Given triangle $ABC$, $D$ is the foot of the external angle bisector of $A$, $I$ its incenter and $I_a$ its $A$-excenter. Perpendicular from $I$ to $DI_a$ intersects the circumcircle of triangle in $A'$. Define $B'$ and $C'$ similarly. Prove that $AA',BB'$ and $CC'$ are concurrent. [i]proposed by Amirhossein Zabeti[/i]

Suppose that each of the diagonals $AD,BE,CF$ divides the hexagon $ABCDEF$ into two parts of the same area and perimeter. Does the hexagon necessarily have a center of symmetry?

Let $n!=n\cdot (n-1)\cdot (n-2)\cdot \ldots \cdot 2\cdot 1$.For example, $5! = 5\cdot 4\cdot 3 \cdot 2\cdot 1 = 120.$ Compute $\frac{(6!)^2}{5!\cdot 7!}$.

With expressions containing the symbol $*$, the following transformations can be performed: 1) rewrite the expression in the form $x * (y * z) as ((1 * x) * y) * z$; 2) rewrite the expression in the form $x * 1$ as $x$. Conversions can only be performed with an integer expression, but not with its parts. For example, $(1 *1) * (1 *1)$ can be rewritten according to the first rule as $((1 * (1 * 1)) * 1) * 1$ (taking $x = 1 * 1$, $y = 1$ and $z = 1$), but not as $1 * (1 * 1)$ or $(1* 1) * 1$ (in the last two cases, the second rule would be applied separately to the left or right side $1 * 1$). Find all positive integers $n$ for which the expression $\underbrace{1 * (1 * (1 * (...* (1 * 1)...))}_{n units}$ it is possible to lead to a form in which there is not a single asterisk. Note. The expressions $(x * y) * $z and $x * (y * z)$ are considered different, also, in the general case, the expressions $x * y$ and $y * x$ are different.

$a_0 = a_1 = 1$ and ${a_{n+1} . a_{n-1}} = a_n . (a_n + 1)$ for all positive integers n. prove that $a_n$ is one integer for all positive integers n.

Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2=3. $ Prove that $$\frac{a^4+3ab^3}{a^3+2b^3}+\frac{b^4+3bc^3}{b^3+2c^3}+\frac{c^4+3ca^3}{c^3+2a^3}\leq4.$$

Given an integer $n>1$ and a $n\times n$ grid $ABCD$ containing $n^2$ unit squares, each unit square is colored by one of three colors: Black, white and gray. A coloring is called [i]symmetry[/i] if each unit square has center on diagonal $AC$ is colored by gray and every couple of unit squares which are symmetry by $AC$ should be both colred by black or white. In each gray square, they label a number $0$, in a white square, they will label a positive integer and in a black square, a negative integer. A label will be called $k$-[i]balance[/i] (with $k\in\mathbb{Z}^+$) if it satisfies the following requirements: i) Each pair of unit squares which are symmetry by $AC$ are labelled with the same integer from the closed interval $[-k,k]$ ii) If a row and a column intersectes at a square that is colored by black, then the set of positive integers on that row and the set of positive integers on that column are distinct.If a row and a column intersectes at a square that is colored by white, then the set of negative integers on that row and the set of negative integers on that column are distinct. a) For $n=5$, find the minimum value of $k$ such that there is a $k$-balance label for the following grid [asy] size(4cm); pair o = (0,0); pair y = (0,5); pair z = (5,5); pair t = (5,0); dot("$A$", y, dir(180)); dot("$B$", z); dot("$C$", t); dot("$D$", o, dir(180)); fill((0,5)--(1,5)--(1,4)--(0,4)--cycle,gray); fill((1,4)--(2,4)--(2,3)--(1,3)--cycle,gray); fill((2,3)--(3,3)--(3,2)--(2,2)--cycle,gray); fill((3,2)--(4,2)--(4,1)--(3,1)--cycle,gray); fill((4,1)--(5,1)--(5,0)--(4,0)--cycle,gray); fill((0,3)--(1,3)--(1,1)--(0,1)--cycle,black); fill((2,5)--(4,5)--(4,4)--(2,4)--cycle,black); fill((2,1)--(3,1)--(3,0)--(2,0)--cycle,black); fill((2,1)--(3,1)--(3,0)--(2,0)--cycle,black); fill((4,3)--(5,3)--(5,2)--(4,2)--cycle,black); for (int i=0; i<=5; ++i) { draw((0,i)--(5,i)^^(i,0)--(i,5)); } [/asy] b) Let $n=2017$. Find the least value of $k$ such that there is always a $k$-balance label for a symmetry coloring.

Suppose that $a$ and $b$ are two distinct positive real numbers such that $\lfloor na\rfloor$ divides $\lfloor nb\rfloor$ for any positive integer $n$. Prove that $a$ and $b$ are positive integers.

Let $ \alpha$ be a rational number with $ 0 < \alpha < 1$ and $ \cos (3 \pi \alpha) \plus{} 2\cos(2 \pi \alpha) \equal{} 0$. Prove that $ \alpha \equal{} \frac {2}{3}$.

Mary chose an even $4$-digit number $n$. She wrote down all the divisors of $n$ in increasing order from left to right: $1,2,...,\tfrac{n}{2},n$. At some moment Mary wrote $323$ as a divisor of $n$. What is the smallest possible value of the next divisor written to the right of $323$? $\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646$

Let $a,b,x,y$ be positive numbers with $a+b+x+y < 2$. Given that $$\begin{cases} a+b^2 = x+y^2 \\ a^2 +b = x^2 +y\end {cases} $$ show that $a = x$ and $b = y$

The positive real $x, y, z$ are such that $x^2+y^2+z^2 = 3$. Prove that$$\frac{x^2+yz}{x^2+yz +1}+\frac{y^2+zx}{y^2+zx+1}+\frac{z^2+xy}{z^2+xy+1}\leq 2$$

For every positive integer $n$, determine the biggest positive integer $k$ so that $2^k |\ 3^n+1$

Three circles are pairwise tangent, with none of them lying inside another. The centres of the circles are the corners of a triangle with circumference $1$. What is the smallest possible value for the sum of the areas of the circles?

Let $ABC$ be a non-isosceles triangle with altitudes $AD$, $BE$, $CF$ with orthocenter $H$. Suppose that $DF \cap HB = M$, $DE \cap HC = N$ and $T$ is the circumcenter of triangle $HBC$. Prove that $AT\perp MN$.

Let $p$, $q$, and $r$ be primes satisfying \[ pqr = 189999999999999999999999999999999999999999999999999999962. \] Compute $S(p) + S(q) + S(r) - S(pqr)$, where $S(n)$ denote the sum of the decimals digits of $n$. [i]Proposed by Evan Chen[/i]