[b]A hexagon is inscribed in a circle of radius $ r$. Find $ r$ if two sides of the hexagon are 7 units long, while the other four sides are 20 units long.[/b]

[quote="buzzer11"][hide]
Draw it out. The two length 7 sides are across from each other.

Call the radius r. drop the height from the length 7 side down to the diameter of the circle formed by the 4 length 20 sides (the one parallel to the length 7 side) and call that h.

This forms 2 right triangles, and 2 variables, r and h. solve for r, and you get $ \boxed{16}$.[/hide][/quote]\

What do you mean by 2 right triangles.  Your solution looks very easy.  I on the otherhand got the same answer as you but I had to use Ptolemy's theorm which is above mathcounts, so can you please explain the two equations you got?

Mine used ptolemys and got a 4th degree polynomial in terms of r., but you could luckily factor it.

[quote="buzzer11"][hide]
Draw it out. The two length 7 sides are across from each other.

Call the radius r. drop the height from the length 7 side down to the diameter of the circle formed by the 4 length 20 sides (the one parallel to the length 7 side) and call that h.

This forms 2 right triangles, and 2 variables, r and h. solve for r, and you get $ \boxed{16}$.[/hide][/quote]
While I know by common sense that the 7's must be across from each other, but how do you know for sure?

actually, you dont know for sure.
now that i look at it, the solutions manual actually has the two sides adjacent to each other.  :maybe:

my solution. it'll be hard to explain (i need to learn asymptote) :)

So i put the two length 7 sides across from each other... and then i cut the hexagon in half with the two 7 sides still across from each other. then two trapizoids are formed, with side lengths 20, 7, 20.

Look at just one trapizoid, drop the height in two places, one from the vertex and one splitting the 7 side in half, draw the line of length r from the bottom of the half-height to the top of the vertex height.

now you have two right triangles. and pythagoreas time. argh thats hard to explain.

Now that I draw the figure closer to scale, it doesn't look cyclic at all!  :o ,
However, the solution manual's figure also looks really wacky. It would really help to know how to see if a polygon is cyclic or not.

[hide="Go Microsoft Paint!"][img]10047[/img][/hide]
Nice solution, buzzer11.
[hide="equations"]
$ (7/2)^{2}+h^{2}=r^{2}$
$ (r-7/2)^{2}+h^{2}=400$
Subtract the second equation from the first.
$ 7r-r^{2}=r^{2}-400$
Factor:
$ 0=(r-16)(2r+25)$
Since $ r$ is positive, $ r=\boxed{16}$.[/hide]

[quote="erdogankerem123"][quote="buzzer11"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=160738]I asked this recently[/url][/quote]
Yes,but that's only for quadrilaterals, though.[/quote]

Oh. oops.

Hm the interior angles of the polygon will always add up to 180... and the exterior angles will always add up to 360...

but im not sure of certain tests that ensure if any polygon is cyclic.

[quote="mihail911"]yeah...in the solutions manual, i was kinda confused on how they went from the two sides with 7 being adjacent to eachother to them being directly across from eachother...something about symmetry.. :huh:[/quote]
I don't posess the manual, but what I think they were saying was:
Take the polygon with the 7s adjacent.  Then split the polygon down the middle into 2 halves.  Flip one half.  Then you get the polygon with the 7s opposite.  Everything else remains the same.

what im confused about, though, archimedes, is are you allowed to just divide the polygon symmetrically and flip it over...?? i mean, it seems kinda weird to me...bear with me, im new to this stuff.... :oops:

Well, making the cut and  flip  doesn't change $ r$, so we can do it and get the same answer.  (Do you see why it doesn't change $ r$?)

yes..i do, but making that cut and flip means the two new resulting polygons are not congruent...which i think they must be? :maybe:

Even though they're not congruent, they both fit the problem description and they both have the same $ r$.


algebra
polynomial
superior algebra
superior algebra unsolved

What is the number of isomorphism class of order n groups?
And Why?

from my best knowledge, there is no formula to find the number of groups of order n

This is not known in general.  The hard cases occur when $ n$ is a power of a prime; explicit formulas are known for $ n \equal{} p^k, k \equal{} 1, 2, ... 7$ but past that there appears to be no hope of explicitly understanding the classification.

If $ n \equal{} p_1^{k_1} p_2^{k_2} ... p_n^{k_n}$, could it be presented by $ f(k_1,k_2,...,k_n)$? Since the abel one is like that...
How much have been done in this problem? Or is it proved impossible to find out?

[quote="jackhui"]If $ n \equal{} p_1^{k_1} p_2^{k_2} ... p_n^{k_n}$, could it be presented by $ f(k_1,k_2,...,k_n)$? Since the abel one is like that...[/quote]

No, there are a different number of isomorphism classes of groups of order 2^4 and 3^4.  It is conjectured that the number of groups of order p^n for fixed n is given by finitely many polynomials fi(p) where the number is fi(p) if i = p mod M for some fixed M.  Various versions of this have been proved.

[quote="jackhui"]How much have been done in this problem? Or is it proved impossible to find out?[/quote]

For numbers other than prime powers, people have been more interested in estimates.  You might check out:

Blackburn, Simon R.; Neumann, Peter M.; Venkataraman, Geetha. "Enumeration of finite groups."
Cambridge University Press, 2007. ISBN: 978-0-521-88217-0
http://www.ams.org/mathscinet-getitem?mr=2382539


Let $ABCDEF$ be a convex hexagon and denote by $A',B',C',D',E',F'$ the middle points of the sides $AB$, $BC$, $CD$, $DE$, $EF$ and $FA$ respectively. Given are the areas of the triangles $ABC'$, $BCD'$, $CDE'$, $DEF'$, $EFA'$ and $FAB'$. Find the area of the hexagon.

[i]Kvant Magazine[/i]

Noting that $[ABC'] = \frac{[ABC] + [ABD]}{2}$, adding up all the areas should give you $\frac{3}{2}$ the total area of the hexagon...


advanced fields
advanced fields unsolved

find neccessary and sufficient condition for a graph to be HAMILTONIAN?

This is most certainly an open question. It's completely intractable.

please tell me where i have to post this problem....

You can post it anywhere you like and get just the same answer. Maybe for graphs with particular restraints it is possible to find necessary and sufficient conditions. But you can't expect people to give you a full discourse on the literature in this field here. Do some research!


Sometimes when writing a post there is a automatic spell check, but not always, why?
How do I get it? Does it have to do with the web browser i use?

Spell checkers are embedded in the browser.  If you use firefox 2, or Internet Explorer 7 (I think on this on, at least), there is spell check enabled by default.  I don't know about other browsers.

If you do not want to upgrade, then there might be extensions you could add on that have spell checks.

Ok, thanks, I think I know what do do now.  :)


trigonometry
function
combinatorics proposed
combinatorics

For the recursive formula $a_{n+1}=4a_n(1-a_n)$, if we choose the initial value
$a_0$ properly,prove that there exist the solution with period of [b]3[/b] which satisfies
\[a_{3k}<a_{3k+1}<a_{3k+2}\ (k=0,1,2,\cdots)\]

Let $f(x) = 4x(1 - x)$, let $h(x) = \sin^2 (\pi x/2)$, and let $T : [0,1] \rightarrow [0,1]$ be the function defined by $T(x) = 2x$ if $0 \le x \le 1/2$ and $2 - 2x$ if $1/2 < x \le 1$, the so-called tent function.

Let $x_0$, $x_1$, $x_2$, $\ldots$, be an orbit under $T$, i.e. $x_{i + 1} = T(x_i)$ for all $i$.  Then it can be shown that $h(x_0)$, $h(x_1)$, $h(x_2)$, $\ldots$, is an orbit under $f$.  Furthermore, any orbit under $f$ is of this form, so orbits under $T$ and orbits under $f$ are equivalent via the transformation $h$.

It is then easy to find an orbit under $T$ with the given property, which can be used to find the desired orbit under $f$.


quadratics
linear algebra
matrix
number theory unsolved
number theory

Give all solutions of $ x^2 \minus{} 2y^2 \equal{} 7$ over the natural numbers.

If $ (x_1^2\minus{}2y_1^2)\equal{}7$ and $ (c^2\minus{}2d^2)\equal{}1$, we have $ (x_1c\plus{}2y_1d)^2\minus{}2(x_1d\plus{}y_1c)^2\equal{}7$. We can find infinitely many values for $ (c,d)$ since $ \mathbb{Q}(\sqrt{7})$, being a real quadratic extension, has its group of units infinite cyclic.

Dario Alpern's [url=http://www.alpertron.com.ar/QUAD.HTM]Quadratic two integer variable equation solver[/url] produces the following recursive series of integer solutions:
\[ \begin{bmatrix} X_{n \plus{} 1} \\
Y_{n \plus{} 1}\end{bmatrix} \equal{} \begin{bmatrix} 3 & 4 \\
2 & 3 \end{bmatrix} \cdot \begin{bmatrix} X_n \\
Y_n\end{bmatrix}
\]
with $ (X_0,Y_0) \equal{} (\pm 3, \pm1).$


arithmetic sequence
number theory unsolved
number theory

The set $ M$ contains all natural numbers from $ 1$ to $ 2007$ inclusive and has the following property: If $ n\in M$, then $ M$ contains all terms of the arithmetic progression with first term $ n$ and difference $ n\plus{}1$. Determine whether there must always exist a number $ m$ such that $ M$ contains all natural numbers greater than $ m$.

Minimal set M is $ M\equal{}\{m|exist \ prime \ p\le 2008, \ suth \ that \ p|m\plus{}1\}$. Obviosly $ p\minus{}1\not \in M$ if p prime $ p>2008$.

[quote="nguyentrang"][quote="Rust"]Minimal set M is $ M \equal{} \{m|exist \ prime \ p\le 2008, \ suth \ that \ p|m \plus{} 1\}$. Obviosly $ p \minus{} 1\not \in M$ if p prime $ p > 2008$.[/quote]

Can you show this $ M$ satisfies the condition?[/quote]

$ n\in \{m|\exists p\ prime\le 2008, \ suth \ that \ p|m \plus{} 1\}$
$ \Rightarrow \forall k\in\mathbb{N}: \ p|[m\plus{}k(m\plus{}1)]\plus{}1$
$ \Rightarrow m\plus{}k(m\plus{}1)\in \{m|\exists p\ prime\le 2008, \ suth \ that \ p|m \plus{} 1\}$
so $ \{m|\exists p\ prime\le 2008, \ suth \ that \ p|m \plus{} 1\}$ checks the property.
and $ \forall p\ prime>2008,\forall q\ prime\le 2008: \ q\not|(p\minus{}1)\plus{}1$ then $ p\minus{}1\not\in \{m|\exists p\ prime\le 2008, \ suth \ that \ p|m \plus{} 1\}$.

$ \{m|\exists p\ prime\le 2008, \ suth \ that \ p|m \plus{} 1\}$ is the mimal set have this poperty because.
if $ M$ cheks then proberty then $ \forall p\ prime\le 2007: \{m\in\mathbb{N}|\ m\equiv p[p\plus{}1]\}\equal{}\{m\in\mathbb{N}|\ p\plus{}1|m\plus{}1\}\subset M$
then $ {\{m|\exists p\ prime\le 2008, \ suth \ that \ p|m \plus{} 1\}\equal{}\bigcup_{p\ prim\le 2008}\{m\in\mathbb{N}|\ p\plus{}1|m\plus{}1\}\subset M}$


Let A,B in M_n(K)
Prove that $r(A)+r(B)\leq n+r(AB)$

We have $corank(AB)\le corank(A)+corank(B)$, thus substracting this from 2n=2n, we get $(n-corank(A))+(n-corank(B)) \le (n)+(n-corank(AB))$ which is simply $rank(A)+rank(B)\le n+rank(AB)$.


Let $ C_1,C_2$ and $ C_3$ be three pairwise disjoint circles. For each pair of disjoint circles, we define their internal tangent lines as the two common tangents which intersect in a point between the two centres. For each $ i,j$, we define $ (r_{ij},s_{ij})$ as the two internal tangent lines of $ (C_i,C_j)$. Let $ r_{12},r_{23},r_{13},s_{12},s_{13},s_{23}$ be the sides of $ ABCA'B'C'$.
Prove that $ AA',BB'$ and $ CC'$ are concurrent.
(see [url=http://img407.imageshack.us/my.php?image=aaaaafz3.png]figure[/url] )

Source: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=229046]Oliforum contest, round 2[/url]

Don't use Imageshack please - this forum should be readable in several years, while Imageshack have a tendency to remove old images with time. You can upload images as attachments to your posts.

Your problem has been posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=140096 . I know a beautiful solution by April. April, have you published it already, or are you going to?

  darij

Did someone solve it in the contest ? If yes please post the solution

[quote="bboypa"]In fact i know that the problem was already posted, but without solution..[/quote]
When will the official solution become available?
I can't wait to see the solution to this one :blush: .

i think this evenig i'll post gradation. you can find solution [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=140096]here[/url](identical to official one (by gabriel)) (another two can be found in relativ topics  :wink: )


Some of the pages on this site have a yellow background, which makes it very uncomfortable to look at. Is there any way we could change it to a less vibrant color. I found that it really strains my eyes to look at it...

Anyone else agree? Or is it just me?     :D

I like the neutral gray of the forum. There is probably a way for the admin to create custom color schemes, and the users to choose which to use. But I agree that the yellow backgrounds can get annoying.

For those that have ever used QBasic, remember the bright white (15 in Screen 12) and just plain white (7 in Screen 12)? I think the second white would be nice if it were a choice.

How about a softer yellow?

We tried whites and greys but they were just horrid for a site with this page structure.


[quote="dvcalculus"]Ok this is a worked out example in the book, but it does not give an explanation on how to do it it just gives the answer

-x +3y=10
x+y=2
Solution: On a single set axes, graph each linear equation

-x+3y=10          x+y=2
X    Y                  X      Y
0    10/3             0      2
-4   2                 2      0
2     4                1      1

Those are the points in the chart and that is where im stuck I NEED TO KNOW HOW THEY GOT THOSE POINTS. The book does not explain how they arrived at those points so can someone please help me out. Thanks in advance.[/quote]
For example, use x = -4, y = -2. -(-4) is 4. Next, 4 + 3y is 10. Thus, 3y is 6. y is 2.

Well, in most solutions by graphing, you just assign random values to x and find the corresponding values for y. Then graph the lines and you get your answer. It's a pretty painful method of solving equations though... I usually prefer solving them algebraically.

Well when you graph the two linear equations they should cross at one point. Now using the Cartesian graphing system the point should be in a $(x,y)$ format, where $x$ is the solution to the x in your equation and $y$ is the solution to the y in your equation.


trigonometry
geometric series
algebra unsolved
algebra

Simplify the following sum-
$ \sum_{i\equal{}1}^{n} \frac{\sin(ix)}{\sin(x)^i}$

Using

$ \sin(x) \equal{} \frac{e^{x\,i}\minus{}e^{\minus{}x\,i}}{2\,i}$

we can write the sum as

$ \frac{1}{2\,i}\sum^n_{k\equal{}1}\left(\frac{2\,i}{1\minus{}e^{\minus{}2\,x\,i}}\right)^k\minus{}\frac{1}{2i}\sum^n_{k\equal{}1}\left(\frac{2i}{e^{2\,x\,i}\minus{}1}\right)^k$

Now just use the expression for the geometric series and that's it.


linear algebra
matrix
vector
function
linear algebra unsolved

Let $ A\in\mathbb{C}^{n\times n}$ and $ \left\Vert\cdot\right\Vert$ be a matrix norm on $ \mathbb{C}^{n\times n}$ such that $ \left\Vert A\right\Vert\equal{}\rho(A)$, where $ \rho(A)$ is the spectral radius of $ A$. How can I show that for each eigenvalue $ \lambda$ of $ A$ such that $ |\lambda|\equal{}\rho(A)$, the geometric multiplicity of $ \lambda$ equals to its algebraic multiplicity?

I'm going to assume that this matrix norm is the norm induced by a vector norm.

WLOG, assume $ \rho(A)\equal{}\|A\|>0.$

Suppose the conclusion does not hold. Then there exists a $ \lambda\equal{}\|A\|$ and vectors $ x,y$ such that $ \{x,y\}$ is independent and $ Ax\equal{}\lambda x\plus{}y,Ay\equal{}\lambda y.$

For $ t\in\mathbb{R},$ define $ z(t)\equal{}x\plus{}\frac{t}{\lambda}\,y.$

Let $ g(t)\equal{}\|z(t)\|.$ We have that $ g$ is a convex function on $ \mathbb{R}$ and that it goes to $ \infty$ as $ t\to\pm\infty.$ Thus is assumes a minimum value, which we call $ m.$ Let $ [a,b]\equal{}\{t\in\mathbb{R}: g(t)\equal{}m\}.$

From the properties of $ x$ and $ y,$ we can compute that $ Az(t)\equal{}\lambda z(t\plus{}1).$ In particular, $ Az(b)\equal{}\lambda z(b\plus{}1).$

But then $ \frac{\|Az(b)\|}{\|z(b)\|}\equal{}\frac{|\lambda|\,\|z(b\plus{}1)\|}{\|z(b)\|}$ is strictly greater than $ |\lambda|,$ contradicting that the norm of the matrix is $ |\lambda|.$


linear algebra
matrix
geometry
geometric transformation
homothety
vector
algebra

find all matrices $M \in M_{n}(\mathbb R)$ such that $Vect( PMP^{-1}: P \in GL_{n}{R}) = M_{n}( \mathbb{R})$

I don't know what the situation is in positive characteristic (and I don't want to worry about that right now :)), but the answer is the same for all fields $K$ of characteristic zero: the only proper, non-trivial linear subspaces $\mathcal L$ of $\mathcal M_{n}(K)$ which are stabilized by all conjugations (i.e. $P\mathcal LP^{-1}=\mathcal L,\ \forall P\in\mbox{Gl}_{n}(K)$) are (a) the space of scalar matrices, and (b) the space $\mbox{sl}_{n}(K)$ of matrices with trace zero. This is a consequence of the following two results, which hold for fields $K$ of characteristic zero (again, I'm not sure what happens for fields of characteristic $p>0$):

$(1)$ If $\mathcal L\le\mathcal M_{n}(K)$ is stabilized by all conjugations, then it's an ideal in $\mathcal M_{n}(K)$, the latter being considered a Lie algebra with the usual Poisson bracket $[A,B]=AB-BA$.

$(2)$ $\mbox{sl}_{n}(K),\ n\ge 2$ is a simple Lie algebra; it follows that the only ideals of $\mathcal M_{n}(K)$ are (a) and (b) in the statement above.

$(2)$ is well-known. I'm only going to prove $(1)$. It's easy to prove it algebraically, but I prefer this:

Let $X$ be an arbitrary matrix, and $M$ an element of $\mathcal L$. We want to prove that $[X,M]$ belongs to $\mathcal L$. Since the bilinear form $f(A,B)=\mbox{trace}(AB)$ is non-degenerate on $\mathcal M_{n}(K)$, it suffices to pick an arbitrary matrix $U$ in the orthogonal complement of $\mathcal L$ with respect to $f$ and prove that $f(U,[X,M])=0$. By working in the field generated by the entries of $U, X, M$, we can assume that $K$ is a subfield of $\mathbb C$.

Now let $\overline K$ be the completion of $K$ with respect to an Archimedean norm (so $\overline K$ is either $\mathbb R$ or $\mathbb C$, and $K$ is dense in $\overline K$). We know that $\mbox{trace}(UPMP^{-1})=0,\ \forall P\in\mbox{Gl}_{n}(K)\ (*)$, so, since $\mbox{Gl}_{n}(K)$ is dense in $\mbox{Gl}_{n}\left(\overline K\right)$, $(*)$ holds for $P\in\mbox{Gl}_{n}\left(\overline K\right)$. Now apply it to $P=\exp(tX),\ t\in\mathbb R$, and differentiate at $t=0$. We get precisely the desired relation $\mbox{trace}(U[X,M])=0$.

A counterexample to the Grobber's questioning about $carac(K)\not=0$.
$K=\mathbb{F}_{2},n=2$ and \[M=\begin{pmatrix}1&1\\0&0\end{pmatrix}\]
$GL_{2}(\mathbb{F}_{2})$ admits 6 elements and $E=\{P^{-1}MP\}$ admits also 6 elements. $E$ generates $\mathcal{M}_{2}(\mathbb{F}_{2})$, $M$ isn't a homothety and $tr(M)\not=0$.

I suggest you look at the question again, because that is certainly [i]not[/i] a counterexample :).

The result, as I formulated it above, is: if $E=\{PMP^{-1}\}$ doesn't generate $\mathcal M_{n}(K)$, then $M$ is either scalar or has trace zero. The contrapositive is: if $M$ is non-scalar and has non-zero trace, then $E$ generates $\mathcal M_{n}(K)$. Your example confirms this rather than disproving it.

Here's my attempt to prove Grobber's claim for fields with at least three elements. It's not complete, but I strongly believe it's not far from that. Suppose $M$ is such that the vector space of its conjugates is not the entire $M_{n}(K)$, thus this vector space is included in some hyperplane of $M_{n}(K)$, which has the form $\{X|Tr(AX)=0\}$ for some nonzero matrix $X$. Therefore $Tr(PAP^{-1}M)=0$ for all invertible $P$. Fix for instant two distinct indices $i,j$ and take the matrices $P=I+aE_{ij}$. Then $P^{-1}=I-aE_{ij}$ and the condition $Tr(P^{-1}APM)=0$ implies that the polynomial $tr(AM)+a(Tr(AE_{ij}M)-Tr(E_{ij}AM)-a^{2}tr(E_{ij}AE_{ij}M)=0$ for all $a\in K$. If $K$ has at least 3 elements, this implies that $tr(E_{ij}AE_{ij}M)=0$. Develop this to get $a_{ji}m_{ji}=0$. BUT (and as an idiot I spent lots of time without noticing this) if $A$ is a matrix satisfying $Tr(PAP^{-1}M)=0$ for all $P$, the same property holds for any conjugate of $A$ (idiot, but essential!). Therefore, for all $i\ne j$ and all matrices $X$ conjugated to $A$ we have $x_{ij}m_{ij}=0$. Suppose that $A$ is not scalar, otherwise $Tr(M)=0$ and it's ok. Then there exists some nonzero vector $v$ with $v,Av$ not collinear. Thus completing this in a basis and puting $v$ as the $i$-th vector of the basis and $Av$ as the j-th vector, we have a conjugate of $A$ which has a $1$ on the $j,i$ position. This implies that $m_{ji}=0$ for all $j\ne i$ and thus $M$ is diagonal. Unfortunately I'm blind and I don't see how to conclude easily from this. I am convinced the argument should be straightforward.


Jim paddles a canoe upstream at a rate of 5 mph but travels a distance of only 9 miles in 3 hours. What was the rate of the current in miles per hour?

Let $ x$ be the rate of the current, in miles per hour.

$ 3(5\minus{}x)\equal{}9$
$ 5\minus{}x\equal{}3$
$ x\equal{}\boxed{2}$ miles per hour is the rate of the current.

I think this is a shorter way.

Without the current, he would have went 15 miles, but he only went 9. 15-9=6, and 6/3 for the mph of the current gives you the current is going at 2 mph.

Edit: Dope!!!! I forgot about that. GJ isabella.

Alternatively, we see that Jim only traveled 3 miles per hour, even though he was going 5 mph. Thus, the current is $ 5 \minus{} 3 \equal{} \boxed{2}$ mph.

I thought that you could do it with a system of equations.

The system of equations is for when you go against the current and with the current. Like say you went 10 miles per hour with the current, and 6 miles per hour against the current.
Thus x+y=10, and x-y=6.


Suppose that $ f: N \rightarrow N$ satisfies $ f(1)=1$ and for all $ n$,
a) $ 3f(n)f(2n+1)=f(2n)(1+3f(n))$
b) $ f(2n)<6f(n)$
Find all solutions to the equation $ f(k)+f(m)=293$


geometry
circumcircle
trapezoid
number theory
least common multiple
geometry proposed

Given quadrilateral $ ABCD$. The circumcircle of $ ABC$ is tangent to side $ CD$, and the circumcircle of $ ACD$ is tangent to side $ AB$. Prove that the length of diagonal $ AC$ is less than the distance between the midpoints of $ AB$ and $ CD$.

$ K, L$ are midpoints of $ AB, CD.$ From circumcircle tangencies, $ \angle ABC \equal{} \angle ACD,$ $ \angle CAB \equal{} \angle CDA$ $ \Longrightarrow$ $ \triangle ABC \sim \triangle DCA$ are oppositely similar $ \Longrightarrow$ $ \angle BCA \equal{} \angle CAD$ $ \Longrightarrow$ $ BC \parallel AD.$ Midline $ KL \parallel (BC \parallel AD)$ of trapezoid $ ABCD$ cuts its diagonal $ AC$ in half at $ M$ $ \Longrightarrow$ $ \triangle AKM \sim \triangle LCM$ are oppositely similar $ \Longrightarrow$ $ AKCL$ is cyclic with diagonal intersection $ M.$ Power of $ M$ to its circumcircle is $ \minus{} KM \cdot ML \equal{} \minus{} AM \cdot MC \equal{} \minus{} \frac {_1}{^4}AC^2$ and $ KL \equal{} KM \plus{} ML \ge 2 \sqrt {KM \cdot ML} \equal{} AC.$ Nothing prevents $ KL \equal{} AC$ when $ AKCL$ becomes a square.


Do there exist 100 positive integers not exceeding 25000 such that the sum of every pair is distinct?

[hide="Hint"]
Every positive integer can be expressed as the sum of distinct fibonacci numbers.
[b]Demonstration:[/b] Note that for j>2, $ F_{j}<F_{j+1}<2F_{j}$. Hence, the difference of any integer and the largest Fibonacci number not greater than that integer is less than that Fibonacci number. This process clearly always yields a positive number (seed integer is not a Fibonacci number) or zero (seed integer is a Fibonacci number), and the output is always decreasing. Eventually, it must reach zero by Pigeonhole.[/hide]


Let a, b be real numbers. And there is at least one real root to the equation \[ 4z^2+(2a+i)z-8b(9a+4)-2(a+2b)i=0  \] Find the maximum of the real root.

Compare imaginary parts of both sides.  Then $z = 2(a+2b)$.

For real parts, $16(a+2b)^2 + 4a(a+2b) -8b(9a+4) = 0$.  It implies $16a^2 + 64ab + 64b^2 + 4a^2 + 8ab - 72ab -32b = 0$

We intend to maximize $K = 2(a + 2b) \Leftrightarrow K - 2a = 4b$, becoming  $20a^2 + 4(K-2a)^2 = 8(K-2a)$, a quadratic in K [we divided by 4],

$4K^2 + (2-4a)K + 9a^2-4a = 0$, which has a maximum of $9a^2 - 4a - \frac{(2-4a)^2}{16} \Leftrightarrow 10a^2 - 4a + 1/4$, which has a maximum of $1/4 - 16/40 = -1/10$.

ps. i didnt check my work, if the ans is wrong, atleast the method is okay


I don't get the upper and lower bounds root location method! Starting from the very beginning
"Suppose we use synthetic division to find $f(x)/(x-c)$ where $f(x)$ has a positive leading coefficient and c>=0."

"Suppose we use synthetic division to find  where  has a positive leading coefficient and c>=0."

so lets create an equation for $f(x)=x^{3}-x^{2}+2x+6$

So lets synthetically divide it by for instance $x-3$ so that:

[code]3|  1  -1   2    6
        3   6   24
    1   2   8   30[/code]
Ok, so you see then when divided by 3 then the coefficients are all positive which means that there is no root greater than 3.  Whenever you synthetically divide by a number and the coefficients and remainder come out positive there cannot be a root that is larger than that number, which is called the upper bound.  This is because the remainder is too large to ever be able to equal zero, and you cannot get it smaller if all your values are positive, and they all will be if the number is larger than the experimented one.  It's basically just logic.

To get the lower bound we once again are going to use $f(x)=x^{3}-x^{2}+2x+6$ as an example and divide this time by $x+3$:

[code]-3|  1  -1   2    6
        -3  12  -42
     1  -4  14  -36[/code]

Ok, so this time the signs in the last one alternate.  If the signs in the answer alternate then you have your lower bound, which in this case is -3, so nothing is smaller than -3.  The reason this works is because, well, try any smaller number and you'll see that the remainder will get closer to -infinity.  It's logic again as well, since obviously you'll end up repeating the same pattern (positive, negative, positive, negative) and cannot get the remainder as 0.

So basically, you can zero in on the answer with this.  So you try c, then -c, each proving to be either too large (if all resulting coefficients are positive) or too small (if resulting coefficients have alternating signs).  You then try d (which is smaller than c) and -d to get possibly a closer upper and lower bound, and do it until you find the number which can be factored from the equation.

I hope all this makes sense...  :D

(thak you solafidefarms for helping me out with this, I actually did my first equations using LaTeX  :lol: )


I'm having trouble seeing how:

$ -\dfrac {1}{2}mv^2 = mgd -\dfrac {1}{2}kd^2$

leads to:

${ d = \dfrac {mg + \sqrt { m^2g^2 + mkv^2}} {k}}$

Can anyone help with the initial step/s please? Thanks!

move everything to the right to get

$ \dfrac {1}{2}kd^2\minus{}mgd\minus{}\dfrac {1}{2}mv^2\equal{} 0$

Then solve as a quadratic of d using the [url=http://en.wikipedia.org/wiki/Quadratic_equation]Quadratic Formula[/url]

Tip: If you're doing physics, unless it is the actual question, it is best not to memorize these formulas. It is often much more convenient to plug in a few numbers first... But yeah, spring problems sometimes require solutions to quadratic equations.

are you having trouble with the other 3 variables?

you can rewrite the equation as
$ (\frac{1}{2}k)d^2\minus{}(mg)d\minus{}(\frac{1}{2}mv^2)$
then set the coefficients equal to a, b, and c

Thanks very much guys! :)
Yeah I solved it by plugging in numbers,
but wanted to see how the algebra worked; it's a physics problem combining
three separate formulas (formulae?).


calculus
derivative
calculus computations

when does $ \frac {d(\frac {\partial L}{\partial \dot{q}})}{dt} \equal{} \frac {\partial \dot{L}}{\partial \dot{q}}$ hold

not sure what dL/dq is but...

anything that follows $ \frac{dy}{dt}\equal{}y$ is of the form $ y\equal{}Ae^t$

Note that there is a dot over the $ L$ in the RHS. Since $ \dot{L}\equal{}\frac{dL}{dt}$, this asks when we can change the order of differentiating with respect to $ t$ and $ \dot{q}$.

This should be when the mixed partials exist and are continuous. Due to the fact that this is probably arising from the Hamiltonian or Lagrangian, there may be less strict requirements due to physical considerations. I don't really remember.


hint:
calcuim carbide in water gives acetylene

[hide="Next"]Then we could use an appropriate catalyst.[/hide]

Just a question: what is the point of this problem?


algebra
polynomial
abstract algebra
vector
calculus
integration
Ring Theory

Let $ I$ be an ideal of the ring $\mathbb{Z}\left[x\right]$ of all polynomials with integer coefficients such that

a) the elements of $ I$ do not have a common divisor of degree greater than $ 0$, and

b) $ I$ contains of a polynomial with constant term $ 1$.

Prove that $ I$ contains the polynomial $ 1 + x + x^2 + ... + x^{r-1}$ for some natural number $ r$.

[i]Gy. Szekeres[/i]

Actually $ I$ contains $ x^r\plus{}x^{r\plus{}1}\plus{}\cdots\plus{}x^s$ for some $ r\le s$, and that's the best you can do.

So you're saying that the statement is wrong? :? Because I'm quite sure it's not. I'll just sketch the proof I had in mind:

I want to show that $ A \equal{} \mathbb Z[x]/I$ is finite. Condition (b) implies that $ x$ is invertible in $ A$, so $ x^t \equal{} 1$ in $ A$ for some positive integer $ t$. Then, if, say, the positive integer $ n$ belongs to $ I$ (there is such a positive integer, for example the order of $ 1$ in the additive quotient group $ (A, \plus{} )$), we have $ 1 \plus{} x \plus{} \ldots \plus{} x^{nt \minus{} 1} \equal{} 0$ in $ A$, i.e. $ 1 \plus{} x \plus{} \ldots \plus{} x^{nt \minus{} 1}\in I$. Hence it suffices to show that $ A$ is finite.

$ I$ has a primary decomposition as $ I \equal{} \bigcap_i\frak q_i$. Since $ A$ injects into $ \prod_i\left(\mathbb Z[x]/\frak q_i\right)$ and the $ \frak q_i$ obviously satisfy conditions (a) and (b), it suffices to prove that a primary ideal $ \frak q$ satisfying (a) and (b) has finite quotient $ \mathbb Z[x]/\frak q$. Let $ \frak p$ be the radical of $ \frak q$. Some power $ \frak p^n$ is contained in $ \frak q$, so it suffices to show that $ \mathbb Z[x]/\frak p^n$ is finite. Finally, since $ \mathbb Z[x]/\frak p^n$ can be written as successive extensions of $ \frak p^k/\frak p^{k \plus{} 1},\ k \equal{} \overline{0,n \minus{} 1}$, and these are finitely-generated modules over $ A/\frak p$, it suffices to show that given a [i]prime[/i] ideal $ \frak p$ with properties (a) and (b), the quotient ring $ A \equal{} \mathbb Z[x]/\frak p$ is finite (in other words, the initial problem has been reduced to the case when $ I$ is prime). This is easy: (a) implies that the ideal contains a positive integer, so, being a prime ideal, it must contain some prime number $ p$. But now (b) shows that $ x\in A$ is algebraic over the field $ \mathbb F_p$ with $ p$ elements, so $ A$ is a finite-dimensional vector space over $ \mathbb F_p$ (since $ A \equal{} \mathbb F_p[x]$).

P.S.

Sometimes I used $ x$ as an indeterminate, and sometimes I used it, by a slight abuse of notation, in order to indicate the image of the indeterminate $ x$ through some canonical map of $ \mathbb Z[x]$ onto a quotient ring; I hope it's clear from the context which is which.

Yes grobber, I misread b) and thought of leading coefficient $ 1$. Sorry.

EDIT
Here is a another proof.

The ideal $ \mathbb{Q}I$ of $ \mathbb{Q}[x]$ must be principal, say $ \mathbb{Q}I \equal{} t(x)\mathbb{Q}[x]$ with some $ t(x)\in\mathbb{Z}[x]$ [url=http://planetmath.org/encyclopedia/PrimitivePolynomial.html]primitive[/url].
Then $ I\subseteq t(x)\mathbb{Z}[x]$ and a) gives $ t(x) \equal{} \pm 1$.
Then $ 1\in \mathbb{Q}I \equal{} \bigcup_{n\in\mathbb{N}}n^{ \minus{} 1}I$ gives some natural number $ n\in I$.
So $ \mathbb{Z}[x]/I$ has some finite characteristic $ n\in\mathbb{N}$ and prime ring $ R\cong\mathbb{Z}/n\mathbb{Z}$.

By b) we find $ f(x)\in I$ of degree $ d$ with $ f(0) \equal{} 1$. Then $ g(x) \equal{} x^df(x^{ \minus{} 1})\in\mathbb{Z}[x]$ is monic.
Let $ y\in \mathbb{Z}[x]/I$ be the inverse of $ x \plus{} I$.
Then $ f(x) \equal{} x^dg(x^{ \minus{} 1})\in I$ gives $ g(y) \equal{} 0$, so $ y$ is [url=http://en.wikipedia.org/wiki/Integral_extension]integral[/url] over $ R$ and $ R[y]$ is finite.
But then $ 1,1 \plus{} y,1 \plus{} y \plus{} y^2,\ldots$ can't be all different
and we find some $ r,s\in\mathbb{N},r < s$ with $ 1 \plus{} y \plus{} \cdots \plus{} y^r \equal{} 1 \plus{} y \plus{} \cdots \plus{} y^s$.
So $ 0 \equal{} y^{r \plus{} 1} \plus{} \cdots \plus{} y^s \equal{} y^{s}(1 \plus{} y^{ \minus{} 1} \plus{} \cdots \plus{} y^{ \minus{} (s \minus{} r \minus{} 1)})$ and $ y$ invertible gives $ 1 \plus{} x \plus{} \cdots \plus{} x^{s \minus{} r \minus{} 1}\in I$.

PS Nice proof grobber! I like yours better than mine. :wink:

This problem can be solved in very easy way.

Easy to see from $(a)$ that $n\in I$. Let $f=1+a_1x+...+a_kx^k\in I$, then if we proved that $x^r(1+x+x^2+...+x^j)\in (n,f)$, then $x^r(1+x+x^2+...+x^j) = ng+fh$ where all coefficients of $h$ not divisible by $n$, so easy to see that $h=x^rh'$, $g=x^rg'$ and $1+x+x^2+...+x^j=ng'+fh'\in I$.

In case where $\gcd(a_k,n)=1$, easy to see that we can take mod $Z_n$ and see that for some $r_1< r_2$ polynoms $1+x+x^2+...+x^{r_1}$, $1+x+x^2+...+x^{r_2}$ will have same remains by $f$ in $Z_n$.

Let polynom $f^*=1+d_1x+...+d_ex^e$ in $I$ with constant term $1$ and minimal value $a=d_e$ of highest term, easy to see that $d_e$ divides every other highest term of any polynom in $I$. Let $a\not=1$.

Fixed number $b$ and take polynom $u=u_1+...+u_{l}x^{l}+u_tx^t$ in $I$ with minimal $\deg\not= 0$ in $Z_n$ and where $t-l=b$ and $u_t=a$. We can see that $u_t=a|n$. Take $\frac{n}{u_t}u=\frac{n}{u_t}u_1+...+\frac{n}{u_t}u_{l}x^{l}+nx^t$ in $Z_n$ his $\deg$ is smaller, so it is $0$ and so $n|\frac{n}{u_t}u_{l}$ and $u_t|u_{l}$, so take $x^{t-l}u-\frac{u_l}{u_t}u\in I$ with bigger value of $t-l$ and highest term $a$.

So we can take polynom $Q=q_1+...+q_gx^g+ax^w$, where $w-g>e$. And from $f^*$ construct polynom in $I$ with smaller highest term. So $a=1$. done


Let $ x,y$ be positive number such that $ x\plus{}y\equal{}\sqrt{10}$. Find the minimum of expression: $ M\equal{}(x^4\plus{}1)(y^4\plus{}1)$

[quote="phuong"]Let $ x,y$ be positive number such that $ x \plus{} y \equal{} \sqrt {10}$. Find the minimum of expression: $ M \equal{} (x^4 \plus{} 1)(y^4 \plus{} 1)$[/quote]
$ M \equal{} (x^4 \plus{} 1)(y^4 \plus{} 1)\equal{}(x^4 \plus{} 1)(1\plus{}y^4) \ge (x^2\plus{}y^2)^2\ge [\frac{(x\plus{}y)^2}{2}]^2\equal{}25$

I think your answer has a little problem. The equality doesn't hold.

Let $ xy\equal{}t$, $ M\equal{}t^4\plus{}2t^2\minus{}40t\plus{}101\ \left(0<t\leqq \frac{5}{2}\right)$.

$ \frac{d}{dt}M\equal{}4(t\minus{}2)\{(t\plus{}1)^2\plus{}4\}$, $ M_{min}\equal{}M_{t\equal{}2}\equal{}\boxed{45}$, when $ (x,\ y)\equal{}\left(\frac{\sqrt{10}\plus{}\sqrt{2}}{2}, \frac{\sqrt{10}\minus{}\sqrt{2}}{2}\right),\ \left(\frac{\sqrt{10}\minus{}\sqrt{2}}{2}, \frac{\sqrt{10}\plus{}\sqrt{2}}{2}\right)$.

Why don't you try to set $ x\plus{}y\equal{}\sqrt{10}\equal{}S$ and $ xy\equal{}P$. After that, you perform the expression by P and S. The maximum value is 101 and the minimum value is done the same way.

[quote="great math"]The maximum value is 101 .[/quote]

Wrong!


function
abstract algebra
group theory
real analysis
real analysis unsolved

Let $ f(x)$ be a nonzero, bounded, real function on an Abelian group $ G$, $ g_1,...,g_k$ are given elements of $ G$ and $ \lambda_1,...,\lambda_k$ are real numbers. Prove that if \[  \sum_{i=1}^k \lambda_i f(g_ix) \geq 0\] holds for all $ x \in G$, then \[  \sum_{i=1}^k \lambda_i \geq 0.\]

[i]A. Mate[/i]

What if $ f \equiv \minus{}1$ and $ \lambda_i \equal{} \minus{}1$?


[quote="Αρχιμήδης 6"]Βρες ολους τους φυσικους $ n$ ωστε ο $ 2^{n}/(n+1)!$ :)[/quote]

Ορίζουμε τη συνάρτηση $ g: \mathbb{N}\to\mathbb{N}$ με τύπο $ g(x)=r \iff 2^{r}\mid x$ και $ 2^{r+1}\not | \ x$
Δηλαδή $ 2^{g(x)}$ είναι η μεγαλύτερη δύναμη του $ 2$ που διαιρεί το $ x$. Μ' άλλα λόγια, η $ g(x)$ μας δείχνει πόσα δυάρια υπάρχουν στο $ x$ όταν το αναλύσουμε σε γινόμενο πρώτων παραγόντων
Προφανώς $ g(ab) = g(a)+g(b), \ \forall a,b\in \mathbb{N}\ \ (1)$

Ορίζουμε επίσης $ f(x)=g(x!)$.
Από τον ορισμό της $ f$ είναι φανερό ότι $ 2^{n}\mid (n+1)! \iff f(n+1) \geq  n$
$ \hline$

Αρχικά θα υπολογίσουμε την $ f(2^{k})$. Είναι $ (2k)! = 1\cdot 2 \cdot 3 \cdot ... \cdot 2^{k}$

Ενδιαφερόμαστε μόνο για τα δυάρια. Έτσι, αγνοούμε τους παράγοντες της μορφής $ 2a+1$ και παίρνουμε μόνο το $ A=2\cdot 4\cdot 6\cdot ...\cdot 2^{k}$

$ A = 2^{2^{k-1}}\cdot (1 \cdot 2 \cdot 3 \cdot ... \cdot 2^{k-1}) = 2^{2^{k-1}}\cdot (2^{k-1})!$

$ g(A) = g( 2^{2^{k-1}})+g((2^{k-1})!) =2^{k-1}+f(2^{k-1})$

Έχουμε λοιπόν $ \boxed{f(2^{k}) = 2^{k-1}+f(2^{k-1})}\ \ (2)$

[color=blue][size=150]Θεώρημα 1[/size][/color] $ \boxed{f(2^{k}) = 2^{k}-1}$

[size=150][hide="Απόδειξη"][/size]
Για $ k=1$ είναι $ f(2^{1})=f(2)=1$, αφού $ 2!=2$

Έστω ότι ισχύει για $ k-1$, δηλαδή $ f(2^{k-1}) = 2^{k-1}-1$. ΘΔΟ ισχύει και για το $ k$

Από τη $ (2)$ έχουμε

$ f(2^{k}) = 2^{k-1}+f(2^{k-1})$
$ f(2^{k}) = 2^{k-1}+2^{k-1}-1$
$ f(2^{k}) = 2^{k}-1$

$ \star$ Το θεώρημα ισχύει και για $ k=0$, αφού $ (2^{0})!=1$ και η μεγαλύτερη δύναμη του $ 2$ που το διαιρεί είναι η $ 2^{0}$ [/hide]
$ \hline$

Επομένως για $ n=2^{k}-1$ έχουμε $ f(n+1) = n$, που σημαίνει ότι $ 2^{n}\mid (n+1)!$.

Δηλαδή οι αριθμοί $ n=2^{k}-1$ είναι λύσεις. Θα δείξουμε ότι είναι και οι μοναδικές

[color=blue][size=150]Θεώρημα 2[/size][/color] Έστω ότι ο αριθμός $ n+1$ δεν είναι δύναμη του $ 2$. Τότε θα είναι $ f(n+1)<n$

[size=150][hide="Απόδειξη"][/size]Έστω $ k$ ο μεγαλύτερος φυσικός τέτοιος ώστε $ 2^{k}<n+1$.

Είναι λοιπόν $ n+1 = 2^{k}+a$ με $ 0<a<2^{k}$

$ (n+1)! = (2^{k})!\cdot (2^{k}+1)(2^{k}+2)...(2^{k}+a) \Rightarrow $
$ (n+1)! = (2^{k})!\cdot B\Rightarrow $
$ f(n+1) = f(2^{k})+g(B)\Rightarrow $
$ f(n+1) = 2^{k}-1+g(B)$

$ f(n+1)<n \iff 2^{k}-1+g(B) <2^{k}+a-1 \iff g(B)<a$

Αρκεί λοιπόν να δείξουμε ότι $ g(B)<a$

Προφανώς κανένας από τους $ 2^{k}+1, ..., 2^{k}+a$ δε διαιρείται με το $ 2^{k}$.

Επίσης $ 2^{k}+i \equiv i \mod 2^{w}, \ \forall w<k$
Δηλαδή ο $ 2^{w}$ θα διαιρεί το $ 2^{k}+i$ τότε και μόνο, όταν διαιρεί το $ i$
Αυτό σημαίνει ότι $ \boxed{g(2^{k}+i) = g(i)}\ \ (3)$

Θα χρησιμοποιήσουμε τη 2η μορφή της μαθηματικής επαγωγής
$ \blacksquare$ Υποθέτουμε ότι το θεώρημα ισχύει για κάθε αριθμό $ m<n$, δηλαδή αν ο $ m+1$ δεν είναι δύναμη του $ 2$, τότε $ f(m+1)<m$
$ \blacksquare$ Θα δείξουμε ότι ισχύει και για το φυσικό αριθμό $ n$

Από τις σχέσεις $ (3)$ και από την ιδιότητα $ (1)$ της $ g$ παίρνουμε
$ g(B) = g((2^{k}+1)(2^{k}+2)...(2^{k}+a))$
$ =g(2^{k}+1)+g(2^{k}+2)+...+g(2^{k}+a)$
$ =g(1)+g(2)+...+g(a)$
$ =g(a!) = f(a)$

Από τον ορισμό του $ a$ είναι $ 0<a<2^{k}$

Αν $ a=2^{q}$ για κάποιο $ q<k$, τότε ισχύει το [b]Θεώρημα 1[/b] $ \Rightarrow f(a)=a-1$
Αν $ \not \exists$ τέτοιο $ q$, τότε σύμφωνα με την υπόθεση ότι το [b]Θεώρημα 2[/b] ισχύει, έχουμε $ f(a)<a-1$

Σε κάθε περίπτωση έχουμε $ g(B)=f(a)\leq a-1<a$ και τελειώσαμε[/hide]

[quote="Αρχιμήδης 6"]Βρες ολους τους φυσικους $ n$ ωστε ο $ 2^{n}/(n+1)!$ :)[/quote]

Αυτή την άσκηση να τη βάλουμε στο βιβλίο. Είναι καλή. Παρακαλώ τον Αρχιμήδη 6 και τον Pontios να τη μεταφέρουν με πλήρη λύση και με τα ονόματά τους στο σχετικό χώρο της θεωρίας αριθμών.

  Καλό δεκαπενταύγουστο !!!

 Μπάμπης

Θα την προσεγγισω διαφορετικα.Εδω φαινεται και η χρησιμοτητα και η δυναμη απλα και κατανοητα της της μεταπιδησης απο το δεκαδικο στο δυαδικο συστημα αριθμησης.Εστω οτι $ 2^{n\minus{}1}/n!$ και εστω οτι $ n \equal{} a_{0}\plus{}a_{1}(2)\plus{}a_{2}(2^{2})\plus{}......\plus{}a_{k}(2^{k})$ η αναπαρασταση του στο δυαδικο συστημα με $ a_{i}\equal{} 0$ η $ a_{i}\equal{} 1$.Eυκολα αποδυκνειεται οτι η μεγιστοβαθμια δυναμη του $ 2$ που διαιρει το $ n!$ ειναι η συναρτηση $ [n/2]\plus{}[n/2^{2}]\plus{}......\plus{}[n/2^{k}]$.Λογω οτι $ 2^{n\minus{}1}/n!$ θα πρεπει να ισχυει οτι $ n\minus{}1\leq [n/2]\plus{}[n/2^{2}]\plus{}......\plus{}[n/2^{k}]$ αρα $ a_{0}\plus{}a_{1}(2)\plus{}a_{2}(2^{2})\plus{}......\plus{}a_{k}(2^{k})\minus{}1\leq [a_{1}\plus{}a_{2}(2)\plus{}....\plus{}a_{k}(2^{k\minus{}1}]\plus{}[a_{2}\plus{}a_{3}(2)\plus{}.....\plus{}a_{k}(2^{k\minus{}2}]\plus{}....\plus{}a_{k}$ αρα
 $ a_{0}\plus{}a_{1}(2)\plus{}a_{2}(2^{2})\plus{}......\plus{}a_{k}(2^{k})\minus{}1\leq a_{1}\plus{}a_{2}(2^{2}\minus{}1)\plus{}.....\plus{}a_{k}(2^{k}\minus{}1)$ δηλαδη $ a_{0}\plus{}a_{1}\plus{}a_{2}\plus{}......\plus{}a_{k}\leq 1$ αρα επειδη $ a_{i}\equal{} 0$ η $ a_{i}\equal{} 1$ σημαινει οτι ολοι ειναι ισοι με $ 0$ και ενας ακριβως ισουται με $ 1$ αρα $ n \equal{} 2^{x}$ με $ 0\leq x\leq k$
ΠΑΡΑΤΗΡΗΣΕΙΣ
-Ευκολα αποδυκνιεται οτι η προταση αποτελει σχεση ισοδυναμιας γιατι ισχυει και η αντιστροφη αυτης.
-Μια γενικευση αυτης ειναι και η εξης:αν $ a^{n\minus{}1}/n!$ τοτε $ a \equal{} 1$ για καθε $ n$ η $ a \equal{} 2$ και $ n \equal{} 2^{k}$.


geometry
geometric transformation
power of a point

Two circles $C_1$ and $C_2$ intersect at two distinct points $P, Q$ in a plane. Let a line passing through $P$ meet the circles $C_1$ and $C_2$ in $A$ and $B$ respectively. Let $Y$ be the midpoint of $AB$ and let $QY$ meet the cirlces $C_1$ and $C_2$ in $X$ and $Z$ respectively. Show that $Y$ is also the midpoint of $XZ$.

The points P, Q are the 2 intersections of the circles $C_1, C_2$, otherwise the problem would not make sense. Let the point A move on the circle $C_1$. The angle $\angle QAB = \angle QAP$ for A on the major arc PQ of $C_1$ and $\angle QAB = 180^\circ - \angle QAP$ for A on the minor arc PQ of $C_1$. Similarly, the angle $\angle QBA = \angle QBP$ for B on the major arc PQ of $C_2$ and $\angle QBA = 180^\circ - \angle QBP$ for B on the minor arc PQ of $C_2$. Thus the angles $\QAB, \angle QBA$ are constant and the triangles $\triangle QAB$ remain similar as the point A moves on the circle $C_1$. If a figure F moves in such a way that it remains similar to a figure $F_0$ and one point Q of the figures $F, F_0$ remains fixed, then the curve $\Gamma_1$ described by the corresponding points A of the figure F is similar to the curve desribed by the corresponding points Y of this figure. Applying this theorem to the triangle $\triangle QAB$ with the fixed point Q and with the vertex A moving a the circle $C_1$, it follows that the locus of midpoints Y of the side AB is a circle (O). When the point A coincides with the point Q, the point B also coincides with the point Q. Hence, the circle (O) passes through the point Q. From the continuity principle, a point $A \in C_1$ exists, such that AP = BP, hence, the circle (O) passes though the point P as well. Consequently, it is centered on the center line $O_1O_2$ of the circles $C_1, C_2$. When the line AB becomes parallel to the center line $O_1O_2$, the angles $\angle QPA = \angle QPB = \angle QPY = 90^\circ$ become all right, which means that the centers $O_1, O_2$ of the circles $C_1, C_2$ become the midpoints of the triangle sides QA, QB and the center O of the circle (O) the midpoint of the median QY of the triangle $\triangle QAB$. This allows to calculate the radius r of th locus circle (O). Let $r_1, r_2$ be the radii of the circles $C_1, C_2$ and $d = O_1O_2$ the distance between their centers. Then $QA = 2r_1, QB = 2r_2, QY = 2r$ and $d = O_1O_2 = \dfrac{AB}{2}$ is the midline of the triangle $\triangle QAB$. The median of this triangle is equal to

$QY = \frac 1 2 \sqrt{2QA^2 + 2QB^2 - AB^2}$

$2r = \frac 1 2 \sqrt{8r_1^2 + 8r_2^2 - 4d^2}$

$r = \frac 1 2 \sqrt{2r_1^2 + 2r_2^2 - d^2}$

Since this is a formula for the median of the triangle $\triangle O_1O_2P$, it follows that the center PO is the median of this triangle and O the midpoint of its side $O_1O_2$. All that was said above applies to the triangle $\triangle PXZ$ subject to the spiral similarity with the center P - the locus of the midpoints of the sides XZ is the same circle (O), hence, these midpoints are identical with the midpoints Y of the sides AB of the triangle $\triangle QAB$ subject to the spiral similarity with the center Q.

APXQ cyclic $\implies$ PYX ~ QYA, PZBQ cyclic $\implies$ PYZ ~ QYB.

Note that $\dfrac{AY}{XY} = \dfrac{QY}{PY} = \dfrac{BY}{ZY}$. So $AY = BY \implies XY = ZY$.

By power of a point $ (PY)(AY)\equal{}(QY)(XY)$ and $ (QY)(YZ)\equal{}(PY)(BY)$.  Since $ AY\equal{}BY$, $ (PY)(BY)\equal{}(QY)(YZ)\equal{}(QY)(XY)$.  Thus, $ YZ\equal{}XY$.

$AXPQ, BPZQ$ are cyclic, hence $\angle XAP=\angle PQX=\angle PQZ=\angle PBZ\implies AX\parallel BZ$; with $AY=BY$ we get $\triangle AYX\cong\triangle BYZ$, done.

Best regards,
sunken rock


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That's what I recieved when I tried to post a quote in HMMT forum. It only included latex and a smile. What happened?

P.S. The latex was in the quote, but the smilie was not.

This is the seventh time someone posts it!
Can you guys not read or what else¿

[quote="Valentin Vornicu"]Sometimes that happens when the load on the server is so big, that it freezes MySQL for a couple of seconds.[/quote]Totally wrong assesment. Bug fixed now.


I am an international student and not exactly sure whether Cambridge would be a better place or Imperial College London to study engineering. All English people out there should have some idea I suppose. Please give some advice.

Thanks.

What kind of engineering? I'm not familiar with the actual universities, but usually one university is better in that research area and the other is better in another area.

I understand your concern. But the fact is that both universities offer slightly different approaches.
Students apply to Cambridge for the course "Engineering" and they specialise afterwards in their 3rd year, i.e., they choose their options and pursue it for 2 years.

In Imperial College London, students apply for more specific engineering courses - Mechanical Engg., Electrical Engg.

What I am then trying to understand is which one is better for engineering in general. THES rankings show Imperial on the 5th and Cambridge on the 6th place for Technology. But then those are just one set of rankings which differ from another set of rankings.

That is why I needed some real opinions.

One of the key points you should consider too is whether you'd prefer life in London, where your living costs are higher and accomodation is more problematic but you will problem encounter a wider range of people or whether you'd prefer a quieter easier life at Cambridge where life is much less hectic but you don't get so many connections as you would studying in the captial.

So, centy, you mean to say that the difference in engineering departments is not that big and I should consider the other factors instead. Right?

Well think about it...how much difference can there be between two engineering departments when it comes to what they teach you. At the end of the day university is more important for the contacts you make and people you meet especially if this leads straight into employment.

You should think about where you would feel happiest and what sort of people you want to spend a good few years of your life with.

Both universities offer great courses and you'll need to take other considerations into account.

The engineering courses are the top two in the country. I would choose on the location, very different places.


I need some help with the following problems, and as far as the last one is concerned, I can only find 14 different line segments, but I am not entirly convinced that this is the answer :?

Compute $1-2+3-4+5-6+...-2002+2003$

Compute the product of $(10101)_2$ and $(101010)_2$ and express the answer in base 2.

Compute the number of different line segments that have their endpoints on two distinct vertices of a cube.

1) [hide="hint"]Look at the terms of the summation two at a time.[/hide]
[hide="answer"]All the terms up to 2002 are basically -1 added 2002/2=1001 times. So the answer is 2003-1001=1002.[/hide]
2) [hide="hint"]Too lazy to actually do it but I would say convert them to base 10, multiply and convert back. Simple hand multiplication in binary would probably be faster, but I've never tried it before so I'm not going around offering advice on it.[/hide]
[hide="possible_answer"]I tried multiplication in binary, so if I did it right, it would be 1101110010_2. I really doubt it is, but whatever.[/hide]
3) [hide="hint"]Think of it in terms of choosing two points to connect with a line segment.[/hide]
[hide="answer"]There are 8 points, and any two points define a segment. So it's just C(8,2)=28 segments.[/hide]

After reading up on changing number bases, I was able to solve the second problem:

The first step is to convert each base 2 number to a base 10 number:

$(10101)_2$ = $1*2^4+0*2^3+1*2^2+0*2^1+1*2^0 = 21$

$(101010)_2$ = $1*2^5+0*2^4+1*2^3+0*2^2+1*2^1+0*2^0 = 42$

Now that the two numbers in base two are converted to base 10, we can multiply them out. Since $21(42) = 882$, we need to convert this back to base 2, which is $(1101110010)_2$.

We might as well check our answer, so:

$(1101110010)_2 =  1*2^9+1*2^8+0*2^7+1*2^6+1*2^5+1*2^4+1*2^3+0*2^2+0*2^1+1*2^0+0*2^0 = 882$

So I was right?! Wow. Double-triple-quadruple-wow.
Anyway, to elaborate on my method for number 2...
[hide]
[color=white]______[/color]1 0 1 0 1 0
[color=white]______[/color]x 1 0 1 0 1
[color=white]______[/color]_________
[color=white]______[/color]1 0 1 0 1 0
[color=white]___[/color]1 0 1 0 1 0
1 0 1 0 1 0
_______________
1 1 0 1 1 1 0 0 1 0
[/hide]
...So it's the same thing as any old multiplication, except that in the adding part, whenever you see two 1's you write down 0 and carry 1, when you see three 1's you write down 1 and carry 1, etc.


AMC
AIME
geometry
3D geometry
octahedron
probability
graph theory

The labels 1, 2, ... , 8 are randomly placed on the faces of an octahedron (one per face). Find the probability that no two adjacent faces (sharing an edge) have adjacent numbers, where 1 and 8 are also considered adjacent.

Copied from sci.math (my favorite because of the comment No reason a bright student couldn't figure

that out with pencil and paper in a few minutes.):

[hide]It's an exercise in elementary graph theory. To get on more familiar

ground, for "faces of octahedron" read "vertices of cube". We have to

count the Hamiltonian circuits in a certain graph, namely, the

complement of the cube Q_3. No reason a bright student couldn't figure

that out with pencil and paper in a few minutes.

The graph consists of two K_4's joined by a matching. The circuit

must use either 2 or 4 edges of the matching, so there are two cases.

In each case you have to decide (a) which edges of the matching to

use, and in which direction, and (b) how to link them up.

In the first case, there are 4 choices for which edge of the matching

to use for the left-to-right crossing, then 3 choices for the

right-to-left crossing, then 2 ways to link then up on the left side

and 2 ways on the right side, for a total of 4*3*2*2 = 48.

In the other case, there are 6 choices for which 2 edges of the

matching will be traversed from left to right, then 2 ways to link

them up on the left side, after which there is only 1 way to link them

up on the right, for a total of 6*2*1 = 12. The grand total is 48+12 =

60 Hamiltonian circuits (with an orientation but no designated

starting point), and the probability is 60/7! = 1/84.[/hide]

neat.  but first things first:
" We have to count the Hamiltonian circuits in a certain graph, namely, the complement of the cube Q_3. "
The graph consists of two K_4's joined by a matching."
What is "the complement of the cube Q_3" and "two K_4"

The complement of a graph \overline{G} of G is the graph on V with edge set [V]^2 \ E. K_4 is the complete graph on 4 nodes. Basically the complement of Q_3 is the cube's diagonals/space diagonals.


If n > 3 setting x1 = x2 = ... = x(n-1) = 1, xn = 1/2 yields a contradiction,
if n = 3 one can set x1 = x3 = 2, x2 = 3 and if n = 2 or n = 1 the inequality becomes obvious.


geometry
3D geometry
Putnam
college contests

A solid has a cylindrical middle with a conical cap at each end. The height of each cap equals the length of the middle. For a given surface area, what shape maximizes the volume?

I don't get it...It seems like they describe the shape pretty much perfectly in the problem.

I agree with [b]Rushil[/b]. The problem statement is not very clear.

[quote="Kent Merryfield"]Let $r$ be the radius of the cylinder and $h$ the "length of the middle portion." Then I interpret this as an object whose overall length is $3h.$

$V=\pi r^2h+\frac23\pi r^2h$

$A=2\pi rh+2\pi r\sqrt{r^2+h^2}$

Now: maximize $V$ for fixed $A.$ (Or minimize $A$ for fixed $V,$ which would be the same problem.)[/quote]

Thanks, I see it now! I made a silly error in thinking the surface area ,$A = 2\pi rh$ (surface of the middle)+ $2\pi rh$(surface of the cone caps)

I guess I need to brush up on my secondary school formulas.


logarithms
inequalities
real analysis
real analysis unsolved

$ \sum_{n\equal{}1}^{\infty}(1\minus{}\frac{ \ln n}{n})^n$

The above series seems convergence....

Use Cauchy's Test , I can't get anything

It seems difficult to work out..... :o

edit: my hint was misleading. Thanks Kent.

It diverges by comparison to the harmonic series.

$ \left(1 \minus{} \frac {\ln n}n\right)^n \equal{} \text{exp}\left(n\ln\left(1 \minus{} \frac {\ln n}n\right)\right)$

$ \equal{} \text{exp}\left( \minus{} \ln n \minus{} \frac {\ln^2n}{2n} \plus{} o(n^{ \minus{} 1})\right)$

$ \equal{} e^{ \minus{} \ln n}\cdot e^{o(1)} \equal{} \frac1n\cdot e^{o(1)}.$

But $ e^{o(1)}$  is the sequence of exponentials of a sequence that tends to zero. It is both bounded above and bounded below away from zero. We conclude that

$ \left(1 \minus{} \frac {\ln n}n\right)^n > \frac cn$ for some positive constant $ c.$

Hence the claim of divergence for the series.

Edit: WWW's inequality points in the wrong direction. $ 1 \minus{} \frac {\ln n}n\to 1$ as $ n\to \infty.$

Thanks very much, Wonderful !

In your reply:

$ \text{exp}\left(n\ln\left(1 \minus{} \frac {\ln n}{n}\right)\right) \equal{} \text{exp}\left( \minus{} \ln n \minus{} \frac {\ln^{2}n}{2n} \plus{} o(n^{ \minus{} 1})\right)$

But I think the right hand of the above equation maybe $ \text{exp}\left( \minus{} \ln n \minus{} \frac {\ln^{2}n}{2n} \plus{} o(\frac {\ln^{2}n}{n})\right)$

because:

$ \ln\left(1 \minus{} \frac {\ln n}{n}\right) \equal{} \minus{} \frac {\ln n}{n} \minus{} \frac {\ln^{2}n}{2n^2} \plus{} o(\frac {\ln^{2}n}{n^2})$

so plus $ n$

it equal:

$ \minus{} \ln n \minus{} \frac {\ln^{2}n}{2n} \plus{} o(\frac {\ln^{2}n}{n})$

and $ o(\frac {\ln^{2}n}{n})$ seems not have the same order as $ o(\frac {1}{n})$

consider: $ \frac {\ln n}{n}$

Use a better series estimate. We have $ \ln(1 \minus{} u) \equal{} \minus{} u \minus{} \frac {u^2}{2} \plus{} O(u^3)$, so that term is $ n\cdot O(\frac {\ln^3 n}{n^3}) \equal{} O(\frac {\ln^3 n}{n^2}) \equal{} o(n^{ \minus{} 1})$, using that logarithms are smaller than powers of $ n$ in the last step.

The whole point of using the big-O, little-o notation is to avoid getting bogged down in details that don't really matter.

In fact, the next term of that Taylor expansion is $ O\left(\frac{\ln^3n}{n^2}\right),$ And it is true that that is smaller than $ \frac1n,$ hence in calling it $ o(n^{\minus{}1})$ I was making a true statement.

I realized a few lines later that even there I was giving more detail than I needed to, and the whole expression $ \minus{}\frac{\ln^2n}{2n}\plus{}o(n^{\minus{}1})$ need only be written as $ o(1).$

In your criticism of my argument, you're trying to be more precise, to give more information. My defense to that is that I didn't need to give that information, and the the effort involved in being more precise is wasted effort.


geometry
rectangle
quadratics
LaTeX
algebra
quadratic formula

Actually, AstroPhys, I think you did the formula for volume when the question is asking for the surface area.

[hide="hide hide hide hide hide hide hide hide...a message brought to you by jli"]Since a cylinder's surface can be "flattened" to a rectangle and two circles, with the rectangle's width being the circles' circumferences, the equation for this problem can be written:

$188.5=2{\pi}r^{2}+7(2{\pi}r)$ or $0=2{\pi}r^{2}=14{\pi}r-188.5$.

Putting this into the quadratic formula, we get:

${(-14{\pi}+/-{\sqrt}{196{\pi}^{2}-8{\pi}+754)}/{4{\pi}}$. This simplifies to $r$ being very close to either $-10$ or $3$. Since the result must be a real answer, $r=3$.[/hide]

EDIT: I'm sorry, I can't quite get the latex on the quadratic formula to work. I hope it's understandable to everyone.

they already explained it.

$Surface area = 2\pi(r^{2}+h)$

[quote="sharkman"]A 39 ounce can of coffee requires 188.5 square inches of aluminum.  If its height is 7 inches, what is its radius?

HINT GIVEN:  Think about a right circular cylinder.[/quote]
[hide="soltuion"]
39 ounces is completely irrelevant.

The surface area of a can is $2\pi rh+2\pi r^{2}=14\pi r+2\pi r^{2}=188.5$

So $r\approx-10$ and $3$ so its $\boxed{3}$.
[/hide]
To igiul, the height needs to be multiplied by r.


logarithms
floor function
number theory proposed
number theory

Prove that for every natural number like $n\geq 2$,

$\sum_{i=1}^{n-1}\lfloor{\sqrt[i+1]n}\rfloor=\sum_{i=1}^{n-1}\lfloor{\log_{i+1}n}\rfloor$

Are you sure this is correct ? Mupad suggests that the statement is false  :?


geometry
ratio
geometric series
area of a triangle
Heron\u0027s formula

The sequence $a_{1}, a_{2}, a_{3}, \ldots$ is a geometric sequence.
The sequence $b_{1}, b_{2}, b_{3}, \ldots$ is a geometric sequence.
$b_{1}= 1$
$b_{2}= \sqrt[4]{7}-\sqrt[4]{28}+1$
$a_{1}= \sqrt[4]{28}$
$ \sum_{n=1}^{\infty}{\frac{1}{a_{n}}}=  \sum_{n=1}^{\infty}{b_{n}}$

Compute the area of the triangle with side lengths $a_{1}, a_{2},$ and $a_{3}$.

[hide]$\sum_{n=1}^{\infty}{\frac{1}{a_{n}}}= \sum_{n=1}^{\infty}{b_{n}}\Rightarrow \frac{1/\sqrt[4]{28}}{1-1/k_{a}}= \frac{1}{\sqrt[4]{28}-\sqrt[4]{7}}\Rightarrow k_{a}= \sqrt{2}\Rightarrow a_{1}= \sqrt[4]{28}, a_{2}= 2\sqrt[4]{7}, a_{3}= 2\sqrt[4]{28}\Rightarrow \boxed{A = \frac{7}{2}}$

$k_{a}$ is the constant ratio of the $a$ geometric series.  I used the formula for the sum of a geometric series as $\frac{t_{1}}{1-r}$ and used Heron's formula to find the area.[/hide]


Using only the numbers 1, 2, 3, 4and 5 (none more than once) and the operations +, -, * and / and brackets, how many of the following numbers is it possible to make: 76, 79, 86, 92, 97 and 98? Is theree a website for this type of question?

Hm... Maybe I could make a computer program for that, but that isn't so easy. Anyway, I'll try.

This link is for 6 numbers.

http://www.ee.unb.ca/cgi-bin/tervo/compte.html


In a community of more than six people each member exchanges letters
with exactly three other members of the community. Show that the com-
munity can be partitioned into two nonempty groups so that each member
exchanges letters with at least two members of the group he belongs to.

EDIT: You are right RealZeta.  That old argument could be modified to show we can partition the group into two groups where each person knows at most 1 person in each group, but for this problem, since the complete community satisfies the condition, I cannot guarantee that one of the groups is non-empty.  I'm working on another approach that I think will work...

Perhaps I am not seeing it, but is it guranteed that both sets remain non-empty?

Ok, lets try this.  It may have a kink or two, so let me know if it does.

First, let $n$ be the number of people in the community.  We have $n\ge 7$ is best possible, since a regular hexagon with the outside edges and $3$ pairs of longest diagnols connected cannot be partitioned correctly.  So $n=6$ does not always work.

Now, let $G=(V,E)$ be the graph of the community, $V$ be the set of people, and $E$ be the edges, such that two vertices are connected if and only if they share a letter.  It is obvious that for $n>6$ that $|E|>n>n-1=|V|-1$.  This means that $G$ cannot be a tree, in other words, $G$ is not acyclic.

Now let $C$ be the cycle in $G$ with the shortest length.  This leaves three cases:

[b]Case I:[/b] $|C|$ is at least $5$.
Now, because the degree of each vertex is $3$, each of these points in the cycle connects to exactly one more point.  However, they cannot connect to any of the other points in the cycle besides their neighbors, for then the cycle would not be the shortest.  Secondly, no two different points in the cycle can connect to the same point outside the cycle because, again, $C$ would not be the shortest.  So each the points in the cycle connect to unique points outside the cycle, and since for all vertices outside $C$ will have degree $2$ in the subgraph of $G$ \ $C$, we have the desired partition.

[b] Case II:[/b] $|C|=3$.
This is the simplest case.  Let $C=C_{1}C_{2}C_{3}$ be the vertices in the cycle.  Then each of the $C_{i}$ connect to one more point outside $C$, and they can make at most one vertex outside $C$ have degree less than $2$ in subgraph $G$ \ $C$.  In this case, include that point with $C$, and since there are at least $3$ more points, we are done.  Otherwise, we are done anyways.

[b] Case III:[/b] $|C|=4$.  This is a bit awkward.  Let $C=C_{1}C_{2}C_{3}C_{4}$ like above.  Now, like in Case I, we cannot connect anymore of the $C_{i}$'s together, otherwise $C$ is not the shortest.  However, we can connect $C_{1}$ and $C_{3}$ to the same point outside and the same goes for $C_{2}$ and $C_{4}$.  Now $n$ cannot be odd, since $\sum_{v\in V}deg(v)=3n=2|E|$ which is impossible.  And if $n=8$, then we cannot have both pairs in the cycle latch to the same points, because then the $deg(v)=3$ for all vertices cannot hold (there are only a finite number of cases to check).  And finally, for $n\ge 10$, there are no problems to partition the groups.  So we are done.

We make a graph $G(V, E)$ with $|V|=6$, we connect two vertice if and only if they communicate with each other.

We call a partition $(A,B), A+B=V$ is good such that the number of edges between $A$ and $B$ reaches the minimum.

It is easy to see a good partition is an answer, because if there is a member doesn't exchanges letters with at least two members in his group, then we put him in the other group, the value I mentioned above will decrease, it is a contradiction with the definition of good partition!

[quote="libra_gold"]
It is easy to see a good partition is an answer.
[/quote]

but, In this situation, A or B can be empty! right?


Παιδιά είμαι καινούριος στο forum κ μ'αρέσει η γεωμετρία.Μ'αρέσει πολύ η παρέα σας.Έχω ένα πρόβλημα που βασανίζομαι 2 μέρες κ θέλω να το δείτε να με βοηθήσετε.Έχουμε ένα ισόπλευρο τρίγωνο κ ένα σημείο που απέχει 1, 3/8 κ 5/8 αντίστοιχα απο τις κορυφές του.Πόση είναι η πλευρά του τριγώνου?

Καλωσήρθες billakos!!! Είναι πολύ ευχάριστο που προστίθενται νέμα μέλη στην παρέα μας :)

Όσον αφορά το πρόβλημά σου σίγουρα αυτό βγαίνει με αναλυτική γεωμετρία (αν και προσωπικά δεν την ξέρω καλά :oops: ).
Πάντως αν θεωρήσουμε $ ABC$ το τρίγωνο, $ M$ το σημείο και $ MA \equal{} 1, MB \equal{} 3/8, MC \equal{} 5/8$, νομίζω μπορούμε να βρούμε τις συντεταγμένες του Μ σε συνάρτηση της πλευράς του τριγώνου ως την τομή δύο κύκλων με κέντρα τα $ A, B$ για παράδειγμα και μετά να πούμε ότι πρέπει να απέχει απ' το τρίτο σημείο την αντίστοιχη απόσταση (π.χ. $ MC \equal{} 5/8$) ώστε να βρούμε μια σχέση με μόνο άγνωστο την πλευρά του τριγώνου.

Επίσης χωρίς να είμαι σίγουρος πιστεύω ότι το συγκεκριμένο πρόβλημα μπορεί να λυθεί και με τριγωνομετρία... με χρήση νόμου συνημιτόνων για παράδειγμα :wink:

Πάντως προσωπικά θα προσπαθήσω να βρω μια λύση καθαρά γεωμετρική που είμαι σίγουρος ότι θα υπάρχει :P

Υπάρχει γεωμετρική λύση . (δες και μια παλαιότερη άσκηση του μήνα είναι ίδια απλά με άλλα νούμερα )
Η βασική ιδέα είναι να πάρεις τα συμμετρικά του σημείου ως προς τις πλευρές του τριγώνου  :wink:
(και μετά να δουλέψεις με εμβαδά)

Ακόμα πιο απλή λύση νομίζω δίνεται αν σκεφτείς ότι το Μ πρέπει να βρίσκεται πάνω στο περιγεγραμμένο του $ ABC$ λόγω του ότι $ MA \equal{} MB \plus{} MC$ και απ' το θεώρημα του Πτολεμαίου.

Μετά απο αυτό και με νόμο συνημιτόνων στο $ MAB$ βρίσκεις αμέσως την ζητούμενη πλευρά (βγαίνει $ \frac {49}{64}$ αν δεν έχω κάνει κανένα λάθος στις πράξεις)

Μήπως έιναι εύκολο να μου πείτε πια είναι η ασκήση του προηγούμενου μήνα που αναφέρατε?

[quote="stergiu"]Όλα σχδόν αυτά τα προβλήματα ανήκουν στην κατηγορία του θεωρήματος Pompeiu.Αν τα μήκη είναι πλευρές τριγώνου συνήθως κάνουμε μια στροφή γύρω πχ στο ΒΜ κατά 60 και  συνεχίζουμε με σύγκριση και νόμο συνημιτόνων. ΑΝ το ένα είναι ίσο με το άθροισμα των δύο άλλων , τότε δουλεύουμε όπως είπαν τα παιδιά.

Μπάμπης[/quote]

Μπορείς να μου πεις φίλε ποιο είναι το θεώρημα του Pompeiu?


I don't know if this the right forum, but what is the Pearson product-moment correlation?  The problem gives a table of x's and y's and asks me to use a graphing calculator to find the Pearson product-moment correlation.

http://en.wikipedia.org/wiki/Pearson_product-moment_correlation_coefficient

This isn't really the right forum for this kind of question.


induction
number theory unsolved
number theory

Given a natural number $n \geq 2$, consider all the fractions of the form $\frac{1}{ab}$, where $a$ and $b$ are natural numbers, relative primes and such that:
$a < b \leq n$,
$a+b>n$.
Show that for each $n$, the sum of all this fractions are $\frac12$.

[quote="Megus"]It is easy to use induction on this one. If by $T_n$ you denote desired sum and by $H_n= \sum_{i=1}^n \frac{1}{i}$ then $T_{n+1}=T_n + \frac{H_n}{n+1} - \frac{H_{n-1}}{n}$[/quote]

But according to the problem, $T_{n+1} = T_n = \frac{1}{2}$, which implies that

$n H_n = (n+1)H_{n-1}$,

which is false for $n>2$.

Does anyone have a valid solution for this problem?

[quote="mavropnevma"]Lemma. $A = \sum_{(a,n+1)=1} \dfrac {1} {a(n+1)} = \dfrac {1} {2} \sum_{(a,n+1)=1} \dfrac {1} {a(n+1-a)} = B$.

Indeed, $A = \dfrac {1} {n+1} \sum_{(a,n+1)=1} \dfrac {1} {a}$, while $B = \dfrac {1} {2(n+1)} \sum_{(a,n+1)=1} \left (\dfrac {1} {a} + \dfrac {1} {n+1-a}\right) = \dfrac {2} {2(n+1)} \sum_{(a,n+1)=1} \dfrac {1} {a}$, since $(a,n+1)=1$ if and only if $(n+1-a,n+1)=1$ if and only if $(a,n+1-a)=1$.
But for $(a,b)=1$, $a+b=n+1$, $1\leq a < b \leq n$ one clearly has $\sum \dfrac {1} {ab} = B$.

Now, when passing from $n$ to $n+1$, the fractions with $b=n+1$ are added, while those with $a+b=n+1$ should be subtracted, hence the value of the sum remains the same, since $A-B=0$. This value is thus that for $n=2$, i.e. $\dfrac {1} {2}$.[/quote]

Great solution!! I knew that after proving that, it solved the problem, but I didn't know how to.


Why is NO3- a stronger oxidizing agent than H+?

edit~if you read what i said before, IGNORE IT!!! it  did not understand your question correctly.

the reason $NO_3$ is a stronger oxidizing agent is that it's reduction half reaction has a higher reduction $E^0$ . the half reactions are as following:

$NO3^- +4H^+ 3e^-\rightarrow NO +2H_2O; E^0=+1.23$

$2H^+ +2e^-\rightarrow H_2; E^0=0.00$


what is the no of solution
           $ 2^x \plus{} 3^x \plus{} 4^x \minus{} 5^x \equal{} 0$ :(

Rewrite your equation as $ \left (\frac {2} {5}\right )^x \plus{} \left (\frac {3} {5}\right )^x \plus{} \left (\frac {4} {5}\right )^x \equal{} 1$. Left-Hand-Side is a strictly decreasing continuous function,
taking value $ 3$ at zero, and with limit $ 0$ at infinity, so it crosses value $ 1$ exactly once.

The approximate solution is 2.373294368404781.


limit
logarithms
calculus
function
calculus computations

Prove that $ \lim_{x\to0}x\ln x\equal{}0.$

Don't apply L'Hôpital's Rule.

thats kind of clear but let $ x\equal{}e^{\minus{}t}$ then we have:

$ L\equal{}\minus{}\lim_{t\rightarrow\plus{}\infty}\frac{t}{e^{t}}\equal{}0$

[quote="sinajackson"]
$ L \equal{} \minus{} \lim_{t\rightarrow \plus{} \infty}\frac {t}{e^{t}} \equal{} 0$[/quote]
Yeah, but, can you explain why $ \lim_{t\rightarrow \plus{} \infty}\frac {t}{e^{t}} \equal{} 0$? I know it's 'cause $ e^t$ grows faster than $ t.$ I'd like another explanation.

At first because $ t>0$ and $ e^t>0$, $ \lim_{t\rightarrow \plus{} \infty}\frac {t}{e^{t}} \ge 0$. At second, for positive $ t$:$ e^t\equal{}1\plus{}t\plus{}\frac{1}{2}t^2\plus{}...>\frac{1}{2}t^2$, so
\[ \lim_{t\rightarrow \plus{} \infty}\frac {t}{e^{t}}\le\lim_{t\rightarrow \plus{} \infty}\frac {t}{\frac{1}{2}t^2}\equal{}\lim_{t\rightarrow \plus{} \infty}\frac {2}{t}\equal{}0\]
 and we have that $ \lim_{t\rightarrow \plus{} \infty}\frac {t}{e^{t}}\equal{}0$

[quote="greenvert"]At first because $ t > 0$ and $ e^t > 0$, $ \lim_{t\rightarrow \plus{} \infty}\frac {t}{e^{t}} \ge 0$. At second, for positive $ t$:$ e^t \equal{} 1 \plus{} t \plus{} \frac {1}{2}t^2 \plus{} ... > \frac {1}{2}t^2$, so
\[ \lim_{t\rightarrow \plus{} \infty}\frac {t}{e^{t}}\le\lim_{t\rightarrow \plus{} \infty}\frac {t}{\frac {1}{2}t^2} \equal{} \lim_{t\rightarrow \plus{} \infty}\frac {2}{t} \equal{} 0
\]
and we have that $ \lim_{t\rightarrow \plus{} \infty}\frac {t}{e^{t}} \equal{} 0$[/quote]

Are you sure we can write: $ e^t \equal{} 1 \plus{} t \plus{} \frac {1}{2}t^2 \plus{} ... > \frac {1}{2}t^2$ ?
remember $ t\rightarrow\infty$!!! and we can write that just in case $ t\rightarrow0$ ...

but we can say that $ e^{t} > t^{2}$ for big $ t$ and continue!

[quote="sinajackson"]remember $ t\rightarrow\infty$!!! and we can write that just in case $ t\rightarrow0$ ...[/quote]

It is an easy exercise in calculus to show that $ e^x > 1 \plus{} x \plus{} \frac{x^2}{2!} \plus{} \cdots \plus{} \frac{x^n}{n!}$ for all $ x > 0$. This even does not need knowledge about power series, and if you know calculus, this becomes much more simple. all terms of $ e^x \equal{} \sum_{n\equal{}0}^{\infty} \frac{x^n}{n!}$ are positive if $ x > 0$, so greenvert's argument is valid.

[quote="sos440"][quote="sinajackson"]remember $ t\rightarrow\infty$!!! and we can write that just in case $ t\rightarrow0$ ...[/quote]

It is an easy exercise in calculus to show that $ e^x > 1 \plus{} x \plus{} \frac {x^2}{2!} \plus{} \cdots \plus{} \frac {x^n}{n!}$ for all $ x > 0$.[/quote]
thats right sos440, i didnt deny this :D , i just said that we cant have the equality for $ x\rightarrow\infty$! but $ >$ doesnt make any problem! :)

Here's an argument. (I don't claim it's a [i]good[/i] argument but at least it's different.)

Let $ s>0$ be arbitrary and consider $ f(x)\equal{}x^s\ln x$ for $ x>0.$

$ f(x)<0$ for $ 0<x<1.$ $ f'(x)\equal{}sx^{s\minus{}1}\ln x\plus{}x^{s\minus{}1}\equal{}x^{s\minus{}1}(s\ln x\plus{}1).$

We have that $ f'(x)<0$ for $ 0<x<e^{\minus{}1/s}.$

Consider the behavior of $ f(x)$ as $ x$ decreases towards zero. The function increases (because we're tracking it backwards; in normal language it is decreasing), and it is bounded above by $ 0.$ Hence $ \lim_{x\to0^\plus{}}x^s\ln x$ exists and is a nonpositive real number for any $ s>0.$

But for any fixed $ s,$ consider $ r$ such that $ 0<r<s.$ Then $ \lim_{x\to0^\plus{}}x^r\ln x\equal{}A$ for some $ A\le0.$ Then

$ \lim_{x\to0^\plus{}}x^s\ln x\equal{}\lim_{x\to0^\plus{}}x^{s\minus{}r}\cdot x^r\ln x\equal{}0\cdot A\equal{}0.$

Just prove that $ \minus{} 2\sqrt {x} < x\ln x < 0\ (0 < x < 1)$. :lol:


Just registered a few days ago but now I've forgotten my password. Can webpage admin send me a new password?

Click on the forgot password link and supply your username and email and a reset password link will be emailed to you. After that you can change your password in the Preferences menu.


How many ten letter codes can you generate(using letters of the english alphabet) such that the letters a and b never occur together.Repetition of letters is allowed. :lol:  :D

Not sure
[hide="Answer"]$ 26 * 25^9$ :?:[/hide]

consider the letters a and b as a single block....find the  no of such permutations....then find the total number of permutations possible with letters of the english alphabet...then subtract the former from the latter

we have to post the answer not theoretical comments on the question
 :maybe:

Yes i asked for an answer not theoretical comments :rotfl:
Somebody in the back of my head called "intuition" is telling me that i should probably use recursion. :rotfl:

[hide="Solution"]$ A(n)$ = number of sequences of length $ n$ that end with a.
$ B(n)$ = number of sequences of length $ n$ that end with b.
$ E(n)$ = number of sequences of length $ n$ that end with some other letter.
$ F(n)$ = number of sequences of length $ n$ that don't have a, b consecutively.

We have the following relations:
$ F(n) = A(n) + B(n) + E(n)$
$ E(n) = 24F(n - 1)$
$ F(n + 1) = 25A(n) + 25B(n) + 26E(n)$

So,
\begin{align*} F(n + 1) & = 25A(n) + 25B(n) + 26E(n) \\
& = 25A(n) + 25B(n) + 26(F(n) - A(n) - B(n)) \\
& = 26F(n) - (A(n) + B(n)) \\
& = 26F(n) - (F(n) - E(n)) \\
& = 25F(n) + E(n) \\
F(n + 1) & = 25F(n) + 24F(n - 1) \end{align*}

We also have intial conditions $ F(1) = 26$ and $ F(2) = 26^2 - 2 = 674$.
Plug and chug using the recursion to find $ F(10)$.

Finally, $ F(10) = \boxed{137565093115874}$.

I hope that's right  :maybe: .[/hide]


Functional Analysis
advanced fields
advanced fields unsolved

Let H be a HILBERT space and let $T: H\rightarrow H$ be linear. If
$\langle Tx,y\rangle = \langle x,Ty\rangle$
for all x,y $\in\ H$
then WITHOUT USING Closed graph theorem show that $T$ IS BOUNDED

Suppose $x_n\to x$. We have to show that $Tx_n\to Tx$. I'll use the following link between weak and strong convergence in a Hilbert space: the sequence $(u_n)_n$ converges strongly to $u$ iff it converges weakly to $u$ and, additionally, $\|u_n\|\to\|u\|$. This is a nice exercise, but it's easy enough, so I won't prove it here. The conclusion is that all we need to do is show that $Tx_n$ converges weakly to $Tx$, and $\|Tx_n\|\to\|Tx\|$,

Fix some $y$. We have $(Tx_n,y)=(x_n,Ty)\to(x,Ty)=(Tx,y)$. Since $y$ is arbitrary, this simply means that $Tx_n$ converges weakly to $Tx$ whenever $x_n\to x$. This is true for any symmetric linear transformation, so it holds for $T^2$ as well.

We have $\|Tx_n\|^2=(Tx_n,Tx_n)=(x_n,T^2x_n)=(x_n-x,T^2x_n)+(x,T^2x_n)\ (*)$. Now, since $(T^2x_n)_n$ is a weakly convergent sequence, it means, in particular, that for every $u\in H$ the sequence $(T^2x_n,u)$ is bounded. By the uniform boundedness principle, this implies the boundedness of $\|T^2x_n\|$. Take some $M$ s.t. $\|T^2x_n\|\le M,\ \forall n$. We have $(x_n-x,T^2x_n)\le M\|x_n-x\|\to 0$, so the left summand of the last expression in $(*)$ goes to $0$. The right summand in that expression, on the other hand, converges to $(x,T^2x)=(Tx,Tx)=\|Tx\|^2$, because $T^2x_n$ converges weakly to $T^2x$. We have thus shown that $\|Tx_n\|\to\|Tx\|$, and, according to the observation in the first paragraph, we're done.

P.S.

Even though I didn't use the closed graph theorem, I did use the uniform boundedness principle (another strong result), but in the form in which I used it (weak boundedness of a sequence of elements of $H$ implies strong boundedness) it has, as I recall, an "elementary" proof for Hilbert spaces, i.e. without any reference to Baire's Theorem.


geometry
3D geometry
number theory solved
number theory

Prove that one can find a $n_{0} \in \mathbb{N}$ such that $\forall m \geq n_{0}$, there exist three positive integers $a$, $b$ , $c$ such that

(i) $m^3 < a < b < c < (m+1)^3$;

(ii) $abc$ is the cube of an integer.

Take $a=[\sqrt{m^3}+1]^2,b=[\sqrt{m^3}+1][\sqrt{m^3}+2],c=[\sqrt{m^3}+2]^2$, where $s,t\in\mathbb{N}$.
Clearly we must have $m^3<a<b<c<(m+1)^3$ for sufficiently large m.

The obvious question --- how did you get this?????

You are looking for $abc=k^3$ for some k, but this is pretty hard to find a,b,c in this case. A natural approach will be to consider $abc=k_1^3k_2^3$ instead.

[hide]I'l be obliged if you can post the entire solution at once. I realise you must be very busy but i'm not really catching your hints. Posting a lot of hints will just be a waste of both of our time and space![/hide]

Nice solution, i did in a different way but either with a similiar kind of idea

First i look for a, b ,c such that each is equal to $2x^3$ for some x, but thats not true so i found that theres is only one equal to $2x^3$, but there s also one equal to $6x^3$ and so on:

$a=2{[\frac{m}{\sqrt[3]{2}}+1]}^3 b=6{[\frac{m}{\sqrt[3]{6}}+1]}^3 c=18{[\frac{m}{\sqrt[3]{18}}+1]}^3$

and for sufficient large m

$m^3 < a < b < c < (m+1)^3$

[quote="Ravi B"]I got the same numbers that mecrazywong did.  I tried to think of 3 expressions that multiplied to a cube.  After a while, I hit upon $x^2$, $xy$, and $y^2$.  To make these three expressions close to each other, I chose $x$ and $y$ to be consecutive integers.

Here is one way to complete the argument.  For $m$ sufficiently large, we have
\[ m^{3/2} + 2 < (m+1)^{3/2}.  \]
Thus there are two integers $x$ and $y$ such that
\[ m^{3/2} < x < y < (m+1)^{3/2}.  \]
Squaring gives
\[ m^3 < x^2 < xy < y^2 < (m+1)^3.  \]
Thus we can choose $a = x^2$, $b = xy$, and $c = y^2$.[/quote]

That is a very beautiful solution! Solving for a bit gives for all $m \geq 3$

take n0=20

for m>=20

m^3 < (m-3)*((m+2)^2) < (m-3)*(m+2)*(m+3) < (m-3)*((m+3)^2)<(m+1)^3

check by expanding..

Trivial solution, although it required me some time to arrive at...


You know the stars in your personal info on the side of every post? How many posts do you need to get
1/2 a star?
1 star?
1 1/2 stars?
etc...

To get 4 stars, you click the search button.

To get 5 stars, you type in what you want to search and hit enter. For example, your question.

Do not pass GO. Do not collect $ \$200$.

Thanks. Also, I found it--the search function works again =]

[quote="Dojo"]Well since you couldn't find it, I found it for you.  :maybe:
0 - 19 New Member (Zero Stars)
20 - 49 P versus NP (Half Star)
50 - 99 Hodge Conjecture (One Star)
100 - 249 Poincare Conjecture (Two Stars)
250 - 499 Riemann's Hypothesis (Two and Half stars)
500 - 999 Yang Mills Theory (Three Stars)
1000 - 2499 Navier-Stokes Equation (Four Stars)
2500 - Birch & Swinnerton Dyer. (Five Stars)[/quote]

I thought it was 0-24 New Member and 25-49 P versus NP...


a) prove that there are no right triangles with all integer sides such that the lengths of both legs are powers of 2.

b) find the smallest integer $k$ such that there exists a right triangle with integer hypotenuse and legs of length $k^{a}$, $k^{b}$ for integers $a,b$.

edit: forgot part b

[quote="chess64"]prove that there are no right triangles with all integer sides such that the lengths of both legs are powers of 2.[/quote]

[hide="Uh?"]$\sqrt{2^{2a}+2^{2b}}$, WLOG $a\ge b$, $\sqrt{2^{2b}(2^{2(a-b)}+1)}=2^{b}\sqrt{2^{2(a-b)}+1}$, $2^{2(a-b)}$ is a square, so $2^{2(a-b)}+1$ is not a square, so $2^{b}\sqrt{2^{2(a-b)}+1}$ is not an integer.[/hide]

oops, i forgot part b... i just added it

It seems like kind of a silly question to me (both parts).  [hide="Because, I mean..."] Obviously if we're looking at primitive triples, one of the legs must be $1$, and $1$ is not part of any integer triples.  [/hide]


geometry
circumcircle
Euler
perimeter
LaTeX
national olympiad

1. http://www.mathlinks.ro/Forum/viewtopic.php?t=137067
Let $ABC$ be a triangle and $M,N,P$ be th midpoints of sides $BC, CA, AB$. The lines $AM, BN, CP$ meet the circumcircle of $ABC$ in the points $A_{1}, B_{1}, C_{1}$. Show that the area of triangle $ABC$ is at most the sum of areas of triangles $BCA_{1}, CAB_{1}, ABC_{1}$.

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=137071
Consider $p$ a prime number and $p$ consecutive positive integers $m_{1}, m_{2}, \ldots, m_{p}$. Choose a permutation $\sigma$ of $1, 2, \ldots, p$. Show that there exist two different numbers $k,l \in \{1,2, \ldots, p\}$ such that $m_{k}m_{\sigma(k)}-m_{l}m_{\sigma(l)}$ is divisible by $p$.

3. http://www.mathlinks.ro/Forum/viewtopic.php?t=6188
Let $ABC$ be a triangle with all angles $\leq 120^{\circ}$. Let $F$ be the Fermat point of triangle $ABC$, that is, the interior point of $ABC$ such that $\angle AFB = \angle BFC = \angle CFA = 120^\circ$. For each one of the three triangles $BFC$, $CFA$ and $AFB$, draw its Euler line - that is, the line connecting its circumcenter and its centroid.

Prove that these three Euler lines pass through one common point.

[i]Remark.[/i] The Fermat point $F$ is also known as the [b]first Fermat point[/b] or the [b]first Toricelli point[/b] of triangle $ABC$.

4. http://www.mathlinks.ro/Forum/viewtopic.php?t=137070
Consider a convex polygon $A_{1}A_{2}\ldots A_{n}$ and a point $M$ inside it. The lines $A_{i}M$ intersect the perimeter of the polygon second time in the points $B_{i}$. The polygon is called balanced if all sides of the polygon contain exactly one of points $B_{i}$ (strictly inside). Find all balanced polygons.

[Note: The problem originally asked for which $n$ all convex polygons of $n$ sides are balanced. A misunderstanding made this version of the problem appear at the contest]
Thanks to iura :)

Here's the PDF version. I'm not too familiar with LaTex (In fact I'm doing one of my first files now!), so could someone give it the OK?


The roots of an equation $ x^{3}\minus{}148x\plus{}672\equal{}0$ are A,B, and C. If A and B are in the ratio of 3 to 4, find AB+C.
A) 34
B) 36
C) 42
D) 62
E) None of the above

did you brute force that or did you set something up using vieta's?

hello
 [hide="solution"]
Let the roots be 3a,4a and -7a
84$ a^3$=672
a=2

AB+C=34  (a)[/hide]

Bachukas  sorry for the previous post.
I have edited my post.

The roots are 3a,4a and -7a, not 3a,4a and 7a, so $ AB \plus{} C \equal{} 6\cdot8 \plus{} ( \minus{} 14) \equal{} 34$


How to do qn 5,6 and 7

http://unitest.acer.edu.au/sample-questions/uniTEST_SampleQuestions.pdf

[hide="Question 5"]
Dale worked for 1 hr. 20 min. or 80 min. 80 min. minus 15 min. (for the initial period) is 65 min. 65 min. divided by 15 is 4.333333... billing sections which rounds up to 5 billing sections of 15 dollars each. 5 times 15 is 75 dollars plus the initial 60 dollars is 135 dollars or answer $ \boxed{D}$.
[/hide]

[hide="Questions 6 and 7"]
The code does not take into account the initial 15 min. work for which Dale charges 60 dollars when Dale works for more than 15 minutes. However, the code does include code to account for 15 dollars worth of the first 15 minutes. Thus, the program will charge 45 dollars less than it should for all jobs exceeding 15 minutes which is answer $ \boxed{B}$. This makes the answer to #7 $ \boxed{B}$ as well.
[/hide]


Let $S$ be the set of all real solutions of the equation
\[ \sec \left[ \arctan \frac{1}{1+e^x} - \arctan (1-e^x) \right] = \frac{\sqrt{5}}{2} \]
Then:

a) $S=\emptyset$
b) $S=\mathbb{R}$
c) $S \subset [1, 2]$
d) $S \subset [-1, 1]$
e) $S = [-1, 2]$

[hide]Let $\arctan \frac{1}{1+e^x}-\arctan (1-e^x)=\theta$.

$\sec \theta=\frac{\sqrt{5}}{2} \Rightarrow \sec^2 \theta=\frac{5}{4}\Rightarrow \tan^2 \theta=\frac{1}{4} \Rightarrow \tan \theta=\pm \frac{1}{2}$.

Use the subtraction formula for tangent to get $\tan \theta=\frac{\frac{1}{1+e^x}-(1-e^x)}{1+\frac{1-e^x}{1+e^x}}$.

Simplify the fraction by multiplying top and bottom by $1+e^x$: $\tan \theta=\frac{1-(1-e^x)(1+e^x)}{2}=\frac{e^{2x}}{2}$.

Since this must equal $\pm \frac{1}{2}$, $e^{2x}=\pm 1$. When -1, no solution and when 1, $x=0\Rightarrow \boxed{D}$.[/hide]

[hide]Let $\theta = \arctan \frac{1}{1+e^x}$ and $\alpha =\arctan (1-e^x)$.

Hence

$\sec (\theta - \alpha) = \frac{\sqrt{5}}{2} \Leftrightarrow \cos(\theta - \alpha)=\frac{2}{\sqrt{5}}\Leftrightarrow \cos \theta \cos \alpha + \sin \theta \sin \alpha = \frac{2}{\sqrt{5}} \Rightarrow$

$\frac{1+e^x}{\sqrt{1+(1+e^x)^2}} \cdot \frac{1}{\sqrt{1+(1-e^x)^2}} + \frac{1}{\sqrt{1+(1+e^x)^2}} \cdot \frac{1-e^x}{\sqrt{1+(1-e^x)^2}}=\frac{2}{\sqrt{5}}\Leftrightarrow$

$\frac{2}{\sqrt{4+e^{4x}}}=\frac{2}{\sqrt{5}} \Rightarrow 4+e^{4x} = 5 \Rightarrow x=0 \Rightarrow \boxed{D}$.[/hide]


Let x, y , z be positive real numbers such that:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$
Show that:
$\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+yx} \ge \sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}$

ahem

al.M.V, the meaning is to have fun solving the problem, not to post a link to the solution :D

well, put a=1/x b=1/y c=1/z, then the inequality becomes:
 :Sigma:  :sqrt: (a+bc)  >= 1+ :Sigma:   :sqrt: ab

now, :sqrt: (a 2 +ab+bc+ca)>=a+ :sqrt: bc because its equivalent to ab+ac>=2a :sqrt: bc. suming the analogous inequalities, and taking into acount that a+b+c=1 and that  a 2 +ab+bc+ca=a(a+b+c)+bc=a+bc one gets the desired result.

I visited the link, but I don't really understand how to do it with a short proof.
I tried a different (longer) method, could anyone comment on my proof?  Maybe there is a flaw to it.  I skipped a few basic steps because it took up too much space and I assume that they are obvious.  This seems to be too long a proof.  If there is a way to write the proof in a similar way and shorter, could anyone comment on it please?

Multiply the first equation by $ \sqrt{abc}$
Substitute the value of $ \sqrt{abc}$ in the inequality with the value of the new equation we got from above.
Then if we prove that
$ \sqrt{a+bc}\ge\frac{\sqrt{bc}}{\sqrt{a}}+\sqrt{a}$ , $ \sqrt{b+ca}\ge\frac{\sqrt{ac}}{\sqrt{b}}+\sqrt{b}$ and $ \sqrt{c+ab}\ge\frac{\sqrt{ab}}{\sqrt{c}}+\sqrt{c}$ the statement holds true.

For $ \sqrt{a+bc}\ge\frac{\sqrt{bc}}{\sqrt{a}}+\sqrt{a}$
$ a+bc\ge\frac{bc}{a}+2\sqrt{bc}+a$
$ bc\ge\frac{bc}{a}+2\sqrt{bc}$
$ bc-\frac{bc}{a}\ge2\sqrt{bc}$
$ bc(1-\frac{1}{a})^{2}\ge4$
$ \frac{c}{b}+\frac{b}{c}\ge\2$ has to be true.

Let $ X=\frac{c}{b}+{b}{c}$
$ \frac{dX}{db}=-\frac{c}{b^{2}}+\frac{1}{c}$
stationary pt when $ -\frac{c}{b^{2}}+\frac{1}{c}=0$
$ \frac{c}{b^{2}}=\frac{1}{c}$
therefore $ b=c$

$ \frac{d^{2}X}{db^{2}}=2\frac{c}{b^{3}}$
When $ b=c$, $ =2\frac{c}{c^{3}}$
As $ c$ is a positive real, $ 2\frac{1}{c^{2}}>0$
Therefore $ b=c$ is a minimum value.
Therefore $ \frac{c}{b}+\frac{b}{c}=\frac{c}{c}+\frac{c}{c}=2$
Therefore $ \frac{c}{b}+\frac{b}{c}\ge\2$ is true.

Prove the same way for the rest of the two inequalities.


Standard trick:
1) Calculate a polynomial $b_{n}$ with $b_{n+1}-2b_{n}=-n^{2}+2n$. Identifying coefficients, $b_{n}=n^{2}+1$, so $b_{1}=2$.
2) Calculate the homegeneous $c_{n}$ with $c_{n+1}-2c_{n}=0$: $c_{n}=2^{n}c_{1}$.
3) $a_{n}=b_{n}+c_{n}$
The only remaining thing to calculate is $c_{1}=a_{1}-b_{1}=-1$. So, $a_{n}=-2^{n}+n^{2}+1$.


Solve the equation
$ \sqrt{x\plus{}\sqrt{x^2\minus{}1}}\equal{}\frac{9\sqrt{2}}{4}(x\minus{}1)\sqrt{x\minus{}1}$

[quote="Algadin"]Solve the equation
$ \sqrt {x + \sqrt {x^2 - 1}} = \frac {9\sqrt {2}}{4}(x - 1)\sqrt {x - 1}$[/quote]
Let $ a=\sqrt{x-1}$, $ b=\sqrt{x+1}$. Then $ b^2-a^2=2$.
There fore :
$ \{ \begin{array}{l} b^2-a^2=2 \\2(a+b)=9a^3 \end{array} $ $ \leftrightarrow \{ \begin{array}{l} b^2=a^2+2 \\2b=9a^3-2a \end{array} $ $ \leftrightarrow \{ \begin{array}{l} b^2=a^2+2 \\81a^6-36a^4-8=0 \end{array} $
$ \leftrightarrow \{ \begin{array}{l} b^2=a^2+2 \\a^2=\frac{2}{3} \end{array} $ $ \leftrightarrow \{ \begin{array}{l} b^2=\frac{8}{3} \\2b=9a^3-2a \end{array} $ $ \leftrightarrow x=\frac{5}{3}$  :)

hello, for $ x \in \mathbb{R}$ we have the range of definition $ x\geq1$. By squaring both sides of the equation we get: $ \sqrt{x^2\minus{}1}\equal{}\frac{81}{8}(x\minus{}1)^3\minus{}x$. The inequality $ \frac{81}{8}(x\minus{}1)^3\minus{}x\geq0$ ist fulfilled for all $ x$ with $ \frac{1}{27}(972\plus{}12\sqrt{6465}^{\frac{1}{3}}\plus{}\frac{8}{9(972\plus{}12\sqrt{6465})^{\frac{1}{3}}}\plus{}1\le x <\infty$.Squaring again and factoring we have
$ \minus{}\frac{1}{64}(3x\minus{}5)(27x^2\minus{}48x\plus{}25)(81x^3\minus{}207x^2\plus{}171x\minus{}53)\equal{}0$. From here we have the solution $ x\equal{}\frac{5}{3}$.
remark: $ x\equal{}\frac{1}{27}(908\plus{}36\sqrt{633})^{\frac{1}{3}}\plus{}\frac{16}{27(908\plus{}36\sqrt{633})^{\frac{1}{3}}}\plus{}\frac{23}{27}$ is not in our range of definition.
Sonnhard.


Let $ az^2 \plus{} bz \plus{} c \equal{} 0$ be a quadratic with complex coefficients
Then after some manipulation we get
$ (2az \plus{} b)^2 \equal{} \Delta$. where $ \Delta \equal{} b^2 \minus{} 4ac$
Now let $ y \equal{} (2az \plus{} b)$ and $ \Delta \equal{} u \plus{} vi$ where $ u,v$ are real
$ y^2 \equal{} u \plus{} vi$
Then $ y \equal{} \pm(\sqrt {\frac {|\Delta| \plus{} u}{2}} \plus{} ($sign[v]$ )\sqrt {\frac {|\Delta| \minus{} u}{2}})$ where sign[v] is the sign of the number v
I don't understand how the above conlusion was reached :?: .

Thanks for your time  :)

EDIT: OOPS :oops:

How do we have $ (2az\plus{}b)^2\equal{}\Delta$ and $ (2az\plus{}b)^2\equal{}y,$ then $ y^2\equal{}\Delta?$

[quote="RoFlLoLcOpT"]How do we have $ (2az \plus{} b)^2 \equal{} \Delta$ and $ (2az \plus{} b)^2 \equal{} y,$ then $ y^2 \equal{} \Delta?$[/quote]
Srry, its edited now

Write in polar form: $ y^2\equal{}u\plus{}vi\equal{}re^{\theta i}$: we have

$ r\cos\theta\equal{}u$, $ r\sin\theta\equal{}v$, $ r\equal{}\sqrt{u^2\plus{}v^2}\equal{}|\Delta|$.

Then, $ y\equal{}r^\frac12e^{(\frac{\theta}2\plus{}\pi k)i}$ for $ k\equal{}0,1$, by DeMoivre's.

That should give you:

$ y\equal{}\pm r^\frac12\left(\cos\frac{\theta}2\plus{}i\sin\frac{\theta}2\right)$

$ \cos^2\frac{\theta}2\equal{}\frac{1\plus{}\cos\theta}2$ and $ \sin^2\frac{\theta}2\equal{}\frac{1\minus{}\cos\theta}2$ follow from the double-angle formulas.

$ \cos\theta\equal{}\frac{u}{r}$, $ \sin\theta\equal{}\frac{v}{r}$ implies $ \cos^2\frac{\theta}2\equal{}\frac{1\plus{}\frac{u}{r}}2$ and $ \sin^2\frac{\theta}2\equal{}\frac{1\minus{}\frac{u}{r}}2$.

$ y\equal{}\pm\left(\sqrt{\frac{|\Delta|\plus{}u}2}\plus{}i\sqrt{\frac{|\Delta|\minus{}u}2}\right)$

Discussion: the 'y' you have written is a [b]real number[/b] as opposed to the 'y' I have found is a [b]complex number[/b]. And the 'y' form that I have written is [u]correct[/u] as it stands when applying DeMoivre's. I triple checked every single detail and researched [b]all[/b] the formulas to make sure I'm not insane, so yea...

Thank you, and my "y" is also a complex number

[quote="Complex_Ninja"]$ y \equal{} \pm(\sqrt {\frac {|\Delta| \plus{} u}{2}} \plus{} ($sign[v]$ )\sqrt {\frac {|\Delta| \minus{} u}{2}})$ where sign[v] is the sign of the number v[/quote]This 'y' is a real number.

If that's the book's answer, it's still wrong anyways, strictly 'speaking' :P


I am not very interest in inequalities so I will post solution with some property of function.
Because $ f(x)\equal{}x^m$ ($ m>1)$ is convect then use Jensen 's inequalities we have :
$ \sum_{i\equal{}1}^n (a_i\plus{}\frac{1}{a_i})^m\geq \ n(\frac{\sum_{i\equal{}1}^n a_i\plus{}\frac{1}{a_i}}{n})^m$
Other :
$ \sum_{i\equal{}1}^n\frac{1}{a_i}\geq \frac{n^2}{\sum_{i\equal{}1}^n a_i}$

If $ m\equal{}1$ $ \sum \left(a_i\plus{}\frac{1}{a_i}\right)\ge n^2\plus{}1$ is an immediate consequence of Cauchy-Schwarz.

If $ m>1$ then
$ \left(\sum\left(a_i\plus{}\frac{1}{a_i}\right)^m\right)^\frac{1}{m}\left(\sum1^\frac{m}{m\minus{}1}\right)^\frac{m\minus{}1}{m}\ge
\sum\left(a_i\plus{}\frac{1}{a_i}\right)$ (Holder).
Raising both sides to the m-th power we obtain
$ \sum\left(a_i\plus{}\frac{1}{a_i}\right)^m\cdot n^{m\minus{}1} \ge \sum\left(a_i\plus{}\frac{1}{a_i}\right)^m$.
Because $ f(x)\equal{}x^m$ is monotonically increasing ($ m>1$) and $ \sum\frac{1}{a_i}\ge \frac{n^2}{\sum a_i}\equal{}n^2$ (CS),
$ \sum\left(a_i\plus{}\frac{1}{a_i}\right)^m\ge \frac{1}{n^{m\minus{}1}}\left(n^2\plus{}1\right)^m\equal{}n\left(n\plus{}\frac{1}{n}\right)^m$
with equality for $ a_1\equal{}a_2\equal{}...\equal{}a_n\equal{}\frac{1}{n}$.


linear algebra
matrix
linear algebra solved

Let be the set $\mathcal M$
\[
\mathcal M=\left\{\left(\begin{array}{lll}
x&0&-x\\
0&y&0\\
-x&0&x
\end{array}\right)\mid x,y\in\mathbb C\right\}
\]
Prove that $(\mathcal M,+,\cdot)$ is a commutative ring and is isomorphic with the set $\mathbb C$X$\mathbb C$

Notice that
$\begin{pmatrix}
x&0&-x\\
0&y&0\\
-x&0&x
\end{pmatrix}\begin{pmatrix}
z&0&-z\\
0&w&0\\
-z&0&z
\end{pmatrix}
=
\begin{pmatrix}
z&0&-z\\
0&w&0\\
-z&0&z
\end{pmatrix}
\begin{pmatrix}
x&0&-x\\
0&y&0\\
-x&0&x
\end{pmatrix}
=
\begin{pmatrix}
2xz&0&-2xz\\
0&yw&0\\
-2xz&0&2xz
\end{pmatrix}
$
Then it is easy to verify that $\mathcal M$ is a commutative ring.

Define $f:\mathcal M\to \mathbb{C}\oplus\mathbb{C}$ as
$f\left(
\begin{pmatrix}
x & 0 & -x\\
0 & y & 0\\
-x& 0 &x
\end{pmatrix}
\right)=(2x,y)$
It is easy to verify that $f$ is an isomorphism of rings.


If a triangle has one side that is 9 cm and a second side that is 3 cm less than twice the third side, what are the possible lengths for the second and thirds sides?

[hide="Answer"]
$ 4 < x < 12$, where x is the shorter side.

If x is the shortest side, 2x-3 is the other unknown side, they add up to more than 9, hence 3x-3>9, x>4.
Also, x+9>2x-3, hence x<12. The inequality (2x-3)+9>x is always true.
[/hide]


1. Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \left( \frac{a}{a\plus{}b}\right)^2 \plus{}\left( \frac{b}{b\plus{}c}\right)^2\plus{}\left( \frac{c}{c\plus{}a}\right)^2 \le \frac{3\plus{}\sqrt{5}}{2}\frac{a^2\plus{}b^2\plus{}c^2}{(a\plus{}b\plus{}c)^2}\]
With equality holds if and only if $ (a,b,c)\equal{}\left( \frac{\sqrt{5}\plus{}1}{2},1,0\right)$.
2. Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \frac{a^3}{a\plus{}b}\plus{}\frac{b^3}{b\plus{}c}\plus{}\frac{c^3}{c\plus{}a} \le \left( \frac{\sqrt[3]{4}}{3}\plus{}1\right) \frac{a^3\plus{}b^3\plus{}c^3}{a\plus{}b\plus{}c}\]
With equality holds if and only if $ (a,b,c)\equal{}\left( 0,1,\sqrt[3]{2}\right)$.
:)

[quote="can_hang2007"]1. Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \left( \frac {a}{a \plus{} b}\right)^2 \plus{} \left( \frac {b}{b \plus{} c}\right)^2 \plus{} \left( \frac {c}{c \plus{} a}\right)^2 \le \frac {3 \plus{} \sqrt {5}}{2}\frac {a^2 \plus{} b^2 \plus{} c^2}{(a \plus{} b \plus{} c)^2}
\]
With equality holds if and only if $ (a,b,c) \equal{} \left( \frac {\sqrt {5} \plus{} 1}{2},1,0\right)$.
2. Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \frac {a^3}{a \plus{} b} \plus{} \frac {b^3}{b \plus{} c} \plus{} \frac {c^3}{c \plus{} a} \le \left( \frac {\sqrt [3]{4}}{3} \plus{} 1\right) \frac {a^3 \plus{} b^3 \plus{} c^3}{a \plus{} b \plus{} c}
\]
With equality holds if and only if $ (a,b,c) \equal{} \left( 0,1,\sqrt [3]{2}\right)$.
:)[/quote]
The general of 2) is
\[ \frac{a^n}{a\plus{}b}\plus{}\frac{b^n}{b\plus{}c}\plus{}\frac{c^n}{c\plus{}a} \le \left( \frac{(n\minus{}1)^{\frac{n\minus{}1}{n}}}{n}\plus{}1\right) \frac{a^n\plus{}b^n\plus{}c^n}{a\plus{}b\plus{}c}\]
for all $ a,b,c \ge 0, k \ge 1$.
I have proved this one for all $ k \ge 3$ and $ k \equal{}2$. But I cannot prove it for all $ 1 \le k \le 3$, I think it is true, too. Could anyone help me prove it or disprove it? Thanks

[quote="Hong Quy"]i  wanna see   solutions[/quote]
Take a look here: http://www.mathlinks.ro/viewtopic.php?t=208794 :)


Let $ f: [0,\plus{}\infty\rangle \rightarrow \langle 0,1],f\in C^{\infty}([0,\plus{}\infty\rangle)$ such that $ \forall k\in\mathbb{N}_0,x\in [0,\plus{}\infty \rangle$
$ (\minus{}1)^k f^{(k)}(x)\geq 0$. Define $ g: \langle 0,\plus{}\infty\rangle \rightarrow \mathbb{R}$, with $ g(x)\equal{}\frac{1\minus{}f(x)}{x}$. Prove that for $ \forall k\in\mathbb{N}_0,x\in \langle 0,\plus{}\infty\rangle$
  $ (\minus{}1)^k g^{(k)}(x)\geq 0$.

I got that $ \frac{d}{dx}( (\minus{}1)^{k\minus{}1}g^{(k\minus{}1)}(x) x^k)\geq 0$, what is not exactly what we're looking for.

Use Leibniz rule:
\[ (f\cdot g)^{(n)} \equal{} \sum_{k \equal{} 0}^{n}\binom{n}{k}f^{(k)}g^{(n \minus{} k)}
\]

Here we have $ f\equal{}\frac1{x}$ and $ g\equal{}1\minus{}f(x)$.


function
logarithms
number theory
totient function
relatively prime
real analysis
real analysis solved

Let f be Euler's totient function. Compute the limit of the Cesaro sequence associated to the sequence $ x_n=f(n)/n$. Find a good asimptotic equivalent of this Cesaro-sequence.

What's the Cesaro sequence associated to another sequence?

And what is totient function? There are many Euler's functions and their names may differ from country to country.

The totient function is $\phi(n)$, the number of numbers $\leq n$ and coprime with $n$.

Is the Cesaro sequence associated to $\frac{f(n)}n$ the sequence $\frac{\phi(n+1)-\phi(n)}{n+1-n}=\phi(n+1)-\phi(n)$?

It is called Euler's function here in Russia.

In Romania too, but if you google "totient function" you'll see that this name is really popular in English.

If Cesaro sequence is $\phi(n+1)-\phi(n)$ then harazi's question is meaningless.

Yeah, I know, so I suppose it was a rather stupid question. Anyway, I think we should wait for his reply :).

[quote="Moubinool"]What about the limit of $\frac{x_1+x_2+...+x_n}{n}$ ?[/quote]
What does it mean? Is it explanation of harazi's question?

Using identities
1) $\frac{\varphi(n)}{n}=\sum_{t|n}\frac{\mu(t)}{t}$
and
2) $\sum_{t=1}^\infty \frac{\mu(t)}{t^2}=\frac{1}{\zeta(2)}=\frac{6}{\pi^2}$
we can easily obtain
\[\frac{1}{n}\sum_{k=1}^{n}\frac{\varphi(k)}{k}=\frac{6}{\pi^2}+O(\frac{\ln n}{n}).\]

Yes harazi very nice result

$\frac{3}{\pi^2}+O(\frac{lnn}{n})$  Walfisz 1963

http://mathworld.wolfram.com/TotientSummatoryFunction.html

Moubinool! What is actually we need to calculate: $\frac{1}{n}\sum_{k=1}^{n}\varphi(k)$ or $\frac{1}{n}\sum_{k=1}^{n}\frac{\varphi(k)}{k}$?

When I saw your anwser myth, I undestood find the limit of $W(n)=\frac{\phi(1)+...+\phi(n)}{n^2}$ it is known  see formulae (3)

http://mathworld.wolfram.com/TotientSummatoryFunction.html

We have to wait harazi to get precision on its question.

Sorry for the late answer. The question is to find an asimptotic behaviour of the sequence $ x_n=\frac{1}{n}*(\frac{f(1)}{1}+\frac{f(2)}{2}+...+\frac{f(n)}{n})$ where f(n) is the number of numbers smaller than n and relatively prime to n.

So I solved this problem. Or you know more precise estimate?

Yes, thanks a lot Myth. Unfortunately, I do not know more precise estimation.


triangle ABC

Squares ABDE and BCGH are constructed on the sides AB and BC of a triangle ABC. Prove the following:

1. The median BM of the triangle formed externally between the squares is also the altitude of triangle ABC.
2. BM = AC/2.

Let $S$ be the midpoint of $AC$ and $N$ be the center of $\Box BCGH$.

From [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=116421[/url],
we know that $\angle SNI=90^\circ$ and $NS=NI$.

So we can rotate $\triangle CNS$ $90^\circ$ with pivot $N$ to be $\triangle BNI$.

From this, we get those 2 properties.