \u03bd\u03b1 \u03c5\u03c0\u03b1\u03c1\u03c7\u03bf\u03c5\u03bd \u03c0\u03bf\u03bb\u03bb\u03b5\u03c2 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2 \u03b1\u03bb\u03bb\u03b1 \u03c0\u03b9\u03c3\u03c4\u03b5\u03c5\u03c9 \u03c0\u03c9\u03c2 \u03b7 \u03c0\u03b9\u03bf \u03b1\u03c0\u03bb\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b7 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03c4\u03c9:\r\n\r\n$ x \\plus{} y \\plus{} z \\equal{} 0 <\\equal{}> x/3 \\plus{} y/3 \\plus{} z/3 \\equal{} 0$\r\n\r\n$ 1/5^x \\plus{} 1/5^x \\plus{} 1/5^z \\equal{} 3 <\\equal{}>\r\n1/5^x \\plus{} 1/5^x \\plus{} 1/5^z \\equal{} 3*1/5^0 <\\equal{}>\r\n1^3/5^{3x/3} \\plus{} 1^3/5^{3y/3} \\plus{} 1^3/5^{3z/3} \\equal{} 3*1/5^{x/3 \\plus{} y/3 \\plus{} z/3} <\\equal{}>\r\n(1/5^{x/3})^3 \\plus{} (1/5^{y/3})^3 \\plus{} (1/5^{z/3})^3 \\equal{} 3*1/5^{x/3}*1/5^{y/3}*1/5^{z/3}$ (1)\r\n\r\n\u0391\u03c0\u03bf \u03c4\u03b7\u03bd \u03c4\u03b1\u03c5\u03c4\u03bf\u03c4\u03b7\u03c4\u03b1 $ a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} 3abc \\equal{} (a \\plus{} c \\plus{} b)((a \\minus{} b)^2 \\plus{} (b \\minus{} c)^2 \\plus{} (c \\minus{} a)^2)$\r\n\u03c0\u03c1\u03bf\u03ba\u03b5\u03b9\u03c0\u03c4\u03b5\u03b9 \u03bf\u03c4\u03b9 \u03b1\u03bd $ a^3 \\plus{} b^3 \\plus{} c^3 \\equal{} 3abc$ \u03c4\u03bf\u03c4\u03b5 $ a\\equal{}b\\equal{}c$ \u03b7 $ a \\plus{} b \\plus{} c \\equal{} 0$ , \u03bf\u03c0\u03bf\u03c4\u03b5 \u03b7 \r\n(1) \u03b3\u03c1\u03b1\u03c6\u03b5\u03c4\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03c5\u03bd\u03b1\u03bc\u03b1: $ 1/5^{x/3} \\plus{} 1/5^{y/3} \\plus{} 1/5^{z/3} \\equal{} 0$ (\u03b1\u03b4\u03c5\u03bd\u03b1\u03c4\u03bf \u03b1\u03c6\u03bf\u03c5 \u03b1\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03b8\u03b5\u03c4\u03b9\u03ba\u03c9\u03bd \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c0\u03b1\u03bd\u03c4\u03b1 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03c2), \u03b7\r\n$ 1/5^{x/3} \\equal{} 1/5^{y/3} \\equal{} 1/5^{z/3} <\\equal{}>\r\n(1/5)^{x/3} \\equal{} (1/5)^{y/3} \\equal{} (1/5)^{z/3} <\\equal{}>\r\nx/3 \\equal{} y/3 \\equal{} z/3 <\\equal{}>\r\nx \\equal{} y \\equal{} z$ (2)\r\n\r\n\u03b1\u03c6\u03bf\u03c5 \u03bf\u03bc\u03c9\u03c2 $ x \\plus{} y \\plus{} z \\equal{} 0$ \u03b1\u03c0\u03bf (2) \u03c0\u03c1\u03bf\u03ba\u03b5\u03b9\u03c0\u03c4\u03b5\u03b9 $ 3x \\equal{} 0 <\\equal{}> x \\equal{} 0$ (3), \u03b1\u03c1\u03b1 \u03b1\u03c0\u03bf (2) \u03ba\u03b1\u03b9 (3) \u03ba\u03b1\u03c4\u03b1\u03bb\u03b7\u03b3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b5\u03bb\u03b9\u03ba\u03b1 \u03bf\u03c4\u03b9 $ x \\equal{} y \\equal{} z \\equal{} 0$", - "Solution_4": "\u039b\u03bf\u03b3\u03bf \u03b2\u03b9\u03b1\u03c3\u03b9\u03bd\u03b7\u03c2 \u03b5\u03c6\u03b1\u03b3\u03b1 \u03b5\u03bd\u03b1 1/2 \u03bc\u03c0\u03c1\u03bf\u03c3\u03c4\u03b1 \u03b1\u03c0\u03bf \u03c4\u03bf 2\u03bf \u03bc\u03b5\u03bb\u03bf\u03c2 \u03c4\u03b7\u03c2 \u03c4\u03b1\u03c5\u03c4\u03bf\u03c4\u03b7\u03c4\u03b1\u03c2, \u03b1\u03bb\u03bb\u03b1 \u03b1\u03c5\u03c4\u03bf \u03b4\u03b5\u03bd \u03b1\u03bb\u03bb\u03b1\u03b6\u03b5\u03b9 \u03c4\u03b9\u03c0\u03bf\u03c4\u03b1 \u03c3\u03c4\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03c4\u03b7\u03c2 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7\u03c2" -} -{ - "Problem": "Four kids (Allie, Bobby, Callie, and David) meet four teletubbies (Dipsy, Laa Laa, Po, and Tinky Winky) and Barney's crew (Barney, Riff, Baby Bop, and BJ). Each person/being shakes hands with every other one once except for those of his/her same species (i.e. a human doesn't shake hands with a human, a teletubby doesn't shake with teletubby, a dinosaur doesn't shake with dinosaur) How many handshakes would take place if Callie wants to shake hands with Laa Laa twice, Bobby refuses to shake hands with Baby Bop, and Allie (being so friendly) wants to shake hands with everyone no matter the species? \r\n\r\nHard but silly problem! Enjoy!", - "Solution_1": "each person shakes hadns with 8 others and we just have to divide by two and aplay all those special conditions\r\n\r\nwe have $ \\frac {12*8}{2} \\minus{} 1 \\plus{} 1 \\plus{} 3 \\equal{} 51$", - "Solution_2": "No, that can't be it. You forgot to consider the exceptions that Allie shakes hands with everyone and Bobby doesn't want to shake hands with Baby Bop. Also Callie wants to shake hands with Laa Laa twice.\r\n\r\nBtw, mods/admins, why did this get moved? I thought this is AMC level.", - "Solution_3": "1. Because the AMC forum is for AMC test problems or issues about the test. Problems on their own go into the high school forums.\r\n\r\n2. Bobby doesn't want to shake hands once: $ \\minus{}1$. Callie wants two handshakes once: $ \\plus{}1$. Allie wants to shake with the three other humans: $ \\plus{}3$. These are accounted for. Why is his solution incorrect?", - "Solution_4": "oh wait didn't see the last few steps in his equation...sorry :oops: \r\n\r\nThis is a public shaming of myself. TYVM" -} -{ - "Problem": "Hi, I was wondering if you guys could help me out with something. I need some interesting tasks(goals, propositions, problems?) with quadratic equations. And I need them fast. Thanks.", - "Solution_1": "Before Arne reads your message and roasts you up like he did to BEE(NO offence Arne,I sort of enjoyed that :D ),i'll tell you that no one is going to post problems \"specially\" for you,at least i wouldn't...I think you should browse through all the questions and read new posts and find what you want.As for goals and propositions,i don't get what you mean...I'm sure there will be some people ready to help you if you be a little more specific.\r\n\r\nThis isn't a homework site as Arne said(I think) but help will be provided if necessary.\r\n\r\nI am really sorry all the mods and administerators if the above things I said were for you to say and I had no right to say them...You might as well delete my post if you feel so :)", - "Solution_2": "Ah... well can't you at least give me a link where I can find some problems, I would truly appreciate it. And what's with this \"specially\"? I thought people love to help each other.", - "Solution_3": "Well, the homework questions that i see posted here are all obvious, (if you looked at the section in the book, you should be able to solve them). Also, those problems give no satisfaction when you solve them, it is not that we don't want to help, but you have to be challenging yourself to get that help, most people won't waste their time with homework problems, but you can always bother your teacher if you need help, in fact, you should if you are having difficulty in math.", - "Solution_4": "But here are a few things that were similar to the problems on my test (maybe a little easier).\r\n\r\nSolve for $x$: $7x^4+8x^2-25=0$\r\n\r\nWrite a quadratic with the solutions $x=5+9i$ and $x=5-9i$\r\n\r\nIn the quadratic $5x^2-7x+85=0$, What is the sum of the roots? What is the product of the roots? (Roots are the same thing as solutions)\r\n\r\nSolve for $x$ in: $x -3\\sqrt{x}-24=0$", - "Solution_5": "[quote=\"falling_cross\"]Hi, I was wondering if you guys could help me out with something. I need some interesting tasks(goals, propositions, problems?) with quadratic equations. And I need them fast. Thanks.[/quote]\r\n\r\nGoogle: quadratic equations. There will be tons. :D", - "Solution_6": "No homework questions? :D \r\n\r\n[quote=\"rrusczyk\"]Actually, I disagree - this is a great place to get homework help. What's the point of a community if people can't get help from it?[/quote]\r\n\r\nMeditate on that ;) \r\n\r\n\r\nWho's BEE?", - "Solution_7": "[quote=\"236factorial\"]No homework questions? :D \n\n[quote=\"rrusczyk\"]Actually, I disagree - this is a great place to get homework help. What's the point of a community if people can't get help from it?[/quote]\n\nMeditate on that ;) \n\n\nWho's BEE?[/quote]\r\n\r\nYeah i read what Mr.rrusczyk wrote befor i gave my post but i also mentioned that help will be providede if necessary and if he was a bit specific(Read my post again ;) )Also BEE didnot even mention any questions he needed help on...first he should have done all the questions he copuld and ask them here if he ahd any doubts.....and yeah ...A+Math is some site with homework help." -} -{ - "Problem": "How much would it cost to cover the entire United States (including Alaska and Hawaii) with dollar bills? \r\n\r\nWhat would be the cost to each person in the United States?", - "Solution_1": "Area of the United States=about $ 3.79*10^6$ square miles\r\nArea of one dollar bill=$ 16.0254$ square inches.\r\n\r\nOne square mile=$ 4.014*10^9$ square inches, so the total area of the U.S. is $ 1.52*10^{16}$ square inches. Dividing this number by 16 gives us approximately $ 9.5*10^{14}$, or $ 950,000,000,000,000$. That's how many dollar bills it would take to cover the entire U.S. \r\n\r\nThere are about $ 300,000,000$ $ (3*10^8)$ people in the U.S. so it would cost each person about $ \\$3,170,000$ $ (3.17*10^6)$ to supply enough money to cover the U.S. with dollar bills.[/url]\r\n\r\nNote: I took the above population and area figures from Wikipedia." -} -{ - "Problem": "Find the number of different right triangles which have these conditions:\r\n1) Each side length is positive integer;\r\n2) Perimeter equals to $ 6^n$ ($ n\\in{\\math{N}}$)\r\nExpress your answer in terms of $ n$", - "Solution_1": "Uh, I got $ \\frac{(n(n\\plus{}2))^2}{8}$ for number of distinct right triangles that satisfy above conditions where $ n$ is even.\r\n\r\nBut I will not post my solution because I highly doubt my formula is correct, since it claims that there are 8 right triangles with perimeter 36. :|", - "Solution_2": "Anybody except FantasyLover tried to solve this problem? \r\nFantasy, try to find mistakes in your solution, and try again :)", - "Solution_3": "[quote=\"FantasyLover\"]Uh, I got $ \\frac {(n(n \\plus{} 2))^2}{8}$ for number of distinct right triangles that satisfy above conditions where $ n$ is even.\n\nBut I will not post my solution because I highly doubt my formula is correct, since it claims that there are 8 right triangles with perimeter 36. :|[/quote]\r\n\r\nIt also claims that there are $ 1/8$ distinct right triangles with perimeter 6... in fact there are 0.\r\n\r\nEuclid's formula is sufficient for producing all Pythagorean triples:\r\n\\[ (a,b,c)\\equal{}(2kmn,k(m^2\\minus{}n^2),k(m^2\\plus{}n^2))\\]\r\nAnd the sum is \\[ k(2mn\\plus{}2m^2)\\equal{}2km(m\\plus{}n).\\]\r\nAnd we want this to equal $ 6^x$ (using a different variable to prevent confusion with the $ n$ above). I don't have time to finish this, hopefully this will be useful.", - "Solution_4": "Why don't you just solve this diophantine equation \r\n$ a^2\\plus{}b^2\\equal{}(6^n\\minus{}(a\\plus{}b))^2$\r\ninstead of thinking hard solutions? :)", - "Solution_5": "$ a^2\\plus{}b^2\\equal{}6^{2n}\\minus{}2*6^n(a\\plus{}b)\\plus{}a^2\\plus{}2ab\\plus{}b^2$\r\n\r\n$ 6^{2n}\\minus{}2*6^n(a\\plus{}b)\\plus{}2ab\\equal{}0$\r\n\r\n$ 6^{2n}\\plus{}2ab\\equal{}2*6^n(a\\plus{}b)$\r\n\r\n$ 3^{2n}2^{2n\\minus{}1}\\plus{}ab\\equal{}6^n(a\\plus{}b)$\r\n\r\nThen plug in Euclid's formula.\r\n\r\nI shall run away now before people tell me that I should have done it.", - "Solution_6": "[hide]$ (a\\minus{}6^n)(b\\minus{}6^n)\\equal{}2^{2n\\minus{}1}3^{2n}$\n$ a\\minus{}6^n$ and $ b\\minus{}6^n$ are divisors of $ 2^{2n\\minus{}1}3^{2n}$\nAlso remember that $ a$ and $ b$ have to satisfy these inequalities:\n$ 0 a^2\\plus{}b^2\\plus{}c^2\\plus{}2ab\\plus{}2bc\\plus{}2ac \\equal{} 6^{2n} \\equal{}> 2(c^2\\plus{}ab\\plus{}bc\\plus{}ac) \\equal{} 6^{2n}$ \r\n\r\n$ \\equal{}> (*) c^2\\plus{}ab\\plus{}bc\\plus{}ac \\equal{} 3^{2n}2^{2n\\minus{}1} \\equal{}> (a\\plus{}c)(b\\plus{}c) \\equal{} 3^{2n}2^{2n\\minus{}1}$\r\n\r\nor find values of n for which:\r\n\r\n$ (a\\minus{}b)^2\\plus{}36^n.2$ is a perfect square since this is the discriminant of equation (*).", - "Solution_9": "go on :) :yup:" -} -{ - "Problem": "Hi! there,\r\n\r\nI use Latex for my thesis. I need to plot some graphics in Latex. May I ask which GUI software is your recommended please? I know Tpx, Matagraf...\r\n\r\nI do not know which one is the best. Thx for your suggestion! :rotfl:", - "Solution_1": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53217]Ultimate guide to inserting Pictures in LaTeX[/url]", - "Solution_2": "[quote=\"stevem\"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53217]Ultimate guide to inserting Pictures in LaTeX[/url][/quote]\r\nAsymptote is best because you can insert it in the forum with [asy]and[/asy] tags.", - "Solution_3": "[quote=\"aopsfan\"][quote=\"stevem\"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53217]Ultimate guide to inserting Pictures in LaTeX[/url][/quote]\nAsymptote is best because you can insert it in the forum with [code][asy]\nand\n[/asy][/code] tags.[/quote]\r\n\r\nAdded code tags.", - "Solution_4": "I believe that you can actually also make images in $ \\text{\\LaTeX}$, but I would suggest Asymptote.\r\n\r\n$ \\text{\\LaTeX}$:\r\n$ \\setlength{\\unitlength}{.5cm}\\begin{picture}(5,4) \\put(.5,.5){\\line(5,0){5}} \\put(.5,.5){\\line(0,3){3}} \\put(.5,3.5){\\line(5,-3){5}} \\put(1,.5){\\line(0,1){0.5}} \\put(.5,1){\\line(1,0){0.5}} \\put(-.1,3.5){A} \\put(-.1,0){B} \\put(5.5,0){C} \\put(2.75,0){a} \\put(3,2.1){b} \\put(0,1.75){c} \\end{picture}$\r\n\r\n[code]\n$\\setlength{\\unitlength}{.5cm}\\begin{picture}(5,4) \\put(.5,.5){\\line(5,0){5}} \\put(.5,.5){\\line(0,3){3}} \\put(.5,3.5){\\line(5,-3){5}} \\put(1,.5){\\line(0,1){0.5}} \\put(.5,1){\\line(1,0){0.5}} \\put(-.1,3.5){A} \\put(-.1,0){B} \\put(5.5,0){C} \\put(2.75,0){a} \\put(3,2.1){b} \\put(0,1.75){c} \\end{picture}$\n[/code]\nAsymptote:\n\n[asy]import olympiad;\ndraw((0,0)--(0,3)--(4,0)--cycle);\nlabel(\"$A$\",(0,3),N);\nlabel(\"$B$\",(0,0),SW);\nlabel(\"$C$\",(4,0),S);\nlabel(\"$a$\",(2,0),S);\nlabel(\"$b$\",(2,1.5),NE);\nlabel(\"$c$\",(0,1.5),W);[/asy]\n\n[code]\n[asy]\nimport olympiad;\ndraw((0,0)--(0,3)--(4,0)--cycle);\nlabel(\"$A$\",(0,3),N);\nlabel(\"$B$\",(0,0),SW);\nlabel(\"$C$\",(4,0),S);\nlabel(\"$a$\",(2,0),S);\nlabel(\"$b$\",(2,1.5),NE);\nlabel(\"$c$\",(0,1.5),W);\n[/asy]\n[/code]\r\nWhich one is more simple? Understandable? :D" -} -{ - "Problem": "This question came up incidentally as part of another problem on my problem set last week, and I'm not entirely convinced that it is very difficult but I wasn't able to solve it:\r\n\r\nGiven a real Banach space [i]E[/i] and [i]n[/i] linearly independent vectors [tex]\\displaystyle{e_1, e_2, \\ldots, e_n \\in E}[/tex], let [tex]\\displaystyle S(M)=\\{(\\alpha_1, \\alpha_2, \\ldots, \\alpha_n)[/tex] such that[tex]\\displaystyle ||\\alpha_1e_1 + \\ldots + \\alpha_ne_n|| \\leq M\\}[/tex].\r\n\r\nProve that there exists [tex]\\displaystyle (a_1, a_2, \\ldots, a_n)[/tex] such that [tex]\\displaystyle \\forall (\\alpha_1, \\alpha_2, \\ldots, \\alpha_n) \\in S(M), |\\alpha_1| \\leq |a_1|, |\\alpha_2| \\leq |a_2|, \\ldots, |\\alpha_n| \\leq |a_n|[/tex]", - "Solution_1": "Is the dimension of your Banach space known? I think I can do it for finite-dimensional Banach spaces.", - "Solution_2": "I don't know too much about Banach spaces. But I figured we could assume the space was finite-dimensional by just focusing on the span of the ei.", - "Solution_3": "That might be true, but the solution I think I have does depend on it being finite dimensional.", - "Solution_4": "I don't know that the entire space is finite-dimensional. As Ravi said, however, we can just consider E' to be our new Banach space with basis e_i. Actually, this question arose in a circumstance where I chose the e_i to be the basis of a finite-dimensional subspace of the Banach space. So, if you can just do it for a Banach space of dimension n, that will take care of it for me.", - "Solution_5": "Can anyone explain what a Banach space is? I got bigheaded reading Mathworld's definition. I took a Linear Algebra course at Mathcamp, so I might be able to give this question a shot.\r\n\r\nThanks!", - "Solution_6": "For these purposes, it's a complete normed real vector space. A vector space has two properties: additivity (if w and v are in the vector space, so is (w + v)) and scalar multiplication -- if r is real and w is in the vector space, rw is in the vector space. Oh, and it's a group. I think that's all the important stuff, anyhow.", - "Solution_7": "I think I understand it. If I do, then here's my proof. If not, try not to make me feel too dumb when you explain why I'm stupid :)\r\n\r\nI'll just prove why [tex]\\alpha_1[/tex] is bounded. The rest follows.\r\n\r\nLet [tex]x_2, x_3, \\ldots x_n[/tex] be the least-squares approximation to\r\n\r\n[tex]x_2 e_2+x_3 e_3 + \\cdots + x_n e_n = -e_1.[/tex]\r\n\r\nSince the e's are linearly independent, this least-squares approximation is not exactly equal to [tex]-e_1[/tex], so we must have\r\n\r\n[tex]\\| e_1+x_2 e_2 + \\cdots + x_n e_n \\| = n > 0.[/tex]\r\n\r\nWe then get\r\n\r\n[tex]\\| \\alpha_1 e_1 + \\cdots + \\alpha_n e_n \\| \\ge \\| \\alpha_1 x_1 + (\\alpha_1 x_2)e_2 + \\cdots + (\\alpha_1 x_n)e_2 \\| = |\\alpha_1| n[/tex].\r\n\r\nBut if [tex](\\alpha_1,\\ldots,\\alpha_n) \\in S(M)[/tex], then this norm must be less than or equal to [tex]M[/tex]. Thus, we must have [tex]|\\alpha_1| n \\le M[/tex], i.e. [tex]|\\alpha_1| < M/n.[/tex] So, [tex]|\\alpha_1|[/tex] is bounded as desired.", - "Solution_8": "Here's a start. I know how to finish, but I'll leave it for you...\n\n[I see that zabelman just posted too. Oh well.]\n\n\n\nSpoiler: [hide]Let a be a vector in Rn. Define f(a) to be the norm of the sum of ai ei. It is easy to see that f is continuous. Let S be the set of vectors in Rn with max norm 1. It is easy to see that S is compact. Thus f achieves its minimum on S. Let c be the minimum value. It is nonzero because the ei are linearly independent.[/hide]\n\n\n\nOkay, can you finish off the proof?", - "Solution_9": "I was just going to pick an orthonormal basis and project all the [tex]e_i[/tex] vectors onto the orthonormal basis since the property clearly holds if the [tex]e_i[/tex]s are basis vectors of an orthonormal basis.", - "Solution_10": "To define orthonormal, don't you need the notion of inner product? I thought Hilbert spaces have an inner product, but Banach spaces don't necessarily have one. But I may not know what I am talking about.", - "Solution_11": "You're right, Ravi -- I thought of doing that, too, Simon, but we don't necessarily know that such a thing would exist.", - "Solution_12": "Don't mind my inquisitiveness: ^_^\"\r\n\r\n1. What's the definition of a compact set that's applicable to this problem?\r\n\r\n2. Why did you choose \"with max norm 1\"?", - "Solution_13": "Bleh. It's not a Hilbert space.", - "Solution_14": "The map he described is linear -- f(l*x) = l*f(x), for l real and x a vector. Thus, how the function operates on the set of all vectors with norm 1 tells us basically everything we could want to know about how the function operates everywhere.\r\n\r\nThe relevant definition of compact is a space in which every sequence of points contains an accumulation point. It's not too hard to show that this causes every continuous function to achieve both its maximum and minimum on the set.", - "Solution_15": "Right. (I think you need an absolute value in one of the l's.)\r\n\r\nAlso, for subsets of Rn, it's nice to know that compact is equivalent to closed and bounded. Without the condition about norm 1, the set S wouldn't be bounded.", - "Solution_16": "zabelman, in your proof, how do you know that the least-squares approximation is actually achieved?", - "Solution_17": "[quote]zabelman, in your proof, how do you know that the least-squares approximation is actually achieved?[/quote]\r\n\r\nI don't understand your question.", - "Solution_18": "In your proof, you say that x2, ..., xn is chosen to be the least-squares approximation. What does that mean? How do you know that such xi exist? In particular, how do you know that the errors can't get arbitrarily close to 0?\r\n\r\nTo put my question in a simpler context, suppose that we wanted to choose a positive number x such that 1/x is as close to 0 as possible. There is no such number. You need a way to rule out that kind of scenario.", - "Solution_19": "Ah, I see. I guess I was just assuming there was a 'best approximation' for e1 in Span(e2,...,en). I thought that was a known property... I guess not. [i]My limited linear algebra knowledge from an audited night class at a local college has failed me! NOOOOO![/i] :? Thanks for the tip... I'll see if I can prove it.", - "Solution_20": "I would venture the thought that proving it is a more significant trick than proving the original problem.", - "Solution_21": "zabelman, the least squares approximation that you used only works when you have an inner-product between vectors. The inner product gives you a notion of angles between vectors. But Banach Spaces are more general. In them you have a notion of distance of vectors (i.e. a norm), but not necessarily a notion of angles between vectors (no inner-product).\r\n\r\nYour proof does work fine for vector spaces with inner products though.", - "Solution_22": "Here is a little simpler proof.\r\n\r\nGiven a vector a in Rn, define ||a|| to be the norm of the sum of ai ei. It's easy to see that this new norm on Rn is indeed a norm. (Because of linear independence, the only vector with norm 0 is 0 itself.) But it is well known that all norms on Rn are equivalent. In particular, this norm and the max norm are within a constant factor of each other. So we're done." -} -{ - "Problem": "hey, every time i swim now, i get a cramp..\r\ni usually go at night around 8 or so..\r\nim talking bout the kinda cramp where ur muscle gets really tight, and u cant move tat part of the body, and it hurts like hell..\r\nhow do they occur and how do you prevent them?\r\nand its so dangerous getting a cramp while u swim..", - "Solution_1": "They usually happen because Your muscles aren't properly warmed up or conditioned for the exercise.", - "Solution_2": "You're more likely to get a cramp in swimming because they happen a lot when your muscles get cold and when you use them too much, like BHorseMath said", - "Solution_3": "lol, I was expecting someone to be like \"oh very simple, it's all due to anarobic (sp?) respiration and the build up of...\" ;)", - "Solution_4": "Something to that effect: A cramp is an involuntary and forcibly contracted muscle which does not relax. Cramps can affect any muscle under your voluntary control. Muscles that span two joints are most prone to cramping. Cramps can involve part or all of a muscle, or several muscles in a group. The most commonly affected muscle groups are: Back of lower leg/calf (gastrocnemius), Back of thigh (hamstrings), and Front of thigh (quadriceps). Cramps in the feet, hands, arms, abdomen and along the rib cage are also very common. Muscle cramps range in intensity from a slight tic to agonizing pain. A cramping muscle may feel hard to the touch and/or appear visibly distorted or twitch beneath the skin. A cramp can last a few seconds to 15 minutes or longer. It might recur multiple times before it goes away. Although the exact cause of muscle cramps is unknown (idiopathic), some researchers believe inadequate stretching and muscle fatigue leads to abnormalities in mechanisms that control muscle contraction. Other factors may also be involved, including exercising or working in intense heat, dehydration and depletion of salt and minerals (electrolytes). Stretching and muscle fatigue: Muscles are bundles of fibers that contract and expand to produce movement. A regular program of stretching lengthens muscle fibers so they can contract and tighten more vigorously when you exercise. When your body is poorly conditioned, you are more likely to experience muscle fatigue, which can alter spinal neural reflex activity. Overexertion depletes a muscles oxygen supply, leading to build up of waste product and spasm. When a cramp begins, the spinal cord stimulates the muscle to keep contracting. Heat, dehydration and electrolyte depletion: Muscle cramps are more likely when you exercise in hot weather because sweat drains your bodys fluids, salt and minerals (i.e., potassium, magnesium and calcium). Loss of these nutrients may also cause a muscle to spasm. Just about everyone will experience a muscle cramp sometime in life. It can happen while you play tennis or golf, bowl, swim or do any exercise. It can also happen while you sit, walk or even just sleep. Sometimes the slightest movement that shortens a muscle can trigger a cramp. \r\n\r\nSome people are pre-disposed to muscle cramps and get them regularly with any physical exertion. Those at greatest risk for cramps and other ailments related to excess heat include infants and young children, people over age 65, and those who are ill, overweight, overexert during work or exercise, or take drugs or certain medications. Muscle cramps are very common among endurance athletes (i.e., marathon runners and triathletes) and older people who perform strenuous physical activities. Athletes are more likely to get cramps in the preseason when the body is not conditioned and therefore more subject to fatigue. Cramps often develop near the end of intense or prolonged exercise, or the night after. Older people are more susceptible to muscle cramps due to normal muscle loss (atrophy) that begins in the mid-40s and accelerates with inactivity. As you age, your muscles cannot work as hard or as quickly as they used to. The body also loses some of its sense of thirst and its ability to sense and respond to changes in temperature. \r\n(from the American Academy of Orthopaedic Surgeons)\r\n\r\nIf you bothered to read that, you would have found out nothing really causes a cramp, or at least we don't know to well about it. So that's probably why know one lived up to JS1527's expectations", - "Solution_5": "LOL. Until the bottom few lines, I thought that you had made all that up by yourself RC-7th.", - "Solution_6": "I thought so too, for the first paragraph or two. Then I wondered who in their right mind could draw so accurate a description off the top of their heads, much less be bothered to type it all rather than summarize.", - "Solution_7": "Eat bananas. Seriously, potassium cures it :)", - "Solution_8": "There is actually a technique for removing cramps; you stretch the affected muscle. After you remove the cramp you should continue to use the affected muscle, but at a reduced pace. This makes the cramp less likely to recur.", - "Solution_9": "My old swim coaches always told me to eat bananas because I kept on getting foot cramps. It works - no doubt about that.", - "Solution_10": "cramps? dont swim that much before exercising and warming up", - "Solution_11": "am i the ONLY one who thought he meant the [i]other[/i] type of cramps? :lol:", - "Solution_12": "[quote=\"bleumoose\"]am i the ONLY one who thought he meant the [i]other[/i] type of cramps? :lol:[/quote]\r\n\r\nI knew he was a guy. And I have always known that Tare is a guy.", - "Solution_13": "agrees with tokenadult", - "Solution_14": "What about dysmenorrhea? It seems like it only happens to girls.", - "Solution_15": "[quote=\"churchilljrhigh\"]What about dysmenorrhea? It seems like it only happens to girls.[/quote]\r\n\r\n :? :? :? :? :? *sigh* It was only a matter of time.... only a matter of time.", - "Solution_16": "[quote=\"tokenadult\"][quote=\"bleumoose\"]am i the ONLY one who thought he meant the [i]other[/i] type of cramps? :lol:[/quote]\n\nI knew he was a guy. And I have always known that Tare is a guy.[/quote]\r\n\r\nJust wondering, what does Tare have to do with this topic?", - "Solution_17": "A while ago, in Fun and Games, there was almost this great uprising of people against Tare being moderator, and threads about Tare's gender, which went on for at least 10 pages. Now when there was a confusion of what types of cramps, which essentially is thinking that some one was female instead of male, and it just spurred the memory of Tare, and I guess that's it...", - "Solution_18": "I still don't get why people had a problem with tare being a mod and why people didn't know he was a guy. I figured that out soon after I joined this forum.", - "Solution_19": "I think t'was a case of people claiming that Tare was abusing his modly powers.", - "Solution_20": "Kentilu implied muscle cramps when he mentioned muscles", - "Solution_21": "[quote=\"RC-7th\"]Just wondering, what does Tare have to do with this topic?[/quote]\r\n\r\nYes, I was only mentioning Tare as an example of a forum participant who, although clearly male (to me), was often thought to be female, presumably because people misinterpret the screen name \"Tare\" as a girl's name. (In fact, in Japanese a girl's name would be very unlikely to take that form.) \r\n\r\nThe screen name \"kentliu\" is clearly a boy's name, and the question that opened the thread clearly refers to muscle cramps that follow exercise.", - "Solution_22": "[quote=\"kentliu\"]im talking bout the kinda cramp where ur muscle gets really tight, and u cant move tat part of the body, and it hurts like -..[/quote]\n\n[quote=\"bleumoose\"]am i the ONLY one who thought he meant the [i]other[/i] type of cramps? :lol:[/quote]\r\n\r\nhmm... how did we know it... \r\n\r\n\r\nhard to say, I think I just knew it from his voice :roll:", - "Solution_23": "[quote=\"theone853\"]A while ago, in Fun and Games, there was almost this great uprising of people against Tare being moderator, and threads about Tare's gender, which went on for at least 10 pages. Now when there was a confusion of what types of cramps, which essentially is thinking that some one was female instead of male, and it just spurred the memory of Tare, and I guess that's it...[/quote]\r\n\r\nAnd why were people making a big deal over wether Tare was a boy or a girl?", - "Solution_24": "Well, it was fun." -} -{ - "Problem": "find all $f(x): R\\to R$ such that : \r\n $f(f(xy)+x+y)=f(x+y)+xy$", - "Solution_1": "I thinh here is right problem\r\nfind all continuous function $f(x): R\\to R$ such that : \r\n $f(f(xy)+x+y)=f(x+y)+xy$ $\\forall x,y\\in R$", - "Solution_2": "[quote=\"quangtrung\"]I think here is right problem\nfind all continuous function $f(x): R\\to R$ such that : \n $f(f(xy)+x+y)=f(x+y)+xy$ $\\forall x,y\\in R$[/quote]\r\n thank you , quangtrung \r\n you are right" -} -{ - "Problem": "Given a matrix $ A$ and a positive integer $ n$, show that there is a monic polynomial $ p$ of degree $ n$ minimizing $ \\|p(A)\\|_2$.\r\n(Also true but much harder to prove is that this polynomial is unique, it's called the Chebyshev polynomial of $ A$.)\r\n\r\n\r\nClearly $ g: P^n\\to\\mathbb{R}^\\plus{}: p\\mapsto\\|p(A)\\|_2$ is continuous, but sadly $ P^n$ is not compact so Weierstrass' theorem can't be applied directly. I'm probably missing something obvious here?", - "Solution_1": "I'm sorry if I'm missing something, but why isn't $ P^n$ compact?", - "Solution_2": "compact = closed and bounded, while $ P^n$ is not bounded? Or by what would it be bounded?\r\n\r\nOf couse, $ \\|p(A)\\|\\to\\infty$ if the coeffient vector grows huge, but how to put this formal? \r\n\r\nProbably I am the one misunderstanding something, it was written here as a \"routine exercice\", but I fail to find it...", - "Solution_3": "I am missing something, too, I guess: how can this polynomial be unique if, for $ n$ large enough, we can take various multiples of the characteristic polynomial as $ p$ and get $ p\\left(A\\right)\\equal{}0$ ?\r\n\r\n darij", - "Solution_4": "Darij - note the word \"monic\" in Peter's definition. That takes care of \"various multiples.\"", - "Solution_5": "[quote=\"Kent Merryfield\"]Darij - note the word \"monic\" in Peter's definition. That takes care of \"various multiples.\"[/quote]\r\n\r\nNot quite - you can multiply the characteristic polynomial with $ x\\minus{}1$ or with $ x\\minus{}2$ and get different monic polynomials both attaining $ 0$ at $ A$.\r\n\r\n darij", - "Solution_6": "They probably assume $ n$ to be no larger than the degree of the minimal polynomial.", - "Solution_7": "Let $ m$ be the minimal polynomial of $ A$. If $ n \\ge \\deg m$, then as darij said, we can just choose a multiple of $ m$. So we may assume that $ n < \\deg m$.\r\n\r\nGiven a polynomial $ p$ of degree less than $ \\deg m$, define the norm $ \\left\\Vert p \\right\\Vert$ by\r\n\\[ \\left\\Vert p \\right\\Vert \\equal{} \\left\\Vert p(A) \\right\\Vert_2 .\r\n\\]\r\nThis is indeed a norm: The only thing interesting to check is that $ \\left\\Vert p \\right\\Vert \\equal{} 0$ implies $ p \\equal{} 0$, which follows because $ \\deg p < \\deg m$.\r\n\r\nBut all norms on a finite-dimensional space are equivalent. So $ \\left\\Vert p \\right\\Vert$ is the same as any standard norm up to a constant factor. \r\n\r\nFrom now on, assume that $ p$ is a monic polynomial of degree $ n$. Then from the paragraph above, the set of $ p$ such that $ \\left\\Vert p \\right\\Vert \\le \\left\\Vert x^n \\right\\Vert$ is a compact set. So we can apply a compactness argument after all." -} -{ - "Problem": "I was studying Tensor Products in Algebra by Hungerford and one question says: Give an example of $u \\in A \\otimes B$ , but $u \\ne a \\otimes b$, for any $a \\in A, b \\in B$. Well I don't know an example for that.\r\n\r\nAnother one: If $A'$ is a submodule of the right $R$-module $A$ and $B'$ is a submodule of the left $R$-module $B$, then there is an isomorphism between $A/A' \\otimes B/B'$ and $(A \\otimes A')/C$, where $C$ is the subgroup of $A \\otimes B$ generated by all elements $a' \\otimes b$ and $a \\otimes b'$ with $a \\in A$ , $a' \\in A'$ , $b \\in B$ , $b' \\in B'$.\r\n\r\nI tried to see if $C$ was the kernel of the homomorphism $\\overline{f}: A \\otimes A' \\to A/A' \\otimes B/B'$ where $a \\otimes a' \\to (a+A') \\otimes (b+B')$, but I couldnt prove kernel of $\\overline{f}$ was contained in $C$.\r\n\r\nIf you have a question about other problems you can consult me and i will try to help you because i have the rest of the problems of that section.", - "Solution_1": "is there someone who knows the answer? i think one of u knows it \r\nc'mon i know u can help me , please", - "Solution_2": "For formulas just use LaTeX .... you can quote my post to see the syntax.\r\n\r\nIf $A,B$ are free modules with bases $X,Y$, then $A \\otimes B$ is also free with the basis $\\{x \\otimes y : x \\in X , y \\in Y\\}$. This follows from elementary properties of tensor product and direct sum.\r\n \r\nNow consider this for $|X| = |Y| = 2$. In this case, \"elementary\" tensors have the form\r\n \r\n$(a x_{1}+b x_{2}) \\otimes (c y_{1}+d y_{2}) =(ac) (x_{1}\\otimes y_{1})+(ad)(x_{1}\\otimes y_{2})+(bc)(x_{2}\\otimes y_{1})+(bd)(x_{2}\\otimes y_{2})$\r\n\r\nWe see that the product of the (unique!) coefficients is a square:\r\n \r\n$(ac)(ad)(bc)(bd) = (abcd)^{2}$\r\n \r\nIn general not every element of $R$ is a square (take $R = \\mathbb{R}$), so that $A \\otimes B$ has elements which don't have the form $a \\otimes b$, $a \\in A, b \\in B$, as desired.\r\n\r\nFor the second question: You mean $(A \\otimes B)/C$ in each case, don't you?\r\n \r\nSo let's consider the epimorphism\r\n \r\n$f : A \\otimes B \\to A/A' \\otimes B/B' , a \\otimes b \\to \\overline{a}\\otimes \\overline{b}$\r\n \r\nwhich is induced by the two projections. Obviously,\r\n \r\n$C : = \\langle a' \\otimes b , a \\otimes b' \\rangle$\r\n \r\nis a submodule of $\\ker(f)$. The other inclusion is not obvious, but can be seen by the following trick, which is - by the way - also useful in other proofs dealing with tensor products:\r\n\r\nBecause of $C \\subseteq \\ker(f)$, we can lift $f$ to an epimorphism\r\n\r\n$g : (A \\otimes B)/C \\to A/A' \\otimes B/B' , \\overline{a \\otimes b}\\to \\overline{a}\\otimes \\overline{b}$\r\n\r\nWe want so show that $g$ is an isomorphism, so let's construct an inverse homomorphism. Of course, this has to be done using the universal property of the tensor product. Consider\r\n \r\n$A/A' \\times B/B' \\to (A \\otimes B)/C , (\\overline{a},\\overline{b}) \\to \\overline{a \\otimes b}$\r\n\r\nTo see well-definedness, write $\\overline{a}= \\overline{x}, ~ \\overline{b}= \\overline{y}$, i.e. $a-x \\in A'$ and $b-y \\in B'$. It follows\r\n\r\n$a \\otimes b-x \\otimes y = a \\otimes b-x \\otimes b+x \\otimes b-x \\otimes y =(a-x) \\otimes b+x \\otimes (b-y) \\in C$\r\n\r\nThus, $\\overline{a \\otimes b}= \\overline{x \\otimes y}$. Therefore, we have indeed a map, which is obviously bilinear, so that it induces a homomorphism\r\n\r\n$h : A/A' \\otimes B/B' \\to (A \\otimes B)/C , \\overline{a}\\otimes \\overline{b}\\to \\overline{a \\otimes b}$\r\n\r\nWe have\r\n\r\n$h(g(\\overline{a \\otimes b})) = h(\\overline{a}\\otimes \\overline{b}) = \\overline{a \\otimes b}$\r\n \r\nSince the elements $\\overline{a \\otimes b}$ generate $(A \\otimes B)/C$, this already shows $h \\circ g = id$. So, we're done.", - "Solution_3": "Thanks dude, as you can see i wrote the post again using LateX, so thanks for that too", - "Solution_4": "Another approach for the first problem:\r\n\r\nLet $V=F^{m}$ and $W=F^{n}$. Write the vectors in these spaces as column vectors.\r\n$V\\otimes_{F}W$ is isomorphic to the space of $m\\times n$ matrices, with the identification $e_{i}\\otimes e_{j}=\\begin{bmatrix}0&0&\\cdots&0&\\cdots&0\\\\ 0&0&\\cdots&0&\\cdots&0\\\\ \\vdots&\\vdots&\\ddots&\\vdots&\\ddots&\\vdots\\\\ 0&0&\\cdots&1&\\cdots&0\\\\ \\vdots&\\vdots&\\ddots&\\vdots&\\ddots&\\vdots\\\\ 0&0&\\cdots&0&\\cdots&0\\end{bmatrix}$. (The 1 is in the $i$th row and the $j$th column.)\r\nThis generalizes to the identification $v\\otimes w=vw^{T}$.\r\nEvery elementary tensor $v\\otimes w$ has rank 1 as a matrix. As long as $m$ and $n$ are each at least 2, there are matrices with rank greater than 1. In fact, the rank of a matrix $M$ is the minimum $k$ for which $M$ can be written as a sum of $k$ elementary tensors." -} -{ - "Problem": "Suppose I have an element (in general some kind of expression containing irrational roots) and I want to find the minimal polynomial over $ \\mathbf{Q}[x]$. Is there any good ways of doing this? For instance, suppose I wish to find the minimal polynomial of, say, $ \\frac{1}{2^{1/5}\\minus{} \\sqrt{3}}$. I guess I could set $ x \\equal{} \\frac{1}{2^{1/5}\\minus{} \\sqrt{3}}$ and then try to multiply/square the different sides until I get rid of the roots, and then check if the polynomial is irreducible. Does this work? Is there any better methods?", - "Solution_1": "Try the technique outlined by jmerry in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=405354180&t=202382]this thread.[/url] (Note that all you need to do is calculate the minimal polynomial of $ \\sqrt[5]{2} \\minus{} \\sqrt{3}$ and reverse the coefficients.)", - "Solution_2": "Thanks, great! Nice solution.\r\n\r\nI dont think, however, this problem was meant to be solved that way (in my case). It would work to set $ x \\plus{}\\sqrt{3} \\equal{} ^5\\sqrt{5}$, raise both sides to the power of 5, then move every term containing $ \\sqrt{5}$ to one side and square both sides, right?\r\n\r\nBtw, how do you write the fifth root nice in Latex (I used ^5\\sqrt.. but it isnt very nice).", - "Solution_3": "try \\sqrt[5]{2}", - "Solution_4": "I guess so; jmerry's technique is best suited to proof, especially when $ \\alpha, \\beta$ are not specified and may themselves have complicated minimal polynomials. In this case, both polynomials are simpler, so the form of the conjugates of both are simple to determine and we can in fact explicitly write the minimal polynomial in factored form (using $ \\omega$ to denote a primitive fifth root of unity) as\r\n\r\n$ \\prod_{0 \\le i \\le 4, 0 \\le j \\le 1} \\left( x \\minus{} \\sqrt [5]{2} \\omega^i \\plus{} \\sqrt {3} ( \\minus{} 1)^j \\right)$.\r\n\r\nThis polynomial is fixed by every member of the Galois group of $ \\mathbb{Q}[ \\sqrt [5]{2}, \\sqrt {3} ]$, so it has rational (in fact integer) coefficients, and no subset of its factors has the same property. It also has the expected degree of $ 10$. Of course, it's not that easy to calculate the coefficients this way.", - "Solution_5": "Isn't the degree $ 40$? You need to attach $ \\zeta_5$ for the extension to be Galois.", - "Solution_6": "I don't think it's true that the minimal polynomial of $ \\sqrt[5]{2} \\minus{} \\sqrt{3}$ is the minimal polynomial of $ \\mathbb{Q}[ \\sqrt[5]{2}, \\sqrt{3}, e^{ \\frac{2 \\pi i}{5} } ]$, but I could be wrong.", - "Solution_7": "What's the minimal polynomial of a field extension\u00bf", - "Solution_8": "The minimal polynomial of $ \\sqrt[5]{2}\\minus{}\\sqrt{3}$ is of degree $ 10$, no doubt, since it has $ 5\\times 2\\equal{}10$ conjugates over $ \\mathbb{Q}$. But how can we talk about the Galois group of $ \\mathbb{Q}(\\sqrt[5]{2},\\sqrt{3})$? I meant that the deg. of the smallest Galois extension that contains it is $ 40$.", - "Solution_9": "Right, right. My apologies. :oops:" -} -{ - "Problem": "So, who's going?", - "Solution_1": "I am!\r\nI'm missing an AP World History exam for this.", - "Solution_2": "Dang, that's pretty intense...lol\r\n\r\nI'm missing a day of French...whoo hoo. Anyone not from Northview's team?", - "Solution_3": "I'm not.\r\nSo why was this thread started ,\r\narvind?", - "Solution_4": "Ummm, who r u???\r\n\r\nI started it to, u know, provide somewhere to talk about it...um...yeah.", - "Solution_5": "i think it's cosmo, b/c i interrogated him yesterday concerning his sn, and he told me something like cbk.", - "Solution_6": "Well then he is going. Ha. Btw I know what cbk stands for XD", - "Solution_7": "ME. I heard the test is made this year by ARML members... this should be interesting :)", - "Solution_8": "oh yeah....varsity was me, eric, and santhosh...i think jv was written by ruby, henry, and sam. hope you guys have fun ;)", - "Solution_9": "Also, most of the harder questions were edited out by teachers. The test was written by \"former students\" aka the people Howard mentioned. \r\n\r\nBTW: The tournament is on the 20th, right? Mr. Hedrick failed to answer that 1 question. \r\nBTW2: Karthik, nice location: 01000111011001010110111101110010011001110110100101100001.", - "Solution_10": "karth's location messes up my window size =(", - "Solution_11": "Yea, I changed it back to USA... I wanted to see how long until someone would complain :P", - "Solution_12": "[quote=\"Xantos C. Guin\"]BTW2: Karthik, nice location: 01000111011001010110111101110010011001110110100101100001.[/quote]\r\n\r\nBut does it actually mean anything? Or is it just a bunch of 1's and 0's. Now if it was binary encoded ASCII, THAT would be good! :-)\r\n\r\nFor instance, USA = 010101010101001101000001", - "Solution_13": "It [i]is[/i] ASCII! (kudos to Karthik) (after Mr. Fulton mentioned that possibility, I fired up python and figured out that it was ascii :))", - "Solution_14": "So it is. Very nice Karthik! I just got lazy and didn't figure it out myself. (And being the old guy that I am, I just read the ascii chart without the electronic help.)", - "Solution_15": "What does Karthik's former location mean? Yeah, I'm technologically challenged, so don't make fun of me... :blush:", - "Solution_16": "Why, it's where he lives of course! :D \r\n\r\nRead it as groups of 8 and use that and this chart:\r\n\r\n[url]http://www.asciitable.com/[/url]", - "Solution_17": "[quote=\"captcha000\"]oh yeah....varsity was me, eric, and santhosh...i think jv was written by ruby, henry, and sam. hope you guys have fun ;)[/quote]\r\n\r\n...crap, i'm dead...i wish i could be a slacker and do jv but no...sigh", - "Solution_18": "Hmm, I liked the test, because even though I couldn't finish, I looked it over and after you get past its original \"omigosh\"ness, it's actually pretty elegant...", - "Solution_19": "Great test! I really commend Santosh, Howard, and Eric who spent much of their time making this test. I love how practically everyone got only 10 or so questions right on the written test and placed :D \r\n\r\nThe ciphering was pretty tricky for me... I knew how to do many of them, but slipped up on some because of going to fast (for example, with the one about the infinite series, I forgot the 'positive' part and put 0 and the negative golden ratio for answers). Great test overall! I feel kinda sad that I won't be able to attend the tournament next year though :(", - "Solution_20": "[quote=\"Karth\"]Great test! I really commend Santosh, Howard, and Eric who spent much of their time making this test. I love how practically everyone got only 10 or so questions right on the written test and placed :D [/quote]\r\n\r\nSeriously? Cause that's not what we were shooting for. Just out of curiosity, what were the top scores like?", - "Solution_21": "Well, I got 15 out of the 18 I answered and got 2nd place. Harrison answered 22 or so, I think, and missed a few.", - "Solution_22": "lol....that's kinda funny...", - "Solution_23": "By the way, for the question on the last page which involved something about maximizing the square root of some algebraic expression, there's a pretty cool way to do it with Cauchy's Inequality... let's see if I can find the question sheet :maybe:", - "Solution_24": "I don't feel I did too well, even though I got 10th...\r\n\r\nI got a 25 on ciphering. A 25. Yes. Count 'em. 25...AARGH!!!\r\n\r\nMy written score is sad, like this smiley. :( Or possibly this one. :stink: \r\n\r\nShould have read AoPS 2 better...lolz\r\n\r\nHoping to do better at GA Tech.", - "Solution_25": "Yeah, haha, it's easy to write a test that's way too hard if you're not careful. You have to have a really light touch or you'll become a mass murderer.", - "Solution_26": "I'm glad you guys liked it. I figured it was kind of too hard; I hope everyone there was able to at least solve a few problems. So what were everyone's favorite problems?", - "Solution_27": "[quote=\"Karth\"]By the way, for the question on the last page which involved something about maximizing the square root of some algebraic expression, there's a pretty cool way to do it with Cauchy's Inequality... let's see if I can find the question sheet :maybe:[/quote]\r\n\r\nI believe the question reads as follows: \r\n\r\nIf $ 3x \\plus{} 7y \\equal{} 42$, find the minimum value of $ \\sqrt {x^2 \\plus{} y^2 \\plus{} 6x \\plus{} 8y \\plus{} 25}$?\r\n\r\nBTW: Where there any errors on the test?", - "Solution_28": "No errors on the test, apart from perhaps a small typo towards the end... when you said $ A_i$ denoted the vertices, I interpreted it (fortunately) as the angles.\r\n\r\nThis was my solution to the above problem:\r\n\r\nWe have that\r\n\r\n\\[ x^2 \\plus{} y^2 \\plus{} 6x \\plus{} 8y \\plus{} 25 \\equal{} (x\\plus{}3)^2 \\plus{} (y\\plus{}4)^2\r\n\\]\r\n\r\nNow, by Cauchy's Inequality,\r\n\r\n\\[ (3^2 \\plus{} 7^2)(x^2 \\plus{} y^2) \\ge (3x \\plus{} 7y)^2 \\iff (3^2 \\plus{} 7^2)[(x\\plus{}3)^2 \\plus{} (y\\plus{}4)^2] \\ge [3(x\\plus{}3) \\plus{} 7(y\\plus{}4)]^2 \\iff (x\\plus{}3)^2 \\plus{} (y\\plus{}4)^2 \\ge \\frac{79^2}{58}\r\n\\]\r\n\r\nTherefore, we have that $ \\sqrt{x^2 \\plus{} y^2 \\plus{} 6x \\plus{} 8y \\plus{} 25} \\ge \\frac{79}{\\sqrt{58}}. \\blacksquare$", - "Solution_29": "SHMOINER SQUARE!!! (Sitan's and my way of saying Halmos) :rotfl: :rotfl: :rotfl: \r\n\r\nok, my fav problem was probably either the one with the cubic roots of , or the hot dog and hamburger one. Either that, or number one (I believe it was the one with the function f(xy)=f(x)+f(y)). I like quick, fun things like that that are easier than they seem, even if they seem easy. Then they are just easier than easy, unless they appear so, in which case they are a bit easier than easier than easy.", - "Solution_30": "Hmm... one thing I must say about the test though was that it was unusually focused on functional analysis. About 3 or 4 problems (I think) were on functional analysis out of the 25... that's more than 12 or so percent. However, for the one about $ f(xy) \\equal{} f(x) \\plus{} f(y)$ a really easy/cheap way to do it is to let $ f(x) \\equal{} \\ln{x}$ and proceed. Or you could do it the rigorous way and factor 2008 and stuff... :roll:", - "Solution_31": "I agree, that's why I liked it, I'm good at that stuffz.\r\n\r\nBut yea, I did it rigorously cuz i'm a nubcake.", - "Solution_32": "[quote=\"Karth\"]Hmm... one thing I must say about the test though was that it was unusually focused on functional analysis. About 3 or 4 problems (I think) were on functional analysis out of the 25... that's more than 12 or so percent. However, for the one about $ f(xy) \\equal{} f(x) \\plus{} f(y)$ a really easy/cheap way to do it is to let $ f(x) \\equal{} \\ln{x}$ and proceed. Or you could do it the rigorous way and factor 2008 and stuff... :roll:[/quote] \r\n\r\nI wrote most if not all of those problems. When writing problems, I wrote them in chunks on the same topic. To choose problems, I categorized everything, set approximate goals for each topic, then we voted on which problems we liked and came to a consensus. After that, the teachers requested changes which we made.\r\n\r\nThat first question was one that I didn't like a lot, but I was willing to put it in. There were some other cool combinatoric/advanced questions that were taken out. One involved a ratio of two integrals that were both relating to the volume of some object. The two integrals themselves were hard to evaluate, but the ratio was easy if you knew that both related to calculating the volume of a certain object.", - "Solution_33": "[quote=\"Arvind_sn\"]the hot dog and hamburger one.[/quote]Interesting - that's one of my least favorite questions that I wrote. It seemed too straightforward and wordy.", - "Solution_34": "I liked the cube root one. I actually already memorized the formula, so it was just plugging in numbers. :lol: I somehow messed up one the hotdog and hamburger one. That was weird.\r\nAnyway, I'm happy with 4th right now." -} -{ - "Problem": "For 0y$, prove that $ a^{x}-b^{x}0.$\r\nThus, $ \\max_{0\\leq x\\leq y\\leq z\\leq1}f=\\max\\{f(0,0,0),f(0,0,1),f(0,1,1),f(1,1,1)\\}=0.$\r\nId est, $ f(x,y,z)<0,$ where $ 0200 or wut? or at least, wut's the average? :) thx!", - "Solution_1": "USAMO index=AMC 10/12 score+10*AIME score\r\n\r\nThe cutoff usually averages around the low 200s.", - "Solution_2": "isnt there supposed to be some conversion between the amc 10/12 cuz then the amc 12 people get a ripoff.", - "Solution_3": "Yes, they do get ripped off. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265771]See this topic[/url] for more discussion on those lines." -} -{ - "Problem": "What if you (mainly your brain) was taken apart (basically you die) and reassembled in the same order (then come back to life) it was originally, when you(reassembled) wake up would you still be yourself(the person youre destined to be)?\r\n\r\nIts hard to explain because we don't know if you wake up as the same conciousness before your brain was reassembled. (For all we know that your old self might have died). \r\n\r\nAnother question: since everything happens once, do you think we are predestined? (IF you are questioning predestination right now, that event might be predestined).\r\n\r\nHeh just a bunch of pretty dumb questions... I hope for interesting discussion...", - "Solution_1": "Another question: If you are \"teleported\" to some other place, is the teleported person still you? :D", - "Solution_2": "[quote=\"jimhu\"]What if you (mainly your brain) was taken apart (basically you die) and reassembled in the same order (then come back to life) it was originally, when you(reassembled) wake up would you still be yourself(the person youre destined to be)?\n\nIts hard to explain because we don't know if you wake up as the same conciousness before your brain was reassembled. (For all we know that your old self might have died). \n\nAnother question: since everything happens once, do you think we are predestined? (IF you are questioning predestination right now, that event might be predestined).\n\nHeh just a bunch of pretty dumb questions... I hope for interesting discussion...[/quote]\r\n\r\n\r\nI think these are great questions! I've been thinking about the former one since I first saw an episode of \"Star Trek\". I have thought about something similar to the latter one a great deal recently. Here's an interesting question along those lines: \"If we live in a deterministic, non-branching universe\", what does it mean to say something is 'possible' or 'necessary' (in the metaphysical, as opposed to epistemological sense). Moreover, what kind of difference is there (if any) between counterfactuals about the past, and those about the future?\r\n\r\nI'm tending towards a radical position (which I would like to call 'modal nihilism') which denies that statements about metaphysical possibility are meaningful (not even logical positists went this far...). For instance, the statements, \"It might rain tomorrow\", or \"If I had only sold it when I had the chance, I'd be rich!\", would be meaningless, or at least mean something quite different from what we'd normally say they do. I may put forth some of my arguments later in this thread if there is interest. I challenge anybody to explain the concept of 'possibility' without resorting to some weird multiple universe theory.\r\n\r\nAny thoughts?", - "Solution_3": "[quote=\"rookworm\"][quote=\"jimhu\"]What if you (mainly your brain) was taken apart (basically you die) and reassembled in the same order (then come back to life) it was originally, when you(reassembled) wake up would you still be yourself(the person youre destined to be)?\n\nIts hard to explain because we don't know if you wake up as the same conciousness before your brain was reassembled. (For all we know that your old self might have died). \n\nAnother question: since everything happens once, do you think we are predestined? (IF you are questioning predestination right now, that event might be predestined).\n\nHeh just a bunch of pretty dumb questions... I hope for interesting discussion...[/quote]\n\n\nI think these are great questions! I've been thinking about the former one since I first saw an episode of \"Star Trek\". I have thought about something similar to the latter one a great deal recently. Here's an interesting question along those lines: \"If we live in a deterministic, non-branching universe\", what does it mean to say something is 'possible' or 'necessary' (in the metaphysical, as opposed to epistemological sense). Moreover, what kind of difference is there (if any) between counterfactuals about the past, and those about the future?\n\nI'm tending towards a radical position (which I would like to call 'modal nihilism') which denies that statements about metaphysical possibility are meaningful (not even logical positists went this far...). For instance, the statements, \"It might rain tomorrow\", or \"If I had only sold it when I had the chance, I'd be rich!\", would be meaningless, or at least mean something quite different from what we'd normally say they do. I may put forth some of my arguments later in this thread if there is interest. I challenge anybody to explain the concept of 'possibility' without resorting to some weird multiple universe theory.\n\nAny thoughts?[/quote]\r\n\r\nThe question stated above is actually linked with predestination... So basically the way it has to do with your question is that if predestination was true, it would make statements like \"I could have been rich\" not meaningful at all...\r\n\r\nAnd plus I don't really understand your vocabulary such as metahpysical lol... I am only a 15 year old with a shallow word bank...", - "Solution_4": "If Chuck Norris is late, time better slow the f*** down.", - "Solution_5": "[quote=\"PenguinIntegral\"]If Chuck Norris is late, time better slow the f*** down.[/quote]\r\n\r\nroundhouse kick'd for blasphemy", - "Solution_6": "Why can't we just treat \"possibility\" (i.e. randomness) as a law of nature? I mean, we make all kinds of other assumptions like gravity and Coulomb's law. With regard to human decision making, that depends on things we can't know by the FWT!", - "Solution_7": "[quote=\"rookworm\"]I challenge anybody to explain the concept of 'possibility' without resorting to some weird multiple universe theory.[/quote]\r\nConsider the set of worlds that would appear identical to the man who makes the statement about a past or future possibility. By this I mean that if he looked around one world for ten minutes, then was picked up and placed in the other world to look around for ten minutes, he wouldn't find any difference. (However, the atoms, bacteria, and dust particles that he can't see may be different.) Then the probability that an event will occur/could have occured is the fraction of these worlds in which the event occurs.\r\n\r\nThis idea is central to the successful theory of [url=http://en.wikipedia.org/wiki/Statistical_mechanics]statistical mechanics[/url]. I would say that any philosophical idea that produces a predictive physical theory is meaningful.", - "Solution_8": "These are hard questions!\r\nI was wondering this when watching the Fly\r\nin this film a lonely scientist builds two machines, one scans the object, the second one completely reconstructs the object\r\nthis isn't actual teleportation but cut and paste\r\nfor some reason the scientist let the first machine destroy the object, which is completely useless to me it seems\r\nwell, i was wondering, when he tried that out on himself, why did you do that? you killed yourself!!\r\nwell he got in trouble when a fly! went along in the machine\r\n\r\nit is also explored in an episode of earth final conflict\r\nan alien robotic explorer scans people and then destroys them\r\na scientist discovers that the machine can also reconstruct and begs his coworkers to force it to rebuild his wife, although they keep telling him it won't be her\r\nin the end, surprisingly, he goes through with it, and lives happily again with the clone\r\n\r\n\r\nwhat is consciousness? i believe in science, but will science ever explain this?", - "Solution_9": "[quote=\"ccy\"]Another question: If you are \"teleported\" to some other place, is the teleported person still you? :D[/quote]Teleportation as you imagine it (as a process of being taken apart, transformed into mere fotons, transported and then reassambled) probably from Star Trek series will not be possible. \r\n\r\nHowever, using a mainfold of the space-time-continuum, bending the subspace field between you and your desired location would allow you to instantly travel to a different place (warp jump) appearing for all others as teleporting. However the amount of energy required to bend space between two locations ... is huuuuge :)", - "Solution_10": "I remember my friend had this obsession with brain transplants and whether that count keep 'you' and your memories and such, when we were in Year 8.\r\n\r\nI think you're eventually touching on the existence of a 'soul'. As this is basically a forum of maths people the very idea would probably earn you ridicule. ;) \r\n\r\nBut such a feat would take very precise nanotechnology and such, and you need to kickstart the brain again, but I guess technically it'd work, hypothetically. It's all in there. :D", - "Solution_11": "for me it is not about a soul\r\ni am a die hard atheist yet i must ask wel : being aware, what is that?", - "Solution_12": "[quote=\"fredbel6\"]for me it is not about a soul\ni am a die hard atheist yet i must ask wel : being aware, what is that?[/quote]\r\n\r\nI'm maintaining that it's all chemical and electrical messages. Nothing happens when those die.", - "Solution_13": "[quote=\"vulgarfraction\"][quote=\"fredbel6\"]for me it is not about a soul\ni am a die hard atheist yet i must ask wel : being aware, what is that?[/quote]\n\nI'm maintaining that it's all chemical and electrical messages. Nothing happens when those die.[/quote]\r\n\r\nMy definition: Being aware is being conscious of your existence.\r\n\r\nPlus, existence of souls cannot be proven or disproven. You cannot just assume that it's not \"scientifically\" present, it is not really present. What's wrong about talking about the concept of souls in a math forum? Mathematicians are supposed to be open to new ideas that are don't necessarily have to make sense.", - "Solution_14": "Your question is a very interesting Jimhu.\r\nI have thinkiking about these kind of questions for a long time. :) \r\n\r\nI just have no idea what the answer is. :oops: \r\n\r\nBut I will ask some few more questions.\r\n\r\nSuppose you take matter(say water) of exactly your own mass and change its subatomic configuration so that the water ends being an exavt copy of you.And now suppose I kill you seeing that there is another 'you'.Does your life still exist?\r\n\r\nYou disintegrate and reintegrate somewhere else.Are you still yourself?\r\n\r\nIf I take the atoms from your body and turn it into a some other creature.then is it still you or some other creature?\r\n\r\nIt all comes down to what exactly is consciousness?\r\nAt present we have very little idea about what conscioussness is and thats the problem.", - "Solution_15": "vulgarfraction, yes i agree I too am an atheist and i believe that it's all chemistry and electric signals , but still how there is a me inside me? I just don't understand, it's so paradoxal, it is for me the single most difficult philosophical question\r\n\r\n\r\nI don't agree with being aware being the same as knowing you exist \r\nbecause what is knowing that\r\na dumb animal like a hamster will not really think about that", - "Solution_16": "[quote=\"fredbel6\"]\na dumb animal like a hamster will not really think about that[/quote]\r\nAre you sure? How do you know this?", - "Solution_17": "well what is knowing that you exist? explicitly sitting down and pondering about it\r\nor just thinking of your next lunch\r\n\r\ndon't get me wrong, i love all rodents! perhaps i should have said guinea pigs, hamsters are a bit more ingenious than guinea pigs in my experience :lol:", - "Solution_18": "What is consciousness? That in itself is probably one of Godel's metastatements.\r\nWell, as study of the brain progresses, we may gain some insight into what consciousness is...but we're nowhere close...", - "Solution_19": "This is just like that book I just read, Teleportation." -} -{ - "Problem": "\"The Surface area, including top and bottom, or a rigth circular cylinder is 950 sq ft. Express the volume of the cylinder as a function of r.\"\r\n\r\nIn this problem you're only suppose to have R's and numbers in the volume of a cylinder right? like the height and pi must be in form of R's?", - "Solution_1": "$\\pi$ can still be $\\pi$, but it just means you can't have any other unknowns in it other than $r$. No heights.", - "Solution_2": "I come up with \r\n\r\npi * r sqaured *((950-2pi rsquared)/2pi r) but it says I am wrong? how could this be I figured out H .", - "Solution_3": "It looks right to me. Just simplify more.", - "Solution_4": "[quote=\"Farcus\"]I come up with \n\npi * r sqaured *((950-2pi rsquared)/2pi r) but it says I am wrong? how could this be I figured out H .[/quote]\r\n\r\nso this would be:\r\n\r\n$\\pi r^{2}(\\frac{950-2\\pi r^{2}}{2\\pi r})$\r\n$r (\\frac{950-2\\pi r^{2}}{2})$\r\n$r (425-\\pi r^{2})$", - "Solution_5": "[quote=\"Farcus\"]\nIn this problem you're only suppose to have R's and numbers in the volume of a cylinder right? like the height and pi must be in form of R's?[/quote]\r\nPi is not a variable, and if you can express it in terms of r, wow, good for you, but you don't have to. pi is like 12, or 23... it's a constant." -} -{ - "Problem": "I haven't got a lot of problems on the math chat today (sadly) and I'm posting some extra problems instead.\r\n\r\nFive animals ---> A,B,C,D, and E --- are either wolves or dogs. Dogs always tell the truth while wolves always tell lies. A claims that B is a dog. B claims that C is a wolf. C claims that D is a wolf. E claims that A is a dog. D claims that B and E are different kinds. The number of wolves is:\r\n\r\nA) 1\r\nB) 2\r\nC) 3\r\nD) 4\r\nE) 5\r\n\r\nThis question was adopted from one of the European Math Contest. Forgot what it was called.", - "Solution_1": "[hide]Either A is a wolf, or A is a dog. From there, we can figure out what the others are. If A is a dog, then D, B, and E are dogs. Since D will tell the truth that B and E are different, this way won't work. That means that A, B, D, and E are all wolves, or (d).[/hide]", - "Solution_2": "Excellent.\r\n\r\n :)" -} -{ - "Problem": "Let [tex]n\\ge2[/tex] be an integer such that [tex]2^n+n^2[/tex] is a prime. Prove that [tex]n\\equiv3 \\mod 6[/tex].", - "Solution_1": "[hide]\n\nfirst of all, n cant be even or the number is even (and greater than 2)\n\nworking in mod 6, 2^n is cyclic...\n\n2^n=2 for n=1,3,5 (mod 6)\n\nso...\n\nif n=1 (mod 6) then n^2 =1 and 2^n+n^2=3 (mod 6) ... divisible by 3\n\nif n=5 (mod 6) ... n=-1 (mod 6) ... then n^2 =1 and 2^n+n^2=3 (mod 6) ... divisible by 3\n\nso if 2^n+n^2 is prime, then n=3 (mod 6)\n\n[/hide]", - "Solution_2": "to be a bit more rigorous you should say that [tex]2^n+n^2 > 3 \\ \\mbox{for} \\ n \\ge 2[/tex] because 3 is prime and [tex]3 \\equiv 3 \\mod 6[/tex]" -} -{ - "Problem": "Caculate lim n->oo from \\sum k=1 to n from 1/C(2k,k).", - "Solution_1": "Any idea how to get started with this one?\r\nI tried guessing the limit via numerical evaluation, but that's not really helpful :huh:", - "Solution_2": "Another example : $a_{n}\\equiv\\sum_{k=1}^{n}\\frac{1}{C_{n}^{k}}\\rightarrow 1\\ .$", - "Solution_3": "[quote=\"xxxxtt\"]Calculate\n$\\lim_{n\\to \\infty}\\sum_{k=0}^{n}\\frac{k!}{(2k)!}$[/quote]\nThe limit is equal to \n$1+\\frac{e^{1/4}\\sqrt{\\pi}}{2}\\text{erf}\\left(\\frac{1}{2}\\right)$\nbut I neither have any idea nor any motivation to try to prove it.\n\nIf you want to approximate it rigorously by hand, Stirling's approximation will work. The sum converges extremely quickly.\n\nedit: \n[quote=\"perfect_radio\"]I tried guessing the limit via numerical evaluation, but that's not really helpful :huh: [/quote]\r\nIf you guessed the answer from the decimal approximation I would be extremely impressed. :P", - "Solution_4": "[quote=\"xxxxtt\"]Calculate ${{\\lim_{n\\to\\infty}\\sum_{k=1}^{n}}\\frac{1}{C_{2k}^{k}}}\\ .$[/quote][size=134][color=darkblue][b]Xevarion[/b], above is another limit, $\\lim_{n\\to\\infty}\\sum_{k=1}^{n}\\frac{k!\\cdot k!}{(2k)!}\\ .$It isn't $\\lim_{n\\to\\infty}\\sum_{k=1}^{n}\\frac{k!}{(2k)!}$ because $C_{2k}^{k}=\\frac{(2k)!}{k!\\cdot k!}\\ .$[/color][/size]", - "Solution_5": "The limits is $\\frac{1}{3}+\\frac{2\\pi}{9\\sqrt{3}}$.\r\nUse the following expansion:\r\n\r\n$\\sum_{k=1}^{\\infty}\\frac{(2x)^{2n}}{nC_{2n}^{n}}=\\frac{2x\\arcsin x}{\\sqrt{1-x^{2}}}$, for $|x|<1$.\r\n\r\nNow take the derivative of the above function and let $x=\\frac{1}{2}$, and we get that $\\sum_{k=1}^{\\infty}\\frac{1}{C_{2n}^{n}}=\\frac{1}{3}+\\frac{2\\pi}{9\\sqrt{3}}$.\r\nDidi", - "Solution_6": "[quote=\"Virgil Nicula\"][quote=\"xxxxtt\"]Calculate ${{\\lim_{n\\to\\infty}\\sum_{k=1}^{n}}\\frac{1}{C_{2k}^{k}}}\\ .$[/quote][size=134][color=darkblue][b]Xevarion[/b], above is another limit, $\\lim_{n\\to\\infty}\\sum_{k=1}^{n}\\frac{k!\\cdot k!}{(2k)!}\\ .$It isn't $\\lim_{n\\to\\infty}\\sum_{k=1}^{n}\\frac{k!}{(2k)!}$ because $C_{2k}^{k}=\\frac{(2k)!}{k!\\cdot k!}\\ .$[/color][/size][/quote]\r\nOops. I always just write the factorials so I forgot which was which $P$ and $C$. :oops:", - "Solution_7": "[color=darkred]I tried to increase this sequence and obtained the identities ${\\frac{4}{C_{2k+2}^{k+1}}-\\frac{1}{C_{2k}^{k}}=\\frac{1}{(2k+1)C_{2k}^{k}}}$ and $\\sum_{k=1}^{n}\\frac{1}{C_{2k}^{k}}=\\frac{1}{3}\\cdot\\left[2-\\frac{4}{C_{2n+2}^{n+1}}+2\\cdot\\sum_{k=1}^{n}\\frac{1}{(k+1)C_{2k+1}^{k}}\\right]\\ .$ Using the [b][u]Didilica's result[/u][/b], we can obtain $\\lim_{n\\to\\infty}\\sum_{k=1}^{n}\\frac{1}{(k+1)C_{2k+1}^{k}}\\ .$ I don't know if it is interesting. [u]The Didilica's proof is proffesionally. I envy him a bit for the his ability ![/u][/color]", - "Solution_8": "[quote=\"didilica\"]Use the following expansion:\n$\\sum_{k=1}^\\infty \\frac{(2x)^{2n}}{n \\binom{2n}{n}}= \\frac{2x\\arcsin x}{\\sqrt{1-x^{2}}}$, for $|x|<1$.\n\nNow take the derivative of the above function and let $x=\\frac{1}{2}$, and we get that $\\sum_{k=1}^\\infty \\frac1{\\binom{2n}{n}}=\\frac{1}{3}+\\frac{2\\pi}{9\\sqrt{3}}$.\nDidi[/quote]\nThank you very much. I tried to do it with \"generating functions\", but I don't believe I would have ever thought to do your nice trick.\n\n[quote=\"Virgil Nicula\"]Another example : $a_{n}\\equiv\\sum_{k=1}^{n}\\frac1{\\binom{n}{k}}\\rightarrow 1$.[/quote]\r\nThis is a piece of cake compared to the limit proposed in this thread.\r\n\r\nPS: The numerical value I obtained with Mathematica was $0.7364$, which indeed approximates didilica's answer (note that asking Mathematica to compute the limit returns hypergeometric stuff). And of course I didn't guess the answer. It's too complicated...", - "Solution_9": "Hi,\r\nAn interesting formula related to Virgil`s sequence, I think it is worth pointing out, is this:\r\n\r\nLet $a_{n}=\\sum_{k=1}^{n}\\frac{1}{C_{n}^{k}}$. Then the following equality holds:\r\n\r\n$\\sum_{k=1}^{n}\\frac{1}{C_{n}^{k}}=\\frac{n+1}{2^{n}}\\sum_{k=0,n-k=odd}^{k=n}\\frac{C_{n}^{k}}{n-k}$.\r\n\r\nIt is interesting to calculate the following limit:\r\n\r\n$\\lim_{n\\rightarrow \\infty}n(1-a_{n})$.\r\nDidi" -} -{ - "Problem": "For an EE class I'm in, we have to find the absolute minimum of a 2D surface given arbitrary parameters. \r\nThe function given is:\r\n\r\n$ Z\\equal{}\\sum_{n\\equal{}1}^{N}{(y_n\\minus{}Bx_n\\minus{}C)}^2$\r\n\r\nWhat values of $ B$ and $ C$ minimize $ Z$ for arbitrary $ x_n$ and $ y_n$?", - "Solution_1": "In other words, you want to find the [url=http://en.wikipedia.org/wiki/Linear_least_squares]linear least-squares[/url] approximation of $ y$ in terms of $ x$ -- that is, you want to find a linear function $ y \\equal{} Bx \\plus{} C$ that is as close to fitting your given data as possible in the least-squares sense.\r\n\r\nOne way to do this is to find the stationary points $ Z(B,C)$, which is a measure of error, by setting the gradient to $ 0$. As the number of parameters increases, this can get annoying. Fortunately, linear algebra gives some insight into how to solve the general problem.", - "Solution_2": "So, to be a bit more mathematical about it, define a function:\r\n\r\n$ f(a,b)\\equal{}\\sum_n \\left(a x_n \\plus{} b \\minus{} y_n\\right)^2$\r\n\r\nThis is the same function you have, just using a nicer form. Now, we find the respective partial derivations of the function:\r\n\r\n$ \\frac{\\partial f}{\\partial a} \\equal{} 2 \\sum_n \\left(a x_n \\plus{} b \\minus{} y_n\\right) x_n$\r\n\r\n$ \\frac{\\partial f}{\\partial b} \\equal{} 2 \\sum_n \\left(a x_n \\plus{} b \\minus{} y_n\\right)$\r\n\r\nSince we wish to find the minimal value of the function, we set the partial derivations to $ 0$ and try to solve for the parameters - in our case $ a$ and $ b$:\r\n\r\n$ \\frac{\\partial f}{\\partial a} \\equal{} 0 \\Rightarrow a \\sum_n x^2_n \\plus{} b \\sum_n x_n \\equal{} \\sum_n y_n x_n$\r\n\r\n$ \\frac{\\partial f}{\\partial b} \\equal{} 0 \\Rightarrow a \\sum_n x_n \\plus{} b \\sum_n 1 \\equal{} \\sum_n y_n$\r\n\r\nWe can use any method to solve this set of equations. I'll multiply each equation so I can cancel out the parameter $ a$ by adding the two equations:\r\n\r\n$ a \\frac{\\sum_n x^2_n}{\\sum_n x_n} \\plus{} b \\equal{} \\frac{\\sum_n y_n x_n}{\\sum_n x_n}$\r\n\r\n$ \\minus{}a \\frac{\\sum_n x_n}{\\sum_n 1} \\minus{} b \\equal{} \\minus{}\\frac{\\sum_n y_n}{\\sum_n 1}$\r\n\r\nNow add them together to get:\r\n\r\n$ a \\left(\\frac{\\sum_n x^2_n}{\\sum_n x_n} \\minus{} \\frac{\\sum_n x_n}{\\sum_n 1}\\right) \\equal{} \\frac{\\sum_n y_n x_n}{\\sum_n x_n}\\minus{}\\frac{\\sum_n y_n}{\\sum_n 1}$\r\n\r\nThis gives us:\r\n\r\n$ a \\equal{} \\frac{\\frac{\\sum_n y_n x_n}{\\sum_n x_n}\\minus{}\\frac{\\sum_n y_n}{\\sum_n 1}}{\\frac{\\sum_n x^2_n}{\\sum_n x_n} \\minus{} \\frac{\\sum_n x_n}{\\sum_n 1}}$\r\n\r\nUsing the same approach again but canceling out $ a$ we get:\r\n\r\n$ b\\equal{} \\frac{\\frac{\\sum_n y_n x_n}{\\sum_n x^2_n}\\minus{}\\frac{\\sum_n y_n}{\\sum_n x_n}}{\\frac{\\sum_n x_n}{\\sum_n x^2_n} \\minus{} \\frac{\\sum_n 1}{\\sum_n x_n}}$\r\n\r\nWe can somewhat simplify this to get:\r\n\r\n$ a\\equal{} \\frac{N \\sum_n y_n x_n\\minus{}\\sum_n x_n\\sum_n y_n}{N \\sum_n x^2_n \\minus{} \\left(\\sum_n x_n\\right)^2}$\r\n\r\n$ b\\equal{} \\frac{\\sum_n x_n \\sum_n y_n x_n\\minus{}\\sum_n y_n \\sum_n x^2_n}{\\left(\\sum_n x_n\\right)^2 \\minus{} N \\sum_n x^2_n}$\r\n\r\nIn general, if you with to find a polynomial of the form $ P(x) \\equal{} \\sum_{i\\equal{}0}^p \\lambda_i\\,x^i$ to approximate a set of points of the form $ (x_n, y_n)$ you need to solve:\r\n\r\n$ \\left[\\begin{array}{cccc}\r\nX_0 & X_1 & \\cdots & X_p \\\\\r\nX_1 & X_2 & \\cdots & X_{p\\plus{}1} \\\\\r\n\\vdots & \\vdots & \\ddots & \\vdots \\\\\r\nX_p & X_{p\\plus{}1} & \\cdots & X_{2p}\r\n\\end{array}\\right] \\left[\\begin{array}{c}\r\n\\lambda_0 \\\\\r\n\\lambda_1 \\\\\r\n\\vdots \\\\\r\n\\lambda_p\r\n\\end{array}\\right]\\equal{}\\left[\\begin{array}{c}\r\nY_0 \\\\\r\nY_1 \\\\\r\n\\vdots \\\\\r\nY_p\r\n\\end{array}\\right]$\r\n\r\nwhere $ X_i\\equal{}\\sum_n x^i_n$ and $ Y_i\\equal{}\\sum_n y_n\\,x^i_n$.", - "Solution_3": "I tend to express the $ a$ and $ b$ in milin's post in a slightly different way, which I find easier to remember. Use a bar over the top of a number to mean the average of that number; for instance, $ \\overline{x}\\equal{}\\frac{1}{N}\\sum_{n\\equal{}1}^Nx_n.$\r\n\r\nThe slope is $ a,$ and $ a\\equal{}\\frac{\\overline{xy}\\minus{}\\overline{x}\\,\\overline{y}}{\\overline{x^2}\\minus{}\\overline{x}^2}.$ In more probabilistic terms, $ a\\equal{}\\frac{\\text{Cov}(X,Y)}{\\text{Var}(X)}.$\r\n\r\nAs for the intercept $ b,$ I could remember that as $ b\\equal{}\\frac{\\overline{x}\\,\\overline{xy}\\minus{}\\overline{x^2}\\overline{y}}{\\overline{x^2}\\minus{}\\overline{x}^2}.$ But there's no good \"hook\" for remembering that formula. The better way of seeing it: the $ y$-intercept is really a nothing-special point, and it has a nothing-special formula. Much easier to remember is the fact that the regression line goes through the center of mass $ (\\overline{x},\\overline{y}).$ So either we write $ y\\minus{}\\overline{y}\\equal{}a(x\\minus{}\\overline{x}),$ or we go ahead and expand that out to $ y\\equal{}ax\\plus{}b,$ this time noting that $ b\\equal{}\\overline{y}\\minus{}a\\overline{x}.$\r\n\r\nI'll add that my answers here are the same as milin's answers; I'm just talking about the form of the answers, and ways to remember them.", - "Solution_4": "Thanks TZF for setting up the general idea for me. Thanks milin for the great post, I was able to follow all your work quite easily and clearly, much appreciated. \r\n\r\nThanks too KM for the tips on putting the solutions in more simple notation. This board is too good." -} -{ - "Problem": "in triangle $ ABC, CC'$ and $ BB'$ are altitudes.$ B'C' and BC$intersect at D.\r\nprove that $ DM(AB^2\\minus{}AC^2)\\equal{}BC^3/2$ where M is the midpoint of BC :ninja:", - "Solution_1": "In this problem, I think shoud be an algebraical prove. And this is my solution\r\nGive DB= x. We'll calculate x follow a,b,c.\r\nFirst, B'C'/BC= HC'/HB= cos A so B'C'= a.cosA.\r\nDC'/DB= DC/DB'= CC'/BB'= b/c. Then DC'= DB.b/c, so DB'= x.b/c +a.cosA, \r\nso DC=(x.b/c+a.cosA).b/c(1) \r\nBut DC= x+a(2)\r\nFrom (1) and (2) we have x=a.(c^2-b^2+a^2)/2(b^2-c^2).\r\nthen DM= x+a/2= a^3/2(b^2-c^2). (which we need prove) :)", - "Solution_2": "[quote=\"ishfaq420haque\"][color=darkred]In triangle $ ABC, CC'$ and $ BB'$ are altitudes.$ B'C' \\cap BC\\equal{}D$.$ M$ is the midpoint of $ BC$.Let, $ DM\\equal{}k$\nProve that, $ k(c^2 \\minus{} b^2) \\equal{}\\frac {a^3}{2}$\n[/color][/quote]\r\n[u][b]\nSolution:[/b][/u]\r\nUsing Menelaus's theorem on $ \\triangle ABC$ and transversal $ B'C'D$ we get,\r\n$ \\frac{k\\minus{}a/2}{k\\plus{}a/2}\\equal{}\\minus{}\\frac{c \\cdot \\cos B}{b \\cdot \\cos C}$\r\nNow using cos law we get,\r\n$ \\frac{k\\minus{}a/2}{k\\plus{}a/2}\\equal{}\\minus{}\\frac{a^2\\plus{}c^2\\minus{}b^2}{a^2\\plus{}b^2\\minus{}c^2}$\r\nwhich implies that,\r\n$ \\frac{k}{\\minus{}a/2}\\equal{}\\minus{}\\frac{a^2}{c^2\\minus{}b^2}$" -} -{ - "Problem": "Prove that there exists a natural number n such that\r\nsin n>2003/2004", - "Solution_1": "More generally, you may prove that the sequence {sin(n)/ n is a positive integer} is dense in [0,1].\r\n\r\nPierre.", - "Solution_2": "Yes, Pierre, how can you prove that the sequence sin n is dense in \r\n0 \\leq x \\leq 1? (Ive asked a teacher who said that it might not be true)", - "Solution_3": "Ok, I give you the steps, I left the details :\r\n\r\nLet a>0 be an irrational number.\r\nLet U(n) = {an} = an - [an], where {.} denotes the decimal part, and [.] denotes the integer part.\r\nLet E = { U(n) / n > 0}.\r\n\r\n1) Use that a is irrational to deduce that if n,m are two distinct integers then U(n) =/= U(m).\r\n\r\n2) Use the pigeon-hole principle to deduce that, for each integer N > 0 there exists p,q in {1,...,N+1} such that p =/= q and 0 < U(p) - U(q) < 1/N.\r\n\r\n3) Deduce from above that there exists a non-zero integer k = p-q (not necessarly positive) such that 0 < U(k) < 1/N.\r\n\r\nNow let x be in [0,1[ and e > 0 be given. Let I = ]x-e,x+e[.\r\nLet N > 0 be an integer such that 1/N < e. Use 3) to get k.\r\n\r\n4) If k > 0.\r\nLet F = { {ka}, 2*{ka}, ...., [1/{ka}]*{ka} }.\r\nProve that I \\cap F contains at least one element, and deduce that I \\cap E contains at least one element.\r\n\r\n5) If k < 0.\r\nLet r = -k. Then 1 - 1/n < U(r) < 1. Let U(r) = 1-d.\r\nLet G = {1,2,..., [1/d]}. For i in G, we have {i*r*a} = 1 - i*d.\r\nAs in 4), deduce that I \\cap E contains at least one element.\r\n\r\n6) Deduce that E is dense in [0,1].\r\n\r\nLet x be in [-1,1], and w in [0,2 \\pi [ such that sin(w) = x. Let t = w/(2 \\pi ), thus t is in [0,1]. Let e > 0 and I = ]x-e,x+e[.\r\n\r\n7) Use 6) for a = 1/(2 \\pi ). Deduce that there exists a positive integer n such that |2 \\pi *{n/(2 \\pi } - w| < e.\r\n\r\n8) Note that, for every real u,v we have |sin(u) - sin(v)| \\leq |u-v|.\r\nUse it to deduce that | sin(2 \\pi *{n/(2 \\pi )}) - x| < e.\r\n\r\n9) Note that 2 \\pi *{n/(2 \\pi )} = n - 2 \\pi *[n/(2 \\pi )], and deduce that |sin(n) - x| < e.\r\nThus { sin(n) / n > 0 is an integer} is dense in [-1;1] (and therefore in [0,1] ).\r\n\r\nPierre.", - "Solution_4": "Let k \\in R\\Q, the sequence (e^(i \\pi nk))_{n \\in Z} is dense in \r\nS^1={ z \\in C; |z|=1}", - "Solution_5": "You are right Moubi, and in fact this latter result is easier to prove.\r\nBut, the original problem ask for a natural number, thus the work is a little bit difficult...\r\n\r\nPierre." -} -{ - "Problem": "Which one of the following statements is false? All equilateral triangles are\n\n$\\textbf{(A)} \\ \\text{ equiangular} \\qquad\n\\textbf{(B)} \\ \\text{isosceles} \\qquad\n\\textbf{(C)} \\ \\text{regular polygons } \\qquad\n\\textbf{(D)} \\ \\text{congruent to each other} \\qquad\n\\textbf{(E)} \\ \\text{similar to each other} $", - "Solution_1": "[hide]\nD becuase the lengths of the sides can be different.[/hide]", - "Solution_2": "[hide]\nD, they can be different side lenghts.[/hide]" -} -{ - "Problem": "There are given $111$ coins and a $n\\times n$ table divided into unit cells. This coins are placed inside the unit cells (one unit cell may contain one coin, many coins, or may be empty), such that the difference between the number of coins from two neighbouring cells (that have a common edge) is $1$. Find the maximal $n$ for this to be possible.", - "Solution_1": "is easy to find one exemple for n = 13.\r\n\r\nto prove that for $n\\geq 15$ is impossible, divide the table in $E(\\frac{n}{2})$ dominoes. every dominoe will have at least one coin! so, for $n\\geq 15$ is impossible, because we have at least 112 coins!\r\n\r\nfor n = 14, we will use other fact.\r\nin any 2x2 table, the sum of the number of coins in your entries is even (is easy to see). so, we can divide one 14x14 table in 49 2x2 tables, and the sum of coins in our table will be even! (impossible, because 111 is odd)", - "Solution_2": "For $n=13$, an example is as follows. Put $1,0,1,\\dots,1$ (total$=7$) coins to odd-indexed rows, and $0,1,0,\\dots,0$ (total$=6$) to even-indexed rows. The total number of coins is $85$. Now, note that, the zeroes are all surrounded by $1$'s, hence, making them $2$ do not change anything. Simply select $13$ cells inside with $0$'s in it, and update their status to $2$ coins.", - "Solution_3": "n=14.parity\nanswer n=13" -} -{ - "Problem": "In a triangle $ \\triangle ABC, BC>AB$, and the angle bisector of $ \\angle ABC$ meets AC at D. The perpendicular from C to BD meets BD at P, and the perpendicular from A to BP is Q. Let M be the midpoint of AC, E be the midpoint of BC. Let the circumcircle of PQM intersect AC at point H (not equal to M). Prove that O, H, E, and M are cyclic.", - "Solution_1": "Dear Mathlinkers,\r\nI suppose that O is the center of PQM.\r\nThis circle is attribuate to a french geometer Calabre.\r\nI need this situation in order to prove synthetically the Feuerbach's theorem.\r\nSee: http://perso.orange.fr/jl.ayme vol. 1 Le th\u00e9or\u00e8me de Feuerbach page 2...(Cons\u00e9quence du lemme 1)\r\nSincerely.\r\nJean-Louis", - "Solution_2": "1.PE\u2225AB,ME\u2225AB P,M,E are collinear\r\n2.let AB/BC=k, we have that QD/DP=AD/DC=AB/BC=k \r\nQD=k*DP BQ/BP=AB/AC=k BQ =k* BP\r\n=> QD/ BQ= DP/ BP\r\nMC/MD=MA/MD=1+AD/DM=1+BD/DP=BP/DP\r\n=> MD/MC=DP/BP\r\n=> QD/ BQ= MD/MC => MQ\u2225BC\r\n3.MQ=MP \u2220QHM=\u2220QPM=\u2220PQM=\u2220PBC=\u2220PBA => B,A,H,Q are concyclic\r\n => BH\u22a5AC\r\n4. \u2220EHO+\u2220OME=\u2220EPO+\u2220OME=\u2220OMP+\u2220OME=180\r\nDone!", - "Solution_3": "Another solution~\r\n\r\nWe shall first prove that $ H$ is the foot of perpendicular from $ B$ to $ AC$. (which is somehow inspired by jayme's link~)\r\nFirst extend $ AQ$ and intersect $ BC$ at $ X$.\r\nSince $ BQ$ bisects $ \\angle ABX$ and $ BQ \\bot AX$, so $ Q$ is the mid-pt of $ AX$.\r\nSince $ M$ is also the mid-pt of $ AC$, so $ QM//BC$. \r\nThus it is not hard to prove that $ H, B, C, P$ concyclic.\r\nSo $ \\angle BHC \\equal{} \\angle BPC \\equal{} 90$ degrees.\r\n\r\nAs $ E$ is the mid-pt of $ BC$, So $ \\angle BEH \\equal{} 2\\angle BCH$\r\n$ \\angle CEM \\equal{} \\angle ABC$ since $ EH//AB$.\r\nSo $ \\angle HEM \\equal{} 180 \\minus{} \\angle HEB \\minus{} \\angle MEC \\equal{} 180 \\minus{} 2\\angle BCA \\minus{} \\angle ABC$\r\n\r\nOn the other hand, since $ M$ is the mid-pt of AC, So the foot of perpendicular of M to PQ is the mid-pt of $ PQ$, denote as $ Y$, i.e. $ MY$ is the perpendicular bisector of PQ.\r\nAs $ OY$ is also perpendicular bisector of PQ, so $ O,Y, M$ collinear.\r\n\r\nAs $ B, H, Y, M$ concyclic ($ \\angle BHM \\equal{} \\angle BYM \\equal{} 90$, from some angle tracing we can get:\r\n$ \\angle HOM \\equal{} 180 \\minus{} 2\\angle OMH \\equal{} 180 \\minus{} 2\\angle YMH \\equal{} 180 \\minus{} 2\\angle HBY \\equal{} 180 \\minus{} 2(0.5\\angle ABC \\minus{} (90 \\minus{} \\angle BAC)) \\equal{} 360 \\minus{} \\angle ABC \\minus{} 2\\angle BAC \\equal{} \\minus{} \\angle ABC \\plus{} 2 (\\angle ABC \\plus{} \\angle ACB) \\equal{} 2\\angle ACB \\plus{} \\angle ABC$.\r\nSo $ \\angle HOM \\plus{} \\angle HEM \\equal{} 180$\r\nSo $ O, H, E, M$ concyclic\r\n\r\nI should've been in this competition... but then i prefer preparing for my public exam to sitting for this. But anyway it's the same coz i can solve the problems at home~\r\n\r\nHope anybody can post the rest of the questions.", - "Solution_4": "[size=117][color=darkblue] [b]I know an old problem which is similarly with this proposed problem.[/b] [/color][/size] \r\n\r\n[quote=\"Virgil Nicula\"] [color=darkred]Let $ ABC$ be an acute and nonisosceles triangle . The bisector of $ \\widehat {BAC}$ meets $ BC$ at $ D$ . Denote the \n\nmidpoints $ M$ , $ N$ , $ P$ of $ [BC]$ , $ [CA]$ , $ [AB]$ respectively and the projections $ \\left\\|\\begin{array}{c}\nH\\in BC\\ ,\\ AH\\perp BC\\\\\\\nR\\in AD\\ ,\\ BR\\perp AD\\\\\\\nS\\in AD\\ ,\\ CS\\perp AD\\end{array}\\right\\|$ . \n\nProve that $ MRHS$ is cyclically and $ S\\in MN$ , $ R\\in MP$ .[/color] [/quote]\r\n\r\n[color=darkblue][b][u]Proof.[/u][/b] The quadrilateral $ ABHR$ is cyclically $ \\implies$ $ \\widehat {MHR}\\equiv \\widehat {BAR}$ . It is well-known that $ AD\\equal{}\\frac {2bc\\cos\\frac A2}{b\\plus{}c}$ \n\nand $ BD\\equal{}\\frac {ac}{b\\plus{}c}$ . Observe that $ AS\\equal{}b\\cos \\frac A2$ and $ \\frac {BD}{BM}\\equal{}\\frac {2c}{b\\plus{}c}\\equal{}\\frac {AD}{AS}$ , i.e. $ MS\\parallel AB$ . Thus, $ \\boxed {\\ S\\in MN\\ }$ and \n\n$ \\widehat {MSR}\\equiv \\widehat {BAR}$ . In conclusion, $ \\widehat {MHR}\\equiv\\widehat {MSR}$ , i.e. [b]the quadrilateral $ MRHS$ is cyclically[/b]. The quadrilateral $ ACSH$ \n\nis cyclically $ \\implies$ $ \\widehat {MHS}\\equiv\\widehat {CAS}$ . Therefore, $ \\widehat {MRS}\\equiv \\widehat {MHS}$ $ \\implies$ $ \\widehat {MRS}\\equiv \\widehat {CAS}$ , i.e. $ MR\\parallel AC$ . Thus, $ \\boxed {\\ R\\in MP\\ }$ .[/color]" -} -{ - "Problem": "for this problem I have the folowing answer:\r\nLet G1 and G2 be groups. \r\na) Define pie : G1XG2 -->G1 by pie 1 (a1,a2)=a1, for all (a1,a2) in G1XG2 and define Pie 2: G1XG2--> G1 by pie 2(a1,a2)=a2, for all (a1,a2) in G1XG2 and. Show that both pie 1 and 2 are Homomorphisms. \r\n\r\nb) Let G be any group, and let Ru: G-->G1XG2 be a function. Show that Ru is a group homomorphism if and only if pie1 Ru and pie2 Ru are both group homomorphism.\r\n\r\nA)I would like to know if I first verify that pie(ab)=pie1(a)pie1(b) for any a,b in G1XG2 if a=(a1, a2) and b=(b1, b2) then, as it is defined ab=(a1b1, a2b2). SO is this correct?\r\nB) Would I do it the same way or how would I do it?", - "Solution_1": "[quote=\"mikejones\"]for this problem I have the folowing answer:\nLet G1 and G2 be groups. \na) Define pie : G1XG2 -->G1 by pie 1 (a1,a2)=a1, for all (a1,a2) in G1XG2 and define Pie 2: G1XG2--> G1 by pie 2(a1,a2)=a2, for all (a1,a2) in G1XG2 and. Show that both pie 1 and 2 are Homomorphisms. \n\nb) Let G be any group, and let Ru: G-->G1XG2 be a function. Show that Ru is a group homomorphism if and only if pie1 Ru and pie2 Ru are both group homomorphism.\n\nA)I would like to know if I first verify that pie(ab)=pie1(a)pie1(b) for any a,b in G1XG2 if a=(a1, a2) and b=(b1, b2) then, as it is defined ab=(a1b1, a2b2). SO is this correct?\nB) Would I do it the same way or how would I do it?[/quote]\r\n\r\nA)You are correct. By definition it suffices to prove that $pie_1(ab)=pie_1(a)pie_1(b)$ for any a,b in ${G_1}\\times{G_2}$, which is obvious by the definition of \"product\" in ${G_1}\\times{G_2}$.\r\n\r\nB)The \"IF\" part is obvious, since the composition of two homomorphisms is also a homomorphism.\r\nThe \"ONLY IF\" part: Notice that $Ru(g)$ can be written $(pie_1\\circ Ru(g),pie_2\\circ Ru(g))$ in ${G_1}\\times{G_2}$, then it is easy to verify that $Ru(g_1g_2)=Ru(g_1)Ru(g_2)$ because $pie_1\\circ Ru$ and $pie_2\\circ Ru$ are both homomorphisms." -} -{ - "Problem": "Prove or disprove:\r\n\r\nGiven three non-collinear points, one and only one circle can be drawn through those three points. \r\n[hide=\"hint\"] Don't use coordinates.... [hide=\"hint\"]equidistance[/hide][/hide]", - "Solution_1": "Maybe I'm missing something, this seems too easy:\r\n\r\n[hide] The points form a triangle. Every triangle has one circumcircle.[/hide]", - "Solution_2": "obviously, you can always show that the perpendicular bisectors of the sides are always concurrent for 3 noncollinear points, furthermore, this is the center of the circle, that will pass through all 3 points", - "Solution_3": "ok yea that was a bit easy..." -} -{ - "Problem": "If one quart of paint is exactly enough for two coats of paint \non a 9-foot by 10-foot wall, how many quarts of paint are \nneeded to apply one coat of paint to a 10-foot by 12-foot wall? \nExpress your answer as a common fraction.", - "Solution_1": "The area of the first wall is 90ft squared.\r\nThe area of the second is 120ft squared.\r\n120/90=4/3, the second is 4/3 the area of the second.\r\nBut u only want 1 coat, so divide by two to get 2/3." -} -{ - "Problem": "If your a good problem solver does that mean your likely to get a good score on the SAT? After all the emphasis is on speed. Did anyone here take it, and if so what were your scores? And do you have any suggestions to keep from making dumb mistakes?", - "Solution_1": "I'm guessing that quite a few of the members on here have perfect 800s.\r\n\r\nThe best thing is if you're fast enough to actually go through the test more than once, reworking every problem.", - "Solution_2": "My impression is that if you are a good problem-solver, you will score very well on the SAT. You may have to take a few practice tests to get a feel for it first, though. If it matters, I got a 1590 when I took it (800 Math, 790 Verbal).\r\n\r\nIt is true there is the speed aspect, but they give you a lot of time for the difficulty (over a minute per question). The most important thing, in my opinion, is to work deliberately. Read the questions SLOWLY, catching every single word, and then just do what you need to do, without rushing your computation. As long as you are focused, you will work very quickly and finish with plenty of time to spare, which you can then use to do the whole section again, checking answers.\r\n\r\nVerbal is different material, but also much the same. As long as you read the questions carefully and think logically, you will do very well if you are a good problem-solver. It's just logic, like the math section, just with a little more fuzzy guess-work. But there is always a clearly correct answer if you know how to catch it." -} -{ - "Problem": "What is the greatest proper factor of 6! ?", - "Solution_1": "A proper factor of a number $ n$ means that it is a factor of $ n$, but is not $ 1$ or $ n$ itself. Therefore, to maximize the proper factor of $ 6!$, we divide out the smallest factor of $ 6!$ (besides $ 1$ of course), which happens to be $ 2$. So we have $ \\frac {6!}{2} \\equal{} \\frac {720}{2} \\equal{} \\boxed{360}$", - "Solution_2": "Actually, 1 is a proper factor of any number (except 1 of course).", - "Solution_3": "According to[url=http://mathworld.wolfram.com/ProperFactor.html]Wolfram Mathworld[/url], $ 1$ does not count as a proper factor, but does count as a proper [i]divisor[/i]. However, [url=http://wiki.answers.com/Q/What_is_a_proper_factor_in_math]WikiAnswers[/url] says that there is debate on it... I'm really not that sure." -} -{ - "Problem": "[img]http://www.dlstudios.net/uploaded/1tri.gif[/img]\r\n\r\nis it even possible to find a, b, c, x, y, and z from the given data?\r\n\r\nI've come up with the equations:\r\na+b+c = 360\r\nx+y+z = 110\r\na+c+x+y = 300\r\na+b+y+z = 330\r\nb+c+x+z = 310\r\na+y = 160\r\nb+z = 170\r\nx+c = 40\r\n\r\nbut none of them give me any unknowns. \r\nI'm writing this problem for a PreCal (algebra 2 and trig) project, if its impossible, what is the least amount of data I can give the person to make it possible, but keep it as hard as possible?", - "Solution_1": "You have eight equations for six unknowns, but they aren't all distinct, i.e. some provide no new information. This system can be reduced to four equations for six unknowns, which has an infinite number of solutions, so it's impossible.\r\n\r\nTo make it possible through a system of equations, you have to provide two new distinct equations involving the angles, but given the triangle in question, I don't think it'd be very difficult to solve, and might seem a little artificial or contrived as a problem. So instead, I would personally suggest giving one additional angle and two side lengths, and ask for all the angles and the remaining side length. It's pretty much a simple law of sines exercise, but it might make the problem slightly more interesting (not to say doable).", - "Solution_2": "Wait, to make sure I understand everything, which 4 equations did you reduce the system to?", - "Solution_3": "Well, there's a couple systems that you could reduce it to, but the one I chose was\r\nx+y+z=110\r\na+y=160\r\nb+z=170\r\nx+c=140\r\ne.g.\r\nx=40, y=50, z=20, a=110, b=150, c=100\r\nand\r\nx=10, y=65, z=35, a=95, b=135, c=130\r\nare two solutions.", - "Solution_4": "[quote=\"Darklightmw\"][img]http://www.dlstudios.net/uploaded/1tri.gif[/img]\n\nis it even possible to find a, b, c, x, y, and z from the given data?\n\nI've come up with the equations:\na+b+c = 360\nx+y+z = 110\n[color=red]a+c+x+y = 300[/color]\na+b+y+z = 330\nb+c+x+z = 310\n[color=blue]a+y = 160[/color]\nb+z = 170\n[color=blue]x+c = 40[/color]\n\nbut none of them give me any unknowns. \nI'm writing this problem for a PreCal (algebra 2 and trig) project, if its impossible, what is the least amount of data I can give the person to make it possible, but keep it as hard as possible?[/quote]\r\nWait a second... your equations don't work... add the two blue equations and you get 200, not 300 for the red equation.", - "Solution_5": "I assumed Darklight meant x+c=140, not x+c=40, since x+c+40=180. If you make the switch, the equations work." -} -{ - "Problem": "Let's discuss this olympiad and its problems.What is your first day results?Of course not official,at least rumours,or your own opinion. \r\nIs here someone who participated at this olympiad,actually,at this time at first day(second day will be tomorrow)?\r\nP.S:\r\nI think that this olympiad problems are much harder than it was at Zimo 2007.What is your opinion?\r\nI hope tomorrow we will continue this topic(of course if i'll not be very busy). :) \r\nI'll post all problems tomorrow. :wink:", - "Solution_1": "What about you Erken? I hope you solved all of them\r\n\r\nAnyway,what about problems? Can anybody post them? :)", - "Solution_2": "[quote=\"cefer\"]What about you Erken? I hope you solved all of them\n\nAnyway,what about problems? Can anybody post them? :)[/quote]\r\nI don't know official results yet,but i'll post all of them,as soon as possible.", - "Solution_3": "Do yow know the results? Thank you!", - "Solution_4": "Please, can you post the results? Thank you!", - "Solution_5": "[quote=\"AdrianDD\"]Please, can you post the results? Thank you![/quote]\r\nThere is a list of results at my school.I'll come back from there soon,and post the results.\r\nGood bye.", - "Solution_6": "1st Day:\r\n[url=http://www.mathlinks.ro/viewtopic.php?p=1013352#1013352]Problem 1[/url] \r\n[url=http://www.mathlinks.ro/viewtopic.php?p=1013354#1013354]Problem 2[/url] \r\n[url=http://www.mathlinks.ro/viewtopic.php?p=1013356#1013356]Problem 3[/url] \r\n\r\n2nd Day:\r\n[url=http://www.mathlinks.ro/viewtopic.php?p=1013358#1013358]Problem 4[/url] \r\n[url=http://www.mathlinks.ro/viewtopic.php?p=1013359#1013359]Problem 5[/url] \r\n[url=http://www.mathlinks.ro/viewtopic.php?search_id=2087670879&t=183916]Problem 6[/url]", - "Solution_7": "You are MONSTERS :lol: \r\n2 gold medals of 4 participants is a very good rezult)))\r\n\r\nps OPUBLIKUYTE REZULTATY plz!!!", - "Solution_8": "[quote=\"A_L_E_N\"]You are MONSTERS :lol: \n2 gold medals of 4 participants is a very good rezult)))\n [/quote]\r\nHI!\r\nBulgarians have 3 gold meals!!!", - "Solution_9": "Gold Medals:\r\n1 \u0412\u043e\u0440\u043e\u0431\u044c\u0435\u0432 \u0418\u043b\u044c\u044f \u0420\u043e\u0441\u0441\u0438\u044f, \u041c\u043e\u0441\u043a\u0432\u0430,\u0421\u0423\u041d\u0426 \u041c\u0413\u0423 \r\n2 \u041a\u043b\u043e\u0447\u043a\u043e\u0432 \u0415\u0433\u043e\u0440 \u041a\u0430\u0437\u0430\u0445\u0441\u0442\u0430\u043d, \u041f\u0430\u0432\u043b\u043e\u0434\u0430\u0440, \u0441\u0431\u043e\u0440\u043d\u0430\u044f \r\n3 \u0414\u0430\u043b\u0438\u0435\u0432 \u0410\u0441\u0435\u0442 \u041a\u0430\u0437\u0430\u0445\u0441\u0442\u0430\u043d, \u0410\u043b\u043c\u0430\u0442\u044b, \u0420\u0421\u0424\u041c\u0421\u0428\u0418 \r\n4 \u0411\u043e\u0440\u0438\u0441\u043b\u0430\u0432 \u0410\u0442\u0430\u043d\u0430\u0441\u043e\u0432 \u0412\u044a\u043b\u043a\u043e\u0432 \u0411\u043e\u043b\u0433\u0430\u0440\u0438\u044f, \u0421\u043e\u0444\u0438\u044f, \u0421\u043e\u0444\u0438\u0439\u0441\u043a\u0430\u044f \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u0447\u0435\u0441\u043a\u0430\u044f \u0433\u0438\u043c\u043d\u0430\u0437\u0438\u044f \r\n5 \u0410\u043b\u0435\u043a\u0441\u0430\u043d\u0434\u044a\u0440 \u041a\u0438\u0440\u0438\u043b\u043e\u0432 \u0414\u0430\u0441\u043a\u0430\u043b\u043e\u0432 \u0411\u043e\u043b\u0433\u0430\u0440\u0438\u044f, \u0421\u043e\u0444\u0438\u044f, \u0421\u043e\u0444\u0438\u0439\u0441\u043a\u0430\u044f \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u0447\u0435\u0441\u043a\u0430\u044f \u0433\u0438\u043c\u043d\u0430\u0437\u0438\u044f \r\n6 \u0422\u0443\u0441\u0443\u043f\u0431\u0435\u043a\u043e\u0432 \u0415\u0440\u043a\u0435\u043d \u041a\u0430\u0437\u0430\u0445\u0441\u0442\u0430\u043d, \u0410\u043b\u043c\u0430\u0442\u044b, \u0420\u0421\u0424\u041c\u0421\u0428\u0418 \r\n7 \u041e\u0440\u0430\u0437\u0431\u0430\u0435\u0432 \u0421\u0430\u043d\u0436\u0430\u0440 \u041a\u0430\u0437\u0430\u0445\u0441\u0442\u0430\u043d, \u0410\u043a\u0442\u043e\u0431\u0435, \u0441\u0431\u043e\u0440\u043d\u0430\u044f \r\n8 Temuge Enkhbaatar \u041c\u043e\u043d\u0433\u043e\u043b\u0438\u044f, \u0423\u043b\u0430\u043d-\u0411\u0430\u0442\u043e\u0440, \u0428\u043a\u043e\u043b\u0430 \u21161 \r\n9 \u041b\u044e\u0431\u043e\u0441\u043b\u0430\u0432 \u041d\u0438\u043a\u043e\u043b\u0430\u0435\u0432 \u041f\u0430\u043d\u0447\u0435\u0432 \u0411\u043e\u043b\u0433\u0430\u0440\u0438\u044f, \u0421\u043e\u0444\u0438\u044f, \u0421\u043e\u0444\u0438\u0439\u0441\u043a\u0430\u044f \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u0447\u0435\u0441\u043a\u0430\u044f \u0433\u0438\u043c\u043d\u0430\u0437\u0438\u044f \r\n10 Deneanu Andrei-Florin \u0420\u0443\u043c\u044b\u043d\u0438\u044f, \u0411\u0443\u0445\u0430\u0440\u0435\u0441\u0442, International Computer Highschool of Bucharest \r\n \r\n \r\nSilver medals: \r\n \r\n1 Bozgan Francisc Teodor \u0420\u0443\u043c\u044b\u043d\u0438\u044f, \u0411\u0443\u0445\u0430\u0440\u0435\u0441\u0442, International Computer Highschool of Bucharest \r\n2 Sarkhan Badirli \u0410\u0437\u0435\u0440\u0431\u0430\u0439\u0434\u0436\u0430\u043d, \u0411\u0430\u043a\u0443, Baku Turkish High School \r\n3 \u0421\u0435\u043c\u0435\u043d\u043e\u0432\u0430 \u0412\u0435\u0440\u0430 \u0423\u043a\u0440\u0430\u0438\u043d\u0430, \u041a\u0438\u0435\u0432, \u0423\u043a\u0440\u0430\u0438\u043d\u0441\u043a\u0438\u0439 \u0424\u041c\u041b \u041a\u041d\u0423 \u0438\u043c. \u0422.\u0428\u0435\u0432\u0447\u0435\u043d\u043a\u043e \r\n4 \u041c\u0443\u043b\u0434\u0430\u0433\u0430\u043b\u0438\u0435\u0432 \u0411\u0438\u0440\u0436\u0430\u043d \u041a\u0430\u0437\u0430\u0445\u0441\u0442\u0430\u043d, \u0410\u043b\u043c\u0430\u0442\u044b, \u0420\u0421\u0424\u041c\u0421\u0428\u0418 \r\n5 \u041a\u0430\u043b\u0438\u043d \u0418\u0432\u0430\u043d\u043e\u0432 \u041b\u0430\u043d\u0434\u0436\u0435\u0432 \u0411\u043e\u043b\u0433\u0430\u0440\u0438\u044f, \u0421\u043e\u0444\u0438\u044f, \u0421\u043e\u0444\u0438\u0439\u0441\u043a\u0430\u044f \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u0447\u0435\u0441\u043a\u0430\u044f \u0433\u0438\u043c\u043d\u0430\u0437\u0438\u044f \r\n6 \u0421\u043c\u0438\u0440\u043d\u043e\u0432 \u0414\u043c\u0438\u0442\u0440\u0438\u0439 \u0420\u043e\u0441\u0441\u0438\u044f, \u0421\u0430\u043d\u043a\u0442-\u041f\u0435\u0442\u0435\u0440\u0431\u0443\u0440\u0433, \u0424\u0422\u0428 \u043f\u0440\u0438 \u0424\u0422\u0418 \u0438\u043c. \u0410.\u0424. \u0418\u043e\u0444\u0444\u0435 \r\n7 \u0421\u0430\u0442\u044b\u043b\u0445\u0430\u043d\u043e\u0432 \u041a\u0430\u043d\u0430\u0442 \u041a\u0430\u0437\u0430\u0445\u0441\u0442\u0430\u043d, \u0422\u0430\u043b\u0434\u044b\u043a\u043e\u0440\u0433\u0430\u043d, \u0441\u0448 \u211620 \r\n8 Bumbacea Radu \u0420\u0443\u043c\u044b\u043d\u0438\u044f, \u0411\u0443\u0445\u0430\u0440\u0435\u0441\u0442, TUDOR VIANU National High School of computer science \r\n9 \u041a\u043e\u0440\u043e\u0431\u0435\u0439\u043d\u0438\u043a\u043e\u0432 \u0418\u0433\u043e\u0440\u044c \u0422\u0430\u0434\u0436\u0438\u043a\u0438\u0441\u0442\u0430\u043d, \u0414\u0443\u0448\u0430\u043d\u0431\u0435, \u0422\u0430\u0434\u0436\u0438\u043a\u0441\u043a\u043e-\u0422\u0443\u0440\u0435\u0446\u043a\u0438\u0439 \u043b\u0438\u0446\u0435\u0439 \r\n10 Byambajav Namsraijav \u041c\u043e\u043d\u0433\u043e\u043b\u0438\u044f, \u0423\u043b\u0430\u043d-\u0411\u0430\u0442\u043e\u0440, \u0428\u043a\u043e\u043b\u0430 \u21161 \r\n11 \u041a\u0430\u0431\u0430\u0441\u043e\u0432 \u0411\u0430\u0433\u0434\u0430\u0442 \u041a\u0430\u0437\u0430\u0445\u0441\u0442\u0430\u043d, \u0410\u043a\u0442\u043e\u0431\u0435, \u0441\u0431\u043e\u0440\u043d\u0430\u044f \r\n12 \u0425\u0430\u0434\u0436\u0438\u043c\u0443\u0440\u0430\u0442\u043e\u0432 \u041d\u0443\u0440\u0441\u0443\u043b\u0442\u0430\u043d \u041a\u0430\u0437\u0430\u0445\u0441\u0442\u0430\u043d, \u0422\u0430\u043b\u0434\u044b\u043a\u043e\u0440\u0433\u0430\u043d, \u0441\u0448 \u211620 \r\n13 \u041a\u0430\u0441\u0435\u043d\u043e\u0432 \u0415\u0441\u043a\u0435\u043d\u0434\u0438\u0440 \u041a\u0430\u0437\u0430\u0445\u0441\u0442\u0430\u043d, \u0410\u0441\u0442\u0430\u043d\u0430, \u0441\u0431\u043e\u0440\u043d\u0430\u044f \r\n14 \u0410\u0442\u043b\u0430\u0441\u043e\u0432\u0430 \u041c\u0430\u0440\u0438\u044f \u0420\u0435\u0441\u043f\u0443\u0431\u043b\u0438\u043a\u0430 \u0421\u0430\u0445\u0430, \u042f\u043a\u0443\u0442\u0441\u043a, \u0420\u0435\u0441\u043f\u0443\u0431\u043b\u0438\u043a\u0430\u043d\u0441\u043a\u0438\u0439 \u043b\u0438\u0446\u0435\u0439-\u0438\u043d\u0442\u0435\u0440\u043d\u0430\u0442 \r\n15 Myagmarjav Bold \u041c\u043e\u043d\u0433\u043e\u043b\u0438\u044f, \u0423\u043b\u0430\u043d-\u0411\u0430\u0442\u043e\u0440, \u0428\u043a\u043e\u043b\u0430 \u21161 \r\n16 \u041a\u0430\u043b\u0434\u044b\u0431\u0430\u0435\u0432 \u041d\u0443\u0440\u0430\u0437\u0435\u043c \u041a\u044b\u0440\u0433\u044b\u0437\u0441\u0442\u0430\u043d,\u0411\u0438\u0448\u043a\u0435\u043a\u041a\u044b\u0440\u0433\u044b\u0437\u0441\u043a\u043e-\u0422\u0443\u0440\u0435\u0446\u043a\u0438\u0439 \u043b\u0438\u0446\u0435\u0439 \"\u0421\u0435\u0431\u0430\u0442\" \r\n \r\n \r\nBronze medals:\r\n \r\n1 \u041b\u0430\u043a\u0438\u0440\u0431\u0430\u044f \u041b\u0430\u0448\u0430 \u0413\u0440\u0443\u0437\u0438\u044f, \u0422\u0431\u0438\u043b\u0438\u0441\u0438, \u0428\u043a\u043e\u043b\u0430 \u2116 199 \"\u041a\u043e\u043c\u0430\u0440\u043e\u0432\" \r\n2 \u041e\u043b\u0438\u043c\u0434\u0436\u043e\u043d\u0438 \u041f\u0438\u0440\u043c\u0430\u0445\u043c\u0430\u0434 \u0422\u0430\u0434\u0436\u0438\u043a\u0438\u0441\u0442\u0430\u043d, \u0414\u0443\u0448\u0430\u043d\u0431\u0435, \u0422\u0430\u0434\u0436\u0438\u043a\u0441\u043a\u043e-\u0420\u043e\u0441\u0441\u0438\u0439\u0441\u043a\u0430\u044f \u0433\u0438\u043c\u043d\u0430\u0437\u0438\u044f-\u0438\u043d\u0442\u0435\u0440\u043d\u0430\u0442 \r\n3 \u041d\u0438\u043a\u0438\u0444\u043e\u0440\u043e\u0432 \u0414\u044c\u0443\u043b\u0443\u0441\u0442\u0430\u043d \u0420\u0435\u0441\u043f\u0443\u0431\u043b\u0438\u043a\u0430 \u0421\u0430\u0445\u0430, \u042f\u043a\u0443\u0442\u0441\u043a, \u0420\u0435\u0441\u043f\u0443\u0431\u043b\u0438\u043a\u0430\u043d\u0441\u043a\u0438\u0439 \u043b\u0438\u0446\u0435\u0439-\u0438\u043d\u0442\u0435\u0440\u043d\u0430\u0442 \r\n4 Ruslan Muslimov \u0410\u0437\u0435\u0440\u0431\u0430\u0439\u0434\u0436\u0430\u043d, \u0411\u0430\u043a\u0443, Baku Turkish High School \r\n5 \u0420\u0430\u0432\u0447\u0435\u0435\u0432 \u041d\u0438\u043a\u0438\u0442\u0430 \u0420\u043e\u0441\u0441\u0438\u044f, \u041a\u0440\u0430\u0441\u043d\u043e\u044f\u0440\u0441\u043a, \u0441\u0431\u043e\u0440\u043d\u0430\u044f \r\n6 \u041a\u0445\u043e\u043c\u0435\u0440\u0438\u043a\u0438 \u0413\u0435\u043e\u0440\u0433 \u0413\u0440\u0443\u0437\u0438\u044f, \u0422\u0431\u0438\u043b\u0438\u0441\u0438, \u0428\u043a\u043e\u043b\u0430 \u2116 199 \"\u041a\u043e\u043c\u0430\u0440\u043e\u0432\" \r\n7 \u0414\u043e\u0441\u044b\u043c\u0431\u0430\u0435\u0432 \u041e\u043b\u0436\u0430\u0441 \u041a\u0430\u0437\u0430\u0445\u0441\u0442\u0430\u043d, \u0410\u043b\u043c\u0430\u0442\u044b, \u0420\u0421\u0424\u041c\u0421\u0428\u0418 \r\n8 \u0413\u0443\u0440\u043e\u044f\u043d \u0412\u0430\u0433\u0430\u043d \u0410\u0440\u043c\u0435\u043d\u0438\u044f, \u0415\u0440\u0435\u0432\u0430\u043d, \u0424\u041c\u0428 \u043f\u0440\u0438 \u0415\u0440\u0413\u0423 \r\n9 Zorigoo Ochirhuyag \u041c\u043e\u043d\u0433\u043e\u043b\u0438\u044f, \u0423\u043b\u0430\u043d-\u0411\u0430\u0442\u043e\u0440, \u0428\u043a\u043e\u043b\u0430 \u21161 \r\n10 Tsogbayar Idertsogt \u041c\u043e\u043d\u0433\u043e\u043b\u0438\u044f, \u0423\u043b\u0430\u043d-\u0411\u0430\u0442\u043e\u0440, \u0428\u043a\u043e\u043b\u0430 \u21161 \r\n11 \u0425\u0430\u0439\u0434\u0430\u0440\u043e\u0432 \u0424\u0430\u0440\u0445\u043e\u0434 \u0423\u0437\u0431\u0435\u043a\u0438\u0441\u0442\u0430\u043d, \u0422\u0430\u0448\u043a\u0435\u043d\u0442, \u0428\u043a\u043e\u043b\u0430 \"\u0418\u043a\u0442\u0438\u0434\u043e\u0440\" \r\n12 \u0428\u0443\u043b\u044c\u0433\u0430 \u041e\u043b\u0435\u0441\u0430\u043d\u0434\u0440 \u0423\u043a\u0440\u0430\u0438\u043d\u0430, \u041a\u0438\u0435\u0432, \u0423\u043a\u0440\u0430\u0438\u043d\u0441\u043a\u0438\u0439 \u0424\u041c\u041b \u041a\u041d\u0423 \u0438\u043c. \u0422.\u0428\u0435\u0432\u0447\u0435\u043d\u043a\u043e \r\n13 Moraru Laurentiu \u0420\u0443\u043c\u044b\u043d\u0438\u044f, \u0411\u0443\u0445\u0430\u0440\u0435\u0441\u0442, TUDOR VIANU National High School of computer science \r\n14 \u0412\u043e\u0439\u0434\u0435\u043b\u0435\u0432\u0438\u0447 \u0410\u043b\u0435\u043a\u0441\u0435\u0439 \u0411\u0435\u043b\u0430\u0440\u0443\u0441\u044c, \u041c\u0438\u043d\u0441\u043a, \u0421\u0440\u0435\u0434\u043d\u044f\u044f \u0448\u043a\u043e\u043b\u0430 \u2116 51 \r\n15 \u0424\u0435\u0440\u0430\u0434\u0437\u0435 \u041b\u0430\u0448\u0430 \u0413\u0440\u0443\u0437\u0438\u044f, \u0422\u0431\u0438\u043b\u0438\u0441\u0438, \u0428\u043a\u043e\u043b\u0430 \u2116 199 \"\u041a\u043e\u043c\u0430\u0440\u043e\u0432\" \r\n16 \u041d\u0443\u043a\u0435\u0436\u0430\u043d\u043e\u0432 \u0415\u0440\u043d\u0430\u0440 \u041a\u0430\u0437\u0430\u0445\u0441\u0442\u0430\u043d, \u0410\u043b\u043c\u0430\u0442\u044b, \u0420\u0421\u0424\u041c\u0421\u0428\u0418 \r\n17 \u041a\u0435\u043d\u0436\u0435\u0442\u0430\u0435\u0432 \u0410\u0437\u0430\u043c\u0430\u0442 \u041a\u0430\u0437\u0430\u0445\u0441\u0442\u0430\u043d, \u041f\u0430\u0432\u043b\u043e\u0434\u0430\u0440, \u0441\u0431\u043e\u0440\u043d\u0430\u044f \r\n18 \u0416\u043e\u043b\u0434\u0430\u0441 \u041c\u0435\u0440\u0435\u0439 \u041a\u0430\u0437\u0430\u0445\u0441\u0442\u0430\u043d, \u0423\u0441\u0442\u044c-\u041a\u0430\u043c\u0435\u043d\u043e\u0433\u043e\u0440\u0441\u043a, \u041a\u0422\u041b \u0434\u043b\u044f \u043e\u0434\u0430\u0440\u0435\u043d\u043d\u044b\u0445 \u0434\u0435\u0442\u0435\u0439 \r\n19 Georgescu Flavian \u0420\u0443\u043c\u044b\u043d\u0438\u044f, \u0411\u0443\u0445\u0430\u0440\u0435\u0441\u0442, International Computer Highschool of Bucharest \r\n20 \u0415\u0440\u043f\u044b\u043b\u0435\u0432 \u0410\u043b\u0435\u043a\u0441\u0435\u0439 \u0420\u043e\u0441\u0441\u0438\u044f, \u041c\u043e\u0441\u043a\u0432\u0430,\u0421\u0423\u041d\u0426 \u041c\u0413\u0423 \r\n21 \u041c\u0438\u0440\u0437\u0430\u0435\u0432 \u0418\u043d\u043e\u043c\u0436\u043e\u043d \u0422\u0430\u0434\u0436\u0438\u043a\u0438\u0441\u0442\u0430\u043d, \u0414\u0443\u0448\u0430\u043d\u0431\u0435, \u0422\u0430\u0434\u0436\u0438\u043a\u0441\u043a\u043e-\u0422\u0443\u0440\u0435\u0446\u043a\u0438\u0439 \u043b\u0438\u0446\u0435\u0439 \r\n22 \u041d\u0443\u0440\u0441\u0443\u043b\u0442\u0430\u043d\u043e\u0432 \u041c\u0435\u0434\u0435\u0442 \u041a\u0430\u0437\u0430\u0445\u0441\u0442\u0430\u043d, \u0410\u0441\u0442\u0430\u043d\u0430, \u0441\u0431\u043e\u0440\u043d\u0430\u044f \r\n23 \u041f\u0440\u0438\u0441\u0438\u0432\u043a\u043e \u0412\u044f\u0447\u0435\u0441\u043b\u0430\u0432 \u0420\u043e\u0441\u0441\u0438\u044f, \u0421\u0430\u043d\u043a\u0442-\u041f\u0435\u0442\u0435\u0440\u0431\u0443\u0440\u0433, \u0424\u0422\u0428 \u043f\u0440\u0438 \u0424\u0422\u0418 \u0438\u043c. \u0410.\u0424. \u0418\u043e\u0444\u0444\u0435 \r\n24 Camburu Oana-Maria \u0420\u0443\u043c\u044b\u043d\u0438\u044f, \u0411\u0443\u0445\u0430\u0440\u0435\u0441\u0442, TUDOR VIANU National High School of computer science\r\nSorry for russian language,cause it is too long to translate all the names into english. :roll:", - "Solution_10": "\u041a\u043b\u043e\u0447\u043a\u043e\u0432 \u0415\u0433\u043e\u0440??? Hmmm... 4th participation in ZIMO)))\r\nCan you post post points nof gold medalists?", - "Solution_11": "I can post the points of all medalists :) :\r\n1 \u0412\u043e\u0440\u043e\u0431\u044c\u0435\u0432 \u0418\u043b\u044c\u044f \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 1RUS 7 7 0 7 7 7 35 GOLD \r\n2 \u041a\u043b\u043e\u0447\u043a\u043e\u0432 \u0415\u0433\u043e\u0440 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 3KAZ 7 3 0 7 7 7 31 GOLD \r\n3 \u0414\u0430\u043b\u0438\u0435\u0432 \u0410\u0441\u0435\u0442 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 9KAZ 5 4 0 7 7 7 30 GOLD \r\n4 \u0411\u043e\u0440\u0438\u0441\u043b\u0430\u0432 \u0410\u0442\u0430\u043d\u0430\u0441\u043e\u0432 \u0412\u044a\u043b\u043a\u043e\u0432 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 BGR 5 3 0 7 7 7 29 GOLD \r\n5 \u0410\u043b\u0435\u043a\u0441\u0430\u043d\u0434\u044a\u0440 \u041a\u0438\u0440\u0438\u043b\u043e\u0432 \u0414\u0430\u0441\u043a\u0430\u043b\u043e\u0432 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 BGR 7 7 0 6 7 1 28 GOLD \r\n6 \u0422\u0443\u0441\u0443\u043f\u0431\u0435\u043a\u043e\u0432 \u0415\u0440\u043a\u0435\u043d \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 9KAZ 7 3 0 7 7 3 27 GOLD \r\n7 \u041e\u0440\u0430\u0437\u0431\u0430\u0435\u0432 \u0421\u0430\u043d\u0436\u0430\u0440 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 2KAZ 7 7 0 6 7 0 27 GOLD \r\n8 Temuge Enkhbaatar \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 MNG 0 7 3 7 7 1 25 GOLD \r\n9 \u041b\u044e\u0431\u043e\u0441\u043b\u0430\u0432 \u041d\u0438\u043a\u043e\u043b\u0430\u0435\u0432 \u041f\u0430\u043d\u0447\u0435\u0432 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 BGR 7 4 0 7 7 0 25 GOLD \r\n10 Deneanu Andrei-Florin \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 2ROM 7 3 2 6 7 0 25 GOLD \r\n11 Bozgan Francisc Teodor \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 2ROM 7 3 0 6 7 1 24 SILVER \r\n12 Sarkhan Badirli \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 1AZE 7 3 0 7 0 7 24 SILVER \r\n13 \u0421\u0435\u043c\u0435\u043d\u043e\u0432\u0430 \u0412\u0435\u0440\u0430 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 UKR 7 2 0 7 7 0 23 SILVER \r\n14 \u041c\u0443\u043b\u0434\u0430\u0433\u0430\u043b\u0438\u0435\u0432 \u0411\u0438\u0440\u0436\u0430\u043d \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 10KAZ 7 0 0 7 7 1 22 SILVER \r\n15 \u041a\u0430\u043b\u0438\u043d \u0418\u0432\u0430\u043d\u043e\u0432 \u041b\u0430\u043d\u0434\u0436\u0435\u0432 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 BGR 7 3 0 5 7 0 22 SILVER \r\n16 \u0421\u043c\u0438\u0440\u043d\u043e\u0432 \u0414\u043c\u0438\u0442\u0440\u0438\u0439 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 3RUS 7 6 2 7 0 0 22 SILVER \r\n17 \u0421\u0430\u0442\u044b\u043b\u0445\u0430\u043d\u043e\u0432 \u041a\u0430\u043d\u0430\u0442 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 4KAZ 0 3 0 3 7 7 20 SILVER \r\n18 Bumbacea Radu \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 1ROM 7 3 0 3 7 0 20 SILVER \r\n19 \u041a\u043e\u0440\u043e\u0431\u0435\u0439\u043d\u0438\u043a\u043e\u0432 \u0418\u0433\u043e\u0440\u044c \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 2TJK 7 0 0 5 7 0 19 SILVER \r\n20 Byambajav Namsraijav \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 MNG 7 1 3 7 1 0 19 SILVER \r\n21 \u041a\u0430\u0431\u0430\u0441\u043e\u0432 \u0411\u0430\u0433\u0434\u0430\u0442 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 2KAZ 7 0 0 5 7 0 19 SILVER \r\n22 \u0425\u0430\u0434\u0436\u0438\u043c\u0443\u0440\u0430\u0442\u043e\u0432 \u041d\u0443\u0440\u0441\u0443\u043b\u0442\u0430\u043d \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 4KAZ 7 3 0 2 7 0 19 SILVER \r\n23 \u041a\u0430\u0441\u0435\u043d\u043e\u0432 \u0415\u0441\u043a\u0435\u043d\u0434\u0438\u0440 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 1KAZ 7 3 0 7 2 0 19 SILVER \r\n24 \u0410\u0442\u043b\u0430\u0441\u043e\u0432\u0430 \u041c\u0430\u0440\u0438\u044f \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 YKT 7 3 0 2 7 0 19 SILVER \r\n25 Myagmarjav Bold \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 MNG 7 4 0 5 2 0 18 SILVER \r\n26 \u041a\u0430\u043b\u0434\u044b\u0431\u0430\u0435\u0432 \u041d\u0443\u0440\u0430\u0437\u0435\u043c \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 3KGZ 7 3 0 2 5 1 18 SILVER \r\n27 \u041b\u0430\u043a\u0438\u0440\u0431\u0430\u044f \u041b\u0430\u0448\u0430 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 GEO 7 3 0 7 0 0 17 BRONZE \r\n28 \u041e\u043b\u0438\u043c\u0434\u0436\u043e\u043d\u0438 \u041f\u0438\u0440\u043c\u0430\u0445\u043c\u0430\u0434 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 1TJK 7 0 0 3 7 0 17 BRONZE \r\n29 \u041d\u0438\u043a\u0438\u0444\u043e\u0440\u043e\u0432 \u0414\u044c\u0443\u043b\u0443\u0441\u0442\u0430\u043d \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 YKT 7 3 0 0 7 0 17 BRONZE \r\n30 Ruslan Muslimov \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 1AZE 7 3 0 7 0 0 17 BRONZE \r\n31 \u0420\u0430\u0432\u0447\u0435\u0435\u0432 \u041d\u0438\u043a\u0438\u0442\u0430 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 5RUS 7 0 0 2 7 0 16 BRONZE \r\n32 \u041a\u0445\u043e\u043c\u0435\u0440\u0438\u043a\u0438 \u0413\u0435\u043e\u0440\u0433 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 GEO 7 0 0 2 7 0 16 BRONZE \r\n33 \u0414\u043e\u0441\u044b\u043c\u0431\u0430\u0435\u0432 \u041e\u043b\u0436\u0430\u0441 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 9KAZ 4 3 0 1 7 1 16 BRONZE \r\n34 \u0413\u0443\u0440\u043e\u044f\u043d \u0412\u0430\u0433\u0430\u043d \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 ARM 7 3 0 6 0 0 16 BRONZE \r\n35 Zorigoo Ochirhuyag \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 MNG 7 0 0 7 2 0 16 BRONZE \r\n36 Tsogbayar Idertsogt \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 MNG 7 2 0 7 0 0 16 BRONZE \r\n37 \u0425\u0430\u0439\u0434\u0430\u0440\u043e\u0432 \u0424\u0430\u0440\u0445\u043e\u0434 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 UZB 7 0 0 2 7 0 16 BRONZE \r\n38 \u0428\u0443\u043b\u044c\u0433\u0430 \u041e\u043b\u0435\u0441\u0430\u043d\u0434\u0440 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 UKR 7 0 0 2 0 7 16 BRONZE \r\n39 Moraru Laurentiu \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 1ROM 0 3 0 6 7 0 16 BRONZE \r\n40 \u0412\u043e\u0439\u0434\u0435\u043b\u0435\u0432\u0438\u0447 \u0410\u043b\u0435\u043a\u0441\u0435\u0439 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 2BLR 7 0 0 1 7 0 15 BRONZE \r\n41 \u0424\u0435\u0440\u0430\u0434\u0437\u0435 \u041b\u0430\u0448\u0430 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 GEO 7 2 0 5 1 0 15 BRONZE \r\n42 \u041d\u0443\u043a\u0435\u0436\u0430\u043d\u043e\u0432 \u0415\u0440\u043d\u0430\u0440 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 11KAZ 0 0 0 7 5 3 15 BRONZE \r\n43 \u041a\u0435\u043d\u0436\u0435\u0442\u0430\u0435\u0432 \u0410\u0437\u0430\u043c\u0430\u0442 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 3KAZ 7 1 0 7 0 0 15 BRONZE \r\n44 \u0416\u043e\u043b\u0434\u0430\u0441 \u041c\u0435\u0440\u0435\u0439 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 5KAZ 7 0 0 1 7 0 15 BRONZE \r\n45 Georgescu Flavian \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 2ROM 0 0 0 2 6 7 15 BRONZE \r\n46 \u0415\u0440\u043f\u044b\u043b\u0435\u0432 \u0410\u043b\u0435\u043a\u0441\u0435\u0439 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 1RUS 7 2 0 5 0 0 14 BRONZE \r\n47 \u041c\u0438\u0440\u0437\u0430\u0435\u0432 \u0418\u043d\u043e\u043c\u0436\u043e\u043d \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 2TJK 7 0 0 7 0 0 14 BRONZE \r\n48 \u041d\u0443\u0440\u0441\u0443\u043b\u0442\u0430\u043d\u043e\u0432 \u041c\u0435\u0434\u0435\u0442 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 1KAZ 0 3 0 7 3 1 14 BRONZE \r\n49 \u041f\u0440\u0438\u0441\u0438\u0432\u043a\u043e \u0412\u044f\u0447\u0435\u0441\u043b\u0430\u0432 \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 3RUS 7 0 0 7 0 0 14 BRONZE \r\n50 Camburu Oana-Maria \u043c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u043a\u0430 1ROM 4 0 0 3 7 0 14 BRONZE", - "Solution_12": "Who can paticipate in the contest?" -} -{ - "Problem": "please, can someone help me..\r\n\r\nIn a big serial 96% of products satisfy to standards. Rude control good product proclaims for ''good'' with probability 0,98 and bad product proclaims for ''good'' with probability 0,05. Find the probability that a chosen product is ''good'' . \r\n\r\nThank you very much", - "Solution_1": "let $ P(G| \\plus{} )$ be probability that a chosen product is ''good'' ( that product proclaimed for \"good\" is really satysfies to standart)\r\n$ P( \\plus{} |G) \\equal{} 0,98$ is probability that good product is proclaimed for \"good\" \r\n$ P( \\plus{} |B) \\equal{} 0,05$ is probability that bad product is proclaimed for \"good\"\r\n$ P(G) \\equal{} 1 \\minus{} P(B) \\equal{} 0,96$ is probability that product satisfies to standards \r\n\r\nso \r\n\r\n$ P(G| \\plus{} ) \\equal{} \\frac {P( \\plus{} |G) \\cdot P(G) }{P( \\plus{} |G) \\cdot P(G) \\plus{} P( \\plus{} |B) \\cdot P(B) } \\equal{} \\frac {0,98 \\cdot 0,96}{0,98 \\cdot 0,96 \\plus{} 0,05\\cdot 0,04}$$ \\equal{} \\frac {2352}{2357} \\equal{}$~$ 0,9979$\r\n\r\n\r\n :roll:", - "Solution_2": "thank you..i've solved problem but differently, maybe more complicated but Bayes is the better solution..tnx very much" -} -{ - "Problem": "An interesting poll. Much logic involved.", - "Solution_1": "Arrg...\r\nThe only thing I thought of was to choose one that didn't have itself in it....", - "Solution_2": "WHAT?!?! I voted for the one that was voted most..., but I got the 2nd and 3rd most voted ones. Meh. Almost had it :D .\r\n[hide=\"It was the \"]2nd one: 1&3.[/hide] Don't look at what I voted if you haven't yet.", - "Solution_3": "What if in the end, it is \"none of the above\"? :huh:", - "Solution_4": "That's not an option. The logic part is in the what to choose part.", - "Solution_5": "It's best to choose 2 - since 3 is expected to be one of the least chosen\r\nbut seems 3 [i]was[/i] the correct option somehow, as things stand now! :o" -} -{ - "Problem": "this is easy...\r\n\r\n$\\sqrt{(x-1)(x-3)} - \\sqrt{(x+1)(x-5)}=4-2\\sqrt{2}$", - "Solution_1": "[hide]\nx=2+root17\nor 2-root17\ni just let 4 = the first surd and 2root2 equal the second.. ><\ni dont know how to do this..\nive tried expanding by squaring both sides.. not going 2 well..\nany hints?\n[/hide]", - "Solution_2": "[hide=\"Solution\"]\\begin{eqnarray*}\\sqrt{(x-1)(x-3)} - \\sqrt{(x+1)(x-5)} &=& \\sqrt{x^2-4x+3} - \\sqrt{x^2-4x-5}\\\\ &=& \\sqrt{(x^2-4x-1)+4} - \\sqrt{(x^2-4x-1)-4}\\end{eqnarray*}\n\nIf we substitute $x^2-4x-1=t$, the equation becomes\n\n$\\sqrt{t+4} - \\sqrt{t-4}=4-2\\sqrt{2}$\n\nand we have \n\n\\begin{eqnarray*}\\sqrt{t+4} - \\sqrt{t-4}=4-2\\sqrt{2} &\\Rightarrow& \\sqrt{t+4} = \\sqrt{t-4}+4-2\\sqrt{2}\\\\ &\\Rightarrow& t+4=t-4+2(4-2\\sqrt{2})\\sqrt{t-4}+(4-2\\sqrt{2})^2\\\\ &\\Rightarrow& 16\\sqrt{2}-16=2(4-2\\sqrt{2})\\sqrt{t-4}\\\\ &\\Rightarrow& \\sqrt{t-4}={{4\\sqrt{2}-4}\\over{2-\\sqrt{2}}}\\\\ &\\Rightarrow& \\sqrt{t-4}={{(4\\sqrt{2}-4)(2+\\sqrt{2})}\\over{2^2-\\sqrt{2}^2}}\\\\ &\\Rightarrow& \\sqrt{t-4}=2\\sqrt{2}\\\\ &\\Rightarrow& t-4=8\\\\ &\\Rightarrow& t=12\\end{eqnarray*}\n\nNow we have $x^2-4x-1=12$, which gives the solutions: $x_{1,2}=2\\pm\\sqrt{17}$.[/hide]" -} -{ - "Problem": "x :^3: + y :^3: = 22z :^3: for x,y,z coprime positive integers.\r\n\r\nI know for a fact that there is only one solution, with x,y,z each having about 5 digits. I'm just wondering if there is a way to find and prove uniqueness without using analytic number theory. I have no clue myself.", - "Solution_1": "Try a google search. This is a very famous problem. I just read something about it recently on the web, but I can't remember where.", - "Solution_2": "Okay I remembered where I saw it. It was mentioned as an aside in this long rambling paper by Louis De Branges about the Riemann Hypothesis. He claimed that it was given to him as a youth and he spent the better part of a year solving it.\r\n\r\nActually, note that there are multiple solutions; but the other ones are just cube multiples of the one you are referring to.\r\n\r\nHere's a site with some information on this general type of equation:\r\nhttp://www.mathsoft.com/mathresources/problems/article/0,,2186,00.html\r\n\r\nIn general these types of diophanitine questions are very hard. I doubt that there is any \"elementary\" solution to it, or even an easy way to show that it has only one solution. I really don't know enough to say for sure though.\r\n\r\nFrom a general (and quite spectacular) result that Gerd Faltings found in 1983, it's known that all such cubic equations have at most a finite number of (independent) solutions in both integers and rational numbers. This is a consequence of very deep results in Algebraic Geometry." -} -{ - "Problem": "learning groups, how do I find all the one dimensional representations of say Z5?", - "Solution_1": "for every positive natural number $ n$ and a group $ G$, the homomorphisms $ \\mathbb{Z}/n \\to G$ correspond to elements $ g \\in G$ with $ g^n\\equal{}1$. apply this to $ n\\equal{}5$ and $ G\\equal{}\\mathbb{C}^*$.", - "Solution_2": "You only need the generators since Z5 is cyclic. Thus you're looking for representations T where T(x)^5=1\r\n\r\n1) T(x)=1\r\n2) T(x) = e^(2pi*i/5)\r\n3) T(x)= e^(4pi*i/5)\r\n4) T(x)=e^(8pi*i/5)\r\n5)T(x)= e^(2pi*i)\r\n\r\nI think that does it? (those are all five one dimensional representations)." -} -{ - "Problem": "Four prime numbers are randomly selected without replacement from the first ten prime numbers. What is the probability that the sum of the four selected numbers is odd? Express your answer as a common fraction.", - "Solution_1": "the sum is prime if and only if one of the numbers is 2 (figure out why this is true youself)\r\n\r\ntherefore, out of the ten numbers, we have to choose a 2, so we need to choose three numbers out of the nine numbers remaining\r\n9C3 = $ \\frac{9\\times8\\times7}{3\\times2} \\equal{} 3\\times4\\times7 \\equal{} 84$\r\n\r\nthere are a total of 10C4 ways to choose 4 numbers out of 10\r\n10C4 = $ \\frac{10\\times9\\times8\\times7}{4\\times3\\times2}\\equal{}10\\times3\\times7 \\equal{} 210$\r\n\r\nso our probability is $ \\frac{84}{210}\\equal{}\\frac{42}{105}\\equal{}\\frac{14}{35}\\equal{}\\boxed{\\frac{2}{5}}$\r\n\r\nhope i didn't make a mistake", - "Solution_2": "As already stated, the sum is odd only if 2 is one of the primes selected. What is the probability that 2 is one of the first 4 primes selected out of 10? $\\frac{4}{10} = \\frac{2}{5}$. Easy peasy.", - "Solution_3": "[font=TIMES NEW ROMAN][hide=Logical Thinking]We can first think about how we can obtain an odd number through addition. We know that for two numbers, whenever we add one odd number and one even number we get an odd number as a result. \n\nSo for four numbers, we need an odd number and an even number, as well as either two odd numbers or two even numbers.\n\nFortunately, there is only one even prime number: two. So we know that one of the numbers that is chosen must be two in order for the two numbers to sum to an odd number. This means that our chosen numbers must include two, and three other odd numbers in order to work.\n\nSo we can solve this problem by supposing we have to choose three numbers from nine - the number two must always be included, so we just choose from prime numbers two to ten.\n\nThis means that there are $\\tbinom{9}{3}=84$ combinations that work. There are $\\tbinom{10}{4}=210$ total possible combinations, so the probability of getting working one is $\\boxed{\\tfrac{84}{210}\\leadsto\\tfrac{2}{5}}.$\n$$\\definecolor{color}{rgb}{0.1, 0.8, 0.1} \\scriptsize{\\colorbox{color}{\\color{white}{A pifinity solution}}}$$\n", - "Solution_4": "just use complimentary counting\n", - "Solution_5": "Yes. Not choosing 2 is 9/10 x 8/9 x7/8 x 6/7, or 3/5, so 1 - 3/5 = 2/5.", - "Solution_6": "Another way that I thought about it was that you must get $2$ in order to fit the solution. You have a $\\frac{1}{10}$ chance of getting it. Since you choose a number $4$ times, you just do $\\frac{1}{10} \\times 4$. This results in $\\frac{2}{5}$" -} -{ - "Problem": "There was some discussion on how to sign up if your school is clueless or reluctant.\r\nSo far I did not see much enthusiasm from neighboring schools to let students from other schools even close to their math teams/ programs/ materials etc. So I do not anticipate any response if I try to sign up a student with another school for AMC.\r\nIs there a list of schools that have room for students from other schools and would be willing to take the other students in for a test? Or homeschools that registered and having room left?\r\nIs it possible to sign up individually? \r\nThank you.", - "Solution_1": "I'm not sure what you mean. The AMC is already done for this year.", - "Solution_2": "I mean for next year. Just doing my homework.", - "Solution_3": "The AMC won't have a list of which schools have registered for the AMC until probably around next November/December. You can start checking up with schools personally right now to see if they'll let you compete next year, though.", - "Solution_4": "The preceding information from Nathan is correct and is good advice.\r\nThroughout this year make contact with local high schools and high school math teachers and math team coaches. \r\nCheck with the AMC office in December 2006 about schools in your region registered for 2007.\r\n\r\nSteve Dunbar\r\nMAA Director, Americna Mathematics Competitions", - "Solution_5": "[quote=\"AMCDirector\"]The preceding information from Nathan is correct and is good advice.\nThroughout this year make contact with local high schools and high school math teachers and math team coaches. \nCheck with the AMC office in December 2006 about schools in your region registered for 2007.\n\nSteve Dunbar\nMAA Director, Americna Mathematics Competitions[/quote]\r\n\r\nSince we have to homeschool, I have had to go through this the last two years. Last year I send email to 3 local private schools. Two got back to me - one to give us the go ahead, and the other to tell us they only offer alternating years. The school we went to last year didn't plan to offer it this year, so I sent email to another private school, and within 15 minutes I had an affirmative. Since my son has a chance of qualifying for the USAMO, I chose among the schools who would probably have a qualifier this year.\r\n\r\nI sign my son up as a guest, which costs $\\$$32, thru the AMC. AMC sent them a copy of the test, so he does not use up one of their tests. He will not be eligible for their site honors if he does well, nor will he affect their statistics if he does poorly.\r\n\r\nLast year we were at a local math competition. I asked someone from a local public school (not our district), if he could take it there. I got a very negative response. This school had once allowed a middle school student from another district to take the test, and they had a bad experience with the mother. He ranted long enough for me to figure out who the family was. I will stick with the private schools!" -} -{ - "Problem": "a boat moves relative to water with velocity which is n=2 times less than the river flow velocity .at what angle to the stream direction must the boat move to minimize drifting?", - "Solution_1": "Unless I am not understanding your English, shouldn't the angle be 180\u00ba?", - "Solution_2": "no .the angle is 120 but l dont know the solution", - "Solution_3": "Two questions:\r\n\r\n1) Doesn't this problem have a picture?\r\n2) What is the meaning of \"minimize drifting\"?" -} -{ - "Problem": "Show that if $ f: R \\rightarrow R$ is a monotone increasing function, then the set where $ f$ is discontinuous has measure zero.", - "Solution_1": "The set where f is dicontinuous is countable, because they are all jump discontinuities. Uncountably many jump discontinuities of positive size on some interval $ (a,b)$ would make $ f$ increase more than $ f(b) \\minus{} f(a)$ on the interval", - "Solution_2": "in fact, I saw what you said, but I don't know how to prove that [i]uncountably many jump discontinuities of positive size on some interval $ (a,b)$ would make increase $ f$ more than on the interval $ (f(a),f(b))$.[/i]", - "Solution_3": "I was just loosely saying that an \"uncountable sum\" of positive numbers is infinite. Index all the discontinuities in $ (a,b)$ as $ x_j$ where $ j$ runs through some index, and let the corresponding jump size at $ x_j$ be $ K_j$. Then the set of $ x_j$ such that $ K_j \\geq 1/n$ is finite. The set of all $ x_j$ is the countable union of the former sets as $ n$ runs through the positive integers. The countable union of finite sets is countable, so the number of $ x_j$ is countable.", - "Solution_4": "[quote=\"Kalle\"]I was just loosely saying that an \"uncountable sum\" of positive numbers is infinite. Index all the discontinuities in $ (a,b)$ as $ x_j$ where $ j$ runs through some index, and let the corresponding jump size at $ x_j$ be $ K_j$. Then the set of $ x_j$ such that $ K_j \\geq 1/n$ is finite. The set of all $ x_j$ is the countable union of the former sets as $ n$ runs through the positive integers. The countable union of finite sets is countable, so the number of $ x_j$ is countable.[/quote]\r\n\r\nThank you Kalle !", - "Solution_5": "[quote=\"Kalle\"]I was just loosely saying that an \"uncountable sum\" of positive numbers is infinite. Index all the discontinuities in $ (a,b)$ as $ x_j$ where $ j$ runs through some index, and let the corresponding jump size at $ x_j$ be $ K_j$. Then the set of $ x_j$ such that $ K_j \\geq 1/n$ is finite. The set of all $ x_j$ is the countable union of the former sets as $ n$ runs through the positive integers. The countable union of finite sets is countable, so the number of $ x_j$ is countable.[/quote]\r\n\r\nwhy is the set of $ x_j$ s.t. $ K_j \\geq \\frac{1}{n}$ finite?" -} -{ - "Problem": "find the limit as x->4 of (4^x - x^4) / (x-4).\r\n\r\nWe are not allowed to use l'hopital's rule.\r\n\r\nAny help is appreciated. Thanks!", - "Solution_1": "Are you allowed to use the definition of the derivative? \r\n(For the given limit this amounts to the same as using L'Hopital.) \r\n\r\nIn any case you should try to break up the fraction into a sum of two terms \r\nthat are difference quotients and deal with them seperately.", - "Solution_2": "yes, we are allowed to use the definition of the derivative.\r\n\r\nI tried using algebra to try and cancel terms so that i could substitute 4 in, but i havent come up with anything. I also tried multiplying by the conjugate, using exponentials, etc.\r\n\r\nAny tips? :)", - "Solution_3": "You're going to have to quote your knowledge of derivatives. $ \\lim_{x\\to 4}\\frac{4^4\\minus{}x^4}{x\\minus{}4}$ is doable algebraically, but you need to know the derivative of the exponential for $ \\lim_{x\\to 4}\\frac{4^x\\minus{}4^4}{x\\minus{}4}$.", - "Solution_4": "is it basically just differentiating $ \\frac{4^{x} \\minus{} 4^{4}}{x \\minus{} 4}$ and then substituting the value 4?\r\n\r\nbut differentiating that would give a denominator of (x-4)^2 which would give a denominator of 0 when substituting 4.\r\n\r\nI'm still stuck :(", - "Solution_5": "$ \\lim_{x\\to 4}\\frac {4^x \\minus{} 4^4}{x \\minus{} 4}$ [i]is[/i] a derivative -- can you recognize the function that it's a derivative of (and the point at which it's the derivative)?", - "Solution_6": "oops, its $ \\lim_{x\\to 4}\\frac{4^{x}\\minus{}x^{4}}{x\\minus{}4}$ not $ \\lim_{x\\to 4}\\frac{4^{x}\\minus{}4^{4}}{x\\minus{}4}$\r\n\r\nif i let y = $ \\frac{4^{x}\\minus{}x^{4}}{x\\minus{}4}$ , and i differentiate w.r.t. x, and substitute it in at point 4, i get 240/0. Does that mean the limit is infinity?", - "Solution_7": "No, it means that you don't know how to recognize a limit as the definition of a derivative. What's the definition of the derivative of the function $ f(x)$ at the point $ x \\equal{} a$?\r\n\r\n(Also, jmerry and I both know that the problem we're suggesting to you is not exactly the same as the problem you asked; but it's an intermediate step in solving your problem. Once you've figured out our problem, you'll begin to see why.)", - "Solution_8": "$ \\frac{240}{0}$? $ 4^4$ is $ 256$, not $ 16$.\r\n\r\nIt has to be a $ \\frac{0}{0}$ form, because it's a derivative.\r\n\r\nAs for the different limits, there are two pieces in my post #4. Their sum is the original limit, and each is simpler than that original.", - "Solution_9": "$ \\lim_{x\\to 0}\\frac{f(a\\plus{}m)\\minus{}f(a)}{m}$ ? I think thats right. Sorry if I'm a bit slow, I'm really bad at limits. Taking calculus in college and learning derivatives without learning limits in high school is a terrible experience.\r\n\r\nI did realise that I could add them together and get the limit, but I can't find the derivative. I'm going around in circles. I can't seem to apply anything I've learnt in my calc class to this. I think I'm missing a huge chunk of the concept here.", - "Solution_10": "For the way you wrote that, you wanted to take $ \\lim_{m\\to 0},$ not $ x.$\r\n\r\nBut you might also want to consider this alternate (but algebraically equivalent) form:\r\n\r\n$ f'(a) \\equal{} \\lim_{x\\to a}\\frac {f(x) \\minus{} f(a)}{x \\minus{} a}.$", - "Solution_11": "[quote=\"tennischan\"]Sorry if I'm a bit slow, I'm really bad at limits.[/quote] No need to apologize :) \r\n\r\nKent made a helpful suggestion for another way to write the same expression. Now, the question becomes: can you recognize the limit you're given as the derivative of some function? This is a pattern-matching exercise: you have to come up with the right choice of $ f(x)$ and $ a$ so that the definition of the derivative of $ f$ at $ a$ ends up looking like the limit you're trying to compute.", - "Solution_12": "Okay, so I tried something I thought of, and I added the limits together by comparing the definition of the derivative for both parts and finding out what f(x) for each part was, and then finding f'(x).\r\n\r\nSo I found the derivative of 4^x and the derivative of x^4, which are 4^x (ln 4) and 4x^3 respectively.\r\n\r\nAdding them together, I got 256 ln 4 + 256 = 256 (ln 4 + 1).\r\n\r\nIs that right?", - "Solution_13": "Almost: are you sure adding the two limits was what you wanted to do? After you fix this, you'll be done; for purposes of your own education, though, you should definitely make sure you understand [i]why[/i] you did each step in this process and practice doing other examples so that you don't forget the important ideas :)", - "Solution_14": "$ \\frac {4^x \\minus{} x^4}{x \\minus{} 4} \\equal{} \\frac {4^x \\minus{} 4^4}{x \\minus{} 4} \\minus{} \\frac {x^4 \\minus{} 4^4}{x \\minus{} 4}$.\r\n\r\nAnd apply definition of derivative.", - "Solution_15": "oops, its 256(ln 4 -1)\r\n\r\nI get it now, thanks so much for all your help! :)" -} -{ - "Problem": "Suppose that $O$ is a point inside a convex quadrilateral $ABCD$ such that \\[ OA^2 + OB^2 + OC^2 + OD^2 = 2\\mathcal A[ABCD] , \\] where by $\\mathcal A[ABCD]$ we have denoted the area of $ABCD$. Prove that $ABCD$ is a square and $O$ is its center.\r\n\r\n[i]Yugoslavia[/i]", - "Solution_1": "From AM-GM we have that $OA^2 + OB^2 \\geq 2OA \\cdot OB$. \r\nOn the other hand we have that $OA \\cdot OB \\geq OA \\cdot OB\\sin \\alpha = 2\\left[OAB\\right]$ because the $\\sin$ is on $(0, 1)$. \r\nSo $OA^2 + OB^2 \\geq 4\\left[OAB\\right]$ and $OA^2 + OB^2 + OC^2 + OD^2 \\geq 2\\left[ABCD\\right]$. \r\nThe equality holds iff the two that we've considered hold, i.e $OA = OB$ for the first and $\\alpha = 90^{\\circ}$ for the second one. \r\nHence $O$ is the center of a square.", - "Solution_2": "Andreas i think your solution is a little bit mistaken \nbecause you can asume that the area of AOB is bigger then the COB or DOB but you cannot asume that the area of triangle AOB and COD are simmultainesly the bigest in the quadrilateral\nP.S.Sorry if i didnt understood smt. wrong", - "Solution_3": "[quote=\"frenchy\"]Andreas i think your solution is a little bit mistaken \nbecause you can asume that the area of AOB is bigger then the COB or DOB but you cannot asume that the area of triangle AOB and COD are simmultainesly the bigest in the quadrilateral\nP.S.Sorry if i didnt understood smt. wrong[/quote]\n\nWell, actually these assumptions haven't been made in the solution proposed by Andreas, which is actually correct.\nWhat he did is the following:\nhe used the inequality that he proved for AOB for the other 3 triangles (AOD, DOC, BOC) and then he added all 4 of them concluding that 2(OA^2 + OB^2 + OC^2 + OD^2) >= 4( [AOB] + [AOD] + [DOC] + [COB] ) = 4 [ABCD]" -} -{ - "Problem": "$ X\\in R^{n}$ with $ X$ the column vector\r\n\r\n(x_1)\r\n(x_2)\r\n.......\r\n(x_n)\r\n\r\nlet $ f(X)\\equal{}max |x_{k}|$ \r\n\r\n$ A\\in M(n,R)$\r\n\r\nsuch that for any $ X\\in R^{n}$, $ f(AX)\\equal{}f(X)$ \r\n\r\nprove that there exist m positive integer such that $ A^{m}\\equal{}I_{n}$", - "Solution_1": "I think we can prove a sharper result than the one you asked for Moubinool.\r\n\r\nWe can show that $ A$ is a signed permutation matrix ([i]i.e.[/i] there is exactly one nonzero entry in each row and each column, that entry being $ \\pm 1$) then your result follows.\r\n\r\nWe have, $ \\forall j \\equal{} 1,\\ldots ,n$, $ \\max_{i \\equal{} 1,\\ldots ,n}|a_{i,j}| \\equal{} 1$.\r\nStart with $ j \\equal{} 1$, there is $ i_{0}$ s.t.$ a_{i_{0},1}\\equal{}\\pm 1$. Now suppose that there is $ j > 1$ s.t. $ a_{i_{0},j}\\not \\equal{} 0$, and consider $ X \\equal{} (a_{i_{0},1}, 0,\\ldots ,0, 1/a_{i_{0},j}, 0,\\ldots , 0)^{T}$ if $ |a_{i_{0},j}|\\geqslant 1$ and $ X \\equal{} (a_{i_{0},1}, 0,\\ldots ,0, a_{i_{0},j}, 0,\\ldots , 0)^{T}$ otherwise. Both cases contradict the asumption that $ f(AX) \\equal{} f(X)$. So row $ i_{0}$ of $ A$ is $ (\\pm 1, 0,\\ldots , 0)$.\r\nContinue the same reasoning for $ j \\equal{} 2,\\ldots,n$.\r\nHence $ A$ is a signed permutation matrix.\r\nFinally there exist a positive integer $ m$ such that $ A^{m}\\equal{} I_{n}$ ($ m\\leqslant 2n!$).", - "Solution_2": "and now, what if $ f(X) \\equal{} (\\sum_{k\\equal{}1}^{n}x_{k}^{p})^{\\frac{1}{p}}$, where $ 1\\leq p\\leq\\infty$ ?" -} -{ - "Problem": "A thin ring is of mass $M$ and radius $R$. \r\nFind the moment of inertia of this ring about its diameter.", - "Solution_1": "My calculation gives me the answer that $\\large I=\\frac{MR^2}{2}$", - "Solution_2": "Prove it !!!", - "Solution_3": "$dM=R\\lambda d\\theta$\r\n\r\n$\\large dI=dMr^2=\\lambda R^3\\cos^2\\theta d\\theta$\r\n\r\n$\\large I=2\\int_{0}^{\\pi}\\lambda R^3\\cos^2\\theta d\\theta=\\pi R^3\\lambda$\r\n\r\nBut $\\large M=2\\pi\\lambda R$\r\n\r\n$\\Longrightarrow I=\\frac{MR^2}{2}$" -} -{ - "Problem": "Hey ppl,\r\n how are all of you? Ppl like varunarasu n all, long time no talk? Ppl, jus get back n do some ob on AOPS as well da..", - "Solution_1": "that wud b best 4 all :rotfl: :oops:", - "Solution_2": "i am waiting for it...........", - "Solution_3": "in fact i d love it if ppl actually spare a few mins from their bz 24x7 studying time and do post here :P\r\nit can be a grand reunion :wink:", - "Solution_4": "yes..but do tell the meganerds :rotfl: :P", - "Solution_5": "Really!!! Grand reunion...", - "Solution_6": "y alternate ID ,Soumya?? :P :rotfl:", - "Solution_7": "forgot password", - "Solution_8": ":rotfl: :rotfl: :P", - "Solution_9": "so i have company :!: (at last)\r\nawesome ,Soumyashant you of all people forgot? :rotfl: :rotfl:", - "Solution_10": "nice 2 see you here soumya :) \r\n\r\n@gaurav:y only soumya could give you company :rotfl: :rotfl:", - "Solution_11": "i tink there wasnt ne girl at inmotc last yr.. :rotfl: :D :P", - "Solution_12": "I was also thinking on same lines. :rotfl: \r\n\r\nRitu and soumya(both girlish names and both brilliant ones) :omighty:", - "Solution_13": "[quote=\"gauravpatil\"]\nawesome ,Soumyashant you of all people forgot? [/quote]\n\nWell its human to forget[hide]but till now i was thinking soum is a beast :rotfl: [/hide]\n\n\n[quote=\"skand\"]Ritu and soumya(both girlish names and both brilliant ones[/quote]\r\n\r\nThats like underestimating SOUMYASHANT NAYAK :o", - "Solution_14": "[quote=\"rituraj007\"]i tink there wasnt ne girl at inmotc last yr.. :rotfl: :D :P[/quote]\r\nThat's because poor Palak was mistaken for a girl :). Curse all uni-sex names...", - "Solution_15": "i said there wasnt ne girl :P" -} -{ - "Problem": "I wonder what Americans call for AM-GM inequality, because I have got a problem about finding minimum for sum of areas which I used AM-GM inequality. However, I was received this statement from my professor \"I really don't find your answer\" :mad: :mad:", - "Solution_1": "Anyone who is familiar with contest math, in the US or otherwise, would know the AM-GM inequality. That's a pretty standard abbreviation, although you could consider writing out the words it stands for.\r\nCollege professors are not all familiar with contest math. You might want to find a reference and provide it to your professor.", - "Solution_2": "Also, the professor may not be the one grading problem sets. They hire undergrads and grads to do such menial labor for them.", - "Solution_3": "Very true; this applies even in graduate classes. I've been burned a few times myself.", - "Solution_4": "It never hurts to show a few more steps and to write out in full words any special terms you use. Good mathematical writing is communicating, so it is important to be clear, not just brief.", - "Solution_5": "As anyone who has taken a course with a truly bad marker knows, there is a big difference between 'good mathematical writing', and 'writing that will be understood by a marker with either a poor grasp of the english language or an insufficient grasp of mathematics to reasonably be a marker'." -} -{ - "Problem": "Prove for all $a_1,a_2,..,a_n>0$ then:\r\n$\\frac{a_1^n}{a_2}+\\frac{a_2^n}{a_3}+...+\\frac{a_{n-1}^n}{a_n}+\\frac{a_n^n}{a_1} \\ge n(\\frac{a_1^n+a_2^n+..+a_n^n}{n})^{\\frac{n-1}{n}}$", - "Solution_1": "Using convexity of $x\\rightarrow x^{\\frac{n-1}{n}}$ we have:\r\n\\[a_1^{n-1}+...+a_n^{n-1}\\geq n \\Bigg( \\frac{a_1^n+...+a_n^n}{n} \\Bigg)^{\\frac{n-1}{n}}\\]\r\nWe need to prove\r\n\\[\\frac{a_1^n}{a_2}+...+\\frac{a_n^n}{a_1}\\geq a_1^{n-1}+...+a_n^{n-1} \\Leftrightarrow a_1^{n-1}\\Big(\\frac{a_1}{a_2}-1\\Big)+...+a_n^{n-1}\\Big(\\frac{a_n}{a_1}-1\\Big)\\geq 0\\]\r\nI don't succeed to continue ...", - "Solution_2": "[quote=\"Rodolphe2005\"]Using convexity of $x\\rightarrow x^{\\frac{n-1}{n}}$ we have:\n\\[a_1^{n-1}+...+a_n^{n-1}\\geq n \\Bigg( \\frac{a_1^n+...+a_n^n}{n} \\Bigg)^{\\frac{n-1}{n}}\\]\nWe need to prove\n\\[\\frac{a_1^n}{a_2}+...+\\frac{a_n^n}{a_1}\\geq a_1^{n-1}+...+a_n^{n-1} \\Leftrightarrow a_1^{n-1}\\Big(\\frac{a_1}{a_2}-1\\Big)+...+a_n^{n-1}\\Big(\\frac{a_n}{a_1}-1\\Big)\\geq 0\\]\nI don't succeed to continue ...[/quote].\r\n\r\nrearrengement with ${a_i^n}$ and ${\\frac{1}{a_i}}$.. it's clear that the worst ordering is $a_1^n\\frac{1}{a_1}+\\dots +a_n^n\\frac{1}{a_n} = \\sum {a_i^{n-1}}$ and the claim follows...", - "Solution_3": "[quote=\"Rodolphe2005\"]Using convexity of $x\\rightarrow x^{\\frac{n-1}{n}}$ we have:\n\\[a_1^{n-1}+...+a_n^{n-1}\\geq n \\Bigg( \\frac{a_1^n+...+a_n^n}{n} \\Bigg)^{\\frac{n-1}{n}}\\]\nWe need to prove\n\\[\\frac{a_1^n}{a_2}+...+\\frac{a_n^n}{a_1}\\geq a_1^{n-1}+...+a_n^{n-1} \\Leftrightarrow a_1^{n-1}\\Big(\\frac{a_1}{a_2}-1\\Big)+...+a_n^{n-1}\\Big(\\frac{a_n}{a_1}-1\\Big)\\geq 0\\]\nI don't succeed to continue ...[/quote]\r\n\r\nhi, i wonder why your claim is true.\r\nBy the famous power mean inequality you have the reverse sign of your inequality.", - "Solution_4": "[quote=\"Rodolphe2005\"]Using convexity of $x\\rightarrow x^{\\frac{n-1}{n}}$ we have:\n\\[a_1^{n-1}+...+a_n^{n-1}\\geq n \\Bigg( \\frac{a_1^n+...+a_n^n}{n} \\Bigg)^{\\frac{n-1}{n}}\\]\n[/quote]\r\n\r\nBut I think \r\n$(\\frac{a_1^{n-1}+a_2^{n-1}+...+a_n^{n-1}}{n})^n \\le (\\frac{a_1^n+a_2^n+...+a_n^n}{n})^{n-1}$", - "Solution_5": "We can write the statement as follows:\r\n\r\n$x_1+...+x_n=n$ implies $x_1*x_2^{-1/n}+...+x_n*x_1^{-1/n} \\ge n$.\r\n\r\nBut, for $n=10$, $x_1=...=x_5=1/2$ and $x_6=...=x_{10}=3/2$, it is false.", - "Solution_6": "$(1/2)^{0.9}=0.535887$\r\n\r\n$4*0.535887=2.143548$\r\n\r\n$(3/2)^{0.9}=1,4404$\r\n\r\n\r\n$4*1,4404=5,7616$\r\n\r\n\r\n$(0.5)/(1.5)^{0.1}=0.5/1.04138=0,48$\r\n\r\nSo finally we have\r\n\r\n$8.38 \\geq 10$\r\n\r\nand the ineq is not true :(", - "Solution_7": "Thanks alot ,Manlio and Vasc!\r\nI'm realize my solution is wrong in general case. :blush: \r\n\r\nI have proof in case n=3\r\nAnd for $n \\ge 4$\r\n$\\sum\\frac{a_{i}^n}{a_{i+1}} \\ge n\\sqrt[n]{\\frac{4}{n}}(\\frac{a_1^n+a_2^n+...+a_n^n}{n})^{\\frac{n-1}{n}}$\r\n\r\nAnd\r\n$\\sum\\frac{a_{i}^k}{a_{i+1}} \\ge n\\sqrt[k]{\\frac{4}{n}}(\\frac{a_1^k+a_2^k+...+a_n^k}{n})^{\\frac{k-1}{k}}$\r\nstill satisfy for all k integer.", - "Solution_8": "Great problem this last one! Taking $ a_1^k+...+a_n^k=1$ we need to prove that $\\sum {\\frac{a_i^k}{a_{i+1}}\\geq 4^{\\frac{1}{k}}}$. Using Cauchy it suffices to prove that $\\sum{x_i\\cdot\\sqrt[k]{4x_{i+1}}\\leq 1}$ if $\\sum {x_i}= 1$. This is true since $\\sqrt[k]{4x_{i+1}}\\leq\\frac{4x_{i+1}+k-1}{k}$ and $ 4\\sum {x_i\\cdot\\ x_{i+1}}\\leq 1$.", - "Solution_9": "Very clearly Solution,Harazi.\r\n :) :) :) :) :)", - "Solution_10": "Here is my solution:\r\n$S=\\sum\\frac{a_i^k}{a_{i+1}}=\\sum\\frac{a_i^{i+1}}{a_{i}a_{i+1}}$\r\n$S^k.(a_1^{k}a_2^k+a_2^{k}a_3^k+...+a_{n-1}^{k}a_n^n+a_n^{k}a_1^{k})\\ge (a_1^k+a_2^k+...+a_n^k)^{k+1}$\r\nBut $4\\sum{b_{i}b_{i+1}}\\le (\\sum{b_i})^2$,It's wellknown.\r\nThen We have result." -} -{ - "Problem": "\"Bella's Lullaby\" from Twilight\r\n\r\n[youtube]4PZb0Ecl1Sw[/youtube]", - "Solution_1": "Nice.. I like this one \r\n\r\nhttp://www.youtube.com/watch?v=dZn_VBgkPNY\r\n\r\n\r\n\r\nThis is such a enchanting piece!", - "Solution_2": "I love the Bernstein conducting Tchaikovsky Sym 5", - "Solution_3": "I also like \"It ends tonight\" by All-American Rejects.", - "Solution_4": "[youtube]lTSr2oz15Xk[/youtube]", - "Solution_5": "[youtube]VK2uTtuI84w[/youtube]", - "Solution_6": "Pirates of the Caribbean : The Black Pearl Played on Piano\r\n\r\n[youtube]itK6FvWde1c[/youtube]", - "Solution_7": "[youtube]Hg8Fa_EUQqY[/youtube]", - "Solution_8": "Why are people posting blank posts?", - "Solution_9": "You must enable BBCode.\r\n\r\nEither that, or your browser just doesn't support youtube embed/your filter blocked youtube.", - "Solution_10": "[color=orange][b]Toccata & Fugue[/b][/color] is one of my favourites, too", - "Solution_11": "I like Moonlight Sonata!", - "Solution_12": "[youtube]ogJFXqYEYd8[/youtube]", - "Solution_13": "If it includes non-classical music, then probably Crazier by Taylor Swift...\r\nhttp://www.youtube.com/watch?v=KfdxeymNwjU&feature=PlayList&p=27886EC594CA51FD&index=5&playnext=6&playnext_from=PL\r\n(sorry it's so long; it's in my playlist) :)", - "Solution_14": "Jonathan Rhys Meyers-Something Inside\r\n\r\nhttp://www.youtube.com/watch?v=LOO2N1mwD8o", - "Solution_15": "[youtube]rEs1wtsw_IA[/youtube]", - "Solution_16": "I like :\r\n\r\nAnimusic\r\n\r\n1. [url]http://www.youtube.com/watch?v=3ahoqR6OGdM[/url] -Future Retro\r\n2. [url]http://www.youtube.com/watch?v=Cgt4DEBQy50&feature=related[/url] -Aqua Harp\r\n3. [url]http://www.youtube.com/watch?v=OROAGNbsikQ&feature=related[/url] -Acoustic Curves\r\n\r\nRuneScape\r\n\r\nI like songs from RuneScape, mainly:\r\n\r\nHarmony, YesterYear, Scape Main, Scape Summon, Start, Doorways, and The Mentor.\r\n\r\nYou can search on Youtube.\r\nSome good channels are Bigstingman, RuneScapeSoundtrack, and RunescapeOST.", - "Solution_17": "I like the Fantasie Impromptu played by Yundi Li. I also like the La Campella (spelling error) also played by Yundi Li.", - "Solution_18": "[youtube]mVq-MU7ojVY[/youtube]" -} -{ - "Problem": "Matt and Claire hiked to a point $\\frac{3}{4}$ the way up the moutain in 43 minutes and 12 seconds. They took the same amount of time to hike each half of the distance from bottom to top. it took them 1 minute and 36 seconds longer to hike the third quarter than the fourth. how many minutges did it take them to hike the entire mountain from bottom to top? express as decimal to nearest tenth.", - "Solution_1": "it takes them 43 min and 12 sec,\r\nwhich is 2592 sec.\r\nsupposing they hiked for equal time for each quater of the \r\nmountain, they would finish the hike in 4x (x standing for the\r\ntime it takes to climb one quarter).\r\nhowever, the 3rd quarter took 96 sec longer, and the first\r\n3 quarters took 2592 sec, you get the equation\r\n3x + 96 = 2592\r\nsolving it gives x= 832\r\nso 4x=3328sec\r\nwhich in min and sec is 55 min and 28 sec", - "Solution_2": "[quote=\"espark52\"]how many minutges did it take them to hike the entire mountain from bottom to top? express as decimal to nearest tenth.[/quote]\n[quote=\"davidlizeng\"]which in min and sec is 55 min and 28 sec[/quote]\r\n[hide]Then, it would be $55 \\frac{28}{60}$ which is $55.4\\bar{6}$ so then the final answer is $55.5$ minutes[/hide]", - "Solution_3": "ooops....\r\nforgot to put ans. into decimal form :D \r\nthanks for the correction quevvy", - "Solution_4": "[hide]Let $x$ be the amount of time they took to climb the third quarter.\nThe first half is $\\frac{216}{5}-x$. And the second is $2x-\\frac{8}{5}$. \n$\\frac{216}{5}-x=2x-\\frac{8}{5}$\nSolve.... and $x=\\frac{224}{15}$\nAnd for the second half they went $\\frac{448}{15}-\\frac{8}{5}=\\frac{424}{15}$.\nBut the problem wants from bottom to top, so multiply $2$, and the answer is $\\frac{848}{15}\\approx56.5$[/hide]", - "Solution_5": "davidli, where did you get all those numbers?\r\nI think davidlizeng explains it the best", - "Solution_6": "I don't think davidlizeng's solution is correct. The third quater is not equal to 96 seconds more than the first quarter. The problem said that the third quater took 96 seconds longer than the fourth quarter.", - "Solution_7": "good point davidli\r\nthen the equation would be\r\n3x +48=2592\r\nso x=848\r\nand in min, the ans. is 56.5\r\nthanks for the correction\r\nstupid me........", - "Solution_8": "[hide=\"probably wrong because I'm dumb\"]\nLet's say it took x minutes to hike the 4th quarter. Then x + (x + 1.6), the time it took to hike the 3rd and 4th quarter, is equal to the time it hiked the first half. The first half is 3/4 minus the 3rd quarter, so it's 43.2 - (x + 1.6), and this is equal to 2x + 1.6 as mentioned above. Now we have 41.6 - x = 2x + 1.6, so 3x = 40. Plug that into (43.2 - (x + 1.6)) and multiply by 2 to get 56.533333... which rounds to 56.5[/hide]" -} -{ - "Problem": "The positive integrer number $n$ has $1994$ digits. $14$ of its digits are $0$'s and the number of times that the other digits: $1, 2, 3, 4, 5, 6, 7, 8, 9$ appear are in proportion $1: 2: 3: 4: 5: 6: 7: 8: 9$, respectively. Prove that $n$ is not a perfect square.", - "Solution_1": "0 -> 14\r\n1 -> 44\r\n2 -> 88\r\n3 -> 132\r\n4 -> 176\r\n5 -> 220\r\n6 -> 264\r\n7 -> 308\r\n8 -> 352\r\n9 -> 396\r\n\r\nThe number of times each digit apears in $n$ is listed above.\r\nTake $n$ mod.$9$ . We get $n\\equiv 1*44+2*88+3*132+...+9*396\\equiv 3$, Which is not a square." -} -{ - "Problem": "Let r(n) be the number of ways n can be written as the sum of two squares of integers that need not be positive, and order is important. For example, r(2)=4 because\r\n\r\n2=\r\n1^2+1^2=\r\n1^2+(-1)^2=\r\n(-1)^2+1^2=\r\n(-1)^2+(-1)^2.\r\n\r\nEvaluate\r\n\r\nsum from n=1 to infinity (-1)^n*r(n)/n.", - "Solution_1": "The solution to this problem can be found on the alumni forum at the Mathcounts website, http://www.mathcounts.org. The idea behind the proof has to do with a graphical representation of what it means for a number to be the sum of squares of other numbers. The answer is [hide]pi*ln(2)[/hide], I believe. As I said, the mathcounts website contains a full explanation.", - "Solution_2": "The explicit answer to this question can be found in an exercise in A Course in Number Theory by H.E. Rose (I highly recommend this book to you Zeta), on page 120, exercise 2.\r\n\r\nThe simplest way to describe it is to take the prime factorization of n and write it as 2^a*(p_1^h_1...p_n^h_n)*(q_1^k_1...q_m^k_m), where the p_i's are primes 1 mod 4 and the q_i's are primes 3 mod 4. Then if any of the k_i's are odd, the number of representations of n as sums of two squares is 0. Otherwise, the number of representations is 4*(h_1 + 1)(h_2 + 1)...(h_n + 1).\r\n\r\nThe sum that you give looks very similar to the average number of representations by sums of two squares (except for the (-1)^n sign). This is known as Gauss's Circle Problem. A discussion about it can be found in just about any good intermediate Number Theory book. A particularly nice and simple proof of it can be found in G.E. Andrews book Number Theory. A more advanced book that covers geometric number theory in detail is Geometric and Analytic Number Theory by Hlawka, et al. I would just recommend the Andrews book though.", - "Solution_3": "Here's another description of the number of ways to represent some number as the sum of two squares taken from Davenport's book, The Higher Arithmatic:\r\n\r\n\"For two squares the rule (due to Legendre) is as follows. Count the number of divisors of n of the form 1 mod 4 and those of the form 3 mod 4. If these numbers are D_1 and D_3 respectively, then the number of representations is 4*(D_1 - D_3).\"\r\n\r\nHe then goes on to give a description of the number of ways to write a number as the sum of 4 squares. He does these without proofs though." -} -{ - "Problem": "What is the constant term in the Laurent polynomial\r\n\\[ \\prod_{1\\le i\\neq j\\le n}\\left(1 \\minus{} \\frac {x_i}{x_j}\\right)^{k_i}\\]\r\nwhere $ k_i$'s are given naturals.", - "Solution_1": "Please continue the discussion in [url=http://www.mathlinks.ro/viewtopic.php?t=76800]Constant term of an expression[/url]" -} -{ - "Problem": "This problem is inspired by AP Government :P.\r\n\r\nSuppose you have a contiguous region $R$ and a subregion $S$ (not necessarily contiguous) such $\\text{area}(S) > \\frac{1}{2}\\cdot \\text{area}(R)$. Can we always partition $R$ into contiguous subregions $P_{1}, P_{2}, \\ldots, P_{n}$ such that\r\n\r\n$\\text{area}(P_{i}\\cap S) > \\frac{1}{2}\\cdot \\text{area}(P_{i})$\r\n\r\nfor all $i$? Basically, can we divide $R$ such that $S$ contains more than half of the area in each division. This would make the process of redistricting very effective for the majority party :wink:.", - "Solution_1": "I'm finding it hard to solve this with any rigor :rotfl:", - "Solution_2": "[quote=\"t0rajir0u\"]I'm finding it hard to solve this with any rigor :rotfl:[/quote]\r\n\r\nSame haha. I think it's true though; at least I can't find any counterexamples...", - "Solution_3": "By \"contiguous,\" I assume you really mean \"connected.\" Also, you really don't have very many constraints ... perhaps you want to require all the $P_{i}$ to have equal areas? In that case, I can generate counter-examples.\r\n\r\nAlthough, on further thought, another reasonable constraint would be that $R$ and the $P_{i}$ be open, in which case the counter-example I imagined wouldn't work.", - "Solution_4": "[quote=\"JBL\"] perhaps you want to require all the $P_{i}$ to have equal areas? In that case, I can generate counter-examples.[/quote]\r\n\r\nHmm. I think this would be more faithful to the motivation for the problem. What's the counterexample?", - "Solution_5": "[quote=\"JBL\"]By \"contiguous,\" I assume you really mean \"connected.\" Also, you really don't have very many constraints ... perhaps you want to require all the $P_{i}$ to have equal areas? In that case, I can generate counter-examples.\n\nAlthough, on further thought, another reasonable constraint would be that $R$ and the $P_{i}$ be open, in which case the counter-example I imagined wouldn't work.[/quote]\r\n\r\nPerhaps, I haven't thought out the problem very clearly, but it would be interesting to see counterexamples for any case that you have one for.", - "Solution_6": "The counter-example is an \"asymmetric bow-tie.\" The top triangle has area between $\\frac{1}{2n}$th and $\\frac{1}{n}$th of the whole thing. The yellow region is $S$.\r\n\r\nAny connected set which contains points from the top half and points from the bottom half must contain the \"knot\" of the bow-tie, the single point in the middle. Thus, we can say about each half of the tie that it must be subdivided into several regions, each with area $\\frac{1}{n}$, except that they are each entitled to one region of area at most $\\frac{1}{n}$. Since the total area of the top half is at most $\\frac{1}{n}$, it must fall entirely within one of the $P_{i}$. But since its area is at least $\\frac{1}{2n}$ and the area of the $P_{i}$ is $\\frac{1}{n}$, this $P_{i}$ shares at most half of its area with $S$.\r\n\r\nAdding openness voids this counter-example, since it's not possible to create a bottleneck in the way I just did. \r\n\r\nIf we don't require the $P_{i}$ to be of equal area but we do require the boundary between $S$ and $R \\setminus S$ to be at least in part a smooth curve, we can always find a solution by taking all but one of the $P_{i}$ to be extremely tiny disks along the boundary and the final $P_{n}$ to be all of $R$ except these disks.", - "Solution_7": "[quote=\"JBL\"]The counter-example is an \"asymmetric bow-tie.\" The top triangle has area between $\\frac{1}{2n}$th and $\\frac{1}{n}$th of the whole thing. The yellow region is $S$.\n\nAny connected set which contains points from the top half and points from the bottom half must contain the \"knot\" of the bow-tie, the single point in the middle. Thus, we can say about each half of the tie that it must be subdivided into several regions, each with area $\\frac{1}{n}$, except that they are each entitled to one region of area at most $\\frac{1}{n}$. Since the total area of the top half is at most $\\frac{1}{n}$, it must fall entirely within one of the $P_{i}$. But since its area is at least $\\frac{1}{2n}$ and the area of the $P_{i}$ is $\\frac{1}{n}$, this $P_{i}$ shares at most half of its area with $S$.\n\nAdding openness voids this counter-example, since it's not possible to create a bottleneck in the way I just did. \n\nIf we don't require the $P_{i}$ to be of equal area but we do require the boundary between $S$ and $R \\setminus S$ to be at least in part a smooth curve, we can always find a solution by taking all but one of the $P_{i}$ to be extremely tiny disks along the boundary and the final $P_{n}$ to be all of $R$ except these disks.[/quote]\r\n\r\nThat's tricky. What if we force $R$ to be convex? Is it still possible to construct a counterexample when the area of the $P_{i}$ are all equal?" -} -{ - "Problem": "find all natural numbers $n$ such that $2^{n}+1$is the power of natural number.", - "Solution_1": "The only solution is 2^3+1=3^2. Assume 2^n+1=x^a where a>1. Then 2^n=(x-1)(x^(a-1)+x^(a-2)+...+x+1) Then x=2^k+1. As a result a must be even for otherwise the second expression would be a sum of an odd number of odd numbers which can't be a power of two. Set a=2b. Then one has 2^n=(x^b-1)(x^b+1). This results in n=3, x=3, a=2 as the only solution for a>1." -} -{ - "Problem": "A sculpture for a math classroom is a figure composed of a 1\" cube stacked upon a 3\" cube which in turn is stacked upon a 5\" cube. The entire sculpture is to be painted. Find the number of square inches of the surface that will be painted.", - "Solution_1": "Quote:A sculpture for a math classroom is a figure composed of a 1\" cube stacked upon a 3\" cube\nwhich in turn is stacked upon a 5\" cube.\nThe entire (which includes the bottom :P ) sculpture is to be painted.\n Find the number of square inches of the surface that will be painted.[hide]\n\nAssuming that the bottom surface will also be painted . . .\n\n\n\nThe 5\" cube has six faces, each with 25 squares inches of surface.\n\nIt has 6 x 25 = 150, minus the 3x3 = 9 covered by the 3\" cube,\n\nor 141 squares inches.\n\n\n\nThe 3\" cube will have five faces exposed, each with 9 square inches of surface.\n\nIt has 5 x 9 = 45, minus the 1x1 = 1 covered by the 1\" cube,\n\nor 44 squares inches.\n\n\n\nThe 1\" cube will have face faces exposed, each with 1 square inch of surface.\n\nSo it has 5 square inches.\n\n\n\nThe total surface area to be painted is: 141 + 44 + 5 = 190 square inches.\n\n\n\n[If the bottom surface is not to be painted, the answer is 165 square inches.][/hide]", - "Solution_2": "A sculpture for a math classroom is a figure composed of a 1\" cube stacked upon a 3\" cube which in turn is stacked upon a 5\" cube. The entire sculpture is to be painted. Find the number of square inches of the surface that will be painted.\r\n\r\n(6*5^2-3^2)+(5*3^2-1)+(5)=141+44+5\r\n =190", - "Solution_3": "[hide]Let's first find the surface area of all the cubes before we subtract. The surface areas are 6,54, and 150. That adds up to 210. Because the 1\" cube touches one squared inch of the 3\" cube, one inch is subtracted from both cubes. The 3\" cube touches nine squared inches on the 5\" cube so nine square inches are subtracted from both cubes. Therefore 9+9+1+1=20 and 210-20=190sq. in.[/hide]" -} -{ - "Problem": "Give example of noetherian closed ring which is not factorial.", - "Solution_1": "What is a closed ring?\r\n\r\nThe ring $\\mathbb Z[x, y]/ (x^2 = y^3)$ is an exaple of a Noetherian ring, which is not factorial.\r\nThe ring is Noetherian by Hilbert's basis theorem. It is not factorial, since the element $x^2 = y^3$ has two non-equivalent factorizations: $x\\cdot x$ and $y \\cdot y \\cdot y$." -} -{ - "Problem": "For every positive integer $ n$, we define $ n?$ as $ 1?\\equal{}1$ and $ n?\\equal{}\\frac{n}{(n\\minus{}1)?}$ for $ n \\ge 2$.\r\n\r\nProve that $ \\sqrt{1992}<1992?<\\frac{4}{3} \\sqrt{1992}.$", - "Solution_1": "[quote=\"moldovan\"]For every positive integer $ n$, we define $ n?$ as $ 1? \\equal{} 1$ and $ n? \\equal{} \\frac {n}{(n \\minus{} 1)?}$ for $ n \\ge 2$.\n\nProve that $ \\sqrt {1992} < 1992? < \\frac {4}{3} \\sqrt {1992}.$[/quote]\r\n\r\nWe have $ \\frac{n?}{(n\\minus{}1)?}\\equal{}\\frac{n (n\\minus{}2)?}{(n\\minus{}1)?(n\\minus{}1)}$\r\n$ \\Rightarrow n?\\equal{}\\frac{n}{n\\minus{}1}(n\\minus{}2)?$\r\n\r\nAlso $ 2?\\equal{}2$, so $ 1992?\\equal{}2 \\cdot \\frac{4}{3} \\cdot \\frac{6}{5} \\cdot ... \\cdot \\frac{1992}{1991}$\r\n\r\n\r\n$ (1992?)^2 > 2^2 \\cdot \\frac{4}{3} \\cdot \\frac{5}{4} \\cdot \\frac{6}{5} \\cdot ... \\cdot \\frac{1993}{1992}$\r\n$ \\equal{}\\frac{4(1993)}{3} > 1992$\r\n\r\n$ (1992?)^2 < (2^2 \\cdot \\frac{4}{3}) \\cdot (\\frac{4}{3} \\cdot \\frac{5}{4} \\cdot \\frac{6}{5} \\cdot ... \\cdot \\frac{1992}{1991})\\equal{}\\frac{16}{9}(1992)$, done." -} -{ - "Problem": "The numbers 1 to 50 are arranged in an arbitrary manner into 5 rows of 10 numbers each. Then each row is rearranged so that it is in increasing order. Then each column is arranged so that it is in increasing order. Are the rows necessarily still in increasing order?", - "Solution_1": "[hide]Haven't found a counterexample yet [/hide]bu[hide]t I'[/hide]m[hide] still looking for a [/hide]p[hide]roof.[/hide]", - "Solution_2": "That was some interesting spoilering you have there.\r\n\r\nI would say assume that there is a possible ending position in which a row is out of order and then work backwards -- you should either get a situation that gives you a contradiction or a situation that works.", - "Solution_3": "If you want to know which direction to pursue, the answer to the original question is [hide]yes[/hide]. This fact is the basis for some sorting algorithms.", - "Solution_4": "no.\n\n\n\n\n\n\n\n\n\n[hide]Do i need to add a solution?[/hide]", - "Solution_5": "A solution is always required.", - "Solution_6": "Here's a hint: [hide]First prove it when the entries are only 0's and 1's.[/hide]", - "Solution_7": "Is this correct?\n\n[hide]I will refer to terms in the form of a_(r,c) where c is the column of the term and r is the row of the term.\n\n\n\nConsider the grid after the rows have been rearranged but before the columns have been.\n\nWe know that for every term of the form a_(r,c) there is another term, namely a_(r,c+1), such that a_(r,c)0; a,b--> + infinity then: \r\n1/(a+b+c)+1/(b+c+d)+1/(c+d+a)+1/(a+d+b) -->0" -} -{ - "Problem": "Find an expression for 3/5 as a finite sum of distinct reciprocals of\r\npositive integers. (For example: 2/7 = 1/7+1/8+1/56.)\r\n\r\nProve that any positive rational number can be so expressed.", - "Solution_1": "Follow these steps to get your proof:\r\n\r\na) Prove that if $ 1/(n\\plus{}1)