geometry
circumcircle
inradius
IMO Shortlist
symmetry
incenter

A circle $ C$ has two parallel tangents $ L'$ and$ L"$. A circle $ C'$ touches $ L'$ at $ A$ and $ C$ at $ X$. A circle $ C"$ touches $ L"$ at $ B$, $ C$ at $ Y$ and $ C'$ at $ Z$. The lines $ AY$ and $ BX$ meet at $ Q$. Show that $ Q$ is the circumcenter of $ XYZ$

Nice problem but not hard. ;)
Assume that C=(O,R),C'=(O_1,R_1) and C"=(O_2,R_2).OX=OY=R,O_1X=O_1Z=R_1 and O_2Y=O_2Z=R_2 =>
 circumcircle of XYZ it is incircle of OO_1O_2.I am prove that BX prpendicular of OO_1 and AY  prpendicular of OO_2=>Q is the circumcenter of XYZ.AY   prpendicular of OO_2 <=>AO^2-(AO_2)^2=OY^2-(O_2Y)^2=R^2-(R_2)^2,AO^2=R^2+4RR_1 and (AO_2)^2=(2(RR_1)^1/2-2(RR_2)^1/2)^2+(2R-R_2)^2=>AO^2-(AO_2)^2=R^2-R_2^2<=> R^2=4R_1R_2 but this is easy question. :oops:  :)

In addition, I proved that:

${\overline {\underline |\ OZ\perp AB\ | }}$ ([u]very nice[/u] !);

$\ 2\rho=\frac{R\sqrt R}{\sqrt R_1 +\sqrt R_2}$, where $\rho$ is the inradius of the triangle $XYZ$ (easily);

$OZ\perp O_1O_2\Longleftrightarrow R_1=R_2$ (easily).

Let $C$ be tangent to $L'$ at $D$. A homothety centered at $Y$ maps $C$ to $C''$. Since $L'$ and $L''$ are parallel, $D$ is mapped to $B$, so $D$, $Y$, and $B$ are collinear. Similarly, $A$, $Z$, and $B$ are collinear.

Let the common external tangents of $C$ and $C'$ intersect at $P$.  is the external homothety center of $C$ and $C'$. $Y$ and $Z$ are the internal homothety centers of $C$ and $C''$ and $C'$ and $C''$, respectively. Hence, by d'Alembert's theorem, $P$, $Y$, and $Z$ are collinear.

Let $PZ$ intersect $C'$ at $Y'$. Since the homothety centered at $P$ maps $C'$ to $C$, $\frac{PY'}{PA} = \frac{PY}{PD}$, so $PY' \cdot PD = PA \cdot PY$. By power of a point on $C'$, $PA^2 = PY' \cdot PZ$, so $PY' = \frac{PA^2}{PZ}$. Substituting this into $PY' \cdot PD = PA \cdot  PY$, so $ADYZ$ is cyclic. Since $B = AZ \cap DY$, $BY \cdot BD = BZ \cdot BA$, so $B$ lies on the radical axis of $C$ and $C'$. By symmetry, $A$ must lie on the radical axis of $C$ and $C'$, so $Q$ must be the radical center of $C$, $C'$, and $C''$. Since $QX$, $QY$, and $QZ$ are tangents and $Q$ has equal power with respect to each of these circles, $QX = QY = QZ$, as desired.

Strangely enough, I found G3 harder than G4.
Let $C$ touch $L'$ and $L''$ at $M$ and $N$ repectively.
Let the centre of $C$ be $O_1$, the centre of $C'$ be $O_2$ and centre of $C''=O_3$.
Join $O_1,X$ and $O_2,X$ as $AO_2\parallel MN\implies \angle AO_2X=\angle XO_1N$
$\implies \Delta XO_1N~\Delta AO_2X\implies \angle AXO_2~\angle O_1XN\implies A,X,N$ are collinear.
Similarly we can prove that $A,Z,B$ are collinear and $B,Y,M$ are collinear.
$\angle AZX=\angle MAN$(alternate segment, $L'$ is tangent to $C'$)
$\angle MAN=\angle ANM$ as $C$ is tangent to $L'$ and $L''$ at diametrically opposite points.
$\implies \angle AZX=\angle XNB\implies XZMN$ is cyclic, $XN\cap ZB=A\implies A$ lies on the radical axis of the circles $C$ and $C''$, but as $C$ and $C''$ touch each other $\implies AY$ is the direct common tangent to $C$ and $C''$.
Similarly, we can prove that $BX$ is the direct common tangent to $C$ and $C'$.
$\implies AY\perp O_1O_3$ and $BY\perp O_1O_2$
$X,Y,Z$ are the points where the incircle of $\Delta O_1O_2O_3$ touches the sides.
This means that the circumcentre of $\Delta XYZ=$ the incentre of $\Delta O_1O_2O_3$
As $AY\perp O_1O_3\implies AY$ passes through the incentre of $\Delta O_1O_2O_3$
$BX\perp O_1O_2\implies BX$ passes through the incentre of $\Delta O_1O_2O_3$
$\implies AY\cap BX=\text{ the incentre of } \Delta O_1O_2O_3=\text{the circumcenter of} \Delta XYZ$


Fix a markov chain on a countable state space and let $ \rho_{xy}\equal{}P_x(T_y<\infty)$; that is, the probability that starting at $ x$, you hit $ y$.  My textbook says that it is a consequence of the strong markov property that for states $ x,y,z$, we have

\[ \rho_{xz}\geq \rho_{xy}\rho_{yz}
\]

Intuitively this is clear; there are more ways to go from $ x$ to $ z$ than there are to go from $ x$ to $ z$ hitting a specific state $ y$ in the middle.  But I cannot see how to prove this :(

Just follow your intuition. We need to consider the paths $ P$ starting with $ x$ and going to $ z$ without hitting $ y$ or $ z$ in the middle, paths $ Q$ going from $ x$ to $ y$ without hitting $ y$ or $ z$ in the middle, paths $ R$ going from $ y$ to $ z$ without hitting $ z$ in the middle, and, finally, paths $ S$ going from $ z$ to $ y$ without hitting $ y$ in the middle. Then $ \rho_{xz}\equal{}\sum_P p(P)\plus{}\sum_{Q,R}p(Q)p(R)$ while $ \rho_{xy}\rho_{yz}\equal{}\Bigl(\sum_Q p(Q)\plus{}\sum_{P,S} p(P)p(S)\Bigr)\Bigl(\sum_R p(R)\Bigr)$ ($ p(A)$ stands for the probability of path $ A$). Now it is enough to open the parentheses and use the estimate $ \sum_A p(A)\le 1$ where $ A$ is any family of paths satrting at the same point to get the result. This may be not the most elegant approach but it is the most natural.


Two consecutive integers have the property that twice the smaller integer is nine more than the larger integer. Find the value of the smaller integer.

Let's call the smaller integer x. The larger integer is x+1.
Therefore, 2x= (x+1)-9
               2x=x-8
                 x=8

It should have been $ 2x\equal{}(x\plus{}1)\plus{}9$.

$ \implies 2x\equal{}x\plus{}10 \implies x\equal{}\boxed{10}$.

letd say that the # is  x therefore the

smaller # = x
and the larger #= x+1

therefore we get the equation

2x=(x+1)+9

solving for x we get the answer is 10 or $\boxed{10}$


Let $ S$ be a set of rational numbers such that:

1. $ \frac{1}{2}\in S$

2. If $ x\in S$ then both $ \frac{1}{(x\plus{}1)}$ and $ \frac{x}{(x\plus{}1)}$ are elements of $ S$.

Prove that all rational numbers on the interval $ 0<x<1$ are elements of $ S$.

[hide] Consider all $ \frac{p}{q}$ with $ 0<p<q$.  We wish to show that $ S$ is this set.  We show this using strong induction on $ q$.  The base case is obvious since it is given that $ \frac{1}{2}\in S$.  Assume that for all $ q< n$ and $ 0<p<q$, $ \frac{p}{q}\in S$.  Then consider all $ \frac{m}{n}$ with $ 0<m<n$.  If $ \frac{m}{n\minus{}m}<1$ then $ \frac{m}{n\minus{}m}\in s$ and using the second operation in (2) yields $ \frac{m}{n}\in S$.  Otherwise, $ 2m\ge n$.  If $ 2m\equal{}n$, then $ \frac{m}{n}\equal{}\frac{1}{2}\in S$.  Otherwise, $ 2m>n$ and thus $ 1>\frac{n\minus{}m}{m}$.  Thus, since $ m<n$, $ \frac{n\minus{}m}{m}\in S$ and using the first operation of (2) on it yields $ \frac{m}{n}\in S$.[/hide]

For reference sake, this is BMO 1 2004/5 Q5


if among of x,y,z have number less than or equal 0 ,assume x \leq 0 .
The Cauchy-Schwartz inequality gives :y+z \leq 2 \sqrt (y 2 +z 2 )
and yz \leq 1/2(y 2 +z 2 ) \leq 1 =>(2-y-z)+x(yz-1) \geq 0 this equivalent to 2+xyz \geq x+y+z
if x,y,z are positive numbers,WLOG assume x \geq y \geq z
 +x \geq 1 we have x+y+z \leq 2 \sqrt (x 2 +(y+z) 2 )=2 \sqrt (yz+1) \leq 2+xyz
 +x<1  2+xyz-x-y-z=(1-x)(1-yz)+(1-y)(1-z) \geq 0
Problem is solved!

[quote=nickolas]Let  x,y,z  reals with  x 2 +y 2 +z 2 =2.Prove that  x+y+z \leq2+xyz
Thanks[/quote]

Generalization (from Romania, 2019 https://artofproblemsolving.com/community/c6t243f6h1924800_n_real_variables_on_a_ball_of_radius_sqrtn1):
Let $n\geq3$  and let $a_1,a_2, \cdots,a_n$  be real numbers such that $a_1^2+a_2^2+\cdots+a_n^2=n-1$ .
(i) Prove that $a_1+a_2+\cdots+a_n-a_1a_2\cdots a_n\leq n-1$ .
(ii) For what values of $k$  the inequality $a_1+a_2+\cdots+a_n-ka_1a_2\cdots a_n\leq n-1$ is always true?


Use the half-angle formulas to simplify the expression.
$ -\sqrt{\frac{1-\cos 8x}{1+\cos 8x}}$.

[hide]

$ \sin\frac{x}{2}= \pm \sqrt{\frac{1-\cos x}{2}}$

$ \cos\frac{x}{2}= \pm \sqrt{\frac{1+\cos x}{2}}$

$ \tan\frac{x}{2}= \pm \sqrt{\frac{1-\cos x}{1+\cos x}}$

so your equation is

$ \tan{4x}$
[/hide]


inequalities
IMO Shortlist
algebra
n-variable inequality
Arithmetic Mean-Geometric Mean

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Its too bad that not all inequalities are like this one...

[b]Solution[/b]

Define $a_{n+1}$ such that $\sum^{n+1}_{k=1} a_k = 1$. Cross multiplication and substitution yields that the desired inequality is equivalent to

\[n^{n+1} \le \frac{(1-a_1)(1-a_2) \dots (1-a_{n+1})}{a_1 a_2 \dots a_{n+1}} \quad (*)\]
By AM-GM we have that

\[n\sqrt[n] {\frac{a_1 a_2 \dots a_{n+1}}{a_i}} \le a_1 + a_2 + \dots + a_{i-1} + a_{i+1} + \dots + a_{n+1} = 1-a_i\]
Multiplying the above for each $i$ and dividing by $a_1 a_2 \dots a_{n+1}$ yields $(*)$ and the proof is complete.

Proceeding in a manner similar to the above posts,

$ n^{n+1}\le\frac{(1-a_1)(1-a_2)\dots (1-a_{n+1})}{a_1 a_2\dots a_{n+1}}$

is easily proven via smoothing with $(a, b) \rightarrow (\frac{a+b}{2}, \frac{a+b}{2})$. The algebra of verifying the smoothing is tedious but straight forward.

$a_1+a_2+...+a_n/n>=( a_1.a_2...a_n)^1/n$. It becomes
$1/n^n>= a_1.a_2...a_n/a_1+a_2+...+a_n$
Take is as (1).
Now there is an inequality known as weitrass inequality. It will give
$1-(a_1+a_2+...+a_n)<=(1-a_1)...(1-a_n)$
Which will become
$1-(a_1+a_2+...+a_n)/(1-a_1)...(1-a_n)<=1$
Take it as (2).
Multiply (1) and(2).you will.get it less than $1/n^n$ which is all less than $1/n^{n+1}$.
Hence shown

More storage
[quote=1998 A1]Let $a_{1},a_{2},\ldots ,a_{n}$ be positive real numbers such that $a_{1}+a_{2}+\cdots +a_{n}<1$. Prove that

\[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \][/quote]

Silly :P

Call $1-(a_1+\dots+a_n)$ as $b$ (it really is decoration). Then we just want $(1-a_1) \dots (1-a_n)(1-b) \ge n^{n+1}a_1\dots a_nb$. However, notice that $1-a_i=a_1+\dots+a_{i-1}+a_{i+1}+\dots+a_n+b \geqslant n\sqrt[n]{a_1\dots a_{i-1}a_{i+1}\dots a_nb}$ and multiplying all of them gives the result. $\blacksquare$

Let $a_i=a,a_j=b$ have fixed sum $c<1$. Note
\[\frac{ab}{(1-a)(1-b)}\le t\iff ab\le t-tc+tab\iff ab(1-t)\le t(1-c)\]
where $t$ is the maximum of the expression.
Clearly $t<1$ and so equality would have to be achieved at $a=b$. Thus by a smoothing argument each $a_i$ is equal. Let $\sum_i a_i=c$ so the desired is
\[\frac{1}{n^{n+1}}\ge \frac{\frac{c^n}{n^n}[1-c]}{c\cdot (1-\frac cn)^n}=\frac{c^{n-1}[1-c]}{(n-c)^n}\iff (n-c)^n\ge n^{n+1}c^{n-1}[1-c].\]
The result is obvious at $n=1$ so we can assume $n>1$.
Let $c=\frac{n}{n+1}-\frac{d}{n+1}$ so we need to check
\[\left(n^2+d\right)^n \ge n^{n+1}\left(n-d\right)^{n-1}\left(1+d\right)\]
and have $-1<d<n$. Since the result is true at the extrema, it suffices to check cases where the derivatives of both sides are equal. In these cases,
\[n(n^2+d)^{n-1}=-(n-1)n^{n+1}(n-d)^{n-2}\cdot (1+d)+n^{n+1}(n-d)^{n-1}.\]
Equivalently,
\[(n^2+d)^{n-1}=n^n(n-d)^{n-2}\cdot [(n-d)-(n-1)(1+d)]=n^n(n-d)^{n-2}\cdot [1-nd].\]
Equality can only hold at $d=0$ because $d>0$ increases the left side and decreases the right side whereas $d<0$ increases the right side and decreases the left side. But at $d=0$ the result is obvious, so we are done.

[size=150]Previous Wrong Solution  :blush: [/size]

Write $a_{n+1}=1-a_1-a_2\ldots a_n >0$. Thus, our expression becomes
\[\prod_{i=1}^{n+1} \frac{a_i}{1-a_i} = e^{\sum \ln(\frac{a_i}{1-a_i})}\]
Note that $\ln(\frac{x}{1-x})$ has a second derivative of $\frac{1}{x(1-x)}$ and is thus convex over $0<x<1$, thus, by jensens
\[\sum \ln(\frac{a_i}{1-a_i}) \geq (n+1)\cdot \ln(\frac{1}{n})\]
Thus, \[\prod_{i=1}^{n+1} \frac{a_i}{1-a_i}\geq e^{(n+1)\cdot \ln(\frac{1}{n})} = \left(\frac{1}{n}\right)^{n+1}\]
and we are done $\blacksquare$.

Edit: The second derivative is in fact
\[\frac{2x-1}{(x-1)^2x^2}\]
[size=150]New Correct Solution[/size]
Write $a_{n+1}=1-a_1-a_2\ldots a_n >0$.
Now, note that
\[\prod_{i=1}^{n+1} \sum_{j\neq i} a_i \leq \prod_{i=1}^{n+1} n\cdot \sqrt[n]{\prod_{j\neq i}a_i} \leq n^{n+1} \cdot \prod a_i\]
But wait! this is the denominator of our new expression. The given expression becomes
\[\frac{\prod a_i}{\prod (1-a_i)}=\frac{\prod a_i}{\prod (\sum_{j\neq i} a_j)}\leq \frac{\prod_{i=1}^{n+1}}{n^{n+1}\prod_{i=1}^{n+1}}=\frac{1}{n^{n+1}}\]
and we are done.

[hide=Solution]
Let $a_{n+1}=1-(a_1+a_2+\ldots+a_n)$. Then, the inequality is equivalent to
$$\frac{a_1a_2\ldots a_{n+1}}{(1-a_1)(1-a_2)\ldots(1-a_n)}\leq\frac1{n^{n+1}},$$
which is the same as
$$\frac{a_2a_2\ldots a_{n+1}}{(a_2+a_3+\ldots+a_{n+1})(a_1+a_3+\ldots+a_{n+1})\ldots(a_1+a_2+\ldots+a_n)}\leq\frac1{n^{n+1}}.$$
However, by AM-GM, we have
$$a_1+a_2+\ldots+a_n\geq n\sqrt[n]{a_1a_2\ldots a_n},$$ so the denominator is greater than or equal to $$n^{n+1}a_1a_2\ldots a_{n+1}.$$ Therefore, the entire fraction is less than or equal to $\frac1{n^{n+1}}$.

We do the trick of letting $a_{n+1}=1-(a_1+a_2+\ldots+a_n)$. Then we must have $a_1+\ldots+a_{n+1}=1$, and we want to prove that $$\frac{a_1\ldots a_{n+1}}{(1-a_1)\ldots(1-a_{n+1})}\le \frac{1}{n^{n+1}}.$$ Now notice that $$1-a_i=(a_1+\ldots+a_{n+1})-a_i\ge n\sqrt[n]{\frac{a_1\ldots a_{n+1}}{a_i}}.$$ Applying this to all the terms in the denominator gives us  \begin{align*} \frac{a_1\ldots a_{n+1}}{(1-a_1)\ldots(1-a_{n+1})} & \le \frac{a_1\ldots a_{n+1}}{n^{n+1}\sqrt[n]{\frac{a_1\ldots a_{n+1}}{a_1}}\ldots \sqrt[n]{\frac{a_1\ldots a_{n+1}}{a_{n+1}}}} \\ &= \frac{a_1\ldots a_{n+1}}{n^{n+1}\sqrt[n]{a_1^n\ldots a_{n+1}^n}} \\ &= \frac{a_1\ldots a_{n+1}}{n^{n+1}(a_1\ldots a_{n+1})} \\ &= \frac{1}{n^{n+1}}.\end{align*} The proof is complete.

[b][color=#f00]Remark:[/color][/b] Try stupid stuff first. Don't be like me who takes two hours to solve the problem because I realize that if we take the natural log of both sides, we essentially need to prove that if $f(x)=\ln\left(\frac{x}{1-x}\right)$, we have $$f(a_1)+\ldots+f(a_{n+1})\le (n+1)f\left(\frac{1}{n}\right).$$ Then I find $$f''(x)=\frac{2x-1}{(x(x-1))^2}$$ so Jensen doesn't work, but I see that there is exactly one inflection point in $(0,1)$, so $n-1$ EV is applicable (you should not need $n-1$ EV on an A1)... You can see how this approach of trying fancy things first fails miserably.

[b]Solution[/b]. Define $a_{n + 1} = 1 - (a_1 + \cdots + a_n)$, and note that $a_{n + 1} > 0$ and $a_1 + \cdots + a_{n + 1} = 1$. Then by AM-GM we have
\begin{align*}
1 - a_i &= a_1 + \cdots + a_{i - 1} + a_{i + 1} + \cdots + a_{n + 1} \\ &\geq n(a_1\cdots a_{i - 1} a_{i + 1} \cdots a_{n + 1})^{\frac{1}{n}}.
\end{align*}
Taking the product over $i$,
$$
 \prod_i (1 - a_i) \geq n^{n + 1} a_1 \cdots a_{n + 1},
$$
that is,
$$
\frac{a_1 \cdots a_n (1 - (a_1 + a_2 + \cdots + a_n))}{(a_1 + \cdots + a_n)(1 - a_1) \cdots (1 - a_n)} \leq \frac{1}{n^{n + 1}}.
$$

So, by GM-HM,
$$\sqrt[n]{\left(\frac{1}{a_1}-1\right) \left(\frac{1}{a_2}-1\right)\ldots \left(\frac{1}{a_n}-1\right)}\geq \frac{n}{\frac{a_1}{1-a_1}+\frac{a_2}{1-a_2}+\ldots+\frac{a_n}{1-a_n}}.$$
Now it is suffices to show that
$$\left(\frac{1}{\frac{a_1}{1-a_1}+\frac{a_2}{1-a_2}+\ldots+\frac{a_n}{1-a_n}}\right)^n\geq n\left(\frac{1}{a_1+a_2+\ldots+a_n}-1\right).$$

By Jensen's inequality, $$\frac{a_1}{1-a_1}+\frac{a_2}{1-a_2}+\ldots+\frac{a_n}{1-a_n}\leq n\frac{\frac{a_1+a_2+\ldots+a_n}{n}}{1-\frac{a_1+a_2+\ldots+a_n}{n}}=\frac{a_1+a_2+\ldots+a_n}{1-\frac{a_1+a_2+\ldots+a_n}{n}}=\frac{n(a_1+a_2+\ldots+a_n)}{n-(a_1+a_2+\ldots+a_n)}.$$
Therefore it is suffices to show that
$$\left(\frac{1}{a_1+a_2+\ldots+a_n}-\frac{1}{n}\right)^{n} \geq n\left(\frac{1}{a_1+a_2+\ldots+a_n}-1\right).$$
Let $t=\frac{1}{a_1+a_2+\ldots+a_n}>1$.
We need
 \[ \left(t-\frac{1}{n}\right)^{n} \geq n\left(t-1\right).\]
Define $f:\mathbb{R}_{>1}\to \mathbb{R}$, such that $f(t)=\left(t-\frac{1}{n}\right)^{n} - n\left(t-1\right)$.

We want to find the minimum of $f$.
We have $f'(t)=n\left(t-\frac{1}{n}\right)^{n-1}-n$ and $f''(t)=n(n-1)\left(t-\frac{1}{n}\right)^{n-2}>0 \forall t>1$.
Also note that $f'(t)=0\Longleftrightarrow t=\frac{n+1}{n}$ and thus the minimum of the function $f$ is
$$f\left(\frac{n+1}{n}\right)=\left(\frac{n+1}{n}-\frac{1}{n}\right)^{n} - n\left(\frac{n+1}{n}-1\right)=1-1=0.$$
Hence, we have proven the inequality  \[ \left(t-\frac{1}{n}\right)^{n} \geq n\left(t-1\right),\]
where the equality holds iff $t=\frac{n+1}{n}$. We conclude that that the equality in our original inequality holds iff $a_1=a_2=\ldots=a_n=\frac{1}{n+1}$. We are done.

It is equivalent to showing that $\left(\frac{1}{a_1}-1\right)\left(\frac{1}{a_2}-1\right)\dots\left(\frac{1}{a_{n+1}}-1\right)\geq n^{n+1}$ for positive reals $a_1,a_2,\dots,a_{n+1}$ with $n\geq 1$ that sum to $1$ (in the problem, $a_{n+1}$ is replaced with $1-(a_1+a_2+\dots+a_n)$).

Then $$\frac{1}{a_1}-1=\frac{a_2+\dots+a_{n+1}}{a_1}\geq \frac{n\sqrt[n]{a_2a_3\dots a_{n+1}}}{a_1}$$ and multiplying together gives the desired result. Equality holds when $a_1=a_2=a_3=\dots=a_{n+1}=\frac{1}{n+1}$.

ISL marabot solve

Let $1-(a_1+a_2+\ldots+a_n)=a_{n+1}$ and take the reciprocal to get that the original inequality is equivalent to \[\left(\frac{1-a_1}{a_1}\right)\cdot \left(\frac{1-a_2}{a_2}\right)\cdots \left(\frac{1-a_{n+1}}{a_{n+1}}\right)\ge n^{n+1}.\]

Now we have $1-a_i=\sum_{k=1, k\ne i}^{n+1} a_k\ge n\sqrt[n]{\prod_{k=1, k\ne i}^{n+1} a_k}$.

So the numerator is greater than or equal to $n^{n+1}a_1\cdot a_2\cdots a_{n+1}$. Dividing by the denominator gives the desired result.

First, throw out the $n = 1$ case and define $a_0 = 1 - \sum\limits_{i=1}^{n} a_i$. Then, we have $\sum\limits_{i=1}^{n} a_i = 1$, and we want to show that \[ \prod\limits_{i=0}^{n} \frac{a_i}{1-a_i} \leq \frac{1}{n^{n+1}} \] and notice that equality holds when $a_0 = a_1 = \cdots = a_n = \frac{1}{n+1}$. Now, we smooth. Say that there is some set of values $(a_0,a_1,\ldots,a_n)$ such that the value of the product is maximal, and assume for the sake of contradiction that there exist some $a_i \neq a_j$. Then, we are done if  \[ \frac{a_i}{1-a_i}\frac{a_j}{1-a_j} \leq \left(\frac{a_i+a_j}{2-a_i-a_j}\right)^2 \] as that would imply that there was some larger value of the product by taking the values \[ \left(a_0,a_1,\ldots,a_{i-1},\frac{a_i+a_j}{2},a_{i+1},\ldots,a_{j-1},\frac{a_i+a_j}{2},a_{j+1},\ldots,a_n\right) \] Now, we prove that this set is actually larger. We use the subsitution $a_i + a_j = 2u$ and $a_i - a_j = 2v$. Then, we can rewrite the equation we want to prove as \[ \frac{a_i}{1-a_i}\frac{a_j}{1-a_j} \leq \left(\frac{a_i+a_j}{2-a_i-a_j}\right)^2 \Rightarrow \frac{u^2 - v^2}{(1-u)^2 - v^2} < \frac{u^2}{(1-u)^2}  \] after some algebra, which further rearranges into  \[ \frac{u^2 - (1-u)^2}{(1-u)^2 - v^2} < \frac{u^2 - (1-u)^2}{(1-u)^2} \] which is true because $2u = a_i + a_j < 1$.

[quote=grobber]Put $a_{n+1}=1-(a_1+a_2+\ldots+a_n)$. The inequality becomes $\frac{a_1a_2\ldots a_{n+1}}{(1-a_1)(1-a_2)\ldots (1-a_{n+1})}\le \frac 1{n^{n+1}}$, where $a_i$ are positive and $\sum_1^{n+1}a_i=1$. This is a mere application of AM-GM to the denominators $A_i=1-a_i=a_1+a_2+\ldots+a_{i-1}+a_{i+1}+\ldots+a_n+a_{n+1}$. We get $A_i\ge n\sqrt[n]{\frac{\prod a_k}{a_i}}$. After multiplying all of these we get the desired result.[/quote]

Very nice idea

Let $a_{n+1}=1-(a_1+a_2+ \cdots a_n)$. We have that the inequality becomes $\left( \frac{1}{a_1}-1 \right) \left( \frac{1}{a_2}-1 \right) \cdots \left( \frac{1}{a_{n+1}}-1 \right) \ge n^{n+1}$. We note that $$\frac{1}{a_1}-1 = \frac{a_2+a_3 + \cdots + a_{n+1}}{a_1} \ge \frac{n \sqrt[n]{a_2a_3 \cdots a_{n+1}}}{a_1}$$ and multiplying similar inequalities gives the result.

Let $a_{n+1}$ be the positive real number such that $a_{1}+a_{2}+\cdots +a_{n+1}=1.$ The desired inequality becomes \[ (na_{1}) (na_{2}) \cdots (na_{n}) (na_{n+1}) \le ( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})(1-a_{n+1}) \]
By AM-GM we have \[1-a_{i}=a_1+a_2+\dots+a_{n+1}-a_i\ge n\cdot \sqrt[n]{\frac{a_1a_2\cdots a_{n+1}}{a_i}}\]
Multiplying the analogous gives the result.

Define $1-(a_1+a_2+\dots+a_n)=a_{n+1}$; the LHS is equal to
\[\frac{a_1a_2\dots a_{n+1}}{\prod_{\text{cyc}}(a_1+a_2+\dots+a_n)}\le \frac{a_1a_2\dots a_{n+1}}{\prod_{\text{cyc}}(n\sqrt[n]{a_1a_2\dots a_n})}=\frac{1}{n^{n+1}}\]
so we are done. $\blacksquare$

ISL Marabot solve

Let, $a_{n+1} = 1-\sum\limits_{i=1}^na_i$. So, $\sum\limits_{i=1}^{n+1}a_i =1$.
Now for each $1\leq k \leq n+1$,
\begin{align*}
\sqrt[n]{\prod_{i\neq k}a_i} &\leq \dfrac{\sum_{i\neq k}a_i}n\\
\implies \sqrt[n]{\prod_{i\neq k}a_i} &\leq \dfrac{(1-a_k)}n
\end{align*}
And, multiplying all of these gives,
\begin{align*}
\prod_{i=1}^{n+1}a_i &\leq \dfrac{\prod\limits_{i=1}^{n+1} (1-a_i)}{n^{n+1}}\\
\implies \dfrac{\prod\limits_{i=1}^{n+1}a_i}{\prod\limits_{i=1}^{n+1} (1-a_i)} &\leq \dfrac1{n^{n+1}} \\
\end{align*}
Which is what we wanted.   $\blacksquare$

Let $a=a_1+a_2+\dots+a_n$. Taking $\ln$ of both sides, we can first optimize $$\sum_{\text{cyc}}\ln\left(\frac{a_1}{1-a_1}\right).$$ Using log properties, we find that the second derivative of the term we are summing is $$-\frac{1}{x^2}+\frac{1}{(1-x)^2}$$ which is zero at $x=\frac12$, so there is exactly one inflection point, meaning that we can use $n-1$ EV to have WLOG $a_1=a_2=\dots=a_{n-1}$. Our wanted inequality is now $$\frac{a_1^{n-1}\cdot a_n\cdot(1-a)}{a\cdot(1-a_1)^{n-1}\cdot(1-a_n)}\le\frac{1}{n^{n+1}}.$$ By AM-GM, we can instead just show that $$(n-1)\cdot\frac{a_1}{1-a_1}+\frac{a_n}{1-a_n}+\frac{1-a}{a}\le\frac{n+1}{n}.$$ Adding $n+1$ to both sides, we want $$\frac{n-1}{1-a_1}+\frac{1}{1-a_n}+\frac{1}{(n-1)a_1+a_n}\le\frac{(n+1)^2}{n}.$$ We can resort to differentiating the LHS with respect to $a_n$ to get $$\frac{1}{(1-a_n)^2}-\frac{1}{((n-1)a_1+a_n)^2}$$ which is zero when $a_n=\frac{1-a_1(n-1)}{2}$. The second derivative is positive for the whole interval, so that root is the global minimum for the interval.  Plugging this back in, we want to show that $$\frac{n-1}{1-a_1}+\frac{4}{1+a_1(n-1)}\le\frac{(n+1)^2}{n}.$$ We can do this using another direct differentiation with respect to $a_1$. We get $$\frac{n-1}{\left(1-a_1\right)^2}-\frac{4\left(n-1\right)}{\left(\left(n-1\right)a_1+1\right)^2}$$ and we have to check $$a_1=0,\frac1{n+1},\frac1{n-1}.$$ It is easy to check that the wanted inequality does hold true for all of these, so we are done.

Let $a_{n+1}=1-\sum_{i=1}^n a_i$, so $\sum_{i=1}^{n+1} a_i = 1$. It then suffices to show $$\prod_{i=1}^{n+1} \left(\frac{1-a_i}{a_i}\right) \geq n^{n+1},$$ which is clearly true by AM-GM on the numerator. $\blacksquare$

[i]This abomination of a solution is caused by the fact that I did not see the clean way.[/i]

[hide=Solution]
Define $S = \sum_{i=1}^n a_i < 1$, and we want
\[
    \frac{a_1}{1-a_1} \cdot \frac{a_2}{1-a_2}
    \cdots \frac{a_n}{1-a_n} \cdot \frac{1-S}{S}
    \leq \frac{1}{n^{n+1}}.
\]

By AM-GM,
\[
    \frac{a_1}{1-a_1} \cdot \frac{a_2}{1-a_2} \cdots \frac{a_n}{1-a_n}
    \leq \frac{1}{n^n} \cdot
    \bigg(\frac{a_1}{1-a_1} + \frac{a_2}{1-a_2} + \cdots + \frac{a_n}{1-a_n}\bigg),
\]
so we want to prove
\[
    \bigg(\frac{a_1}{1-a_1} + \frac{a_2}{1-a_2} + \cdots + \frac{a_n}{1-a_n}\bigg)^n
    \cdot \frac{1-S}{S} \leq \frac1n.
\]

Note that if $f(x) = \tfrac{x}{1-x}$, then $f''(x) = 2(1-x)^{-3} > 0$ so $f$ is convex. Sort the sequence $a$ in nondecreasing order, so that $a_i \leq a_{i+1}$. Define a sequence $(b_i)_{i=1}^n$ as $b_1 = S-(n-1)t$ and $b_i = t$ for $i \geq 2$ where $t = \tfrac{S-a_1}{n-1}$, and note that $(b_i) \succ (a_i)$.
We apply Karamata, which says
\[
    \frac{a_1}{1-a_1} + \frac{a_2}{1-a_2} + \cdots + \frac{a_n}{1-a_n}
    \leq \frac{b_1}{1-b_1} + \frac{b_2}{1-b_2} + \cdots + \frac{b_n}{1-b_n}.
\]

We want to prove
\[
    \bigg(\frac{S-(n-1)t}{1-S+(n-1)t} - (n-1) \cdot \frac{t}{1-t}\bigg)^n
    \cdot \frac{1-S}{S} \geq \frac1n.
\]
Define
\[
    g(t) = (n-1) \cdot \frac{t}{1-t} + \frac{S-(n-1)t}{1-S+(n-1)t}
         = (n-1) \cdot \frac{1}{1-t} + \frac{1}{1-S+(n-1)t} - n,
\]
and note that
\[g'(t) = \frac{n-1}{(1-t)^2} - \frac{n-1}{(1-S+(n-1)t)^2}\]
is $0$ when $1-t = 1-S+(n-1)t \Longleftrightarrow t = S/n$, which means $(a_i)$ is constant.
Also $t=0$ and $t=1$ are impossible since $S < 1$ and $a_1 > 0$.

We want to prove
\[
    \bigg(n \cdot \frac{S/n}{1-S/n}\bigg)^n \cdot \frac{1-S}{S}
    = \frac{S^{n-1}(1-S)}{(1-S/n)^n} \leq \frac1n \tag{$*$}
\]
Define
\[
    h(S) = \frac{S^{n-1}(1-S)}{(1-S/n)^n},
    \qquad h'(S) = \frac{(n-1)S^{n-2}}{(1-S/n)^{n+1}}
             \cdot \bigg(1 - \frac{n+1}{n} \cdot x\bigg),
\]
which is only $0$ when $x = \tfrac{n}{n+1}$.
Plugging this in, we get equality in ($*$) so we are done. \qed
[/hide]

Let $a_{n+1} = 1-a_1-a_2-\cdots-a_n$. Then, the inequality is equivalent to $$\frac{a_1a_2\cdots a_{n+1}}{(1-a_1)(1-a_2)\cdots (1-a_{n+1})} \leq \frac 1{n^{n+1}}.$$ However, observe that $$1-a_1 = a_2+a_3+\cdots+a_{n+1} \geq n\sqrt[n]{a_2a_3\cdots a_{n+1}},$$ so multiplying this inequality cyclically in the denominator yields the result.

[hide=Solution]
Let $a_{n+1} = 1-a_1-a_2-\cdots - a_n$. Then, $a_{n+1} > 0$, and the inequality is equivalent to $$\frac{a_1a_2\cdots a_{n+1}}{(1-a_1)(1-a_2)\cdots (1-a_{n+1})} \leq \frac{1}{n^{n+1}}.$$ Now, note that by AM-GM
\begin{align*}
\frac{a_1a_2\cdots a_{n+1}}{(1-a_1)(1-a_2)\cdots (1-a_{n+1})} &= \prod_{\text{cyc}} \frac{a_1}{1-a_1} \\ &= \prod_{\text{cyc}} \frac{a_1}{a_2+a_3+\cdots + a_{n+1}} \\ &\leq \prod_{\text{cyc}} \frac{a_1}{n\sqrt[n]{a_2a_3\cdots a_{n+1}}} \\ &= \frac{1}{n^{n+1}},
\end{align*}
as desired. $\blacksquare$

Let $a_{n+1} = 1- \sum a_i$. Then it remains to show $$\prod \frac{a_i}{1-a_i} \leq \frac 1{n^{n+1}}.$$ This is easy with AM-GM, as $$\prod \frac{a_i}{1-a_i} \leq \prod \frac{a_i^{(n+1)/n}}{n \cdot \prod a_i} = \frac 1{n^{n+1}}.$$

Let $a_{n+1}=1-\sum_{i=1}^na_i$, thus the inequality is equal to: $\frac{\prod_{i=1}^{n+1}a_i}{(a_1+... a_n)\cdots (a_1+...a_{n+1})}$
Furthermore notice that:\begin{align*} a_1+\dots +a_n\ge n\sqrt[n]{\prod_{i=1}^n a_i}\end{align*}$
$ \begin{align*}\vdots\end{align*}
\begin{align*}a_1+\dots +a_{n-1}+a_{n+1}\ge n\sqrt[n]{a_1\cdots a_{n-1}a_{n+1}} \end{align*}
By multiplying the inequalities: $LHS\ge n^{n+1}\prod_{i=1}^{n+1}a_i$
Thus $LHS \le\frac{\prod_{i=1}^{n+1}a_{n}}{n^{n+1}\prod_{i=1}^{n+1}a_n}=\frac{1}{n^{n+1}}$. QED
And we are done! :play_ball:

Let $a_{n+1} = 1-\sum_{i=1}^n a_i$. We claim the maximum can be achieved when all the variables are equal. Assume not. WLOG, $a_1 \neq a_2$. Let $x=a_1, y=a_2$. But consider
\[
\frac{\left( \frac{x+y}2 \right)^2}{\left(1 - \frac{x+y}2 \right)^2} - \frac{xy}{(1-x)(1-y)} = \frac{(x-y)^2 (1-x-y)}{(1-x)(1-y)(2-x-y)^2} \ge 0,
\]
so making them equal (with the same sum) results in a product at least as big as the original one. The conclusion follows.

[quote=orl]Let $a_{1},a_{2},\ldots ,a_{n}$ be positive real numbers such that $a_{1}+a_{2}+\cdots +a_{n}<1$. Prove that \[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \][/quote]

Putting $a_{n+1}=1-\sum_{k=1}^{n}a_k$, it suffices to prove that $\prod_{i=1}^{n}\frac{(1-a_i)}{a_i}\ge n^{n+1}$. Note that 1-a_i=the sum of all a_j's from 1 to n excluding a_i. Then this inequality follows immediately from $\prod_{i=1}^{n}\frac{(1-a_i)}{a_i}=\frac{\prod_{i}(\sum_{j\ne i}a_j)}{\prod_{i}a_i}\ge \frac{(n-1)^n\prod_{i}(\prod_{j\ne i}a_j)^{\frac{1}{n-1}})}{\prod_{i}a_i}=(n-1)^n$, where the greater or equal to is just the AM-GM. Each of these steps are reversible, so we're done. $\blacksquare$

@above, I think the product should go from $i=1$ to $i=n+1$, no? And then you would get $\prod_{i=1}^{n+1} \geq n^{n+1}$ instead of $\prod_{i=1}^{n}\frac{1-a_i}{a_i} \geq (n-1)^n$.

Sorry, yeah. The Latexing took a long time so I couldn't notice what was wrong. The index would be changed over by 1 (1->n+1) and like you said.


Cool, Tiks! :ninja:
My  proof:
$\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)+27\geq4\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}c^{2}(a-b)^{4}\geq0.$ :)

we can use the inequality$(a_{1}+...+a_{n})(\frac{1}{a_{1}}+...+\frac{1}{a_{n}})\ge n^{2}$
adn I didn't understand your solution. arqady

ha ok.
שמתי לב שאתה כל הזמן שם אי שיוויונים, למה?
וגם ראיתי שאתה התותח בהייד פארק  :)  תבקר בתפוז נחמד שם גם!
we also see that $(a^{2}+b^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}})\ge (a+b)(\frac{1}{a}+\frac{1}{b})$(generalisation) and than we can say LHS$\ge 3^{2}+27=36$ and RHS $\ge 4\cdot 3^{2}=36$

[quote="srulikbd"]
שמתי לב שאתה כל הזמן שם אי שיוויונים, למה?
[/quote]
כיוון שאני הכי אוהב את זה במתמטיקה של בית הספר.
[quote="srulikbd"]
תבקר בתפוז נחמד שם גם...
[/quote]
בשמחה, תן לינק  :)
[quote="srulikbd"]...and than we can say LHS$\ge 3^{2}+27=36$ and RHS $\ge 4\cdot 3^{2}=36$[/quote]
But why $LHS\geq RHS$ $?$ :wink:

[quote=arqady]Let's $a,$ $b$ and $c$ are different from zero real numbers. Prove that:
$\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)+27\geq4\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).$
Maybe it was already? Thank you.[/quote]
Let $a,$ $b$ are different from zero real numbers. Prove that:
$$\left(a^{2}+b^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)+4\geq2\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right).$$

Let $x=\frac ab+\frac bc+\frac ca$ and $y=\frac ba+\frac cb+\frac ac$, then the inequality can be rewritten as
$$x^2+y^2-2x-2y+3+27\ge 4x+4y+12\Longleftrightarrow (x-3)^2+(y-3)^2\ge0$$

Let $a_1,a_2,\cdots,a_n  (n\ge 2)$are different from zero real numbers. Prove that:$$\left(a_1^2+a_2^2+\cdots+a_n^2\right)\left(\frac{1}{a^2_1}+\frac{1}{a^2_2}+\cdots+\frac{1}{a^2_n}\right)+3n^2\geq4\left(a_1+a_2+\cdots+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right).$$
[hide=Origin. ]Let $a,b,c,d$  and $a_1,a_2,\cdots,a_n  (n\ge 2)$ are different from zero real numbers.
$$\left(a^{2}+b^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)+12\geq4\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\iff \left(\frac{a}{b}+\frac{b}{a}-2\right)^2\ge0.$$
$$\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)+27\geq4\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\iff \left(\frac ab+\frac bc+\frac ca-3\right)^2+\left(\frac ba+\frac cb+\frac ac-3\right)^2\ge0.$$
$$\left(a^{2}+b^{2}+c^{2}+d^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}\right)+48\geq 4\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\iff \left(\frac ab+\frac bc+\frac cd+\frac da-4\right)^2+\left(\frac ba+\frac cb+\frac dc+\frac ad-4\right)^2\ge0?$$
$$\left(a_1^2+a_2^2+\cdots+a_n^2\right)\left(\frac{1}{a^2_1}+\frac{1}{a^2_2}+\cdots+\frac{1}{a^2_n}\right)+3n^2\geq4\left(a_1+a_2+\cdots+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)\iff \left(\frac{a}{b}+\frac{b}{a}-2\right)^2\ge0 \iff \left(\frac {a_1} {a_2}+\frac {a_2} {a_3}+\cdots+\frac {a_{n-1}} {a_n}+\frac {a_n} {a_1}-n\right)^2+\left(\frac {a_2} {a_1}+\frac {a_3} {a_2}+\cdots+\frac {a_n} {a_{n-1}}+\frac {a_1} {a_n}-n\right)^2\geq 0?$$[/hide]

[quote=sqing]
Let $a,$ $b$ are different from zero real numbers. Prove that:
$$\left(a^{2}+b^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)+4\geq2\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right).$$[/quote]
Let $a,b,c,d$ are different from zero real numbers. Prove that:
$$2\left(a^{2}+b^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)+8\geq\left(a^{2}+b^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)+12\geq4\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right).$$

[quote=sqing]Let $a_1,a_2,\cdots,a_n  (n\ge 2)$are different from zero real numbers. Prove that:$$\left(a_1^2+a_2^2+\cdots+a_n^2\right)\left(\frac{1}{a^2_1}+\frac{1}{a^2_2}+\cdots+\frac{1}{a^2_n}\right)+3n^2\geq4\left(a_1+a_2+\cdots+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right).$$[/quote]
Proof of Zhangyanzong:


A thin plastic rod bent into a semicircle of radius R has a charge of Q, in esu, distributed uniformly over its length.  Find the strength of the electric field at the center of the semicircle.

It's not too hard but I'm messing something up somewhere.  The length of the semicircle is $R\pi$ (half the circumference), so the linear charge density, $\lambda = \frac{Q}{R\pi}$.  But how exactly do you set up the integral, is it $dE = \frac{dq}{R^2} = \frac{Qds}{R^3\pi}$?

you are forgetting the $\frac{1}{4\pi\epsilon_0}$ in the $dE$ expression.

also, don't forget to consider the direction of the $dE$ vectors.

The charge is measured in esu, meaning that we're talking in cgs not SI.  In cgs, $k = 1$, so it makes basic electricity calculations easier.

But what would be the bounds of the integral?  If it was with $d\theta$ then I'd say from $0$ to $\pi$, but that's not exactly the case here.

by symmetry, for every charge element on the rod, there will be another charge element on the rod such that the net electric field of the two charges will be entirely in the horizontal direction.  thus, when integrating, we consider only the horizontal component of the electric field of the charge elements.

so consider some infinitesimally small arc of the bent rod subtending an angle $d\theta$.  we can basically treat it as a point, so suppose the angle the line connecting the center and this point makes with the horizontal is $\theta$.  then, the charge contained on this element is $dq = \lambda r d\theta$.  the horizontal electric field generated by this element is then $dE = \frac{\lambda cos(\theta) d \theta }{r}$.  by symmetry, the contribution from the top half of the semicircle is the same as the bottom half, so we take $\theta$ from $0$ to $\frac{\pi}{2}$ but multiply the whole integral by 2.

Since $\lambda = \frac{Q}{r\pi}$, $E = 2\int^\frac{\pi}{2}_0 \frac{Q cos(\theta) d \theta }{\pi r^2} = \frac{2Q}{\pi r^2}$.

Thank you very much; I understand how to involve $\theta$ now.  Shouldn't the final answer be $\frac{2Q}{R\pi}$, though?

i'm fairly certain the denominator should involve the squared term because the dimensions don't work out otherwise.

i think you may be forgetting that the r in the dq differential will cancel out with one of the r's in the denominator of the dE differential.


i cant figure out how to prove this:

$\text{div}(fF)=f\text{div}(F)+f\cdot F$

$f$ is a scalar and $F$ is vector

edit:oops...forgot the upwards trinagle thing.  it goes before the small f in the third term.

The proof is just a straightforward application of the product rule. Assume that we are in $\mathbb{R}^{3}$ (other dimensions would be similar) and that $\vec{F}=\langle F_{1},F_{2},F_{3}\rangle.$

Then $\nabla\cdot\vec{F}=\frac{\partial}{\partial x}(fF_{1})+\frac{\partial}{\partial y}(fF_{2})+\frac{\partial}{\partial z}(fF_{3})$

$=f\,\frac{\partial F_{1}}{\partial x}+\frac{\partial f}{\partial x}\,F_{1}+f\,\frac{\partial F_{2}}{\partial y}+\frac{\partial f}{\partial y}\,F_{2}+f\,\frac{\partial F_{3}}{\partial z}+\frac{\partial f}{\partial z}\,F_{3}$

$=f\left(\frac{\partial F_{1}}{\partial x}+\frac{\partial F_{2}}{\partial y}+\frac{\partial F_{3}}{\partial z}\right)+\left\langle \frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right\rangle\cdot\langle F_{1},F_{2},F_{3}\rangle$

$=f\,\nabla\cdot\vec{F}+\nabla f\cdot \vec{F}$

As I tell my class (I am teaching Calculus III this semester), treat the symbol $\nabla$ as having two natures, both of which must be satisfied. It is a vector, and its interaction with other vectors or with scalars must be consdided one of the possible products, and it is a differential operator which is "hungry" and must be fed a function. As a differential operator, it obeys all reasonable product rules. In many cases, we can follow this to obtain the useful formulas.

[quote="jli"]i cant figure out how to prove this:

$\text{div}(fF)=f\text{div}(F)+f\cdot F$
[/quote]
This formulation is incorrect a $div$ is missing
it should be
$\text{div}(fF)=f\text{div}(F)+div \,f\cdot F$

That should be $\nabla f$, not $\mathrm{div}\,f$.
For bonus points: prove that $\nabla\times (f\vec F)=f(\nabla\times \vec F)+(\nabla f)\times\vec F$. Is it true that $\nabla\times (\vec F\times \vec G)=\vec F\times (\nabla\times \vec G)+(\nabla \times\vec F)\times\vec G$?


Determine the most integer $ i$ such that

$ n\minus{}\[\sum_{k\equal{}2}^n \frac{k}{\sqrt{k^2\minus{}1}}\geqq \frac{i}{10}$

holds for all integer $ n\geqq 2$

[hide="Hint"]
\[ n \minus{} \sqrt{n^2\minus{}1}\minus{}\frac{2}{\sqrt{3}}\plus{}\frac{5}{\sqrt{8}} \le S \minus{} \sum_{k\equal{}2}^n \frac{k}{\sqrt{k^2\minus{}1}} \le n \minus{} \sqrt{n^2\plus{}2n}\minus{}\frac{2}{\sqrt{3}}\plus{}\sqrt{8}\][/hide]


Japan,China,Taiwan(Chinese Province),Koreans,and
Malaysia are only top five,and known "tiger economic"nation in asian continent.Is asian 'region'with all its insecurities,political
turmoils,terrorisms,and continous regional conflicts ready for their possible "Asian Regional Currency?" Just how the asian
be fair to the Western dominating currency such as the American Dollar,and the Euros? any comment?

                                                     Respectfully,
                                                     A.Eddington

hmm, this is a good question. One i have never thought of. I was born in Pakistan, and i hear that it is rapidly developing to become an 'Asian tiger'. I'm not sure if a singe currency will do more good than bad for Asia.
It has a lot of good aspects, and i can't thing of any bad aspects... :) I don't know, i'm just typing my thoughts. :)
Also, you didn't consider Russia, aren't they a tiger too.(unless Russia isn't part of Asia...)

[quote="Ignite168"]It will never happen considering China's control on their currency and their bad relations. It also isn't good economically due to the vast differences of development between countries and the standard cost of living as well as wages in proportion to other countries.[/quote]
I agree with the last part, and hopefully in the future we'll get there. But China doesn't have bad relations, infact they have very good realtions. I guess it depends what countries you're refering to . :)

I agree, it would probably not happen for a very long time. However if it were to happen I'm pretty sure the currency would be bigger than EUR. For instance right now, JPY/USD trades more than USD/EUR. However EUR overall trades more than JPY. However could it eventually become bigger than USD?


Let $ a\neq 1$ be a positive constant. Given inequalities with respect to $ x$,
\[ 6x^2 \minus{} 5x \plus{} 1 < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots [1]\]

\[ \log_a x \plus{} 2\log_x a < 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots [2]\]
Find the range of $ a$ such that $ [2]$ holds for any real number $ x$ satisfying $ [1]$.

[1]
$ 6x^2 \minus{} 5x \plus{} 1 \equal{} (3x \minus{} 1)(2x \minus{} 1) \equal{} (x \minus{} \frac {1}{3})(x \minus{} \frac {1}{2}) < 0$

$ \frac {1}{3} < x < \frac {1}{2}$

[2]
$ \frac {1}{\log_x a} \plus{} 2 \log_x a < 3$

$ 1 \plus{} 2 \log_x^2 a < 3 \log_x a$

$ (2 \log_x a \minus{} 1)(\log_x a \minus{} 1) < 0$

$ \frac {1}{2} < \log_x a < 1$

$ x^{\frac {1}{2}} < a < x$

I'm a little confused with including the intervals of x with this last inequality.

[quote="kunny"]Let $ a\neq 1$ be a positive constant. Given inequalities with respect to $ x$,
\[ 6x^2 - 5x + 1 < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots [1]\]

\[ \log_a x + 2\log_x a < 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots [2]\]
Find the range of $ a$ such that $ [2]$ holds for any real number $ x$ satisfying $ [1]$.[/quote]

$ 6x^2 - 5x + 1 < 0\implies (3x - 1)(2x - 1) < 0\implies x\in\left(\frac13,\ \frac12 \right)$

${ \log_a x + 2\log_x a < 3\implies \frac {1}{\log_x a} + 2\log_x a} < 3$

Case 1: $ a > 1\implies \log_x a < 0\ \forall\ x\in\left(\frac13,\ \frac12 \right)\implies [2]$ is true

Case 2:$ a < 1\implies \log_x a > 0\ \forall\ x\in\left(\frac13,\ \frac12 \right)\implies 2(\log_x a)^2 - 3\log_x a + 1 < 0$

$ \implies (2\log_x a - 1)(\log_x a - 1) < 0\implies \frac 12 < \log_x a < 1$

$ \implies x < a < x^{\frac 12}\ \forall\ x\in\left(\frac13,\ \frac12 \right)\implies \frac 12 < a < \left(\frac 13\right)^{\frac 12}$

$ \implies \boxed{\boxed{a\in \left(\frac 12 ,\ \left(\frac 13\right)^{\frac 12}\right)\bigcup(1,\ \infty)}}$


[i]Αφού "περνάνε" τελευταία τα διανύσματα δείτε και ένα ακόμη ωραίο θεματάκι![/i]

Δίνεται τρίγωνο $ABC$ στο οποίο είναι $\left|\overrightarrow{AB}\right|=4, \left|\overrightarrow{AC}\right|=6$ και η γωνία των $\overrightarrow{AB}$ και $\overrightarrow{AC}$ είναι $60^\circ$. Αν $M$ το μέσο της πλευράς $BC$ τότε:

α) Να υπολογίσετε το μέτρο του διανύσματος $\overrightarrow{AM}$

β) Να υπολογίσετε το συνημίτονο της γωνίας των διανυσμάτων $\overrightarrow{AM},\overrightarrow{AB}$

γ) Να αποδείξετε ότι η προβολή του διανύσματος $\overrightarrow{AB}$ πάνω στο διάνυσμα $\overrightarrow{AM}$ είναι το διάνυσμα $\frac{14}{19}\overrightarrow{AM}$

Sto teleutaio erwtima mipws einai $\frac{14}{19}$ kai oxi $\frac{14}{9}$ ?

Alexandros

Y.G. An kai de nomizw oti tha meinei aliti parapanw apo 1 imera (kai auto einai poli epeidi thelei vareto grapsimo se LaTeX, entoutois kalitera na asxolithoun osoi kanoun dianismata sto sxoleio. Ean den lithei se 2-3 meres tha valw lisi... Egw eksetastika se auti (To 99 edwsa Panellinies).

Αυτή είναι πολύ καλή άσκηση με την έννοια της εξάσκησης κι όχι με την έννοια του πρόβληματος (που συνηθίζουμε όσοι πάμε μαθηματικούς διαγωνισμούς) είναι προφανές σχεδόν ο τρόπος λύσης
0(Ασχετο Ζήτω η ΑΕΚΑΡΑ 1-0 την Μίλαν)  καλή συνέχεια σε όλους

Αλέξανδρε έχεις δίκαιο, αν και την εκφώνηση εγώ την είχα βρει ως $\frac{14}{9}\overrightarrow{AM}$ κάνοντας τις πράξεις βγαίνει $\frac{14}{19}\overrightarrow{AM}$ !

PS: Νομίζω ότι στο τότε βιβλίο σας είχε στα διανύσματα και τύπο που συνέδεε $\tan$ με συντελεστή διεύθυνσης;

[quote="mostel"]
PS: Νομίζω ότι στο τότε βιβλίο σας είχε στα διανύσματα και τύπο που συνέδεε $\tan$ με συντελεστή διεύθυνσης;[/quote]

Ennoeis kapoio tipo pou sindeei tin efaptomeni me tis sintetagmenes twn dianismatwn? Ean nai, tote den ipirxe kati tetoio sto vivlio tis 1hs Desmis para mono o klassikos tipos pou sinedee to sinimitono tis gwnias me tis sintetagmenes twn dianismatwn (auto pou exei kai sto twrino vivlio tis B Likeiou).

Ektos an ennoeis oti an $\vec{a}=[x_{1},y_{1}]$ tote $\lambda = \frac{y_{1}}{x_{1}}=\tan{\omega}$, opou $\omega$ einai i gwnia pou sximatizei to dianisma $\vec{a}$ me ton aksona $x'x$ (to opoio to vivlio tis 1hs Desmis den to leei fanera alla to iponoei).

Apo ekei kai pera an ennoeis kati allo dieukrinise mou! (Genika omws, twra pou eriksa mia matia den anaferei kati gia efaptomeni).

Alexandros

Όχι, δεν εννοώ αυτόν τον τύπο. Αν θυμάμαι καλά είχε το $\lambda$ στον παρανομαστή! Θα το ψάξω αν είναι και θα σου πω!

Για το πρώτο βρίσκουμε το ΑΜ διάνυσμα πάντα συναρτήσει των ΑΒ και ΑΓ και υψώνουμε στο τετράγωνο .

β) Βρίσκουμε τη BΓ και μετά νόμο συνημιτόνων
γ)δύο πυθαγόρεια θεωρήματα ή θεώρημα προβολών ,,,, :lol:

Είναι δύσκολο να ταγράψω σε LateX


[quote="Drunken_Math"][quote="nat mc"]what is the radius of the incircle of a right triangle with legs $a,b$ and hypotenuse $c$[/quote]
[hide]
$\frac{ab}{2}=r*\frac{a+b+c}{2}$

$ab=r*(a+b+c)$

$r=\frac{ab}{a+b+c}$[/hide][/quote]
simplify please

By Pythagorean Theorem,

$a^2+b^2=c^2$

$(a+b)^2=c^2+2ab$

$2ab=(a+b)^2-c^2$

$ab=\frac{(a+b+c)(a+b-c)}{2}$

Plug it into Drunken_Math's equation:

$r=\frac{(a+b+c)(a+b-c)}{2(a+b+c)}$

$r=\boxed{\frac{a+b-c}{2}}$

[b][u]Another approach[/u][/b]
[hide]Let ABC be the triangle with right angle at C and incentre I
Let also the incircle touches sides BC, CA, and AB respectively at P, Q and R
It is easy to verify that CPIQ is a square with sidelength r
then AQ=AC-QC=b-r, BP=BC-PC=a-r
so AB=AR+BR=AQ+BP
or c=(b-r)+(a-r)
$r=\frac{a+b-c}{2}$[/hide]


inequalities
function
triangle inequality
real analysis
real analysis unsolved

Let $f,g: [a,b]\to \mathbb{R}$. How to show shat
1. if $f,g$ is of a bounded variation then so is $fg$?
2. if $f,g$ is of an absolutely continuous then so is $fg$?

I tried to use identity $fg=\frac{1}{4} ((f+g)^{2}-(f-g)^{2})$ but I don't see how
\[\sup \sum_{k=1}^{n}|f(x_{k})-f(x_{k-1})|<\infty\] implies
\[\sup \sum_{k=1}^{n}|f(x_{k})^{2}-f(x_{k-1})^{2}|<\infty\]

Use $|f(a)g(a)-f(b)g(b)| \leq |f(a)\parallel g(a)-g(b)|+|g(b)\parallel f(a)-f(b)|$
This way of using the triangle inequality is sort of the standard way to prove things like if $a_{n}\to a$ and $b_{n}\to b$ then $a_{n}b_{n}\to ab$. You should have seen the "trick" even in the very beginning of your education  :wink:

I think you should use the fact that any function with bounded variation is bounded.


function
limit
calculus
derivative
calculus computations

[quote="countdownmath"]Let $f$ and $g$ be $C^{3}$ functions, defined in a neighborhood of $x_{0}\in R$, s.t. $f(x_{0})=g(x_{0})=0$, but $g'(x_{0}) \neq 0$.

[b](part a)[/b] Show that $\lim_{x\to x_{0}}\frac{f(x)}{g(x)}$ exists.

[b](part b)[/b] Show that the derivative of the function $f/g$ exist at $x_{0}$. To be absolutely precise, we have to say that the derivative of a function $F$ defined by

$F(x) = f(x)/g(x), (x\neq x_{0})$

and

$F(x) = \lim_{x\to x_{0}}\frac{f(x)}{g(x)}, (x= x_{0})$

exists at $x_{0}$.

[b](part c)[/b] Show that the derivative of $f/g$ is continuous at $x_{0}$.[/quote]
obviusly by Hospital theorem. I don't know how it;s called in English.
but I know next: if $lim_{x_{0}}f,g=0$ then $\lim{f/g}=\lim{f'/g'}$ in point $x_{0}$

Thanks, I was thinking of the same method.
It's actually called the L'Hopital's rule.

i still cannot get how you can get part b from the results of part a.
can anyone show me how this works?

[quote="blahblahblah"]By part a), the derivative of $F$ at $x_{0}$ (if it exists) is given by
\[\lim_{h\to 0}\frac{\frac{f(x_{0}+h)}{g(x_{0}+h)}-\frac{f'(x_{0})}{g'(x_{0})}}{h}\]
which is just
\[\lim_{h\to 0}\left( \frac{g(x_{0}+h)f'(x_{0}+h)-g'(x_{0}+h)f(x_{0}+h)}{g^{2}(x_{0}+h)}\right) \]
by L'hopital's rule.[/quote]

at the first step, where did the $h$ in the denominator go?

what's the derivative of $h$ with respect to $h$?

o, my mistake... :p
i didn't really think about that ,,,
thanks for the input.

It's very possible that I made a mistake, and you're probably right.  In any event, now that you understand the method, there's probably no value in correcting my earlier post.


Solve for real $ x,y,z$
$ x\plus{}y\minus{}z\equal{}4$
$ x^2\minus{}y^2\plus{}z^2\equal{}\minus{}4$
$ xyz\equal{}6$

We have three equations and three variables so it can be done.  Basically, all there is to it is defining one variable in terms of the other two (maybe $ x$ works best, I don't know for sure).  Now since these equations aren't the simplest in the world, there's going to be some algebra, but none of it too difficult.  Thus keep at it and try different substitutions and eventually you'll get it.  You'll know you're right because the answer is $ \boxed{x\equal{}2}$, $ \boxed{y\equal{}3}$, and $ \boxed{z\equal{}1}$.  Good luck.

[quote="hell_ever"] [color=darkred]Solve for real $ x,y,z$ the system $ \{\begin{array}{c} x + y - z = 4\\\
x^2 - y^2 + z^2 = - 4\\\
xyz = 6\end{array}$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Observe that $ z\ne - 4\ .$ Thus, $ \{\begin{array}{c} x + y = z + 4\\\
x^2 - y^2 = - z^2 - 4\end{array}\|$ $ \implies$

$ \{\begin{array}{c} x + y = z + 4\\\\
x - y = \frac { - z^2 - 4}{z + 4}\end{array}\|$ $ \implies$ $ \{\begin{array}{c} \boxed {x = \frac {2(2z + 3)}{z + 4}}\\\\
\boxed {y = \frac {z^2 + 4z + 10}{z + 4}}\end{array}\ .$ Thus, $ xyz = 6$ $ \implies$

$ z(2z + 3)(z^2 + 4z + 10) = 3(z + 4)^2$ $ \implies$ $ (z - 1)(z + 2)(2z^2 + 9z + 24) = 0$ .

Thus, $ z\in \{1, - 2\}$ . Therefore, the solutions are $ (\begin{array}{ccc} \boxed x & \boxed y & \boxed z\\\
\Downarrow & \Downarrow & \Downarrow\\\
2 & 3 & 1\\\
- 1 & 3 & - 2\end{array})$ .[/color]

[quote="hell_ever"]Solve for real $ x,y,z$
$ x \plus{} y \minus{} z \equal{} 4$
$ x^2 \minus{} y^2 \plus{} z^2 \equal{} \minus{} 4$
$ xyz \equal{} 6$[/quote]

Hmmmmm.........

[hide="Solution?"]

My idea is to rearrange everything to obtain a function with one variable.

First notice $ x,y,z$ cannot be $ 0$.

$ x \minus{} z \equal{} 4 \minus{} y \Rightarrow (x \minus{} z)^{2} \equal{} (4 \minus{} y)^{2}$

Expanding we get $ (x^{2} \plus{} z^{2}) \minus{} 2zx \equal{} (y^{2} \minus{} 4) \minus{} 2zx \equal{} 16 \minus{} 8y \plus{} y^{2} (*)$

Since $ zx \equal{} \frac {6}{y}$, substituting this into (*) and multiplying both side by $ y$ we simplify it down to a simple quadratic $ 2y^{2} \minus{} 5y \minus{} 3 \equal{} 0$.

$ y \equal{} 3$ or $ y \equal{} \frac { \minus{} 1}{2}$

If $ y \equal{} \frac { \minus{} 1}{2}$ then we get $ x^{2} \plus{} z^{2} < 0$ clearly invalide over $ \mathbb{R}$.

So $ y \equal{} 3$ giving $ x \minus{} z \equal{} 1$, $ x^{2} \plus{} z^{2} \equal{} 5$ and $ zx \equal{} 2$ which we find $ (x \plus{} z)^{2} \equal{} 9$ and hence $ (x \plus{} z) \equal{} \pm 3$.

Hence $ (x,y,z) \equal{} (2,3,1), ( \minus{} 1,3, \minus{} 2)$[/hide]

@Virgil: when I took $ z\equal{} \minus{}2.5$ it didn't work for the third criteria. Did I do something wrong?


geometry
geometric transformation
reflection
conics
ellipse

Let $ B,C$ be $ 2$ fixed points on the plane. Let $ A$ be moving point on the plane such that $ AB\plus{}AC\equal{}c$, for $ c$ is constant. The circles with centers $ B,C$, radius $ BA,CA$ respectively intersect at the $ D$- as a second point. Find the locus of $ D$.

[quote="Criticalline1859"]Isn't this just the locus of A, D is the reflection of A over the major axis of the ellipse?[/quote]
[quote="mathVNpro"]Geometry problem
for the beginner one[/quote]Certainly not hard for an experienced math avid, but it's not obvious at all for beginners.


group theory
abstract algebra
superior algebra
superior algebra unsolved

suppose $G$ is a group and $H$ is a subgroup such that $[G: H]=n$is there any Homomorphism from $G$ to $S_{n}$?

Yes !
If we consider the left cosets xH as the elements for permutation then for every y in G , we established an element of the symmetric group in the way
xH->(yx) H. We map y to this element when we obtain a homomorphism. OK!


:)  It's an old problem.
Hint: Using order

[quote="apollo"]Determine all $ m,n \in \mathbb{N}$ such that $ m,n \ge 2$ and
$ n | m^{2\cdot 3^{n}} \plus{} m^{3^n} \plus{} 1$[/quote]
[hide="Solution"]
Note that $ (x^2 \plus{} x \plus{} 1)|(x^3 \minus{} 1)$, so $ n|(m^{3^{n \plus{} 1}} \minus{} 1)$, implying that $ m^{3^{n \plus{} 1}}\equiv 1\bmod n$.  Letting the order of $ m\bmod n$ be $ k$, we have that $ k|3^{n \plus{} 1}$, meaning that $ k$ is a power of $ 3$.  Let $ k \equal{} 3^l$.  Since $ k\le \phi (n) < n$, we have that $ l < 3^l < n$.  Notice that $ m^{3^n} \equal{} (m^{3^l})^{3^{n \minus{} l}}$, so we  have that $ m^{2\cdot 3^n} \plus{} m^{3^n} \plus{} 1\equiv 1 \plus{} 1 \plus{} 1\equiv 3\bmod n$.  Yet, it is also congruent to $ 0\bmod n$, implying that $ n \equal{} 1,3$.  Clearly, $ (m,n) \equal{} (m,1)$ is a solution for natural $ m$.  If $ n\equal{}3$, then $ 3$ cannot be a factor of $ m$.  We then get solution $ (m,n) \equal{} (m, 3)$ for $ m$ natural and not divisible by $ 3$.  We now prove that this works for only $ m$ congruent to $ 1\bmod 3$.  If $ m\equiv 1\bmod 3$, then $ m^{2\cdot 3^{n}} \plus{} m^{3^n} \plus{} 1\equiv 1 \plus{} 1 \plus{} 1\equiv 0\bmod 3$.  If $ m\equiv 2\bmod 3$, then $ m^{2\cdot 3^{n}} \plus{} m^{3^n} \plus{} 1\equiv 1 \plus{} 2 \plus{} 1\equiv 1\bmod 3$, which is a contradiction.  In conclusion, the solutions are $ (m,n) \equal{} (m,1), (m,3)$, and in the latter case, $ m\equiv 1\bmod 3$.  [/hide]


[quote="wiki"]Let $ a,b,c \in {[0, \frac {1}{2}]}$ and $ a \plus{} b \plus{} c \equal{} 1$.
Prove that : $ \frac {a}{\sqrt {a^2 \plus{} \frac {4}{3}bc}}\plus{}\frac{b}{\sqrt {b^2 \plus{} \frac {4}{3}ac}}\plus{}\frac {c}{\sqrt {c^2 \plus{} \frac {4}{3}ab}} \geq \frac {3\sqrt {21}}{7}$[/quote]

$ a,b,c \in \left[0,\frac 12\right]$ indicates $ 0 \le a,b,c \le \frac 12$. :wink:


algebra
polynomial
Vieta
number theory proposed
number theory

[i]Let $ m ; n$ be odd integers such that :

   $ ( m^2 \minus{}n^2 \plus{}1) | (n^2 \minus{}1)$

   Prove that  :   $ | m^2 \minus{}n^2 \plus{}1|$       is a perfect square  [/i]

We can expand the condition of this problem. Do not need $ m,n$ be odd integers , we can change to $ m,n$ with the same parity.

The main idea to solve this problem : WLOG , we may assume that $ m \geq n$. Existence $ a,b \in N$ and $ a,b>0$ with $ a > b$ such that $ m \equal{} a \plus{} b$ and $ n \equal{} a \minus{} b$.
Hence, this problem becomes $ 4ab \plus{} 1|(a \plus{} b)^2$

Now solving this problem is much easier by vieta-jumping

i think we can also try proving by infinite descent


The solution to this one would interest me too!

I think x=2.

1) Suppose x>0.
Let A=(x+2)2-1, B=(x+2)2+1 ==> (2(x+2))x=(B2-A2)x/2. So one thing we need to prove is (rx-1)1/x is increasing function when x>0 where r>1.
OK. ((rx-1)1/x)'=(rx-1)1/x*(rxln(r)/x(rx-1)-ln(rx-1)/x2) > (rx-1)1/x*(rxln(r)/x-ln(rx)/x2) = 0.

2) Suppose -1<x \leq 0. Then Ax \geq Bx, (2(x+2))x>0.

3) Must we consider x \leq -1 ?

[quote="Myth"]Is it an irony, siuhochung?[/quote]

nono..
at least i can't solve the part for x>0
somebody asked this question in mathdb too, and another guy used numerical analysis and proved that x=2 is the only solution.
I dunno if there's an elementary method to show that the only solution is x=2.
And now you show me x=2 for x>0, and the method is quite elementary~
 :)

((x+2)2-1)x+(2(x+2))x=((x+2)2+1)x
if we denote T=x+2 then we'll have:
(T^2-1)^x+(2T)^x=(T^2+1)^x
dividing the members with (T^2+1)^x:
( (T^2-1)/(T^2+1) ) ^ x  +  (2T/(T^2+1))^x = 1
if i'm not mistaken we can use T=tg(a/2),  then we'll have

(sin(a))^x + (cos(a))^x = 1 with the solution 2
 but i can't prove that is the only one... maybe someone else can help


geometry
circumcircle
rotation
geometric transformation
reflection
perpendicular bisector
geometry proposed

Let P a point on side BC of triangle ABC. An arbitrary secant through P meets the circumcircle in N and X. The lines AN and AX intersect BC in M and Y. Prove that when the secant rotates around P the circumcircle of triangle AMY passes through a fix point (different from A of course :P )

Nice problem, I like it!  :)

Here is my solution:

Call $N'$ the reflection of $N$ wrt the perpendicular bisector of $BC$. It's obvious that $N'$ lies on the circumcircle of $\Delta ABC$.
We define $Z$ as the intersection of $N'Y$ and the circumcircle of $\Delta ABC$.

I will prove that $Z$ always lies on the circumcircle of $\Delta AMY$.
First, note that
\[\angle YPX = \angle NPB = \angle NXB+\angle CBX = \angle N'ZC+\angle CZX = \angle YZX \]
We can conclude that $YPZX$ is a cyclic quadrilateral. $(1)$

Now, we make the following observation:
\[\angle PZN = \angle XZN-\angle XZP = \pi-\angle XAN-\angle AYM = \angle NMP \]
Hence we can conclude that $NMZP$ is a cyclic quadrilateral. $(2)$

Now, out of $(1)$ and $(2)$, we deduce:
\[\angle YZM = \angle YZP+\angle PZM = \angle AXN+\angle XNA = \pi-\angle XAM \]
All we have to show now is that $Z$ is fixed. This is very easy:
Call $A'$ the second intersection of $PZ$ with the circle. Clearly, $\angle N'ZA' = \angle YXP$ (follows from $(1)$)
Therefore, $A'$ is the reflection of $A$ wrt the perpendicular bisector of $BC$. Since $P$ and $A$ are fixed, $Z$ must be fixed too.

$\mathbb{QED}$

[img]http://img161.imageshack.us/img161/7633/ciprianmathlinks1iz7.jpg[/img]

How do u get that Z is fiz?N is a mobile point. Maybe I have missed something in your solution :maybe:  I can't say you are wrong beacause the point Z you describe is the desired point. Observe that if L is the reflection of A wrt the perpendicular bisector of BC than L, P, Z are colinnear. Moreover, if AZ intersect BC in O, than O is the central point of the homography described by (B,C) and (Y,M) (you can use Fregier theorem).From here you get that OM*OY=OB*OC=OA*OZ so A,Y,Z,M are on the same circle :)

You are right, I forgot to tell why it is fixed.
It's very obvious anyway, but I should have mentioned it.
I'll add it to my solution, thanks for pointing out!  :wink:

Well, Jan, please tell how did you get the desired point?  :)

No special trics, I just made a very accurate drawing and I saw that the fixed point lies on the circumcircle.
When a problem asks to prove that there is a fixed point somewhere, you can almost be sure that the fixed point will be a 'special' point.
So I went looking for a way to construct point $Z$ given all the other points, and I found what you can read above in my proof.
Once you find a construction which you think will give a fixed point (this is not hard in this problem) you just have to prove your conjecture. If your construction towards the point $Z$ is easy, the proof will often be easy too.


The circumference of a circle is increased by $ 10\%$. By what percent will the area of the circle increase?

Original circumference: $ 2\pi r$ radius is $ r$

New circumference: $ \frac{110}{100} 2\pi r \equal{} 2 \pi (\frac{110r}{100})$ radius is $ \frac{110r}{100}$.

Original area: $ \pi r^2$

New area: $ \pi (\frac{110r}{100})^2 \equal{} \pi \frac{121r^2}{100}$

Hence, it is increased by $ \boxed{21 \%}$


search
Euler
modular arithmetic
algebra
function
domain
Ring Theory

(x^3)+(y^3)=3(z^3)

show this has no integer solutions

(0,0,0) is an ordered triplet satisfying the above equation

See Theorem 232 in Hardy, G. H.; Wright, E. M.
An introduction to the theory of numbers.
Fourth edition. The Clarendon Press, Oxford University Press, New York, 1975.

[hide="hint"]Infinite descent  :D [/hide]

[quote="apollo"]Infinite descent  :D[/quote]

the proof given in the book (mentioned by re'born) isn't elementry at all,so I wouldn't consider this as a hint...

anyway,I'll rewrite the proof in the book here (but as you will see,it's hard to understand the whole proof,because it uses some other theorems,I'll write the toher ones at the end of this post without a proof):

[quote="THE BOOK"]
[b]theorem 232. The equation $ x^3 \plus{} y^3 \equal{} 3z^3$ has no solution in integers,excepr the trivial solutions in which $ z \equal{} 0$.[/b]

The proof is as might be expected,substantially the same as that of theorem 227,since $ 3$ is an associate of $ \lambda^2$.we again prove more,viz.that there are no solutions of:

$ \xi^3 \plus{} \eta^3 \plus{} \epsilon\lambda^{3n \plus{} 2}\gamma \equal{} 0$   (13.5.1)

where

$ (\xi,\eta) \equal{} 1$,   $ \lambda\not |\gamma$,
in integers of $ k(\rho)$,and again we prove the theorem by proving two propositions,viz.

(a) if there is a solution, then $ n > 0$;
(b) if there is a solution for $ n \equal{} m\geq 1$, then there is a solution for $ n \equal{} m \minus{} 1$;

which are contradictory if there is a solution for any $ n$.

we have  $ (\xi \plus{} \eta)(\xi \plus{} \rho\eta)(\xi \plus{} \rho^2\eta) \equal{} \minus{} \epsilon\lambda^{3m \plus{} 2}\gamma^3$.

hence at least one factor on the left,and therefore every factor,is divisible by $ \lambda$; and hence $ m > 0$.it then follows that $ 3m \plus{} 2 > 3$ and that one factor is divisible by $ \lambda^2$,and (as in 13.4) only one. we have therefore

$ \xi \plus{} \eta \equal{} \lambda^{3m}\kappa_1$,   $ \xi \plus{} \rho\eta \equal{} \lambda\kappa_2$,   $ \xi \plus{} \rho^2\eta \equal{} \equal{} \lambda\kappa_3$,

the $ \kappa$ being coprime in pairs and not divisible by $ \lambda$.

hence as in 13.4,    $ \minus{} \epsilon\gamma^3 \equal{} \kappa_1\kappa_2\kappa_3$,
and $ \kappa_1,\kappa_2,\kappa_3$ are the associates of cubes,so that

$ \xi \plus{} \eta \equal{} \epsilon_1\lambda^{3m}\theta^3$,   $ \xi \plus{} \rho\eta \equal{} \epsilon_2\lambda\phi^3$,   $ \xi \plus{} \rho^2\eta \equal{} \epsilon_3\lambda\psi^3$.

it then follows that

$ 0 \equal{} \xi \plus{} \eta \plus{} \rho(\xi \plus{} \rho\eta) \plus{} \rho^2(\xi \plus{} \rho^2\eta) \equal{} \epsilon_1\lambda^{3m}\theta^3 \plus{} \epsilon_2\rho\lambda\phi^3 \plus{} \epsilon_3\rho^2\lambda\psi^3$

$ \phi^3 \plus{} \epsilon_4\psi^3 \plus{} \epsilon_5\lambda^{3m \minus{} 1}\theta^3 \equal{} 0$;

and the remainder of the proof is the same as that of theorem 227.[/quote]

[b]theorem 227:there are no solutions of $ \xi^3 \plus{} \eta^3 \plus{} \zeta^3 \equal{} 0$    $ (\zeta\neq 0,\eta\neq 0, \zeta\neq 0)$

in integers of $ k(\rho)$.in particular,there are no solutions of $ x^3 \plus{} y^3 \equal{} z^3$ in rational integers,except the trivial solutions in which one of $ x,y,z$ is $ 0$.[/b]

also

[b]13.4. The equation $ x^3 \plus{} y^3 \equal{} z^3$.if Fermat's theorem is true for some n,it is true for any multiple of $ n$,it is true for any multiple of $ n$,since $ x^{ln} \plus{} y^{ln} \equal{} z^{ln}$ is

$ (x^l)^n \plus{} (y^l)^n \equal{} (z^l)^n$.

the theorem is therefore true generally if it is true (a) when $ n \equal{} 4$ (as we have shown) and (b) when $ n$ is an odd prime. the only case of (b) which we can discuss here is the case $ n \equal{} 3$.

the natural method of attack,is to write Fermat's equation in the form

$ (x \plus{} y)(x \plus{} \rho y)(x \plus{} \rho^2y) \equal{} z^3$,

and consider the structure of the various factors in $ k(p)$.[/b]

I've just realized that a similar problem was posted before at [url]http://www.mathlinks.ro/viewtopic.php?search_id=657786750&t=28102[/url]

I have also found a proof to this (somhow) general problem:

[b]Theorem.[/b] let $ \varepsilon$ be a unit in $ \mathbb{Z}[\omega]$.the equation $ x^3 + y^3 = \varepsilon z^3$

has no solution $ \{x_1,y_1,z_1\}$ where $ x_1,y_1$ and $ z_1$ belong to $ \mathbb{Z}[\omega]$ and $ x_1y_1z_1\neq 0$.

this theorem is due essentially to Euler; note that Fermat's conjecture for $ n = 3$ follows. we begin with a definition and two lemmas.

[b]DEFINITION.[/b] if $ \alpha\in\mathbb{Z}[\omega]$ and $ \alpha = \varepsilon\lambda^n\beta$ where $ \varepsilon$ is a unit and $ \lambda\not |\beta$ we define $ ord\alpha = n$.

[b]LEMMA 1.[/b] let $ \varepsilon$ be a unit in $ \mathbb{Z}[\omega]$.if $ x^3 + y^3 = \varepsilon z^3$,$ \lambda\not |xy$, and $ \lambda\mid z$, where $ x,y,z\in\mathbb{Z}[\omega]$,then $ ord z\geq 2$.

[b]Proof.[/b]we show first that if $ \alpha\in\mathbb{Z}[\omega]$,

$ \alpha\equiv 1\pmod\lambda$ implies $ \alpha^3\equiv 1\pmod{\lambda^4}$.

for,if $ \alpha = 1 + \lambda\beta$,

\begin{eqnarray*}\alpha^3 - 1 & = & (\alpha - 1)(\alpha - \omega)(\alpha - \omega^2) \\
& = & \lambda\beta(1 - \omega + \lambda\beta)(1 - \omega^2 + \lambda\beta) \\
& = & \lambda^3\beta(\beta + 1)(\beta + \omega + 1) \\
& = & \lambda^3\beta(\beta + 1)(\beta - \omega^2)\end{eqnarray*}

(in above equalities we have used the facts that $ 1 - \omega = \lambda$ and $ \omega^2 + \omega + 1 = 0$).

but $ \omega^2\equiv 1\pmod\lambda$ and also we know that $ \alpha\in\mathbb{Z}[\omega]$ implies $ \alpha\equiv - 1,0,\textrm{ or } 1\pmod\lambda$ so our claim is prooved.

reducing the equation in the lemma we have,by the above claim that:

$ \pm 1\pm 1\equiv \varepsilon z^3\pmod{\lambda^4}$

if $ \varepsilon z^3\equiv\pm 2\pmod{\lambda^4}$ then $ \lambda\mid 2$, which is impossible.hence $ \varepsilon z^3\equiv 0\pmod{\lambda^4}$ giving $ 3ord z\geq 4$ and so finally $ ord z\geq 2$.

the next lemma provides the descent step in the main proof.

[b]LEMMA 2.[/b] suppose $ x^3 + y^3 = \varepsilon z^3$ where $ \varepsilon$ is a unit in $ \mathbb{Z}[\omega]$, $ (x,y) = 1$,$ \lambda\not | xy$,and $ ord z\geq 2$,then we can find $ x',y',z'$ and a unit $ \varepsilon'$ in $ \mathbb{Z}[\omega]$ to satisfy

$ x'^3 + y'^3 = \varepsilon'z'^3$ , $ \lambda\not | x'y'$, and $ ord z' = ord z - 1$.

[b]Proof.[/b] as

$ (x + y)(x + \omega y)(x + \omega^2 y) = \varepsilon z^3$ (*)

and $ ord\varepsilon z^3\geq 6$,at least one of the factors on the left-hand side of the last equation is divisible by $ \lambda^2$.so,substituting $ \omega y$ or $ \omega^2 y$ for $ y$ if necessary, we have $ ord (x + y)\geq 2$.as $ ord(1 - \omega)y = ord\lambda y = 1$ by assumption,we have

$ ord (x + \omega y) = ord(x + y - (1 - \omega)y) = \min\left(\ord (x + y),\ord\lambda y\right) = 1$.

using a similar argument we also have $ ord (x + \omega^2 y) = 1$.combining these three equations with (*)  we derive

$ ord (x + y) = 3ord z - 2$. (**)

we show next that $ (x + y,x + \omega y) = \lambda$.for,if $ \pi$ is an irreducible element not equal to $ \lambda$,$ \pi\mid x + y$ and $ \pi\mid x + \omega y$,then $ \pi\mid (1 - \omega)y$; that is,$ \pi\mid y$,and similarly $ \pi\mid x$,contrary to assumption.similar arguments show that $ (x + y,x + \omega^2 y) = (x + \omega y,x + \omega^2 y) = \lambda$.thus as $ \mathbb{Z}[\omega]$ is a unique factorization domain we have by (**)

$ x + y = \varepsilon_1\alpha^3\lambda^{3ord z - 2}$

$ x + \omega y = \varepsilon_2\beta^3\lambda$

$ x + \omega^2 y = \varepsilon_3\gamma^3\lambda$,

where $ \lambda$ does not divide $ \alpha,\beta,$ or $ \gamma$; $ (\alpha,\beta) = (\beta,\gamma) = (\gamma,\alpha) = 1$; and $ \varepsilon_1,\varepsilon_2$ and $ \varepsilon_3$ are units.now according to $ \omega^2 + \omega + 1 = 0$

$ \varepsilon_1\alpha^3\lambda^{3ord z - 2} + \omega\varepsilon_2\beta^3\lambda + \omega^2\varepsilon_3\gamma^3\lambda = 0$

or,dividing by $ \lambda\omega\varepsilon_2$,

${ \beta^3 + \varepsilon_4\gamma^3 = \varepsilon_5\alpha^3\lambda^3(ord z - 1)}$  (***)

where ${ \varepsilon_4 = \omega\varepsilon_2^{ - 1}\varepsilon_3}$ and $ \varepsilon_5 = - \omega^{ - 1}\varepsilon_1\varepsilon_2^{ - 1}$ are both units.hence as $ \lambda$ does not divide $ \alpha,\beta,$ or $ \gamma$,

$ \pm 1\pm\varepsilon_4\equiv 1\pmod{\lambda^2}$

further,$ \lambda^2$ is an associate of $ 3$ and so this congruence is only possible if $ \varepsilon_4 = \pm 1$.thus if we let $ x' = \beta,y' = \pm\gamma$,where the sign depends on the sign of $ \varepsilon_4$, and $ z' = \alpha\lambda^{ord z - 1}$ in (***) the lemma follows.

returning to the proof of theorem,there are four cases to consider.

[i]Case 1.[/i] our equation has a solution $ \{x_1,y_1,z_1\}$ with $ \lambda\not | x_1y_1z_1$.this gives

$ \pm 1\pm 1\equiv \pm\varepsilon\pmod{\lambda^4}$

which is impossible as $ 9$ is an associate of $ \lambda^4$.

[i]Case 2.[/i] our equation has a solution $ \{x_2,y_2,z_2\}$ with $ \lambda\not | x_2y_2$ and $ \lambda\mid z_2$.in this case we cal apply Lemmas 1 and 2 to obtain a new solution $ \{x_3,y_3,z_3\}$ with $ ord z_3 < ord z_2$ and $ \lambda\not | x_3y_3$.repeating this rgument $ k$ times,where $ k = ord z_2 - 1$,we obtain a contradiction because the resulting solution would satisfy the conditions of Case1.

[i]Case 3.[/i]our equation has a solution $ \{x_4,y_4,z_4\}$ with $ \lambda\mid x_4$ and $ \lambda\not | y_4z_4$.in this case we have $ \varepsilon\equiv\pm 1\pmod{\lambda^4}$ and so $ \varepsilon = \pm 1$ hence the equation can be rewritten as $ (\pm z_4)^3 + ( - y_4)^3 = x_4^3$, and Case 2 now applies.

[i]Case 4.[/i] our equation has a solution $ \{x_5,y_5,z_5\}$ with $ \lambda\mid y_5$ and $ \lambda\not | x_5z_5$.this is similar to Case 3.

as Cases 1-4 cover all possibilities,the theorem is proved.


It's true, we are getting married - she told me!

 - Alex says:
so when are we getting married?
Sarah says:
hahaha
Sarah says:
whenever you can find okotoks on the map of canada
 - Alex says:
okay

Good luck. lol.

Okotoks is so small...hahahaha.

Three cheers for randomness!

So Alex and I were thinking, this forum is DEAD.  Where is everyone?!

this fourm is dead, becuase I killed it. Just to watch it die.

I am so very very tired, and I've already stoped doing all my homework for the year.

[quote="rzsolt"]this fourm is dead, becuase I killed it. Just to watch it die.
[/quote]
If I'm still here, this forum is still alive.  Nice try  :P

WHERE IS EVERYONE?

  We were out getting lives.  Sorry for leaving you behind.  :rotfl:   :rotfl:   :rotfl:

[quote="G-UNIT"]
  We were out getting lives.  Sorry for leaving you behind.  :rotfl:   :rotfl:   :rotfl:[/quote]

For someone who has more than 1000 posts on this website?  Hahahahaha. that's kinda ironic  :D

Lol sarah you aren't that far behind in posts .. like you should talk

thats right, dont talk :) D:

[quote="Singular"]
whenever you can find okotoks on the map of canada
[/quote]
:rotfl:

ooo, i sense a level of animosity in here :? ....alex wice: first stop dreaming, then watch out for Shaun!!!!  :D  :D  :D

[quote]ooo, i sense a level of animosity in here Confused ....alex wice: first stop dreaming, then watch out for Shaun!!!! Mr. Green Mr. Green Mr. Green[/quote]

 :rotfl:  :rotfl:  :rotfl:  :rotfl:

Ya Alex, you'd better listen to Allen.. It's a good advice..

[quote="lightrhee"]I found Okotoks on a Canada map. :rotfl:
[url=http://local.google.ca/local?q=okotoks&sll=50.746884,-113.829346&spn=51.603254,112.552734&sspn=1.682310,3.702393&hl=en]Google Local[/url][/quote]

Hahahaha that so doesn't count, there's zoom in it.  For Okotoks to show up on your regular map of Canada, I'd figure it needs to have about...a million people.  More maybe.  That'll NEVER happen, hehehe, cuz they're putting a population restriction on Okotoks, hehehe.

Population restriction?  So how does that work..you have to get interviewed before you can move into the city? t.t

[quote="Sparky"]Alex, even though okotoks isn't on the map of Canada, I still love you with all my heart and I want to marry you...
We can live on a beach house in Argentina; after our honeymoon throughout Europe!  xoxo[/quote]

aww thats so sweet thank you

Oh man.  I miss this forum!

Come on Sarah... you know I'm the only one for you... all those late nights we stayed up eating fudge! ;)

Heheheh... unfortunately I already have a girlfriend :(.  But yeah, when you get your ass down to Waterloo, we're gonna party it up like you've never done before!  (You're still planning on coming to Waterloo, right?)

Sarah, you're my favourite person from Okotoks ever!  We should turn this into a Sarah Sun appreciation thread.

Hopefully I'll see you at seminar this year!

[quote="Singular"]you mean this was all just a joke to you sarah?  im hurt

:)[/quote]

Haha, Shaun's post actually made me believe that he was being serious for a second (yes i'm gullible :(), until i realized "Prime power factorizing you do death" isn't a serious way of killing someone  :rotfl:

[quote="Elyot"]Oh man.  I miss this forum!

Come on Sarah... you know I'm the only one for you... all those late nights we stayed up eating fudge! ;)

[/quote]

Ah hahahaha, how can I forget?  And the poker?  lol.  You owe me Microsoft.

Sweet!  Elyot's joined us once again.

[quote="TheDreamer"]
Haha, Shaun's post actually made me believe that he was being serious for a second (yes i'm gullible ), until i realized "Prime power factorizing you do death" isn't a serious way of killing someone  [/quote]

Hey!  You never know!  Prime power factorizing sounds it could kill people!  At least it sounds scary to my friends, lol.

Yea, actually they got me back to the site when they warned me that you were getting married!  I got down on me knees and cried "NOOOOOOO"!!!!!!!!  How could she?!?!?!?!?!  I had to make sure it wasn't actually true!  All my hopes and dreams would have been shot down!  lol ;)

and yea...
I owe you Microsoft...
but you owe me some pretty big stuff too...
I forget what all of it was...
It was too late in the morning to remember!

And you know, there are other number systems where unique factorization isn't possible!  They're not necessarily rings like the integers are, but they DO EXIST.

[quote="Elyot"]And you know, there are other number systems where unique factorization isn't possible!  They're not necessarily rings like the integers are, but they DO EXIST.[/quote]

$\mathbb{Z}[\sqrt{-5}]$ is a ring without unique factorization.

I said not necessarily... not that they couldn't be rings.

Your example is boring.

Mine is better... you can't uniquely factor polyominoes.  ;)

[quote="Elyot"]I said not necessarily... not that they couldn't be rings. [/quote]

I know that you know that :)

I like providing examples for the sake of completeness.

[quote="blahblahblah"][quote="Singular"]talk about dorky[/quote]

Hey, I go to AoPS to do math, not to profess my e-love to female members of the site.  But different strokes for different folks, I guess.[/quote]

wasnt talking to you sry i didnt make that clear

[quote="Singular"][quote="lotus87"]My top 5 favourite things..
5 ddr
4 sleep
3 prime numbers
2 sarah
1 chocolate milk

Death by inversion would be so much more interesting than death by prime factorization...
Ok... all you sarah lovers back off... I'll take it from here...   :P[/quote]

what about prime numbers covered in chocolate milk?  or....     :huh:  :idea:

 :lol:  :lol:  :lol:  :lol:  :lol:[/quote]

LMFAO!!!!!   You're such a dirty b@stard!

...

But I think fudge > chocolate milk.

list didnt give me much to work with

ps. you can make more than one interpretation


[size=150][b][i][color=green]Prove that in an acute angled $ \Delta$ $ ABC$
A) $ \frac{tanA}{A} \plus{}\frac{tanB}{B}\plus{}\frac{tanC}{C}$ $ >$ $ \frac{36}{\pi^2}$
B) $ AcotA \plus{} BcotB \plus{} CcotC$ $ <$ $ \frac{\pi^2}{4}$
where $ A,B,C$ denote the angles of the triangle.[/color][/i][/b][/size]

The first,by AM-GM we have:
$ \frac{tanA}A\plus{}\frac{tanB}B\plus{}\frac{tanC}C\ge 3\sqrt[3]{\frac{tanAtanBtanC}{ABC}}\ge \frac{9\sqrt{3}}{\pi} > \frac{36}{\pi^2}$
It's true because $ tanAtanBtanC\ge 3\sqrt{3}$ and $ ABC\le(\frac{A\plus{}B\plus{}C}3)^3\equal{}({\frac{\pi}3})^3$

I think the second's not true.What happen if $ A\to 0?$

@duongptnkvn:The inequality is absolutely true ...I can't understand ur point

I don't know how to prove $ cotA\plus{}cotB\plus{}cotC < \frac{3\pi}4$?!!


number theory
greatest common divisor
search
number theory unsolved

Consider the following sequence

$ a_1\equal{}1001$, $ a_2\equal{}1004$, $ a_3\equal{}1009$, $ \ldots, a_n\equal{}1000\plus{}n^2$

where $ n$ is a positive integer. Denote by $ d_n$, the Greatest Common Divisor of $ a_n$ and $ a_{n\plus{}1}$. Find the greatest possible value of $ d_n$


algebra
polynomial
Vieta
pigeonhole principle
algebra unsolved

Polynomial p(x) = x^3 + ax^2 + bx + c has all real roots.

Prove that polynomial q(x) = x^3 - bx^2 + acx - c^2  has at least one nonnegative root.

Let $ r_{1}, r_{2}, r_{3}$ be the roots of $ R(X)$, then $ \minus{}r_{1}r_{2}r_{3} \equal{} c$ is a real number since $ r_{i}$ so is and $ c^{2} \geq 0$. Let $ x_{1}, x_{2}, x_{3}$ be the roots of $ Q(X)$, then if $ x_{i} < 0$, $ i \equal{} 1,2,3$ then we have $ x_{1}x_{2}x_{3} \equal{} c^{2} < 0$. A contradiction!. Then at least, one of the $ x_{i}$ must be nonnegative.

in my opinion You have assumed that q has exactly 3 real roots but I think You should prove it.

If $ P$ has roots $ r_1, r_2, r_3$, then it's trivial to show with Vieta's formulas that $ Q$ has roots $ r_1 r_2, r_1 r_3, r_2 r_3$. There necessarily exist two $ r_i$ that are either both nonpositive or both nonnegative by pigeonhole, finishing the proof.

matex(L)(L)(L) Yes, I assumed that it has three real roots but I didn't prove it becuase it is a immediate consequence of Vieta as MellowMelon proved, but, instead of writting the roots of $ Q$ as the products of roots of $ P$ I just renamed that.  :blush: sorry about that.


Well, I'm working on the first one.  As for the second one,

[hide](2, 2, 2) gives us

(x - x + 1)(y - y + 1)(z - z + 1) = 27

(xyz) - xyz + 1 = 57, so that one doesn't work.  It also took me a while to find an example that did work, but I finally stumbled across (2, 2, 1/2) which gives you 27/8 > 3/2.[/hide]

The third one [hide]follows directly from the fact that x, y, z > 0.  All of the terms on the right side are contained in the expansion of the left side, with a lot of stuff left over.[/hide]

Yeah, I also found out that the second and the third one are easy.
You can find an obvious counterexample for the second one.
For the third one, expanding is indeed sufficient.

But I don't know how to tackle the first one. Just an idea : maybe the inequality
a - a + 1 >= (a + a + 1)/3
(which is very easy to prove) can be helpful ?
Then the problem reduces to proving that
(x + x + 1)(y + y + 1)(z + z + 1) >= 9((xyz) + xyz + 1).
Do you have another idea ?

[color=cyan]A nice thought, but if you push x, y, and z close to zero you get
1 >= 9.  In fact, if you take them all less than 1, I think your suggested inequality fails.  [/color]

[color=cyan]I want maybe to consider what happens around the points for which xyz = 1 -- you can make it look nicer when you multiply by
(x + 1), (y + 1), (z + 1), and (xyz - 1), which might help.  But you have to break it into 2 cases, then.
I was also thinking we might try to prove you force it closer by forcing xyz to 1, then showing that when xyz = 1 equality is only at (1, 1, 1).

This of course wouldn't work if (1, 1, 1) is the only equality point, but it was the only one I could find.

Where did these questions come from?[/color]

Now I think if we can do the case xyz >= 1 then we're done.
Since if xyz > 1, then we can divide the given inequality by (xyz).
Then putting x' = 1/x, y' = 1/y, z' = 1/z we get x'y'z' >= 1.
And then we can apply the result of the case xyz >= 1.

So : is it true that if xyz >= 1 then
(x + x + 1)(y + y + 1)(z + z + 1) >= 9((xyz) + xyz + 1)) ?
Who knows ? Maybe you can use Minkowski ?

(If xyz >= 1 then your argument of pushing x, y, z to 0 does not work.)

I think it's an olympiad problem but I don't know of which olympiad anymore. It was proposed at IMO training but I don't know how to solve it, neither do the other ones who participated in the training sessions. I'll look up from which olympiad this problem comes. I'll try to tell you this tomorrow ... Bedtime (its 22.30 pm in Belgium) :wink:

[color=cyan]No, I didn't suggest pushing x, y, and z to zero to prove it.  I was just saying that your transformation creates an inequality that isn't true over the entire domain.  Anyhow, your dividing-through idea takes care of that problem and looks promising.  That doesn't mean I know what to do, though.  I'll keep thinking.[/color]


linear algebra
matrix
linear algebra unsolved

It is not difficult to prove that if $A$-is a matrix over an algebraically closed field $F$ then $A$ is similar to it's transpose-we can use there a Jordan form.
Is it true in a general for any field $F$ ?

If two matrices with entries in a field $k$ are similar over an extension field $K$ of $k$, then they're similar over $k$ as well. A characterisation of similarity says that two matrices are similar iff they have the same rational canonical form, which is a generalization of the Jordan form, and this is true if they have the same Jordan form in $K$. Try searching for "rational canonical form" (both with Google and on the forum). I'm sure you'll find tons of relevant results.


What is a matrix?  Does anybody have a link to a site that explains what a matrix is?

[quote="now a ranger"]Yah. I still don't get how to find the determinant and sum and stuff.

But then again I've never tried to learn about them. :([/quote]
Me too, but AoPS Volume II has a lot about matrices and vectors.
Sum? You just add each element. Although, multiplying did take me a while. :D
The determinant: Multiply the numbers diagonally down and to the right. Add all of these products. Then multiply diagonally down and to the left and subtract all these products. :D
Can you take the determinant of a non-square matrix? :maybe:

[quote="lotrgreengrapes7926"]Can you take the determinant of a non-square matrix? :maybe:[/quote]
Nope, coz dets are defined only for square matrices. You can take a det of any square SUBmatrix of a given matrix, though, and that practice is crucial for definition of what's called a [i]rank[/i] of a matrix. (Won't go into details on that here, though.)

If you want alot of basic stuff about matrices,

ALGEBRA II is the way to go. Algebra I has some stuff, but not the best. AoPS 2 has some advanced stuff involving matrices, like vectors and stuff.

haha im only in geometry. perhaps doing something else my benefit me :roll:


Never mind. It's more of an AMC 8.9 as I look over the questions.

I think he has a new Mock AMC, one with easier questions.

But I would like to take it, even if it was more of an AMC 9.

I still think you should put up the questions.  Just make open to people who didn't qualify for the mock AIME though.

an easier contest would be great - one where you can answer a question with a number higher than 13. it should be open to everyone, though.

If it's a Mock AMC 9, it'll probably come down to who can make the least mistakes

Perhaps you will want to throw in one difficult problem or two just to make sure no two people will get a full score - though I doubt I will even come close because I make so many dumb mistakes every single time.

Samus, are you still going to do this?  If so do you have a time in mind?

Is this a viable project, a work in progress, a suggestion for someone else to create a contest, or something else?  Explain or this gets deleted in the name of organization.

Hes already written it mathfanatic.  Hes just wondering if anyones interested in it.


calculus
integration
search
derivative
logarithms
calculus computations

These are very interesting problems. Find
$\int^{1}_{0}\frac{1-x}{lnx}dx$

$\int^{\inf}_{0}\frac{sin(x)}{x}dx$

inf= infinity
 :D  :D

Both problems have been discussed many times on the forums, and the latter is particularly classical.

For the first one, search for 'parameter differentiation'.

I still can't find it Blalala. The only useful link I find is http://www.mathlinks.ro/Forum/viewtopic.php?highlight=differentiation+parameter&t=41288
but I don't know how to prove the last one.

Any helps will be appreciated. :)

For the first one, put
\[I(t)=\int^{1}_{0}\frac{x^{t}-1}{\log x}dx \]
and evaluate
\[I'(t)=\int^{1}_{0}\frac{\partial}{\partial t}\frac{x^{t}-1}{\log x}dx \]
You want $I(1)$, of course.

The second one is more involved and is easiest if you know some complex analysis.  Do you know any?

No, I didn't. Someone from my class gives me this problem. I solved it for 3 days already but got stuck :blush:I know only a few about triple integral. That is all.
  BTW I can prove that $\int^{\inf}_{0}\frac{sin^{2}(x)}{x^{2}}dx=\int^{\inf}_{0}\frac{sin(x)}{x}dx$
so the answer is $\frac{\pi}{2}$
Can you help me?
 :lol:

There was a mistake in my post on how to evaluate the first integral; it's been fixed now.


Let a,b,c>0 such that     $ a^2\plus{}b^2\plus{}c^2\equal{}3$                              Prove that
$ \frac{9}{2(a\plus{}b\plus{}c)}\leq\sum\frac{a}{b\plus{}c}$

[quote="tranquoc"]Let a,b,c>0 such that     $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$                              Prove that
$ \frac {9}{2(a \plus{} b \plus{} c)}\leq\sum\frac {a}{b \plus{} c}$[/quote]
it's very easy for you

we have $ \sum\frac {a}{b \plus{} c}$ = $ \sum \frac {a^2}{ab \plus{} ac} \ge \frac { (a \plus{} b \plus{} c)^2 }{2(ab \plus{} bc \plus{} ac)}$  then in view the problem :$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3 \minus{} > s^2 \equal{} 2p \plus{} 3$ . we obsever $ s^3 \plus{} s^3 \plus{} 27 \ge 9s^2$

[color=green]Hey friend, you can use $ \longrightarrow$ or $ \rightarrow$ for $ \minus{} >$,  :) .

Code:

[code]\longrightarrow[/code] $ \longrightarrow$.
[code]\rightarrow[/code] $ \rightarrow$.

[/color]

<=>$ (3 \minus{} s)(9 \plus{} 3s \minus{} 2s^2) \ge 0$ (Finish your solution :D)
 .

we have
$ \sum\frac{a\plus{}b}{c} \geq  \frac{18}{a\plus{}b\plus{}c}$
    IT very EASY


geometry
perimeter
inequalities
triangle inequality
geometry unsolved

Hi all - anyone got an idea how this can be proved:

DEF is a triangle enclosed within triangle ABC. Show that the perimeter of DEF is always smaller  than (or equal to) the perimeter of ABC.

I guess it can be extended to similar polygons in general.

Cheers
J

If by "enclosed" you mean that the vertices of triangle $DEF$ belong to each side of triangle $ABC$, then the problem is solved by using the triangle inequality. Still, I am not quite sure that this is what you mean by "enclosed".

Hi andyciup... thanks for the feedback.

By 'enclosed', I mean that the 'smaller' triangle (DEF)  is within the perimeter of the larger one (ABC). Also, it follows that DEF's area is smaller than ABC's.

The case you mentioned - where each vertices of DEF lie on each side of ABC - is a special case of the above.

Thanks.

Ok, I have solved the problem. So we have that the points $D,E,F$ are inside the interior of triangle $ABC$. The line $DE$ can be parallel to at most one of $AB, BC,CA$, and also, the line $DE$ intersects two of the lines $AB,BC,CA$ in the segment $[AB]$,  $[BC],$ or $[CA]$. Assume WLOG $DE\cap [BC]=\{P\}$. From the triangle inequality, we have $EF+EP\leq FP$, which means $DE+EF\leq PD+PF$. $(1)$

If the line $PF$ has a common point with the segment $[AB]$, let $PF\cap [AB]=\{O\}$. Then $PF+FD\leq PO+OD$, which means, taking into account the relation $(1)$,  $DE+DF+EF\leq PO+OD+DP$, with $P\in[BC]$ and $O\in [AB]$.
          IF $OD$ and $[AC]$ have a common point, let it be $Q$. Then $PO+OD+DP\leq PQ+QO+OP$, and thus $DE+EF+DF\leq PQ+QO+OP$. And now by applying the triangle inequality, \[PQ\leq PC+CQ\] \[QO\leq AQ+AO\] \[OP\leq OB+BQ\], we sum these last three relations to obtain $PQ+QO+OP\leq AB+BC+CA$, which is what we wanted to prove.
             IF $OD$ and $[AC]$ don`t have a common point, then $OD$ and $[BC]$ have a common point.Let it be $R$. We may assume $B-R-P-C$ are positioned in this order on the line $BC$.  Then $PO\leq AC$, $OR\leq BR+BO\leq BR+BA$, and $RP\leq RC$. By summing up the last three relations we obtain $PO+OR+RP\leq AB+BC+CA.$ But because $DE+EF+FD\leq PO+OD+DP\leq PO+OR+RP\leq AB+BC+CA$, be obtain what we wanted to prove :)

If the line $PF$ doesn`t have a common point with $[AB]$, then it has a common point with $[AC]$, and the argument is similar.

Great. Thanks a bunch andyciup. I look forward to the next one  :)

Incidentally, I needed this to show that: For any point D inside triangle ABC, (AB+BC) > (AD+DC),, which now becomes trivial.

Thanks again
J


Let $ ABC$ be a non-isosceles and acute triangle. $ (O)$ is the circumcircle $ (ABC)$.The tangent line through $ A$ of $ (O)$ meets $ BC$ at $ D, OD$ meets $ AB,AC$ at $ E,F$ respectively. Denote $ M,N$ be the midpoints of $ AB,AC$ respectively. Prove that $ AO,FM,EN$ are concurrent.

Posted on this forum before!

Best regards,
sunken rock

This is probably way overkill, but I found [hide=this solution.]
First, we consider all the important points under a dilation with center $A$ and factor $2$. Thus, $M, N$ map to $B, C$; $O$ maps to the $A$-antipode $A'$; and line $EF$ maps to a line going through $A'$ and perpendicular to the $A$-symmedian - let it intersect $AB, AC$ at $S, T$. (The fact that line $EF = $ line $DO$ is perpendicular to the $A$-symmedian can be proven as follows: Let the tangents from $D$ to the circumcircle be tangent at $A$ and $A_1$. By La Hire's theorem, $BC$ going through the intersection of the tangents at $A$ and $A_1$ ($D$) implies that $AA_1$ goes through the intersection of the tangents at $B$ and $C$, thus making $AA_1$ a symmedian.) Thus, the problem statement is equivalent to showing that $AA'$, $TB$, and $SC$ are concurrent.

Now, we invoke Desargues' theorem. Letting $A'C \cap BS = A'C \cap AB = X$, $A'B \cap CT = A'B \cap AC = Y$, and $BC \cap XY = Z$, we now need to show that $Z$ is on $ST = A_1A'$ ($A_1$ is on line $ST$ since $\angle AA_1A' = 90$). Notice that $A'C \perp AC$ and $A'B \perp AB$, so $A'$ is the orthocenter of $\triangle AXY$. Thus, by a well-known lemma, the intersection of the tangents at $B$ and $C$ is the midpoint of $XY$ - call it $M'$. Now, since $A_1$ is on $AM'$, the problem is reduced to showing that $A'Z$ is perpendicular to the $A$-median of $\triangle AXY$, but this is immediate by Brocard's theorem on $XBCY$.
[/hide]

Let $ED \cap MN = Q$. Let $X = BB \cap CC$. We know that $AX$ is the polar of $D$ so we have $AX \perp DO$ which means the $A$-symmedian of $\triangle AMN$ is perpendicular to $OQ$. It suffices to show that $A(E,F;O,Q) = -1$, by reflecting through the angle bisector of $\angle BAC$ this becomes $A(M,N;R,Q') = -1$ where $R$ is the projection of $A$ on $MN$ and $Q'$ is the reflection of $Q$ over the angle bisector of $\angle BAC$, which is equivalent to $Q'$ being on the line through the feet of altitude of $M,N$.

Working with respect to $\triangle AMN$ the problem becomes: Given $\triangle ABC$ with altitudes $AD,BE,CF$, $AA_1$ diameter and $AX$ the symmedian chord. Prove that $AT,AS$ are isogonal where $T = EF \cap BC$ and $S \in BC$ satisfies $A_1Q \perp AX$. Since $\angle AXA_1 = 90^{\circ}$ this is just $S = BC \cap XA_1$. It suffices to show that after we perform an inversion with center $A$ and radius $\sqrt{AB \cdot AC}$ composed with a reflection through the angle bisector of $\angle BAC$, $S$ ends up on line $AT$. It's clear that $X$ goes to the midpoint $M$ of $BC$, and since $S \in BC$ we have $S' \in (ABC)$ such that $\angle AS'X' = \angle AXS = 90^{\circ}$. Therefore $S' = A_1M \cap (ABC)$, it is well known that $A1M$ passes through the orthocenter $H$ of $ABC$ which means $S'$ is on $(AEFH)$. Therefore by radical axis on $(AS'FHE),(BFEC),(AS'BC)$ we get $AS',FE,BC$ concur which means $A,T,S'$ are collinear as desired.

Here is my solution for this problem
[b]Solution[/b]

Let $I$ $\equiv$ $OM$ $\cap$ $AC$, $J$ $\equiv$ $ON$ $\cap$ $AB$, $G$ be orthogonal projection of $A$ on $OD$
Since: $AG$ $\perp$ $OD$, we have: $AG$ is $A$ - symmedian line of $\triangle$ $ABC$
But: $BC$, $IJ$ are antiparallel so: $AG$ is $A$ - median line of $\triangle$ $AIJ$
Then: $I$, $J$, $G$, $O$ lie on a circle
We see that: $MN$, $OG$, $IJ$ are radical axises of (($AMN$); ($INMJ$)), (($INMJ$); ($IOGJ$)), (($IOGJ$); ($AMN$))
Hence: $MN$, $OG$, $IJ$ concurrent at $K$
But: $O$ is orthocenter of $\triangle$ $AIJ$ so $A(IJOK)$ = $-$ 1 or ($FEOK$) = $-$ 1
Then: $AO$, $EN$, $FM$ concurrent

[quote=khanhnx]
Then: $I$, $J$, $G$, $O$ lie on a circle
[/quote]
Can anyone tell me why we have it?

Let $H$ be the projectionf of $A$ on $EF$ and $X$ the intersection of $EF$ with $MN$. Clearly $AH$ is the pole of $D$, thus $AH$ is the $A$-symmedian in both $\triangle{ABC}$ and $\triangle{AMN}$. Note that $AMHON$ is cyclic.
$$-1=O(A,H;M,N)=A(AO\cap MN , X;M,N)=(OX,EF) \ \ \ \blacksquare$$


[quote="t0rajir0u"]$(3, 4, 5)$.  But I have no proof that there are no other solutions.  =/[/quote]
Well, since all terms are similar, there are at least $3!=6$ solutions, being:
$\\(3,4,5)\\ (3,5,4)\\ (4,3,5)\\ (4,5,3)\\ (5,3,4)\\ (5,4,3)$

Masoud Zargar

Well, it's implied I meant $(3, 4, 5)$ and permutations.   :lol:

[quote="pkothari13"][quote="jimhu"][hide]
Let $a=1+\frac{1}{x}, b=1+\frac{1}{y}$, and $c=1+\frac{1}{z}$
If $x,y,z$ are positive integers, then $a,b,c>1$.
$a,b$ and $c$ are represented as $\frac{n+1}{n}$, where n is an arbitrary number.
Say the numbers are being multiplied in an order, $n_1, n_2$ and $n_3$ being the 1st,2nd and 3rd multiples, respectively.
For the product of all $n_1,n_2$ and $n_3$ to be 2, the the denominator of $n_3$ has to be equal to the numerator of of $n_1*n_2$ and the numberator of $n_3$ has to be 2 times the denominator of $n_1*n_2$.

Because $n+3=2n$, -> $n=3$,

The only set of rational numbers ${n_1,n_2,n_3}$ that can work is $(\frac{4}{3},\frac{5}{4},\frac{6}{5})$.

Therefore we have solution set: $(3,4,5),(3,5,4),(4,3,5),(4,5,3),(5,3,4),(5,4,3).$

QED
[/hide][/quote]

When you do this, aren't you assuming that $x,y,$ and $z$ are consecutive integers?[/quote]

I'm not sure your meaning. The thing is that $a,b, c$ don't have to be consecutive integers, but that was was the case that i investigated. There were 5 other cases, which I can also find by rearranging the terms in $(3,4,5)$.

[quote="pkothari13"]
When you do this, aren't you assuming that $x,y,$ and $z$ are consecutive integers?[/quote]

Don't they have to be consecutive since the numerators and denominaters must cancel? And since the answer must be $2$, the largest of the integers plus $1$ must go into the smallest twice, so that has to be the only solution set...

[quote="nawram88"]sorry for you all
 the solutions are permutations of

(2,4,15) and (2,5,9) and (2,6,7)

if you want i give you the solution[/quote]

Please do so (using Latex)

[quote="jimhu"][quote="nawram88"]sorry for you all
 the solutions are permutations of

(2,4,15) and (2,5,9) and (2,6,7)

if you want i give you the solution[/quote]

Er.. I dont think those work..[/quote]

Oh, they work alright. Be careful with your arithmetic.

[quote="pkothari13"]Find all solutions in positive integers $(x,y,z)$ to the expression

$\displaystyle \left (1+\frac{1}{x} \right ) \left (1+\frac{1}{y}\right ) \left (1+\frac{1}{z}\right ) = 2$[/quote]
WLOG x>=y>=z.
Thus x must be 4 or less since the cbrt of 2 is greater than 5/4. By the same token z must be greater than 4.Thus, if x=4, then the maximum value of the expression is 1.875

Thus x cant equal 4
now if x =3, then the largest possible sums are (x=3,y=3,z=3), (x=3,y=3, z=4),  (x=3, y=3, z=5), (x=3,y=4,z=4), (x=3, y=4, z=5) in order of smallest to largest
every other possible value of the expression when x=3 is larger.
if x=2, then (1+1/z)(1+1/y)=4/3
and y and z are greater than 3, since 3/2*4/3 is already 2.
or 3(y+1)(z+1)=4yz.
3zy+3y+3z+3=4yz
yz-3y-3z-3=0
Thus (y-3)(z-3)=12.
since y-3 and z-3 must be positive
we can factor 12 into 2 factors in 3 ways
(1, 12)(2,6)(3,4).
Thus if x=2, then (2, 4, 15)(2,5,9) (2, 6, 7)
thus our solutions are (2,4,15)(2,5,9)(2,6,7) and (3,4,5)
and those must be the only ones

problem was on an England national olympiad 1995, wasn't it?
classical problem for estimations.
I'll post another one


On the trip home from the meeting where this AMC$ 10$ was constructed, the Contest Chair noted that his airport parking receipt had digits of the form $ bbcac$, where $ 0 \le a < b < c \le 9$, and $ b$ was the average of $ a$ and $ c$. How many different five-digit numbers satisfy all these properties?

$ \textbf{(A)}\ 12\qquad
\textbf{(B)}\ 16\qquad
\textbf{(C)}\ 18\qquad
\textbf{(D)}\ 20\qquad
\textbf{(E)}\ 24$

[hide]
You can choose $ a$ from 0 to 7.
As $ c$ has to have the same parity as $ a$ (so that the average is an integer), there are 4 + 4 + 3 + 3 + 2 + 2 + 1 + 1 possibilities for C. That amounts to $ \boxed{\textbf{(D)}\ 20}$ numbers.
[/hide]

[hide="Click for solution"]
Once we pick $ a$ and $ c$, $ b$ is uniquely determined.  Note that $ a$ and $ c$ must be of the same parity.  $ a\equal{}0 \implies c\equal{}2,4,6,8$, $ a\equal{}2 \implies c\equal{}4,6,8$, $ a\equal{}4 \implies c\equal{}6,8$, and $ a\equal{}6 \implies c\equal{}8$.  This totals to $ 4\plus{}3\plus{}2\plus{}1\equal{}10$ possibilities where $ a$ is even.  We do the same thing when $ a$ is odd, and it is still $ 4\plus{}3\plus{}2\plus{}1\equal{}10$, so our answer is $ 10\plus{}10\equal{}20$, or $ \boxed{\textbf{(D)}}$.
[/hide]

[hide=" a different solution"]

find a value of b instead of a, and c.

let b= 1 then a must be 0.
let b = 2, then a must equal 0,1. (we dont worry about c because it will "automatically choose itselfe" :)  if you know what i mean)

let b = 3, a = 0,1,2
.      .
.      .
.      .

b= 8, a =7

we notice that b=1 is similar to b=8, because there is only 1 solution for a, similarly so are (7,2) ,(6,3), (5,4). so we just find b= 1, b=2, b=3, and b=4, find the total and multiply by 2
we get our answer to be (1+2+3+4)*2=20

[/hide]

(takes a while to explain, but is fast with paper and pencil)

[hide="non-bashing solution"]
If we pick numbers $ a,c$ then there will only be an intiger average if $ a,c$ are both odd or both even.
So there are $ \binom{10}{2}$ ways to pick two numbers $ a$ and $ c$. and the probability that they will both be odd or even is $ 2(\frac {5}{10}\cdot \frac {4}{9}) \equal{} \frac {4}{9}$

So our answer is $ \binom{10}{2}\cdot \frac {4}{9} \equal{} 20 \longrightarrow \boxed{D}$[/hide]


This was inspired by another recent post on perfect numbers:
Prove that if $n$ is of the form $(2k+1)^{2}$ for $k = 1,2,3,\cdots$, then $n$ cannot be a perfect number.

[hide]
Perfect squares have an odd number of divisors, and thus an even number of proper divisors. Each divisor must be odd, since $(2k+1)^{2}$ is odd. Thus, the sum of the proper divisors of the number is even, and thus is not equal to $(2k+1)^{2}$.
[/hide]


modular arithmetic
function
algorithm
logarithms
algebra
polynomial
number theory

So we know by Fermat's Little Theorem and Euler's extension to it that
$ a^{\phi(p)} \equiv 1 \pmod p$
However, under what conditions do we have $ a^m \equiv 1 \pmod p$ for $ a < \phi(p)$? Hopefully this question isn't completely trivial as it doesn't seem difficult and I'm sure there have been results on it... but I wasn't able to come up with anything, other than that $ m|\phi(p)$.

I think its true for all $ m$ such that $ m \equal{} k \cdot O(p)$ where O(p) is the Eulerian function of p i.e. the smallest value for which $ a^{O(p)} \equiv 1 \mod p$ :maybe:

$ O(p)$ is usually called the order of $ a$ modulo $ p$. m is indeed a multiple of the order. However, m does not need to divide $ \phi(p)$  - simple counterexample - it can exceed $ \phi(p)$.

The order of $ a \mod n$ always divides $ \varphi(n)$. In fact it divides the even smaller number $ \lambda(n)$ (Carmichael function).

It's surely not written $ O(p)$ by the way, because it strongly depends on $ a$. Something like $ \mathrm{ord}_p(a)$ or $ o_p(a)$ would be more usefull.
You can find it by "trying" all divisors of $ \varphi(p)$, or by iteratively calculating $ a,a^2,a^3, ...$. There are not very much general results, just some partial properties.

Algorithmic methods - [url=http://en.wikipedia.org/wiki/Exponentiation_by_squaring]exponentiation by squaring[/url], for example (for large $ p$) - are available, but a general method would probably imply [url=http://en.wikipedia.org/wiki/Artin%39s_conjecture_on_primitive_roots]Artin's conjecture[/url] (I think it was you who told me that, ZetaX :) ).

This algorithm is very useful (and up to constant the best way) to check whether $ a^m \equiv 1 \mod n$ for some given $ m$. The problem is, that we want to find the smallest positive such $ m$, so we have to do this very often.
The problem itself is called "finding discrete logarithms" and no polynomial time algorithm is known (as exponentiation can be done much faster, it gives raise to some cryptosystems).