In triangle $ABC$, $M$ is midpoint of $AC$, and $D$ is a point on $BC$ such that $DB=DM$. We know that $2BC^{2}-AC^{2}=AB.AC$. Prove that \[BD.DC=\frac{AC^{2}.AB}{2(AB+AC)}\]
2007 Iran Team Selection Test
Day 1
Let $A$ be the largest subset of $\{1,\dots,n\}$ such that for each $x\in A$, $x$ divides at most one other element in $A$. Prove that \[\frac{2n}3\leq |A|\leq \left\lceil \frac{3n}4\right\rceil. \]
Find all solutions of the following functional equation: \[f(x^{2}+y+f(y))=2y+f(x)^{2}. \]
Day 2
In an isosceles right-angled triangle shaped billiards table , a ball starts moving from one of the vertices adjacent to hypotenuse. When it reaches to one side then it will reflect its path. Prove that if we reach to a vertex then it is not the vertex at initial position By Sam Nariman
Find all monic polynomials $f(x)$ in $\mathbb Z[x]$ such that $f(\mathbb Z)$ is closed under multiplication. By Mohsen Jamali
Let $\omega$ be incircle of $ABC$. $P$ and $Q$ are on $AB$ and $AC$, such that $PQ$ is parallel to $BC$ and is tangent to $\omega$. $AB,AC$ touch $\omega$ at $F,E$. Prove that if $M$ is midpoint of $PQ$, and $T$ is intersection point of $EF$ and $BC$, then $TM$ is tangent to $\omega$. By Ali Khezeli
Day 3
Does there exist a a sequence $a_{0},a_{1},a_{2},\dots$ in $\mathbb N$, such that for each $i\neq j, (a_{i},a_{j})=1$, and for each $n$, the polynomial $\sum_{i=0}^{n}a_{i}x^{i}$ is irreducible in $\mathbb Z[x]$? By Omid Hatami
Suppose $n$ lines in plane are such that no two are parallel and no three are concurrent. For each two lines their angle is a real number in $[0,\frac{\pi}2]$. Find the largest value of the sum of the $\binom n2$ angles between line. By Aliakbar Daemi
$O$ is a point inside triangle $ABC$ such that $OA=OB+OC$. Suppose $B',C'$ be midpoints of arcs $AOC$ and $AOB$. Prove that circumcircles $COC'$ and $BOB'$ are tangent to each other. Click to reveal hidden textwording fixed
Day 4
Find all polynomials of degree 3, such that for each $x,y\geq 0$: \[p(x+y)\geq p(x)+p(y)\]
Triangle $ABC$ is isosceles ($AB=AC$). From $A$, we draw a line $\ell$ parallel to $BC$. $P,Q$ are on perpendicular bisectors of $AB,AC$ such that $PQ\perp BC$. $M,N$ are points on $\ell$ such that angles $\angle APM$ and $\angle AQN$ are $\frac\pi2$. Prove that \[\frac{1}{AM}+\frac1{AN}\leq\frac2{AB}\]
Click for solution I have a nice and short metric solution, so here it goes: We have that $AN=\frac{AQ}{\cos{QAN}}$ and $AM=\frac{AP}{\cos{PAM}}$. But $AQ=\frac{x}{\cos{QAC}}$ and $AP=\frac{x}{\cos{PAB}}$, where $x=\frac{AB}{2}$. So $AM=\frac{x}{\cos{PAM}\cdot \cos{PAB}}$ and $AN=\frac{x}{\cos{QAN}\cdot\cos{QAC}}$, so the inequality rewrites as: $\cos{QAC}\cdot \cos{QAN}+\cos{PAM}\cdot \cos{PAB}\leq 1$ $\iff \cos{B}+\cos{(QAP-\frac{A}{2})}+\cos{BAM}+\cos{(QAP-\frac{A}{2})}\leq 2$. But $\angle{B}=180-\angle{BAM}$, therefore $\cos{B}=-\cos{BAM}$, thus the inequality rewrites as $\cos{(QAP-\frac{A}{2})}\leq 1$, which is true.
Let $P$ be a point in a square whose side are mirror. A ray of light comes from $P$ and with slope $\alpha$. We know that this ray of light never arrives to a vertex. We make an infinite sequence of $0,1$. After each contact of light ray with a horizontal side, we put $0$, and after each contact with a vertical side, we put $1$. For each $n\geq 1$, let $B_{n}$ be set of all blocks of length $n$, in this sequence. a) Prove that $B_{n}$ does not depend on location of $P$. b) Prove that if $\frac{\alpha}{\pi}$ is irrational, then $|B_{n}|=n+1$.