This problem is generalization of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5918]this one[/url].
Suppose $G$ is a graph and $S\subset V(G)$. Suppose we have arbitrarily assign real numbers to each element of $S$. Prove that we can assign numbers to each vertex in $G\backslash S$ that for each $v\in G\backslash S$ number assigned to $v$ is average of its neighbors.

I wanted to post this generalization after seeing these problems:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=468314#p468314[/url]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=468314#p468314[/url]

If $S$ does not meet every connected component of $G$, then it's easy to take care of the components disjoint from $S$: just assign the value $0$ to each point in such a component. This means that it suffices to restrict our attention to the case when $S$ does meet every component, and in this case the proof in that first link works, with only the obvious modifications (that last argument in that topic basically says that in that particular case, $S$ meets every component; since here we're already assuming this, no such argument is needed).

Let $T = V(G) / S.$ We interpret the desired labelling as a system of equations. Consider the adjacency matrix $A$ of the vertices in $T$, with the one difference that we write $-deg(v_i)$ along the main diagonal. In this way, the problem's condition can be interpreted as:

$$A \cdot v = v',$$

where $v$ is the vector of the labellings of the elements of $T$, and $v'$ is some fixed vector. We need to show that this has a solution.

First of all, as [b] grobber [/b] did above, we can WLOG assume that there is no connected component in which $S$ contains no elements.

With this assumption, we're ready to show that the rows of $A$ are linearly independent. Indeed, suppose that they were not for contradiction. Then, if we let $r_i$ be the vector corresponding to the $i$th row of $A$, then there exist real numbers $x_i$ so that
$$\sum x_ir_i = 0.$$
To see that this cannot occur, consider the largest $x_i.$ Then since $-x_ideg(v_i)$ must be summed with other terms to make the $i$th coordinate zero, we know that all neighbors $v_j$ of $v_i$ must be in $T$, and have $x_j = x_i.$ If we continue this logic, we get that the connected component of the subgraph of $G$ spanned by $T$ which contains $x_i$ is adjacent to no vertices of $S$, i.e. it should've been removed by the above assumption, contradiction.

$\square$

Another approach. The idea is at first to assign some number to each vertex of $V(G)\setminus S$ and then iterate assigning to each vertex of $V(G)\setminus S$ the average of its neighbours. If the corresponding real sequences, obtained for each vertex, converge then their limits will do the job.
	It's not certain that each initial assigning will bring convergence, it's even not true, but we show it holds for a specific initial assignment.

	To begin with, it's enough to prove the claim when all numbers assigned to $S$ are non negative. Indeed, in general case, we can represent each number in $S$ as difference of non negative numbers, then get missing numbers to both configurations and finally subtract vertex-wise corresponding values. So, we may assume the assigned numbers to $S$ are non negative.

	Let $M$ be the maximal number assigned to $S$. By $v^{k}_j$ we denote the number assigned to the vertex $v_j\in V$ at step $k$. We set the initial assigning as $v^{(0)}_j:=M\,,\,\forall j,v_j\notin S $. We construct inductively the sequence $v^{k}_j, k=1,2,\dots$
	$$ (*)\,\,\,\,\,\,\,\, v^{(k+1)}_j:=
	\begin{cases}
	\displaystyle \frac{1}{|N(v_j)|} \sum_{v_i\in N(v_j)} v^{(k)}_i\,,& v_j\notin S\\
	v^{(k)}_j\,,& v_j\in S
	\end{cases}
	$$
	where $N(v_j)$ denotes the set of verticies adjacent to $v_j$ . Clearly $v^{(1)}_j\leq v^{(0)}_j\,,\,\forall v_j\in V$.
	Now, by induction, applying $(*)$, we conclude $v^{(k+1)}_j\leq v^{(k)}_j\,,\,\forall v_j\in V;k=0,1,\dots$ . It means the sequence $v^{(k)}_j$ for fixed $j$ and $k=0,1,\dots$ is monotonically decreasing and non negative, hence it converges to $v^*_j$.
	Finally, taking the limit in $(*)$, we conclude the assignment $v^*_j$ satisfies the requirement.


trigonometry
function
trig identities
Law of Cosines

[quote="Hikaru79"]Find the length of a side of a regular polygon of $24$ sides, incribed in a circle of unit radius.[/quote]
[hide] Draw 2 radii which intercept one side of the 24-gon. This forms an isoceles triangle with vertex angle $\frac{360}{24}=15^\circ$. With the Law of Cosines, $x^{2}=1^{2}+1^{2}-2\cdot1\cdot1\cdot\cos15=2-2\cos15=2-2(\cos45\cos30+\sin45\sin30)=2-2(\frac{\sqrt6+\sqrt2}{4})=\boxed{2-\frac{\sqrt6+\sqrt2}{2}}$ [/hide]

You've found $x^{2}$. Still have to take a root of that.

I think there's an easier way.
[hide]
We know that the angle from vertice $v_{1}$ to vertice $v_{2}$ is $15^\circ$ (because $\frac{360}{24}=15$) Lucky for us, two of the well known trig functions are $15$ apart -- $30$ and $45$! So we know that there is a vertice at $(\cos{30},\sin{30})$ and one at $(\cos{45},\sin{45})$. The length of the vertice is just the distance between those two. Plug those two points into a distance formula and you're done. The choice of the points is important because we know exact values for 30 and 45.
[/hide]


topology
induction
geometry
3D geometry
sphere
function
combinatorial geometry

It is a well-known old Kvant problem that any $n$ red and $n$ blue points in the plane, no three of which are collinear, can be grouped into $n$ pairs in such a way that every pair consists of points of different colors, and no two of the $n$ segments formed by the pairs intersect (to show this, just consider the grouping that minimizes the sum of the lengths of these segments). Here is an analogous problem for three dimensions:

[b]Problem. [/b] Consider any set of $3n$ points in $\mathbb{R}^3$, no four of which are coplanar. Color the points in three colors (red, green, blue) arbitrarily, with each color used exactly $n$ times. Is it always possible to partition the points into triples $(A_i,B_i,C_i)$ with $A_i$ red, $B_i$ green, $C_i$ blue, in such a way that the $n$ triangles $A_iB_iC_i$ are pairwise disjoint?

[Edited. ]

--Vesselin

I think this problem is also rather well known. Take a line not parallel to any of the lines determined by the triangles "far away" from the set of points, and then start moving it (keep its direction fixed) towards the set of points. It will meet our points one by one. We form our triangles from the first three points it meets, then the next three, then the next three, and so on.

I didn't say that the problem is difficult  :) . (Only you have to take a [i]plane[/i] instead of a line: the points are in $\mathbb{R}^3$).

Oh, sorry about that. I missed that $\mathbb R^3$ there :).

Oh, silly me, I knew that I had missed something  :blush:  :blush:  :blush:. Now, this will make the problem non-trivial; look back at my edited post above  :) .

I have seen a proof that uses topology, namely the Ham Sandwich theorem.

another general result:Let $A_1,\dots,A_d$ be d pairwise disjoint subsets of $\mathbb{R}^d$,each containing n points ,and suppose that no hyperplane contains $d+1$ of the points in the union of all the sets $A_j$,then there is a partition of $\cup A_j$ into disjoint sets $S_1,\dots S_n$ each containg pricisely one point from each $A_i$,such that the n simplices $conv(S_1),\dots ,conv(S_n)$ are pairwise disjoint.

Can someone prove Vess's edited problem. :(
please and thanks   ;)

Man, is this going to be long.. :)

Assume we prove the following result:

[b]Theorem 1[/b]

Given $3n$ points in space together with another point $P$, no four coplanar, s.t. among the first $3n$ points there are $n$ of each color (red, green, blue), we can find a plane $\pi$ through $P$ s.t. the complement of $\pi$  in $\mathbb R^3$ consists of two disjoint open half-spaces, each one containing equal numbers of points of the three colors.

Then the positive answer to the problem in post #1 would follow by induction on $n$:

Take the point $P$ inside the convex hull of the $3n$ points, not lying on any of the planes determined by the $3n$ colored points. Take a plane $\pi$ as in the statement of Theorem 1. One of the open half-spaces it determined contains $3u$ points, $u$ of each color, while the other one contains $3v$ points, $v$ of each color, with $u+v=n,\ u,v>0$. $u$ and $v$ must be positive because $P$ is inside the convex hull of the $3n$ points, so there are no half-planes through $P$ which have all the $3n$ points on the same side. By the induction hypothesis, we can construct $3$-colored disjoint triangles on the $3u$ points and also on the $3v$ points, since $u,v<n$, so that's that.

In order to prove Theorem 1, we first prove a topologial result (as Ravi suggested in post #6 above):

[b]Theorem 2[/b]

Given three bounded measurable subsets $S_1,S_2,S_3$ of $\mathbb R^3$ with equal volumes and a point $P\in\mathbb R^3$, we can find a plane $\pi$ through $P$ s.t. both half-spaces determined by it contain equal measures of $S_1,S_2,S_3$ (or, in other words, this plane is supposed to divide the measures of $S_i$ in the same fraction).

Theorem 2 follows immediately from the [url=http://en.wikipedia.org/wiki/Borsuk-Ulam_Theorem]Borsuk-Ulam Theorem[/url]:

To each point $X$ on the unit sphere centered at $P$ we associate the plane $\pi_X$ passing through $P$ and orthogonal to $PX$. For a given $X$, let $S_i(X)$ be the measure of the part of $S_i$ that lies on the same side of $\pi_X$ as $X$, and define $f(X)=(S_3(X)-S_1(X),S_3(X)-S_2(X))$. Clearly, $f$ is a continuous function from the unit sphere centered in $P$ to the plane. By the cited theorem, there are two antipodal points $X,X'$ for which $f(X)=f(X')$. However, for every two antipodal points $U,U'$ we have $f(U)=-f(U')$, which means that for $X$  and $X'$ we have $f(X)=f(X')=0$, or, in other words, $S_1(X)=S_2(X)=S_3(X)$, as desired.

The only thing left is to prove Theorem 1 using Theorem 2.

Let the red points be $R_1,\ldots,R_n$. Similarly, we have $G_1,\ldots,G_n$ and $B_1,\ldots,B_n$. Given $m\in\mathbb N$, let $\mathcal S_r(m),\mathcal S_g(m),\mathcal S_b(m)$ be the unions of the balls of radius $\frac 1m$ centered in the red, green, and blue points respectively. Also let $\pi_m$ be the plane through $P$ whose existence is stated in Theorem 2. Finally, let $X_m$ be one of the points on the unit sphere around $P$ s.t. $PX_m$ is orthogonal to $\pi_m$, and $\mathcal H_m$ the half-space determined by $\pi_m$ which contains $X_m$.

Because the $3n$ points together with $P$ are in general position, for large enough $m\ \pi_m$ cuts at most two spheres of radius $\frac 1m$ cetnered at $R_i,G_i,B_i$ This means that for large $m\ \pi_m$ intersects such spheres centered in point of at most two colors. We assume WLOG that $\pi_m$ never cuts spheres centerd in red points, and we also assume WLOG that $(X_m)_m$ is convergent (we can extract a convergent subsequence, but adding extra indices would be too nasty at this point). Let $X$ be the limit of $(X_m)_m$, and let $\pi$ be the plane through $P$ orthogonal to $PX$. $\cal H$ is the half-space determined by $\pi$ which contains $X$.

For large $m$ the difference bwteen the number of spheres of radius $\frac 1m$ centered around red points contained in $\mathcal H_m$ and those contained in the other open half-space becomes stationary: a certain fixed $k$. The spheres centered around green points can be divided into three classes: those contained in $\mathcal H_m$, those contained in the other open half-space, and those cut by $\pi_m$. The difference between the numbers in the first two classes differes from $k$ by at most the number of spheres in the third class. This happens for all large $m$, so when we pass to the limit, the conclusion is that the difference between the number of red points in $\cal H$ and those in the other half-space is $k$, while the difference between the number of green points in $\cal H$ and those in the other half-space differs from $k$ by at most the number of green points contained in $\pi$. The same holds for the blue points. There are at most $2$ colored points on $\pi$, so we can move $\pi$ slightly to get another plane, $\pi'$ say, which still passes through $P$ but satisfies the requirements stated in Theorem 1.

That's about it, I think. I hope it's Ok.

P.S.

The argument above could easily be extended to get a proof of the generalization posted by sam-n in message #7.


calculus
derivative
inequalities
quadratics
algebra
quadratic formula

show that the system

$xe^{x^{2}}+ye^{y^{2}}=3$
$x^{2}+y^{2}=1$

hasn't solution

[quote="ehehheehee"][hide="stuck here"]x^2+y^2=1 is the equation to a unit circle with the origin as the center.

So, we can also express that equation as
x=cos(a)
y=sin(a)

sin(a)e^(sin^2(a))+cos(a)e^(cos^2(a))=3

Now all that is left is to maximize the left side and prove that the maximum is still less than 3.  I think it looks like the max is e, because if cos(a) or sin(a) is not equal to one that small number times e to the power of an even smaller number won't raise it as much as the other term falls[/hide][/quote]
The maximum value is higher than $e$, because if you take derivative of left side and put there $a=0$ it doesn't equal $0$.

I'll take your word for it, I don't know calculus...

We will prove that if $x^{2}+y^{2}=1$, then $xe^{x^{2}}+ye^{y^{2}}<3$. When $x<0$ we have $xe^{x^{2}}+ye^{y^{2}}<(-x)e^{(-x)^{2}}+ye^{y^{2}}<$ (analogous for $y$), so we can assume that $x,y\geq 0$. Applying Cauchy-Schwarz we get $xe^{x^{2}}+ye^{y^{2}}\leq\sqrt{x^{2}+y^{2}}\sqrt{e^{2x^{2}}+e^{2y^{2}}}=\sqrt{e^{2x^{2}}+e^{2y^{2}}}$, so we have to prove that $e^{2x^{2}}+e^{2y^{2}}<9$. Let $a=x^{2}$. Then $0\leq a \leq 1$ and we have to prove that $e^{2a}+e^{2(1-a)}<9$. Let $t=e^{2a}$, $1\leq t\leq e^{2}$. Then the inequality simplifies to $t+\frac{e^{2}}{t}<9\iff t^{2}-9t+e^{2}<0$. By the quadratic formula it is equivalent to $t\in\left( \frac{9-\sqrt{81-4e^{2}}}{2}, \frac{9+\sqrt{81-4e^{2}}}{2}\right)$. We have to prove that the interval $<1,e^{2}>$ enclosed in the interval $\left( \frac{9-\sqrt{81-4e^{2}}}{2}, \frac{9+\sqrt{81-4e^{2}}}{2}\right)$. It is equivalent to two inequalities:
1. $\frac{9-\sqrt{81-4e^{2}}}{2}<1$
2. $e^{2}<\frac{9+\sqrt{81-4e^{2}}}{2}$.
Proof 1. The inequality is equivalent to $\sqrt{81-4e^{2}}>7\iff e^{2}<8$ which is obvious.
Proof 2. It is equivalent to $\sqrt{81-4e^{2}}>2e^{2}-9$. Both sides are positive, so squaring we get $81-e^{2}>4e^{4}-36e^{2}+81 \iff$ $4e^{4}-35e^{2}<0\iff 4e^{2}-35<0\iff e^{2}<\frac{35}{4}$ which is obvious because $e^{2}<8$ and $8<\frac{35}{4}$. The proof of $xe^{x^{2}}+ye^{y^{2}}<3$ is ended.  Of course $xe^{x^{2}}+ye^{y^{2}}<3$ implies $xe^{x^{2}}+ye^{y^{2}}\ne 3$ so we are done.


limit
trigonometry
irrational number
continued fraction
real analysis
real analysis unsolved

Do you know any problems or references on the sequence $\sin(n)^{n}$ ?
thank's

I made the following problem.

Evaluate $\lim_{n\to\infty} \frac{1}{(n+1)^n} \sum_{k=1}^n \sin ^{-1} (\sin k^k).$

[quote="amcavoy"]kunny, is that $\sin\left(k^{k}\right)$ or $\sin\left(k\right)^{k}$?[/quote]

It's $\sin\left(k^k\right).$

I wonder if someone can explain how we might be sure that $\lim_{n\to\infty}\sin^n(n) = 0$. I mean, obviously the terms will be very small for most $n$, but for some $n$, we will have $\sin(n)$ very close to 1. Can we prove that we will never have some $n$ so close to a $k\pi$ that $\sin(n)^n$ still is close to 1? It seems to me that this should never happen ($n$ will be too large) but I don't know how to make a formal argument.

It doesn't go to zero. There exists a $c$ and infinitely many pairs $m,n$ such that $|n-m\pi-\frac\pi2|<\frac cn$, whence $|\sin n|>1-\frac{c^2}{2n^2}$ and $|\sin n|^n>1-\frac{c^2}{2n}$.

This would be much easier to prove for $\cos n$, but I'm sure it's true for $\sin n$ as well.

Oh, right! Now I recall something, is it Hurwitz' irrational number theorem?
If $x$ is irrational, then there are infinitely many $p, q$ so that
$|x - \frac{p}{q}| < \frac{c}{q^2}$ for some $c$ (which I think is even less than 1).
Will that give what we want? Maybe it doesn't, or at least not immediately. But it's the same idea, I guess.

You can take $c=1$ in that theorem- every convergent of the continued fraction is that good. The problem we're looking at is harder, since it's shifted by $\frac12$.

[quote="jmerry"] The problem we're looking at is harder, since it's shifted by $\frac12$.[/quote]
This is to formally justify jmerry's claim. The statement we need is the following: if $x$ is irrational, then there exist infinitely many [b]odd[/b] $n$ such that $\text{dist}(nx,\mathbb Z)\le \frac Cn$ where $C$ is some absolute constant. By the theorem mentioned by Xevarion, we can find infinitely many irreducible fractions $\frac pq$ such that $\left|x-\frac pq\right|<\frac 1{q^2}$. If $q$ is odd, take $n=q$. If $q$ is even, find $n,m\in \mathbb Z$ such that $0<n<q$ and  $np+mq=1$. Note that $n$ must be odd and that $\text{dist}(nx,\mathbb Z)\le \frac 1q+\frac n{q^2}\le \frac 2q<\frac 2n$.


Let $n\geq 2$ be a positive integer and $p$ be a prime divisor of $n$. Prove that there exists a set $A=\{a_{1},a_{2},...,a_{n}\}$ of positive integers such that the product of every two numbers from $A$ is divisible by the sum of every $p$ numbers from $A$.

My solution:
let $x_{1}<\dots< x_{n}$ be postive integers.
Let $X=\prod_{\{i_{1},i_{2},\dots,i_{p}\}\subseteq \{1,\dots,n\}}{(i_{1}+\dots+i_{p})}$
then Let $a_{i}=Xx_{i}$ satisfied the required condition.


It's true that if E is measurable and m(E) is finite then for every positive number e,there exists some opem set G such that m(G\E) <e.

Is it also true withouth the assumption tha m(E) is finite,where m is the lebesgue measure?

vacously true if G is not required to be a superset of E.

I'll assume G is required to be a superset of E.

If E is measurable, m*(E) = m(E) where m*(E) is the outer measure of E. In particular if m*(E) = p < infinity, according to the definition of outer measure and the definition of infimum, there must be a countable collection of open intervals with their sum of length arbitrary close to the number p. So in finite measure case, let G be the union of that collection of open intervals.

In the case that p = infinity, the usual trick of decomposing the set E into countably many pieces of finite measure and decomposing epsilon into a sum of countably many positive numbers, for example, epsilon/2, epsilon/4, epsilon/8, ... ...
That trick works.


Jones covered a distance of $ 50$ miles on his first trip. On a later trip he traveled $ 300$ miles while going three times as fast. His new time compared with the old time was:

$ \textbf{(A)}\ \text{three times as much} \qquad\textbf{(B)}\ \text{twice as much} \qquad\textbf{(C)}\ \text{the same}$
$ \textbf{(D)}\ \text{half as much} \qquad\textbf{(E)}\ \text{a third as much}$

[hide]
For simplicity's sake let Jones' speed on the first trip be 50 mph. So on the first trip, he covered 50 miles at 50 mph in 1 hour. On the second trip, he covered 300 miles at 150 mph in 2 hrs. So, his new time compared to his old tim is $ \frac{2}{1}\equal{}2$, which is twice as much. $ \text{(B)}$
[/hide]


symmetry
inequalities unsolved
inequalities

a,b,c,d are different real numbers. a/b+b/c+c/d+d/a=4 and ac=bd
(a/c+b/d+c/a+d/b)max =?

don't worry positive or negative real numbers...

we can show that exactly $2$ of them are negetive
well , they all can not be positive cause if they be :
we have $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \geq 4\sqrt[4]{1}
$ then we will have $a=b=c=d$ and since they are all different it can not be true.
and since $ac=bd$ even number of $a,b,c,d$ should be negetive ,It means $2$ of them or all of them ,  but simply they all can  not be negetive (this will lead us to , that they are all the same, so we will have that $2$ of them should be negetive.

cause of the symmetry we should only examine the cases that $a,c <0$ or $a,b<0$

To be countinud :)


I'm having trouble getting the answers (maybe the answer key is wrong?)

1. The social convenor has  volunteers to work at a school dance. Each dance requires  volunteers at the door,  volunteers on the floor, and  floaters. Joe and Jim have not volunteered before, so the social convenor does not want to assign them to work together. In how many ways can the volunteers be assigned?

Answer: [hide][/hide]

2. Jeffrey is a DJ at a local radio station. For the second hour of his shift he must choose all his music from the top  songs for the week. Jeffery will play exactly  songs during this hour.

a) How many different stacks of discs could Jeffery pull from the station's collection if he chooses at least  songs that are in positions 15 to 40 on the charts?

[hide][/hide]

b)Jeffrey wants to start his second hour with a hard rock song and finish with a pop classic. How many different play lists can Jeffrey prepare if he has chosen  rock songs,  soul pieces, and  pop classics?

[hide][/hide]

c)Jeffrey has  favorite songs currently on the top  list. How many different subsets of these songs could he choose to play during his shift?

[hide][/hide]

k i got most of the solutions to the problems

Solutions:

[hide]

1.

Very direct way. I'm trying to find another solution to this.

here, you assume that joe is at the door, so you only have 10 other people to choose from to be his partner

you also assume that jim is on the floor, so you have 9 people to choose 3 from

\dbinom{6}{6} is the rest

here, joe is at the door

jim is a floater, so there are 9 people to choose from (you cant choose jim)

 is the rest

 because you cant pick either jim or joe for this case

 because only joe is on the floor,so you have 8 others

 is the rest, including jim

Multiply by  because Joe and Jim can flip flop

2. a)

My mistake first was assuming there  songs to choose from to be at least ... but its  (position  to ).  choose , you have  left. We need exactly 12 songs, so we choose  from the . Continue the pattern and add the possibilities.

b)

Still doesn't match the answer key. Anything wrong? You can choose  songs to be the Hard Rock and  songs to be the pop classics. You have  songs left and order matters.

c)

should be self-explanatory... These are the ways to choose  songs in the different subsets.

[/hide]


Let a,b,c$ \geq 0$ such that a+b+c=6.Prove that:
$ a^2\plus{}2b^2\plus{}3c^2\plus{}abc \geq 24$

Can you show me when does equality occur ?
I don't think it's an old problem. Is it yours ? If it's trues, it's very intersting, my brother  :)

Equality occurs when $ (a,b,c) \equal{} (4,2,0)$.

[quote="tran300491"]Let a,b,c$ \geq 0$ such that a+b+c=6.Prove that:
$ a^2 \plus{} 2b^2 \plus{} 3c^2 \plus{} abc \geq 24$[/quote]

Use lagrange multiplier
\[ \frac {\partial}{\partial a} (a^2 \plus{} 2b^2 \plus{} 3c^2 \plus{} abc) \equal{} \lambda
\]

\[ \frac {\partial}{\partial b} (a^2 \plus{} 2b^2 \plus{} 3c^2 \plus{} abc) \equal{} \lambda
\]

\[ \frac {\partial}{\partial c} (a^2 \plus{} 2b^2 \plus{} 3c^2 \plus{} abc) \equal{} \lambda
\]
we have $ 2a \plus{} bc \equal{} 4b \plus{} ac \equal{} 6c \plus{} ab$   ---- (1)
              and   $ a \plus{} b \plus{} c \equal{} 6$           ----(2)

substitute $ a \equal{} 6 \minus{} b \minus{} c$ into (1)
              we have $ c \equal{} \sqrt {b^2 \minus{} 2b}$
substitute $ c \equal{} 6 \minus{} a \minus{} b$ into (1)
              we have $ b^2 \minus{} 2b \equal{} a^2 \minus{} 4a$
hence $ c \equal{} \sqrt {b^2 \minus{} 2b} \equal{} \sqrt {a^2 \minus{} 4a}$

we need $ b\ge 2$ and $ a\ge 4$ in the square root

it follows that the only extreme value occurs when $ (a,b,c) \equal{} (4,2,0)$
and it can be shown to be a minimum  :)

is there any another approach  :?:


geometry
geometric transformation
reflection
geometry proposed

Let $ ABC$ be a triangle and $ d$ is a line pass through orthocenter $ H$. $ d_a$ is reflection of $ BC$ through $ d$ and $ d_a$ intersect the line through $ H$ and parallel to $ BC$ at $ A'$. Defined similarly we have $ B',C'$. Prove that $ A',B',C'$ are collinear.

This nice problem has been solved by a friend of mine [b][size=100]Dimitris Papadimitriou[/size][/b] and I will post here later, the translation of his solution which has been published [url=http://www.mathematica.gr/forum/viewtopic.php?f=50&t=2890][b][color=red]Here[/color][/b][/url].

Kostas Vittas.

I expected a quick replies to this particular topic from jayme and Darij
just because they adored the Swiss pair of Arnold Droz and Lina Farny
and published pretty good papers on the subject of that  tremendous theorem.

Clearly, the collinearity here is on the reflection of Droz-Farny line in the
given one through the orthocenter.

I quickly thought of providing a related historical note on Droz-Farny, but
decided against it.

M.T.

@Vitasko:  the proof on the site you recommended although seemed Greek to me,
possibly can be interpreted as another proof of D-F theorem.

[quote="armpist"] seemed Greek to me,
possibly can be interpreted as another proof of D-F theorem.[/quote]

I think Insland is quite far with USA to know about google translate. Well, Droz-Farny is my way in this problem.

But there are many things that's more important than a problem!

I want to say many thank to vittasko who is always spend his enthusiasm for geometry - the great science. And now I feel happy to see another solution for Droz-Farny line. I like and admire vittasko with his aestheticism, he make geometry become art. thank you again vittasko.

E.C.

@armpist I adored the Swiss pair of Arnold Droz and Lina Farny, don't forget me!


Hello, I have saved some web pages by clicking Save As and saving them in .html format. The web pages contain weblogs. Does anyone know how I can write a program which formats the .html files so that a new file can be created from each .html file which contains only the pure text in the weblog? (i.e.The text which the weblog user has written, not the tags and code)

For Example the file below:

The new file should print from Al Jazeera... up to hyenas (End of Post) without any of the code for links or '/' character in between.

Notify Blogger about objectionable content.
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Blogger <http://www.blogger.com/>
Get your own blog <http://www.blogger.com/>Flag Blog
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BlogThis! <javascript:BlogThis();>

The Daily Demarche <http://dailydemarche.blogspot.com>
Thursday, July 21, 2005
Al Jazeera supporters are right.
As London recovers from the latest act of terror to strike the city a
terrorist and mass murderer in Iraq begins his march to justice. Saddam
has begun his day in court, unfortunately media coverage is minimal due
the continued threat to the west posed by al Qaeda and its sympathizers.
Luckily al Jazeera is covering the trial
<http://www.aljazeera.com/cgi-bin/news_service/middle_east_full_story.asp?service_id=9291>:

/Long-bearded Saddam, who was wearing a white shirt, a suit jacket and
glasses, appeared to be defiant and stared directly into the judge's
eyes, who in turn seemed uncomfortable and avoided eye contact with the
ousted president./

I'd be uncomfortable too, looking into the eyes of a man who ordered a
province of his own country to be gassed, although I assume al Jazeera
finds the judge's revulsion to be a victory of will for Hussein. Not
surprisingly, some of their readers feel the same way (from the comments
on the AJ site):

/he lived like a lion and even living like lion. these pupets don't dare
to look into his eyes. they know he can not do a thing but their guilty
and fear don't let them to look into his eyes. salute to a lion heart./

and
//
/saddam husseins trial should be open so that the whole world can
understand the history of these past crimes and that there are many
others deeply implicated, particularly the usa. /

Luckily, there are other sources of news to refer to- the Internet
allows anyone interested in learning more to do so, easily. Take this
view <http://www.thestandard.com.hk/stdn/std/World/GG21Wd03.html> of
Saddam, for example:

/A man describes how Saddam Hussein's secret police shoved a dissident's
baby into a sack with a vicious cat that scratches it. /

/Undercover agents throw a man to his death from the roof of a building./

/Iraqiya state television is reviving images of life under Saddam as a
court prepares to announce his trial date./

/``I wish they were here to see the day when Saddam is finished,'' a
tearful woman who lost her relatives under Saddam tells viewers of
Iraqiya, which broadcast footage of abuses filmed by members of Saddam
security forces as they committed them./
//
/Grainy footage of senior officials, including Ali Hassan al-Majid -
nicknamed Chemical Ali because his men allegedly gassed 5,000 Kurds in
1988 - shows them questioning Shiites after a failed rebellion in 1991./

/One official calmly smokes a cigarette and then kicks one of them in
the face. The bound men were later executed. Other reminders include a
blindfolded man with his hands bound behind his back being pushed off
the roof of a building./

/Another scene captured on video shows a man being held on the ground
with an arm extended. The arm is beaten with a club until the bone breaks./
//
So, for perhaps the first and last time, I agree, at least in part, with
sentiments expressed on the al Jazeera site- the trial of Saddam Hussein
should be open and transparent, it should be broadcast on the satellite
channels and the Internet, the entire world should have the
opportunityto view the process of the rule of law in action. The lion of
the desert will quickly be exposed as the hyena he is, or was- with full
apologies to hyenas.

(End of post.)

  7/21/2005 09:52:00 PM posted by Dr. Demarche
  <http://dailydemarche.blogspot.com/2005/07/al-jazeera-supporters-are-right.html>

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So basically you have to read from files and reprint them in form you want.

Deleting tags is simple. When you encouter opening of tag '<' then don't print things untile you will encounter closing bracket.

In that way you can eliminate whether you want.

Hope that helped.

I don't know the syntax to do this.
I have a basic algorithm:

-read in file character by character
-print it to an output file character by character.
-if '<' is encountered, don't print it
-don't print anything else until '>' is encountered.
-keep printing characters to the output file.

But I don't know how to tell Java not to print stuff.

bool inside = false;
char c;

for(i = 0; i < textsize; i++){
      fscanf(fin, "%c", c);
      if(isalpha(c) && !inside)fprintf(fout, "%c", c);
      else if(c=='<')inside = true;
      else if(c=='>')inside = false;
}

Just translate it to java and it's functions. (I could write it in java but it looks like you need some practice ;-) )


Determine Whether or Not There are exist a set of infinity integer points S such that for every pair(A,B,C) lies on S then the orthocenter of triangle ABC lies on S,too.

Wikipedia has the following theorem: [i]An orthocentric system is a set of four points in the plane one of which is the orthocenter of the triangle formed by the other three. If four points form an orthocentric system, then each of the four points is the orthocenter of the other three.[/i] This may help.

uhm.May be.But It's My Home Work, My teacher give me last week.U Just trying to help me^^^


modular arithmetic
number theory
prime numbers

[hide="i guess you could do it this way"]do a lot of casework!
assume that some number $ n\equiv0,1,2...15\pmod{16}$ take the 4th power of each of them and show that each of those is $ \equiv 1/0\pmod{16}$[/hide]

is there a faster/more creative way?

Maybe you could find what $ x^{4}$ is congruent to when $ x$ is congruent to 0, 1, and prime numbers that are less than 16, which are 2, 3, 5, 7, 11, 13. If $ x^{4}$ is congruent to $ 1$ or $ 0$ when $ x \equal{} 0, 1, 2, 3, 5, 7, 11, 13$, then you know that all other numbers that x is congruent to that are below 16 can either be congruent to 0 or 1 using modular arithmetic rules. This isn't that clever, but it saves you from checking 8 cases.  :wink:

That works... basically an odd number to the fourth is 1 mod 16 (An odd number squared is 1 mod 8, which is a far more useful conclusion).

Do I even need to explain for an even number?

[quote="i_like_pie"][quote="mdk"]show that any number raised to the 4th power is either 0 or 1 mod 16[/quote]
[hide]$ n\equiv0,1\bmod2\,\forall\,n\in\mathbb{N}\Longrightarrow n^{4}\equiv0,1\bmod2^{4}\,\forall\,n\in\mathbb{N}$[/hide][/quote]

This is a restatement of the question; the lemma you are implicitly trying to assert is not, in general, true.  $ \bmod 2^{k}$ is a special case, and similar reasoning does not hold in any other cases.  For example,

$ n\equiv 0, 1, 2\bmod 3$

but

$ n^{2}\equiv 0, 1, 4, 7\bmod 9$

[hide="Straightforward solution"] $ (2k\plus{}1)^{4}\equal{} 16k^{4}\plus{}32k^{3}\plus{}24k^{2}\plus{}8k\plus{}1\equiv 16\frac{k(3k\plus{}1)}{2}\plus{}1\equiv 1\bmod 16$ [/hide]
[hide="How this fact relates to general theory"] Let $ \lambda(m)$ denote the smallest exponent $ e$ such that $ a^{e}\equiv 1\bmod m$ for all $ gcd(a, m) \equal{} 1$.  For an odd prime $ p$ and for $ m \equal{} 1, 2, 4, p^{k}, 2p^{k}$ we have $ \lambda(m) \equal{}\varphi(m)$.

[b]Lemma:[/b]  $ \lambda(2^{k\plus{}2}) \equal{} 2^{k}, k\in\mathbb{N}$ [/hide]

[quote="i_like_pie"][quote="mdk"]show that any number raised to the 4th power is either 0 or 1 mod 16[/quote]
[hide]$ n\equiv0,1\bmod2\,\forall\,n\in\mathbb{N}\Longrightarrow n^{4}\equiv0,1\bmod2^{4}\,\forall\,n\in\mathbb{N}$[/hide][/quote]

You are implicitly assuming that

$ a\equiv b\pmod{p}$

implies

$ a^{k}\equiv b\pmod{p^{k}}$,

which is false. For example,

$ 3\equiv 1\pmod{2}$

$ 3^{3}\not\equiv 1\pmod{2^{3}}$

[quote="i_like_pie"][quote="vishalarul"]$ 3^{3}\not\equiv 1\pmod{2^{3}}$[/quote]
But $ 3^{4}\equiv 1\pmod{2^{4}}$, and the problem involves $ 2^{4}$, not $ 2^{3}$.[/quote]

$ 10\equiv 1\pmod{9}$

$ 10^{4}\not\equiv 1\pmod{9^{4}}$

Either way, it doesn't work. And don't tell me that the problem involves $ \pmod{2}$, because the problem is equivalent to proving:

$ x\equiv y\pmod{2}\rightarrow x^{4}\equiv y\pmod{2^{4}}$, which you implicitly assumed.


geometry
circumcircle
geometric transformation
reflection
perpendicular bisector
geometry proposed

In an acute triangle ABC, AC > BC and M is the
midpoint of AB. Let altitudes AP and BQ meet at H, and let lines
AB and PQ meet at R. Prove that the two lines RH and CM are
perpendicular.

Note that $ CPHQ$ and $ BPQA$ are cyclic quadrilaterals. Let $ C_1,C_2,C_3$ be the circumcircles of $ CPHQ,BPQA,BHA$ respectively. Let $ C_1$ and $ C_3$ intersect at $ H$ and $ T$. Let $ O_1$ and $ O_3$ be the centers of $ C_1$ and $ C_3$ respectively. Let $ O$ be the circumcenter of triangle $ ABC$. Since $ \angle CPH\equal{}90^{\circ}$, then $ CH$ is the diameter of $ C_1$, so $ O_1$ is the midpoint of $ CH$. It is well-known that $ CH\equal{}2OM$, therefore $ CO_1\equal{}OM$. Note that $ \angle BOA\equal{}2\angle BCA\equal{}2(180^{\circ}\minus{}\angle PHQ)\equal{}2(180^{\circ}\minus{}\angle BHA)$. But $ \angle BO_3A\equal{}2(180^{\circ}\minus{}\angle BHA)$, so $ \angle BOA\equal{}\angle BO_3A$. Since $ O_3A\equal{}O_3B$ and $ OA\equal{}OB$, $ O_3$ and $ O$ lie on the perpendicular bisector of segment $ AB$. Thus $ O$ is the reflection of $ O_3$ with respect to $ AB$. Hence $ O_3M\equal{}OM\equal{}CO_1$. Consider $ CO_1O_3M$. We have $ CO_1\equal{}O_3M$. Also $ CH\perp BC,BC\perp O_3M$, so $ CH\parallel O_3M$. So $ CO_1O_3M$ is a parallelogram. Since $ HT\perp O_1O_3$ and $ HT\perp CT$, so $ O_1O_3\parallel CT$. Therefore $ CT\parallel CM$, and hence $ CTM$ is a straight line. Consider $ C_1,C_2,C_3$. The three radical axes (for each pair of circles) are concurrent. So they concur at $ R$. We now have $ RH\parallel RT\perp CM$, as desired.


geometry
rectangle
search
combinatorics proposed
combinatorics

In how many ways can you tile a $ 4\times n$ by $ 3\times 1$ dominoes ? Justify fully your answer

I guess we can proceed similarly as in http://www.mathlinks.ro/viewtopic.php?search_id=1699599031&t=54081

Yes: a quick examination revealed five possible variables: $ a_n$ for a $ 4 \times 3n$ block, $ b_n$ for such a $ 4 \times 3(n \minus{} 1)$ block with a $ 3\times 3$ block glued on one end, $ c_n$ for a $ 4 \times 3(n \minus{} 1)$ block with a $ 3 \times 2$ block glued on, $ d_n$ for a $ 4 \times 3(n \minus{} 1)$ block with a $ 3 \times 1$ block glued on, and $ e_n$ for a $ 4 \times (3n \minus{} 1)$ block with a $ 1 \times 1$ block glued on the end in the top- or bottom-most row.  Then we have equations like
$ a_n \equal{} b_n \plus{} e_n$
$ b_n \equal{} c_n \plus{} a_{n \minus{} 1}$
$ c_n \equal{} d_n \plus{} e_{n \minus{} 1}$
$ d_n \equal{} a_{n \minus{} 1} \plus{} d_{n \minus{} 1}$
and $ e_n \equal{} c_n$, or something close to this.  There's lots of simplification that can be done, computation of the actual answer, etc. -- if I have time later, I may come and complete this argument.


calculus
integration
logarithms
trigonometry
function
geometric series
calculus computations

Evaluae $ \int_{0}^{1} \left( \frac{1}{1\minus{}x} \plus{} \frac{1}{\ln x} \minus{} \frac{1}{2}\right) \frac{1}{\ln x} \, dx$

Of course it would be easy for some people here...

somewhat of a long solution...  :|   :(

we are given,
$ I \equal{} \int_{0}^{1} \left( \frac {1}{1 \minus{} x} \plus{} \frac {1}{\ln x} \minus{} \frac {1}{2}\right) \frac {1}{\ln x} \, \textbf dx$

let us transform it using $ x \equal{} e^{ \minus{} y}$ to get

$ I \equal{} \int_0^\infty\left(\frac {1}{2} \plus{} \frac {1}{y} \minus{} \frac {1}{1 \minus{} e^{ \minus{} y}}\right)\;\frac {e^{ \minus{} y}}{y}\;\textbf dy$

noting that we want $ I(1)$, let us define
$ I(z) \equal{} \int_0^\infty\left(\frac {1}{2} \plus{} \frac {1}{y} \minus{} \frac {1}{1 \minus{} e^{ \minus{} y}}\right)\;\frac {e^{ \minus{} yz}}{y}\;\textbf dy$

then,
$ \frac {d}{dz}\;I(z)\; \equal{} \; \minus{} \int_0^\infty\left(\frac {1}{2} \plus{} \frac {1}{y} \minus{} \frac {1}{1 \minus{} e^{ \minus{} y}}\right)\;e^{ \minus{} yz}\;\textbf dy$

using mittag-leffler, we get
$ \frac {1}{2} \plus{} \frac {1}{y} \minus{} \frac {1}{1 \minus{} e^{ \minus{} y}}\; \equal{} \; \minus{} 2\sum_{k \equal{} 1}^\infty\frac {y}{y^2 \plus{} 4\pi^2\,k^2}$

so,  $ \frac {d}{dz}\;I(z)\; \equal{} \;2\int_0^\infty\sum_{k \equal{} 1}^\infty\frac {y}{y^2 \plus{} 4\pi^2\,k^2}\; e^{ \minus{} yz}\;\textbf dy$

now using the fact
$ \int_0^\infty e^{ \minus{} au}\;\sin\,(bu)\;\textbf du\; \equal{} \;\frac {b}{b^2 \plus{} a^2}\qquad \ldots (1)$

we have
 $ \frac {d}{dz}\;I(z)\; \equal{} \;2\int_0^\infty\sum_{k \equal{} 1}^\infty\int_0^\infty e^{ \minus{} 2\pi\,k\,u}\;\sin\,(yu)\;\textbf du\; e^{ \minus{} yz}\;\textbf dy$

we know, since it is a geometric series,
 $ \sum_{k \equal{} 1}^\infty\;e^{ \minus{} 2\pi\,k\,u}\; \equal{} \;\frac {1}{e^{2\pi\,u} \minus{} 1}$

so, $ \frac {d}{dz}\;I(z)\; \equal{} \;2\int_0^\infty\int_0^\infty\frac {\sin\,(yu)}{e^{2\pi\,u} \minus{} 1}\;\textbf du\; e^{ \minus{} yz}\;\textbf dy$

again, using the identity in $ (1)$, this time w.r.t. $ \textbf dy$,

$ \frac {d}{dz}\;I(z)\; \equal{} \;2\int_0^\infty\frac {u}{(u^2 \plus{} z^2)\; (e^{2\pi\,u} \minus{} 1) }\;\textbf du$

integrating w.r.t. $ z$ between limits $ 1$ and $ \infty$ gives

$ \Rightarrow 0 \minus{} I(1)\; \equal{} \;2\int_0^\infty\int_1^\infty \frac {u}{(u^2 \plus{} z^2)\; (e^{2\pi\,u} \minus{} 1) }\;\textbf dz\;\textbf du$

since we know $ \int\;\frac {u}{u^2 \plus{} z^2}\;\textbf dz \equal{} \tan^{ \minus{} 1}\,\frac {z}{u}\qquad \text{and}\qquad \frac {\pi}{2} \minus{} \tan^{ \minus{} 1}\;\frac {1}{u} \equal{} \tan^{ \minus{} 1}\;u$,

we have
$ I(1)\; \equal{} \; \minus{} 2\int_0^\infty \frac {\tan^{ \minus{} 1}\,u}{e^{2\pi\,u} \minus{} 1}\;\textbf du$

the integral appearing on the right hand side of this equation was asked by me on this forum before,
don't know if anyone showed it, but i can show it equals $ \frac {1}{2}\left(1 \minus{} \ln\,\sqrt {2\pi}\right)$

hence, $ I(1)\; \equal{} \;\boxed{\ln\,\sqrt {2\pi} \minus{} 1}$

sos, i would like to ask this question on the next exam in my class, hope that is OK with you. i am sure most of the students will like this question.  :)

[quote="misan"]sos, i would like to ask this question on the next exam in my class, hope that is OK with you. i am sure most of the students will like this question.  :)[/quote]

Sure. -o-;;

Your method is very impressive to me, since it involves some neat techniques which are new to me. Anyway, here is my solution:

[b]Step 1) Preliminary[/b]

One can show the following equalities. You may use the infinite product formula for Gamma function to prove these.

$ (1) \quad \ln \Gamma(1 + x) = - \gamma x + \sum_{n = 2}^{\infty} \frac {( - 1)^{n}}{n}\zeta(n) \, x^{n}\quad \quad (|x| < 1),$

$ (2) \quad \int_{0}^{1} \frac {1 - t^{x}}{1 - t} \, dt = \gamma + \frac {d}{dx} \ln \Gamma(1 + x).$

[b]Step 2) Evaluation of $ \int_{0}^{1} \ln \Gamma(1 + x) \, dx$[/b]

One can confirm the equality $ \left( 1 + \frac {1}{n} \right)^{n} = \frac {n!}{n^n} \prod_{k = 1}^{n} \left( 1 + \frac {1}{k}\right)^{k}.$ Taking $ \ln$ to both side,

\begin{eqnarray*} \ln \left( 1 + \frac {1}{n} \right)^{n} & = & \ln \left( \frac {n!}{n^n} \right) + \sum_{k = 1}^{n} k \ln \left( 1 + \frac {1}{k}\right) \\
& = & \ln \left( \frac {n!}{n^n} \right) + \sum_{k = 1}^{n} \sum_{j = 0}^{\infty} \frac {( - 1)^{j}}{j + 1} \frac {1}{k^{j}} \\
& = & \ln \left( \frac {n!}{n^n} \right) + n - \frac {1}{2} \mathrm{H}_n + \sum_{j = 2}^{\infty} \sum_{k = 1}^{n} \frac {( - 1)^{j}}{j + 1} \frac {1}{k^{j}}. \end{eqnarray*}

Taking $ n \to \infty$ to both side, Stirling formula yields $ 1 = \ln \sqrt {2\pi} - \frac {\gamma}{2} + \sum_{j = 2}^{\infty} \frac {( - 1)^{j}}{j + 1} \zeta(j).$ Together (1), this confirms

$ (3) \quad \int_{0}^{1} \ln \Gamma(1 + x) \, dx = \ln \sqrt {2\pi} - 1.$

[b]Step 3) An integral formula for $ \ln \Gamma (1 + x)$[/b]

Integrating both side of (2) on $ [0, x]$ and changing the order of integration immediately yields

$ (4) \quad \int_{0}^{1} \left( \frac {x}{1 - t} + \frac {1 - t^x}{1 - t}\frac {1}{\ln t} \right) \, dt = \gamma x + \ln \Gamma (1 + x).$

Especially,

$ (5) \quad \gamma = \int_{0}^{1} \left( \frac {1}{1 - t} + \frac {1}{\ln t} \right) \, dt.$

Combining (4) and (5), we obtain an integral formula

$ (6) \quad \ln \Gamma (1 + x) = \int_{0}^{1} \left( \frac {1 - t^x}{1 - t} - x \right) \frac {dt}{\ln t}.$

[b]Step 4) Final Result[/b]

Now integrate both side of (6) on $ [0, x]$ again, then we have

$ (7) \quad \int_{0}^{x} \ln \Gamma (1 + t) \, dt = \int_{0}^{1} \left( \frac {x}{1 - t} + \frac {1 - t^x}{1 - t}\frac {1}{\ln t} - \frac {x^2}{2} \right) \frac {dt}{\ln t}.$

This completes the evaluation by plugging $ x = 1$.

Wow.  Those are some amazing solutions.

Unfortunately, I'm not familiar with the gamma and zeta functions because I've never used them in classes.

about 2 step:
$ G: \equal{} \int_{0}^{1}\ln \Gamma(1 \plus{} z)dz \equal{} \int_{0}^{1}\ln z dz \plus{} \int_{0}^{1}\ln \Gamma(z)dz \equal{} \minus{} 1 \plus{} \int_{0}^{1}\ln \Gamma(z)dz$ since $ \Gamma(1\plus{}z)\equal{}z\Gamma(z)$
$ I \equal{} \int_{0}^{1}\ln \Gamma(z)dz$ say.
put $ 1 \minus{} z\equiv y$ then
$ I \equal{} \int_{0}^{1}\ln \Gamma(1 \minus{} y)dy$
then
$ 2I \equal{} \int_{0}^{1}\ln \Gamma(1 \minus{} z)dz \plus{} \int_{0}^{1}\ln \Gamma(z)dz \equal{} \int_{0}^{1}\ln \Gamma(z)\Gamma(1 \minus{} z)dz \equal{} \int_{0}^{1}\ln \frac {\pi}{\sin\pi z}dz$ since
 $ \Gamma(z)\Gamma(1 \minus{} z) \equal{} \frac {\pi}{\sin\pi z}$
thus:
$ 2I \equal{} \ln \pi \minus{} \int_{0}^{1}\ln \sin \pi z \equal{} \ln \pi \plus{} \ln 2$
hence:
$ I \equal{} \ln \sqrt {2\pi}$ and $ G \equal{} \minus{} 1 \plus{} \ln \sqrt {2\pi}$

[quote="Carcul"]Look also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148672]here[/url].[/quote]

I forgot that post... :blush: It's surprising that there are lots of (seemingly) different ways to show something. This is amusing.


Find the greatest constant k such that the following inequality holds for any positive real numbers $ a, b, c$
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\le k$

[quote="vo thanh van"]Find the greatest constant k such that the following inequality holds for any positive real numbers $ a, b, c$
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\le k$[/quote]
I think U want find the Minimum of k .I sure $ k\equal{}2$. And I have seen this problem on forum  :)

[quote="onlylove_math"][quote="vo thanh van"]Find the greatest constant k such that the following inequality holds for any positive real numbers $ a, b, c$
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\le k$[/quote]
I think U want find the Minimum of k .I sure $ k \equal{} 2$. And I have seen this problem on forum  :)[/quote]
if $ a,b,c$ are side of a triangle,$ k\equal{}2$ is true  :)but in here $ a,b,c$ be positive real numbers

Dear,anh Van! Can you post your solution,I want to see it. :lol:

[quote="vo thanh van"][quote="onlylove_math"][quote="vo thanh van"]Find the greatest constant k such that the following inequality holds for any positive real numbers $ a, b, c$
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\le k$[/quote]
I think U want find the Minimum of k .I sure $ k \equal{} 2$. And I have seen this problem on forum  :)[/quote]
if $ a,b,c$ are side of a triangle,$ k \equal{} 2$ is true  :)but in here $ a,b,c$ be positive real numbers[/quote]
Oh no I think with $ a,b,c$ be positive real numbers . Also $ k\equal{}2$

[quote="vo thanh van"]Find the greatest constant k such that the following inequality holds for any positive real numbers $ a, b, c$
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\le k$[/quote]

Let $ c \equal{} 0,b \equal{} 1$ and $ a \to \plus{} \infty$

Clearly we have the expression $ \frac {{a^2 \plus{} b^2 \plus{} c^2 }}{{ab \plus{} bc \plus{} ca}} \to \plus{} \infty$ and hence there is no $ k$ which satisfy that inequality.  :|

For onlylove_math, we know $ \frac{{a^2  \plus{} b^2  \plus{} c^2 }}{{ab \plus{} bc \plus{} ca}} \ge 1$ is true. With the $ \frac{{a^2  \plus{} b^2  \plus{} c^2 }}{{ab \plus{} bc \plus{} ca}} \ge 2$ you can try with $ b\equal{}c\equal{}1$ and $ a \le 2$.  :)


Assume that $ P(A) \equal{} 0.6$ and that $ P(B) \equal{} 0.3$ and A and B are independent.
Find $ P(A\cup \overline{B})$.

I've tried with the following $ P(A) \equal{} P(A\cup E) \equal{} P(A\cap (B\cup \overline{B})) \equal{} P(A\cap B) \plus{} P(A\cap \overline{B})$
Now from $ P(A\cup B) \equal{} P(A)P(B) \equal{} 0.18$ because A and B are independent we find the probability of
intersection of A and B's complementary so $ P(A\cap \overline{B}) \equal{} 0.42$
Using this and the relation above we get $ P(A\cup \overline{B}) \equal{} P(A) \plus{} P(\overline{B}) \minus{} P(A\cap\overline{B}) \equal{} 0.6 \plus{} 0.9 \minus{} 0.42 > 1$ so there is a problem with this solution.
Any comments and suggestions are much appreciated.

EDIT:corrected a typo

$ P(\overline{B})\equal{}0.7,$ not $ 0.9.$ Fix that and your difficulties will go away.

As a check or as an alternative computation, try finding $ P(\overline{A}\cap B)$ and then subtracting that from $ 1.$

thanks Kent Merryfield,now it makes sense and the solution coincides with the one in the book.


quadratics
number theory unsolved
number theory

Decide det irrational integer that has the continued fractions [9; 9,18]!

I know that the answer is sqrt(83), but how do i get there if i don't that is a quadratic irrational?

big thanks for any solutions!

You have (neglect that $9$ at the beginning):
Let $x=[0;9,18,9,18,...]$
$x=\frac{1}{9+\frac{1}{18 + x}}$ (since $x$ stands just for the infinite chain of fractions, thus we can 'get out' the first period of $x$ without changing $x$).
But this equation gives $x=\frac{18 + x}{163 + 9x} \iff 9x^2 + 163x = 18+x \iff x^2+18x-2=0$.
So solving this gives $x= -9 +\sqrt{83}$ (and another root we can exclude to be correct since its $<0$, or just by testing) as it also should be, since than the desired number is $x+9= \sqrt{83}$.


Hello,
I'm reading a book about Number Theory. And its presenting an example to work through it.

Let $ n$ be an integer greater than $ 1$. Prove that $ 2^n$ is the sum of two odd consecutive integers.

For me I just tried it in a piece of paper but how can I prove it? I never proof anything before. When I see the solution I get  :huh: .

This was the solution: the relation $ 2^{n} \equal{} (2k \minus{} 1) \plus{} (2k \plus{} 1)$ implies $ k \equal{} 2^{n \minus{} 2}$ and we obtain $ 2^n \equal{} (2^{n \minus{} 1} \minus{} 1) \plus{} (2^{n \minus{} 1} \plus{} 1)$.

but from where he bring all that relations. How he get $ 2^n \equal{} (2k \minus{} 1) \plus{} (2k \plus{} 1)$ and so on  :( .

What is a sum of two odd consecutive integers?  How would you write down an odd integer?  How would you write down the next odd integer?

OK first of all how he come up with $ 2^n\equal{}(2k\minus{}1)\plus{}(2k\plus{}1)$  :?: and as I understand that after this he implied that $ k\equal{}2^{n\minus{}2}$ OK before that how he know what is value of $ k$. There is something I didn't get it maybe if I do everything will be OK with me.

I think the solution assumes that it exists and then finds it. (If it didn't exist, some contradiction would occur.) Is this book by any chance 104 NT Problems?

Yes it is 104 NT problems but I'm still in the beginning. So do you mean that you should try it in small numbers and draw a proof  :huh:

"$ 2^n$ is the sum of two odd consecutive integers" is equivalent to "there exists an integer $ k$ such that $ 2^n \equal{} (2k \minus{} 1) \plus{} (2k \plus{} 1)$."  Do you understand this?  That's all the solution is.

OK I prefer to work with an example:
lets say that $ 2^5$ . I know that the numbers are $ 15$ and $ 17$ but the problem with the proof thing. Is the proof need you to come to the solution or a method to apply with every case?

And how did you know that the numbers are $ 15$ and $ 17$?

By: $ 2^5\equal{}32$ so if we say $ \frac{32}{2}\equal{}16$ then $ 16\plus{}1\equal{}17$ and $ 16\minus{}1\equal{}15$

So I can call it: $ 2^n\equal{}S$, so the answer is $ \frac{S}{2} \pm 1$ from my view.

Two consecutive odd integers always add up to what?

I see: I understand now. $ 2k \plus{} 1$ and $ 2k \minus{} 1$ are all ways to make odd numbers. Which makes an even number when added.
Thanks to "t0rajir0u" and "sunehra" and also for "Ihatepie" for his post.  :)


geometry
geometric transformation
rotation

In how many ways can we seat 6 people around a table if Fred and Gwen insist on sitting opposite each other?  (Two seatings are considered equivalent if one is a rotation of the other.)

Isn't there 6 ways they can sit opposite from each other?

Yes, but there are $ 6$ rotations which we consider the same, so you over count and need to end up dividing it by $ 6$ anyways.
[hide="Solution"]
First, we place Fred.  Fred has $ 6$ different places to sit.  Gwen has no choice but the seat directly opposite Fred.  The rest of the people can be seated in $ 4!$ ways.  So we have $ 6\cdot4! \equal{} 144$.  However, there are $ 6$ different rotations, so to correct for our over counting, we do $ \frac {144}6 \equal{} \boxed{24}$
[/hide]

Or, we can just say that Fred is somewhere, then Gwen is directly opposite, and we have $ 4!\equal{}24$ ways to arrange the rest.

6 ways to seat Fred and only 1 way to seat Gwen. Then the rest is just 4 factorial. $\frac{6 \cdot 1 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{6}=24$


What do you do if someone locked a topic just to shut you up? :mad: And especially gave no explanation!!

[quote="jtw"]What do you do if someone locked a topic just to shut you up?[/quote]There's a reason someone wants you to shut up.

Because I'm right and he/she can't prove me wrong.

First of all you're not right. Secondly, there is already a thread discussing (the science of) evolution versus (the myth of) creationism. All discussions about that will take place there not in new threads.


function
induction
geometry
geometric transformation
abstract algebra
combinatorics proposed
combinatorics

Let $C = \{1, 2, 3, \cdots, p\}$ where $p$ is a prime number.
Let $r : C \to C$ be a function defined by
$\forall i \in C, ~ r(i) \equiv i+1 \mod{p}$
Let $h : C \to C$ be a function that is not onto.
Prove that there exists a constant function $f : C \to C$ that can be written as $f = h_{1}\circ h_{2}\cdots \circ h_{k}, \quad h_{i}\in \{r,h\}$.

For a function $f$, we denote it's range by $f(C)$. We will apply induction on the size of $h(C)$.

If the range has one element then there's nothing to prove. Let $|h(C) | = k < p$. Suppose that the statement is true for all functions $f$ such that $|f(C)| < k$. Let  $h(C) = \{h_{1}, \ldots,h_{k}\}$ and $S_{i}= h^{-1}(h_{i})$. We will show that there exists some $c \in C, n \in [k]$, such that \[|\,(h(C)+c) \cap S_{i}\,| \geq 2 .\] Let us first see that the existence of $c, n$ as above is enough to solve the problem. Let $f_{1}= h\circ r^{c}\circ h$ ($r^{0}$ is the identity function). Note that $(r^{c}\circ h) (C) = \{c+h_{1},\ldots, c+h_{k}\}$. So there are $k$ elements in $(r^{c}\circ h)(C)$ and two of them belong to $S_{i}$ for some $i$. Therefore the range of $f_{1}= h \circ r^{c}\circ h$ has size at most $k-1$. So that the statement of the problem holds for $f_{1}$ by induction, which in turn implies that it holds for $h$.

[b]Proof of existence of $c, n$:[/b] Suppose that there exists no cuch $c, n$. We will show that in this case $k | p$, which is a contradiction (as $1 < k < p$). We have that for all $c, i$, $|(h(C)+c ) \cap S_{i}| < 2$. This implies that for all $i, c$, we have $|(h(C)+c) \cap S_{i}| = 1$. So each translation of $h(C)$ has one element in each of the $S_{i}$'s. Hence, for a fixed $i$, (call this equation (1)) \[\sum_{c \in C}|(h(C)+c)\cap S_{i}| = p .\] Further, each element $a \in S_{i}$ lies in $h(C)+c$ for exactly $k$ values of $c$ (these are $a-h_{1}, a-h_{2}, \ldots$). Therefore (call this equation (2)) \[k \cdot |S_{i}| = \sum_{c \in C}|(h(C)+c) \cap S_{i}|.\] From (1) and (2) we have $k \cdot |S_{i}| = p$, a contradiction as $1 < k < p$.

Nice solution.

in fact, this is a lemma that G. L. O'Brien used in a proof of partial result of Road Coloring Conjecture. (G. L. O'Brien. The road coloring problem, Israel J. Math., 39 (1981) )

O'Brien also tried to generalize the lemma (G. L. O'Brien. Zero-inducing functions on finite abelian groups.  PACIFIC JOURNAL OF MATHEMATICS, 1982 )

"n fact, this is a lemma that G. L. O'Brien used in a proof of partial result of Road Coloring Conjecture. (G. L. O'Brien. The road coloring problem, Israel J. Math., 39 (1981) )

O'Brien also tried to generalize the lemma (G. L. O'Brien. Zero-inducing functions on finite abelian groups. PACIFIC JOURNAL OF MATHEMATICS, 1982 )"

Could you give some details of these?


Let be given a real number $k$ in the interval $(-1;2)$ and three pairwise distinct real numbers $a;b;c$.Prove that:
$[a^{2}+b^{2}+c^{2}+k(ab+bc+ca)].[\frac{1}{(a-b)^{2}}+\frac{1}{(b-c)^{2}}+\frac{1}{(c-a)^{2}}] \geq \frac{9}{4}(2-k)$

The problem was proposed by Nam Dung in [b]Mathematics and Youth Mgazine"[/b] in Viet Nam. It's hae still in period dicussing. I think we hadn't better post discussion here


The numbers 1, 2, 3 and 4 are written on four cards and randomly placed in the boxes below. What is the probability, expressed as a common fraction, that no card number appears in the box of the same number?
[asy]draw((0,0)--(2,0)--(2,1)--(0,1)--cycle);
draw((3,0)--(5,0)--(5,1)--(3,1)--cycle);
draw((6,0)--(8,0)--(8,1)--(6,1)--cycle);
draw((9,0)--(11,0)--(11,1)--(9,1)--cycle);
label("1",(1,.5));
label("2",(4,.5));
label("3",(7,.5));
label("4",(10,.5));[/asy]

Even before we start, we can determine there are $ 4!$ or $ 24$ possibilities.

1234, 1243, 1324, 1342, 1423, 1432, 2134, [color=green]2143[/color], 2314, [color=green]2341[/color], [color=green]2413[/color], 2431, 3124, [color=green]3142[/color], 3214, 3241, [color=green]3412[/color], [color=green]3421[/color], [color=green]4123[/color], [color=green]4132[/color], 4213, 4231, [color=green]4312[/color], [color=green]4321[/color]

The answer is $ \frac{10}{24}$ or $ \boxed{\frac{5}{12}}$.

The answer 3/8 is correct.  The answer 5/12 was suggested, but 4231 was mistakenly counted as a possibility.

There are 24 possible ways to fill all four boxes, if we ignore the restrictions, and there are 9 ways to fill the boxes with non-matching card numbers.
-- 3 options to fill the first box, then
-- 3 options for card number 1 to go into one of the remaining boxes, then
-- 1 option for the remaining two cards, since one of the two remaining cards must match the number of an unfilled box, leaving only one option for placement of the two cards

3*3*1/24 = 9/24 = 3/8

You can use the dearrangement theorem, without having to make cases.
It says that there are n![1/0!-1/1!+1/2!-1/3!+1/4!......1/n!] ways to arrange objects such that no object goes in the right place.

So 4![1/0!-1/1!+1/2!-1/3!+1/4!] = 12-4+1=9

Solution: Let's start with the card with #1 on it.
We can put it in 2,3, or 4. Let's call the box we put it in $n$ [b]3 possibilities[/b]

now we find a place for the card $n$. that goes in 3 places as well. assume we don't pick the first. there's 2 possibilities. And that leaves the last card (not the one that's same as the box just picked by $n$) one spot. That's [b]2 ways[/b] And if we put $n$ in box 1, there's 1 possibility for last 2 cards as well. So that's [b]1 way[/b]

So $3\cdot(2+1)=9$

basically using whatever the derangement theorem is

And then $\frac9{24}=\frac38$

You could also use casework:

We can either have $2$ cards being flipped (like $1$ with $3$ and $2$ with $4$) or all $4$ cards being flipped ($1$ with $2$, $2$ with $3$, $3$ with $4$ and $4$ with $1$ for example).  (Odd number cards don't work because if $3$ cards are switched then one card remains the same, and if we have "$1$ card switched" that just means no cards are switched, since you can't switch a card with itself.)

So if we have $2$ cards flipped, we have $\binom{4}{2}$ to choose a pair of cards to be flipped, and the other two will be flipped no matter what.  Hence there are $6$ ways for this case.

If we have $4$ cards flipped, then there are just $3$ ways since we have $\binom{4}{1} = 4$ ways to arrange this but of course we can't have $1,2,3,4$.

So $6 + 3 = 9$, and out of $4! = 24$ ways the probability is $\boxed{\frac{3}{8}}$.

Phew longest FTW solution I ever typed :)

Whoever does not understand this should look at the first solution, but don't count 4132, because 3 is in the box number 3.

ya 4!=24
9 possiblilities so 9/24 or 3/8

Just a quick note: the number of ways you can place the numbers into a different position is called a [b]derangement[/b]. it's represented with a $!$ before the number.

$!1=0$
$!2=1$
$!3=2$
$!4=9$
$!5=44$

just to name the first five ($!0=1$, but that's confusing)

knowing this, it's easy to see that the answer is $\frac{!4}{4!} \implies \boxed{\frac{3}{8}}$


MATHCOUNTS
analytic geometry
FTW
email
quadratics
probability
percent

Hello!

Many of you are aware that I recently hosted a contest for 5th-6th graders. This 10 Round Tournament is open to all grade levels. Each Round will have 10 questions, each one worth 10 points. So the highest possible score is 100 for each round, and the highest for the whole Tournament is 1000. See if you can reach 1000!

This Tournament is similar the Wickedestjr's tournament in that people who move on to the next round are a few short. (In other words, for each round, there will be a few people who don't make it. The number depends on how many sign up.)

I will give you all until Friday, June 13th, midnight forum time to sign up. Once everyone who wishes to do so signs up, I will give more information.

So please, sign up!  :)

[b][u]The Sign-Up List[/u][/b]

1. 12markkram34

2. mewto55555

3. BOGTRO

4. dynamo729

5. D3m0n Shad0w

6. shentang

7. Urc

8. hunter34

9. Canton

10. Bacteria

11. scast

12. AIMEman

13. moogra

14. 7h3.D3m0n.117

I will update this list as more people join.

I'll join..... I have never done MOEMS so what is required??

I will join except I am attending a sleep-over on Friday, June 13th to June 14th at 10:00 am.

@D3m0n Shad0w

I'll go into more detail later, but that shouldn't be a problem.  :)

I guess I will join...but I don't know whats in my plans.

By the way, conflicts don't really ruin this, because you have 3 days for each round.

Can you post all the questions and answers somewhere?

[quote="isabella2296"]Sure. Advice: It's beastly hard. Good luck![/quote] I'll try my best...here goes...

izzy-can you post full solutions to each of the problems?

i understna dif you cant :P

To fishy the fish:what kind of excuse is that???
to isabella:when do the answers come out, and theres no round ten???

crazy4math, try actually reading some of the posts.  :P

:o I DO
do u know HOW hard it is to read 15 pages of posts i have been here since july and have trying to read all the posts and thats way to much, because i don't know abut u but i cant't spend my whole day at the computer reading posts of which some have unimportant answers. i read the last few pages so, i found out.
1)there's no round ten because too many people dropped out. so that answers one of my questions but as hard as i look i cannot find where the answers are posted,
u have offended me  :mad:
i really am trying hard to keep up with all of this.

crazy-this tournament is OVER.  Don't post here anymore.  You're just adding to the spam (i guess this post is also...)

Mods-lock this please

sorry. :blush:
fine, but i'm blaming it on fishy

and i'm blaming it on you, all the :spam:

not relaly lol  :rotfl:

whoops
more :spam:

This tourneys over, now we just have 1000000000 pages of spam, please let this die.

anyway, can you post the answers to the problems?

That's 90 problems....anyway my answer key disappeared, sorry. Ask Yongyi, he got almost all perfects.

I'll be posting answers to my problems!  and thats' more than 90 for sure...


(page 32, number 35)

"Mike and Joey bought identical loaves of bread and packages of bologna. Mike made sandwiches with 5 slices of bologna and had 4 slices of bread left when he ran out of meat. Joey made sandwiches with 4 slices of bologna and had 4 slices of meat when he ran out of bread. How many slices of bread were in each loaf?"

This is in the linear equations section. Don't know if that helps...

I never liked this problem either. In fact, most people don't... Anyway, it was discussed before: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47721

[quote="mathnerd314"]I did the problem on a long car trip, checked the solution manual months later, and found that I got the wrong answer. I hated that problem.

Speaking of solution manuals, do you not have a solution manual, imber.matutinis? Because you can always peek at that and get the hints you want.[/quote]

I do not have a solutions manual...because someone told me that I shouldn't get one. o.O I thought they had the answers at least in the book, but apparently they don't. >.<

Who told you that you shouldn't get a solutions manual? You should get one. It's not too costly anyway.


calculus
integration
trigonometry
calculus computations

[quote="xirti"]
2)$\int_1^\infty sin(x^2)dx$[/quote]

Could the check the 1 in the integral

This one is known as Fresnel integral $\int_0^\infty sin(x^2)dx$

Actually, this was the form in my book.. I looked up frensel integrals and found a proof. Thank you! What about the first?

[quote="xirti"]1)$\int_{0}^\infty \frac 1{(1+x^{a})(1+x^{2})}dx,a>0$
2)$\int_{1}^\infty sin(x^{2})dx$[/quote]
This had been lying around for quite sometime.

Solution to [b]problem (1)[/b]:

Let $x = \tan\theta$.

Then, $I = \int_{0}^{\frac{\pi}{2}}\frac{\,d\theta}{1+{\tan}^{a}\theta}= \int_{0}^{\frac{\pi}{2}}\frac{{\cos}^{a}\theta \,d\theta}{{\cos}^{a}\theta+{\sin}^{a}\theta}$
$\Rightarrow I = \int_{0}^{\frac{\pi}{2}}\frac{{\sin}^{a}\theta \,d\theta}{{\cos}^{a}\theta+{\sin}^{a}\theta}$ ... [using the identity $\int_{0}^{a}f(x)\,dx = \int_{0}^{a}f(a-x)\,dx$]
$\Rightarrow 2I = \int_{0}^{\frac{\pi}{2}}\,d\theta$
$\Rightarrow I = \boxed{\frac{\pi}{4}}$ ... [Not so hard after all!]


algebra
polynomial
group theory
abstract algebra
invariant
Ring Theory
Galois Theory

Hi, I am interested in the folowing theorem but I do not know where to find information on it.

Let $ F$ be a polinomial with integer coeficients that and $ k$ irreducible components. Let $ s(p)$ be the number of solutions to the congruence $ F\equiv 0\mod p$, for each prime $ p$. Prove that the average of $ s(p)$ over all primes is $ k$.

By average it seems that one the folowing may be understood.
-- $ \displaystyle\frac {1}{\pi(x)}\sum_{p\le x}{s(p)}$ is asymptotically equal to $ k$

-- $ \displaystyle\sum_{p\le x}{\frac {s(p)}{p}\log(p)}$ is asymptotically equal to $ k\log(x)$

and so on
.
.
.

Does the theorem have a name? Has anybody a literature suggestion? I am interested in the proof or related developments which might have inspirede the theorem.

Thanks in advance.

You should alsy assume that the irreducible components are pairwise different. This means, that for almost all $ p$, they have no common root $ \mod p$. Thus the problem reduces to the case when $ F$ is irreducible.

Now if $ F$ is irreducible:
Let $ K$ be splitting field of $ F$ over $ \mathbb Q$ and let $ G$ be the corresponding Galois group. If $ \mathcal O_K$ is $ K$'s number ring and $ \mathfrak p$ is a prime ideal over $ p$, then the map $ \mathcal O_K / \mathfrak p \to \mathcal O_K / \mathfrak p$ with $ x \mapsto x^p$ is called the Frobenius; it permutes the roots of $ F \mod \mathfrak p$, and a fixed point corresponds to a solution of $ F(x) \equiv 0 \mod p$, so we want to count those. Note that a $ \sigma \in G$ with $ \sigma \mathfrak p \equal{} \mathfrak p$ (these $ \sigma$ build a subgroup $ G_{\mathfrak p}$) induces some automorphism $ \mod \mathfrak p$, which by elementary field theory is a power of the Frobenius. For unramified primes $ p$ (those that are "nice"), this induces an isomorphism between $ G_{\mathfrak p}$ and the Galois group of $ \mathcal O_K / \mathfrak p$ over $ \mathbb Z /(p)$.

Now we can write $ G$ as a union of it's conjugacy classes $ G_i$ and apply Chebotarev's theorem, which states that for $ \sigma \in G$, the density of the primes $ p$ such that the Frobenius of $ p$ is of type $ \sigma$ (as a permutation of the roots of $ F$) is $ \frac{|G_\sigma|}{|G|}$, where $ G_\sigma$ is $ \sigma$'s conjugacy class. Note that the number of fixed points of $ \sigma$ is invariant under conjugation, so $ |G_\sigma| \cdot \mathrm{Fix}(\sigma) \equal{} \sum_{\tau \in G_\sigma} \mathrm{Fix}(\tau)$. This gives us that the average of $ s(p)$ over $ p$ is equal to $ \frac{1}{|G|} \sum_{\sigma \in G} \mathrm{Fix}(\sigma)$, which equals $ 1$ by elementary group theory. This is what we wanted to prove.

As a literature suggestion, most texts on Chebotarevs theorem should do (but most will not mention this application). I can't give a good reference, but most good books on algebraic number theory should have a reference or a proof for Chebotarev's theorem.


192. 1
193. a.6.24
       b.6.24
[hide="."]  Btw, hi strangers from antarctica and germany
:D  :rotfl:   :P  [/hide]

let's wait for a few more people to reply, which i very much doubt will happen :D

mr antarctica, cant you say whether its right or wrong atleast? :huuh: :D

1.infinity
2.--6.24(both)

i m pretty sure 1 is infinity :D

it has 2 b infinity
no. of eggs is fixed--u can't write as 20.00000000000000000000000eggs

What eggs? :D :D Well i dont think that might be the case :maybe:

that's the case  :w00t:   :whistling:  :sleeping:

thats what shrids says is written in the 11th chem text book, but the book in which i read abt it (some guide i dont remember) says that its only 1 :maybe:

the guide guided u wrong!!!!!!!!!!! :rotfl:  :rotfl:  :D

i dont know which one man but ive seriously read that somewhere :D

ok i made a mistake bcos i didnt see that he was counting a definite quantity like egs or sth :D :D :o

ritu, you're spot on  :thumbup:
looks like you've nv'd chem text book very well :D

[quote="hash_include"]looks like you've nv'd chem text book very well
[/quote]
that was a  wild guess :rotfl:


function
trigonometry
algebra proposed
algebra

Determine all monotonous functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfy \[ f^3(x)\minus{}3f^2(x)\plus{}6f(x)\equal{}4x\plus{}3,\] for all $x \in \mathbb{R}$, where $ f^3\equal{}f \circ f \circ f$ and $ f^2\equal{}f \circ f$.

[quote="moldovan"]Determine all monotonous functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfy: $ f^3(x) - 3f^2(x) + 6f(x) = 4x + 3, (\forall) x \in \mathbb{R}$, where $ f^3 = f \circ f \circ f$ and $ f^2 = f \circ f.$[/quote]

Let $ u\in\mathbb R$. Consider the sequence $ a_n=f^{[n]}(u)$

We get $ a_n=3a_{n-1}-6a_{n-2}+4a_{n-3}+3$ with $ a_0=u,a_1=f(u)$, and $ a_2=f(f(u))$

The equation $ x^3-3x^2+6x-4$ has 3 roots $ 1,2e^{\frac{i\pi}3}$ and $ 2e^{-\frac{i\pi}3}$ and so we get :

$ a_n=u+n+r2^n\cos(\frac{n\pi}3+\theta)-r\cos(\theta)$ for some $ r>0,\theta$ which may be be computed from $ a_1$ and $ a_2$

Since $ f(x)$ is monotonous, we need to have $ u_n=\frac{a_{n+2}-a_{n+1}}{a_{n+1}-a_n}$ $ =\frac{f(a_{n+1})-f(a_n)}{a_{n+1}-a_n}$ with a constant sign.

But $ u_n=\frac{1-r\sqrt 32^{n+1}\sin(\frac{(n+1)\pi}3+\theta)}{1-r\sqrt 32^{n}\sin(\frac{n\pi}3+\theta)}$ $ =\frac{1-r\sqrt 32^{n+1}v_{n+1}}{1-r\sqrt 32^{n}v_n}$ where $ v_n=\sin(\frac{n\pi}3+\theta)$

And, obviously, when $ n\to +\infty$, we always have some $ n$ where $ v_{n+1}$ and $ v_n$ have the same sign and some $ n$ where $ v_{n+1}$ and $ v_n$ have opposite sign.

So $ u_n=\frac{1-r\sqrt 32^{n+1}v_{n+1}}{1-r\sqrt 32^{n}v_n}$ can have a constant sign when $ n\to +\infty$ only if $ r=0$

So $ a_n=u+n$ and $ f(u)=u+1$ which indeed is a solution.

So the unique monotonous solution to this equation is $ \boxed{f(x)=x+1}$


n boys are sitting in a circle. Each has an even number of sweets. Each then gives half his sweets to the boy on his left. The teacher then gives any boy who ends up with an odd number of sweets an extra one. Show that after a sufficient number of repetitions every boy has the same number of sweets.

I'm not sure if this is correct. I just tried with the following example. Here there are 3 boys having 2, 4 and 6 sweets and the boy who has 6 sweets starts giving his sweets to the boy in left with 4 sweets and so on.

2	4	6
2	7	3
2	8	4
6	4	4
3	4	7
4	4	8
4	8	4
8	4	4

after this the same 8, 4 and 4 is going to repeat.

Sorry if I have understood the problem wrongly. Correct me if I'm wrong.

i think all of them pass on the sweets instantly :D  :lol:

[hide="Solution"]Suppose at some point define $ D \equal{} \max \minus{} \min$
Clearly the $ \max$ can never increase(even if the teacher gives the extra), nor the $ \min$ can decrease. If, after some turn, $ D$ remains same, then The $ \min$ must have remained same. But in that case the number of minimum values must have decreased unless $ \min \equal{} \max$. Thus after finitely many turns everyone will have same number of sweets.[/hide]

Why is everybody assuming that $ n\equal{}3$? Consider the general case please.

[quote="Johan Gunardi"]Why is everybody assuming that $ n \equal{} 3$? Consider the general case please.[/quote]

I had the same idea that is in the link that dbgreen posted. Let the largest number of sweets any one person has be l, and the smallest be m. Then after each step i.e. passing on half and giving an extra one to the odd ones if any, l cannot increase and m cannot decrease. The number of people with m sweets must decrease because there must be someone with m sweets to the left of someone with > m sweets, otherwise we are done. So the number of people with m sweets goes to 0, and then m to goes to at least m+2 etc.

[quote="Johan Gunardi"]Why is everybody assuming that $ n \equal{} 3$? Consider the general case please.[/quote]
I have been so stupid.
Now I have edited.


Let circle O1 and circle O2 be tangent internally to circle O at A.B, respectively. Line g1 and line g2 are internal tangent of circle O1 and O2. g1 intersects g2 at C. Circle O1, cirlce O2 touches line g3 at D,E, respectively. M is midpoint of the line segment DE. N is foot of perpendicular from C to g3. AM intersects  circle O again at P. BN intersects circle O again at Q. Prove that PQ is parallel to g3.

g3 is external tangent of circle O1 and O2

please send your solution Fang_jh(preferably hidden) i wanna know how did you solve it.....
how can you call such a problem EASY :maybe: ???????????!!!!!!!!!!!!

Let line g3 intersect center line of circle O1,O2 at K.
The three points K,A,B are collinear (by Monge's theorem).
Let the circle O1 tangent to g1 at H, the circle O2 tangent to g2 at I.
 DH intersects EI at T.  The three points T,C,N are collinear (?).
Let DH, EI intersect center line of the circle O1,O2 at S,R, respectively.
ES is perpendicular to DT(?)
DR is perpendicular to TE(?)
M,N,S,R are concyclic (NPC wrt triangle TDE)
KN*KM=KS*KR
E,D,S,R are concyclic
KS*KR=KD*KE
E,D,A,B are concyclic(?)
KD*KE=KA*KB
yielding KN*KM=KA*KB
as a result M,N,A,B are concyclic
∠AMN=∠ABQ=∠APQ
yielding PQ ∥g3

My solution may be similar to your solution, but it is simpler
Let line g3 intersect center line of circle O1,O2 at K.
The three points K,A,B are collinear (by Monge's theorem).
All we have to do is prove M,N,A,B are concyclic
Call $r$ and $R$ are the radius of (O1) and (O2) ($r < R$)

We have
$\frac{r}{R}= \frac{{ND}}{{NE}}= \frac{{KD}}{{KE}}= \frac{{KN}}{{KE+NE}}$ (Because K,C are the external and internal center of similitude of the circumcircle (O1) and incircle (O2).And $CN//O1D//O2E$)
$\begin{array}{l}\Rightarrow \frac{{KN}}{{KD}}= \frac{{KE+NE}}{{KE}}= 1+\frac{{NE}}{{KE}}\\ \Leftrightarrow \frac{{KN}}{{KD}}+\frac{{KN}}{{KE}}= 1+\frac{{NE}}{{KE}}+\frac{{KN}}{{KE}}= 2 \\ \Leftrightarrow KD \times KE = KN \times \left({\frac{{KD+KE}}{2}}\right) = KN \times KM \\ \end{array}$
We also have $KD \times KE = KA \times KB$
So $KA \times KB = KN \times KM$

So that M,N,A,B are concyclic


this is weaker than maclaurin's symmetric mean ineq
\[
\sqrt {\frac{{\sum {a^2 } }}
{4}}  \geqslant \sqrt {\frac{{ab + bc + cd + da + ac + bd}}
{6}}  \geqslant \sqrt[3]{{\frac{{\sum {abc} }}
{4}}}
\]

Yeah, I knew it , I wanted to post it after this one. :D

It also (of course) has an easy Muirhead solution.  Raising to the sixth power, we get

$\frac {1}{4^3}\left(\sum a^6+3\sum a^4b^2+6\sum a^2b^2c^2\right) \geq \frac {1}{4^2}\left(\sum a^2b^2c^2+2\sum a^2b^2cd\right)$

which is pretty obvious since all of the sums on the LHS majorize the sums on the RHS.


Like, say I want the margin to be different on one page than the other ones, what do I do?
And say I want to end a line, but I don't want to have the huge space that appears when you have an empty line between them, what do I do?
And uhh, say I want to have a vertical space, but not a line skipping space, why doesn't it work when i do something like this?:
blah blah blah
\vspace{10mm}
blah blah blah
\vspace{10mm}
blah blah blah
It still either all end up on one line for some reason, or the space only happens to one of the lines, and the space created becomes too big.

I'm not sure for your first question, but I can answer your second and third.

For your second question, use \\ at the end of a line and then hit Enter once, then begin typing your next line. (Do not skip lines by hitting Enter twice.)

For your third question, you should set \vspace off in a separate line; skip lines, basically.

[quote="ZhangPeijin"]Like, say I want the margin to be different on one page than the other ones, what do I do?[/quote]
Try the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=chngpage]chngpage package[/url]. The documentation is inside the file itself. It will do a lot of things but for changing margins use the adjustwidth environment:
[quote] [b]adjustwidth[/b]

    Within an adjustwidth environment the left and right margins can be
 adjusted. The environment takes one optional argument and two required
 length arguments:

 \begin{adjustwidth}[]{leftmargin}{rightmargin}

     A positive length value will increase the relevant margin (shortening
 the text lines) while a negative length value will decrease the margin
 (lengthening text lines). An empty length argument means no change
 to the margin. At the end of the environment the margins revert to
 their original values.

    For example, to extend the text into the right margin:
 \begin{adjustwidth}{}{-8em}[/quote]


$ abc\equal{}216$
$ a \plus{} b \plus{} c \equal{}19$

$ a,b,c \in \mathbb{Z}$

[quote="mathwizarddude"]$ abc \equal{} 216$
$ a \plus{} b \plus{} c \equal{} 19$

$ a,b,c \in \mathbb{Z}$[/quote]
Note that $ 216 \equal{} 6^3$..
Case $ a,b,c>0$:
We can't have $ 3 \mid a,b,c$ and on the other hand we can't have $ 27 | a$ or $ 27 | b$ or $ 27 | c$. So assume $ 9 \mid a$, since $ a \equal{} 18$ is not a solution: $ a\equal{}9$. Let $ b \equal{} 3k$ So $ abc \equal{} 9bc \equal{} 27kc\equal{}27k(10\minus{}k)$ $ \iff$ $ k(10\minus{}3k) \equal{} 8 \iff k \equal{} 2$. So $ (a,b,c) \equal{} (9,6,4)$ or any permutation of that is a solution.
Case $ a>0$ and $ b,c<0$.
Obviously $ a>19$ and $ a \mid 2^3 \cdot 3^3$. If $ a > 27$ we have that $ abc \ge 28(\minus{}8)(\minus{}1) \equal{} 224 > 216$. So $ a \in [20;27]$.
So we can split into cases:
case $ a \equal{} 24$: $ b\plus{}c \equal{} \minus{}5$ and $ bc \equal{} 9$ but $ |bc| \le (\frac{b\plus{}c}{2})^2 < 9$
case $ a \equal{} 27$: $ b\plus{}c \equal{} \minus{}8$ and $ bc \equal{} 8$, but this also impossible.

So the only solution is $ (9,6,4)$ and it's permutations..


Let a+b+c = pi/2 Write cos(a)cos(b)cos(c) as a sum of sines.

Let x=a+b, then:
cos(a)cos(b)cos(c)=
cos(a)cos(x-a)cos(pi/2-x)=
cos(a)*sin(x)*[cos(x)cos(a)+sin(x)sin(a)]=
cos(a)cos(x)sin(a)sin(x)+cos(a)^2*sin(a)^2=
cos(a)cos(x)sin(a)sin(x)+sin(a)^2-sin(a)^4=
cos(a)cos(x)sin(a)sin(x)-(sin(4a)-1)/8=
???
...I don't know where to go from here.

@lokito: I don't think that's the best way to go.

[hide="Solution"]\begin{eqnarray*}\cos a\cos b\cos c &=& \cos a\cos b\cos (\pi/2-a-b)\\ &=& \cos a \cos b \sin(a+b)\\ &=& \left[ \frac{\cos(a+b)+\cos(a-b)}{2}\right]\sin(a+b)\\ &=& \frac{\cos(a+b)\sin(a+b)}{2}+\frac{\cos(a-b)\sin(a+b)}{2}\\ &=& \frac{\sin(2(a+b))}{4}+\frac{\sin(2a)+\sin(2b)}{4}\\ &=& \frac{\sin(2a+2b)+\sin(2a)+\sin(2b)}{4}\\&=& \frac{\sin(2c)+\sin(2a)+\sin(2b)}{4}\end{eqnarray*}

I think that's good enough. I have repeatedly made use of the product-sum identities.[/hide]


Find $ m$ such that $ x^2\plus{}3xy\plus{}x\plus{}my\minus{}m$ be the product of two factors of degree $ 1$ and with integral coefficients.

[hide][b]Form:[/b] $ (x \plus{} ay \plus{} b)(x \plus{} cy \plus{} d)$

$ \blacktriangleright$ There is no $ y^2$ term, so $ c \equal{} 0$.
$ \blacktriangleright \; xy$ coeff. is $ 3$, so $ a \plus{} c \equal{} a \equal{} 3$
$ \blacktriangleright \; x$ coeff. is $ 1$, so $ b \plus{} d \equal{} 1 \implies b \equal{} 1 \minus{} d$

$ (x \plus{} 3y \plus{} 1 \minus{} d)(x \plus{} d) \equal{} (x^2 \plus{} 3xy \plus{} x) \plus{} d \minus{} d^2 \plus{} 3dy$
$ \therefore my \minus{} m \equal{} d \minus{} d^2 \plus{} 3dy$

Plugging in $ y \equal{} 1$ gives $ d(1 \minus{} d) \plus{} 3d \equal{} 0 \implies d \equal{} 0,4$

[b]Answer:[/b]
$ (x \plus{} 4)(x \plus{} 3y \minus{} 3) \implies m \equal{} 12$
$ x(x \plus{} 3y \plus{} 1) \implies m \equal{} 0$[/hide]


geometry
function
trapezoid
algebra
system of equations
national olympiad

1. Solve the system of equations:
$(1+\frac{12}{3x+y})\sqrt{x}=2\\(1-\frac{12}{3x+y})\sqrt{y}=6$

2. Given $x,y$ are integer number and different to -1, satisfying:
$\large \frac{x^{4}-1}{y+1}+\frac{y^{4}-1}{x+1}\in Z$ . Prove that $x^{4}y^{44}-1 \vdots (x+1)$.

3. A triangle ABC with BC are fixed. H and G are the orthocenter and centroid of triangle ABC, respectively. A is moving such that the midpoint of HG belongs to the line BC. Find the locus of A.

4. Given a regular 2007-gon.  Find the minimal number $k$ such that: Among every $k$ vertexes of the polygon, there always exists 4 vertexes forming a convex quadrilateral such that 3 sides of the quadrilateral are also sides of the polygon.

5. Given $b$ is a positive integer. Find all the function $f : R \to R$ such that:

$f(x+y)=f(x)3^{b^{y}+f(y)-1}+b^{x}(3^{b^{y}+f(y)-1}-b^{y})$

6. Given a trapezoid ABCD, BC is the bigger base, is incribed by (O). . The point P varies on the line BC and lies outside the segment BC such that: PA is not tangent to BC. The circle having PD as diameter cuts (O) at E ( $E\neq D$). M is the interection point of BC and DE. N ($N \neq A$)is the intersection point of PA and (O). Prove that the line MN goes through a fixed point.

7. Given a number $a > 2$. Denote $f_{n}(x)=a^{10}x^{n+10}+x^{n}+...+x+1$, $n$ is positive integer number.

Prove that $f_{n}(x)=a$ has only one solution $x_{n}\in (0,+\infty)$. Prove that the sequence $(x_{n})$ has  finite limit

Very long! :D
What about on students of you? :P

Our students solved 5 to 6 problems. They waste time on the analytical geometry problem.

So if, Our students only sloved at most four problem :D I think :P

Problem 1 http://www.mathlinks.ro/Forum/viewtopic.php?t=132535
Problem 2 http://www.mathlinks.ro/Forum/viewtopic.php?t=132536
Problem 3 http://www.mathlinks.ro/Forum/viewtopic.php?t=132547
Problem 4 http://www.mathlinks.ro/Forum/viewtopic.php?p=750364#750364
Problem 5 http://www.mathlinks.ro/Forum/viewtopic.php?t=132556
Problem 6 http://www.mathlinks.ro/Forum/viewtopic.php?t=132551
Problem 7 http://www.mathlinks.ro/Forum/viewtopic.php?t=132554

This is result

Total: 41 ( Have all 399 candidate)   :(

Rank 1: 2
Rank 2: 7
Rank 3: 7
Rank 4: 25


When Babe Ruth hit a homer over the 12-m-high right-field fence 100 m from home plate, roughly what was the minimum speed of the ball when it left the bat? assume the ball was hit 1.0 m above the ground and its path initially made a 40 degree angle with the ground.

Well, I used the following formula, where $ R$ is range, $ v_0$ is initial velocity, and $ \theta$ is the angle. I'm assuming that if everything started on the ground instead of one meter up, then the minimum height the ball would have to go to clear the fence is 11 meters. Since the fence is 100 meters away from the home plate, and it is the maximum height, then the ball will land 100 meters away (neglecting air resistance). So the $ R$ is 200 m.

$ R \equal{} \frac {(v_0^2)(\sin{2\theta})}{g}$.

Plugging in values gives:

$ 200 \equal{} \frac {(v_0^2)(\sin{80})}{9.8}\Leftrightarrow$
$ v_0 \equal{} 44.6 \frac {m}{s}$

The book says the answer is roughly 34, so there is some flaw in my logic. I tried various ways of solving this, and gotten a variety of answers, but never 34. How did the book get this answer?

Also, in real life, why isn't the ball at its maximum height at $ \frac {t}{2}$, where $ t$ is the time its in the air? Wouldn't the air resistance affect it the same way on the journey up and down?

EDIT: Another problem...

Agent Tim, flying a constant 200km/hr horizontally in a low-flying helicopter, wans to drop an explosive onto a master criminal's automobile traveling 150 km/hr on a lvel highway 88 m below. At what angle (with the horizontal) should the car be in his sights when the bomb is released?

Well, the first think I did was calculate the time it would take for the bomb to fall 88 m:
$ 88 \equal{} \frac {1}{2}(9.8)(t^2)\Leftrightarrow$
$ t \equal{} 4.24 \sec$

I'm not sure if this is allowed, but from the frame of reference of the car, the plane is moving 50 km/hr. Since the plane is dropping a bomb, not shooting it, that means the bomb's initial horizontal velocity is going to be 50 km/hr as well. So I converted that to m/s:

$ \frac {50 km}{hr}\cdot\frac {1hr}{60min}\cdot\frac{1min}{60sec}\cdot\frac{1000m}{1km} \equal{} 13.9 \frac {m}{s}$

Since the bomb will be traveling 4.24 seconds, I multiplied the rate I just found with the time to find the distance the bomb travels, giving me 58.9 meters. So...
$ \tan^{ \minus{} 1}\theta \equal{} \frac {88}{58.9}\Leftrightarrow$
$ \theta \equal{} 56.2$

Is this feasible?

Start by writing the parametric equations for the ball motion:

$ x \equal{} v_o \cos \theta t$
$ y \equal{} v_o \sin \theta t \minus{} \frac{1}{2}gt^2$.

According to the data of the problem, we want that, when x = 100 m we must have y = 11 m. Substituting in the above two equations (with $ \theta \equal{} 40^o$) we get $ v_o \equal{} 33.8\, m/s$. Formulas should be used with care.


The roots of $ x^3$ $ \plus{}$ $ ax^2$ $ \plus{}$ $ bx$ $ \plus{}$ $ c$ $ \equal{}$ $ 0$ are $ \alpha$, $ \beta$ and $ \gamma$.Find the cubic whose roots are $ \alpha$^3, $ \beta$^3 and $ \gamma$^3

Hello
[hide="hint"]
Let assume that the required eq. be $ x^3 \plus{} px^2 \plus{} qx \plus{} r \equal{} 0$ whose roots are $ \alpha^3,\beta^3,\gamma^3$.
now
from the given eq. we have
$ \alpha \plus{} \beta \plus{} \gamma \equal{} \minus{} a$
$ \alpha\beta \plus{} \gamma\beta \plus{} \alpha\gamma \equal{} b$
$ \alpha\beta\gamma \equal{} \minus{} c$

In our eq.
$ p \equal{} \alpha^3 \plus{} \beta^3 \plus{} \gamma^3 \equal{} (\alpha \plus{} \beta \plus{} \gamma)((\alpha \plus{} \beta \plus{} \gamma)^2 \minus{} 3(\alpha\beta \plus{} \alpha\gamma \plus{} \gamma\beta)) \minus{} 3\alpha\beta\gamma$
 where r=$ (\alpha\beta\gamma)^3$=$ \minus{} c^3$
find q  by using same method that we have used for p.
[/hide]

Thank u

so we have:
[hide="answer"]$ x^3\plus{}(a^3\plus{}3c\minus{}3ab)x^2\plus{}(b^3\minus{}2abc\plus{}3c^2)x\plus{}c^3\equal{}0$
[/hide]


Accordingly,there are 2,279,184 solutions to any N-Queen problems for n=15.Can anyone show me how such calculations happened?How about 14,772,512 solutions to N-Queens problem for n=16;And 95,815,104 solutions to n-queens problem for n=17? I'm just intrigue about how such calculating possibilities happened.

                                                                   Respectfully,
                                                                   A.Eddington

       [/b]

See http://oeis.org/A000170 and references there.


geometry
geometric transformation
reflection
circumcircle
Euler
power of a point
radical axis

Let $ ABC$ be a triangle and let $ M, N, P$ be the midpoints of sides $ BC, CA, AB$ respectively. Denote by $ X, Y, Z$ the midpoints of the altitudes emerging from vertices $ A, B, C$ respectively. Prove that the radical centre of the circles $ AMX, BNY, CPZ$ is the centre of the nine-point circle of triangle $ ABC$.

This is from Math Reflections 2008 Issue 1, and since deadline is over for those problems I think I can post them here.

Let $ G$ be the centroid and $ O$ the circumcenter of the $ \triangle ABC.$ $ G$ is also the centroid of the medial $ \triangle MNP$ and $ O$ is its orthocenter. Let $ U, V, W$ be midpoints of $ NP, PM, MN.$ Let $ \mathcal Q_m, \mathcal Q_n, \mathcal Q_p$ be circles internally tangent to the circles $ \odot (AMX), \odot (BNY), \odot (CPZ)$ at $ M, N, P$ and with similarity coefficients $ 1/3.$

Let $ D, E, F$ be the altitude feet of the medial $ \triangle MNP.$ Since the circles $ \odot (AMX), \odot (BNY), \odot (CPZ)$ pass through the reflections $ A', B', C'$ of $ M, N, P$ in the perpendicular bisectors of $ AX, BY, CZ$ and $ \overline {MA'} \equal{} 3\ \overline {MD},$ $ \overline {NB'} \equal{} 3\ \overline {NE},$ $ \overline {PC'} \equal{} 3\ \overline {PF},$ the circles $ \mathcal Q_m, \mathcal Q_n, \mathcal Q_p$ pass through the altitude feet $ D, E, F$ of the medial $ \triangle MNP$ and they are centered on the sidelines $ VW, WU, UV$ of its medial $ \triangle UVW.$ Thus the powers of the orthocenter $ O$ of the medial $ \triangle MNP$ to the circles $ \mathcal Q_m, \mathcal Q_n, \mathcal Q_p$ are all equal to half the power of orthocenter $ O$ to the circumcircle $ (K)$ of this medial triangle, identical with the 9-point circle of the $ \triangle ABC$:

$ \overline{OD} \cdot \overline {OM} \equal{} \overline{OE} \cdot \overline {ON} \equal{} \overline {OF} \cdot \overline {OP} \equal{} \frac {_1}{^2}\ p(O, (K)).$

Since the circles $ \odot (AMX), \odot (BNY), \odot (CPZ)$ pass through the reflections $ A, B, C$ of $ M, N, P$ in $ U, V, W$ and $ \overline {MA} \equal{} 3\ \overline {MG},$ $ \overline {NB} \equal{} 3\ \overline {NG},$ $ \overline {PC} \equal{} 3\ \overline {PG},$ the circles $ \mathcal Q_m, \mathcal Q_n, \mathcal Q_p$ all pass through the centroid $ G.$ Thus the powers of the centroid $ G$ of the medial $ \triangle MNP$ to the circles $ \mathcal Q_m, \mathcal Q_n, \mathcal Q_p$ are all equal to zero.

Consequently, the circles $ \mathcal Q_m, \mathcal Q_n, \mathcal Q_p$ are coaxal and the common Euler line $ GO$ of the $ \triangle MNP, \triangle ABC$ is their common radical axis. The circumcenter $ K$ of the medial $ \triangle MNP,$ identical with the 9-point center of the $ \triangle ABC,$ then lies on their common radical axis. Let the lines $ MK, NK$ cut the circles $ \mathcal Q_m, \mathcal Q_n$ again at $ M_q, N_q.$ Since $ \overline{KM} \equal{} \overline{KN}$ and $ \overline{KM} \cdot \overline{KM_q} \equal{} \overline{KN} \cdot \overline {KN_q},$ it follows that $ \overline {KM_q} \equal{} \overline {KN_q}$ and $ \overline {MM_q} \equal{} \overline {NN_q}.$

Let the lines $ KM, KN$ cut the circles $ \odot (AMX), \odot (BNY)$ again at $ M_0, N_0.$ Since $ M, N$ are similarity centers of $ \mathcal Q_m \sim \odot (AMX), \mathcal Q_n \sim \odot (BNY),$ we get $ \overline {MM_0} \equal{} 3\ \overline {MM_q} \equal{} 3\ \overline {NN_q} \equal{} \overline {NN_0}.$ Using $ \overline{KM} \equal{} \overline{KN}$ again, this yields $ \overline{KM_0} \equal{} \overline{KN_0}$ and  $ \overline{KM_0} \cdot \overline{KM} \equal{} \overline{KN_0} \cdot \overline{KM}.$ In conclusion, $ K$ is on the radical axis of the circles $ \odot (AMX), \odot (BNY)$ and similarly, $ K$ is on the pairwise radical axes of $ \odot (BNY), \odot (CPZ)$ and $ \odot (CPZ), \odot (AMX).$ As a result, $ K$ is (at least) the radical center of these 3 circles.

Let $D$ be the foot of the A-altitude and $ H,N$ the orthocenter and the 9-point center of $ \triangle ABC.$ Let $ L$ denote the midpoint of $AH$ (orthocenter of APN) and let the ray $ ML$ cut the circle $ \mathcal{G}_a \equiv \odot(AMX)$ at $ T.$ Since $ LM$ is a diameter of the 9-point circle $ (N),$ we have that $ p(N,\mathcal{G}_a) \equal{} NM \cdot NT$ and from the power of $ L$ WRT $ \mathcal{G}_a$ we get  $ R \cdot TL \equal{} AL \cdot LX$ since $ \triangle ABC \sim \triangle APN$ with similarity coefficient $ 2.$ Then the power $ AL \cdot LX$ equals a quarter of the power $ AH \cdot HD.$

$ p(N,\mathcal{G}_a) \equal{} \frac {R}{2} \left (\frac {R}{2} \plus{} TL \right ) \equal{} \frac {R}{2} \left (\frac {R}{2} \plus{} \frac {AH \cdot HD}{4R} \right )$

$ p(N,\mathcal{G}_a) \equal{} \frac {R^2}{4} \plus{} \frac {AH \cdot HD}{8} \equal{} \frac {R^2}{4} \plus{} \frac {k^2}{8}$

$ AH \cdot HD$ equals the power of inversion $ k^2$ that takes the circumcircle $ (O)$ of $ \triangle ABC$ into its 9-point circle $(N),$ thus we conclude that $ p(N,\mathcal{G}_a) \equal{} p(N,\mathcal{G}_b) \equal{} p(N,\mathcal{G}_c)$ $ \Longrightarrow$ $ N$ is the radical center of $ \mathcal{G}_a,\mathcal{G}_b,\mathcal{G}_c.$


While going about trying to figure out a problem about the static equilibrium of a rigid beam, I found the following equations after setting the sum of the forces equal to zero and setting them to zero and setting the sum of the torques equal to zero:

$ T\cos{\theta} \minus{} F\cos{\phi} \equal{} 0$
$ T\sin{\theta} \plus{} F\sin{\phi} \minus{} mg \equal{} 0$
$ \minus{}\frac{1}{2}lmg \plus{} lT\sin{\theta} \equal{} 0$

where $ m$, $ l$, and $ \theta$ are knowns.  (Of course, $ g$ is known as well.)  The $ l$ can be divided out of the last equation.  This leaves three equations in three unkowns: $ T$, $ F$, and $ \phi$.  Now how do I get $ F$ in terms of the knowns?  In particular, I'm having trouble with the $ \phi$ and not getting something nasty......

Any help would be greatly appreciated.

Making $ (T\cos{\theta} \minus{} F\cos{\phi})^2 \plus{} (T\sin \theta \plus{} F\sin{\phi})^2$ gives cosine rule.