In triangle $\triangle ABC$ , $M, N$ are midpoints of $AC,AB$ respectively. Assume that $BM,CN$ cuts $(ABC)$ at $M',N'$ respectively. Let $X$ be on the extention of $BC$ from $B$ st $\angle N'XB=\angle ACN$. And define $Y$ similarly on the extention of $BC$ from $C$. Prove that $AX=AY$.

Does there exist bijections $f,g$ from positive integers to themselves st: $$g(n)=\frac{f(1)+f(2)+ \cdot \cdot \cdot +f(n)}{n}$$holds for any $n$?

For each $k$ , find the least $n$ in terms of $k$ st the following holds: There exists $n$ real numbers $a_1 , a_2 ,\cdot \cdot \cdot , a_n$ st for each $i$ : $$0 < a_{i+1} - a_{i} < a_i - a_{i-1}$$And , there exists $k$ pairs $(i,j)$ st $a_i - a_j = 1$.

For any function $f:\mathbb{N}\to\mathbb{N}$ we define $P(n)=f(1)f(2)...f(n)$ . Find all functions $f:\mathbb{N}\to\mathbb{N}$ st for each $a,b$ : $$P(a)+P(b) | a! + b!$$

There is $n$ black points in the plane.We do the following algorithm: Start from any point from those $n$ points and colour it red. Then connect this point to the nearest black point available and colour this new point red. Then do the same with this point but at any step , but you are never allowed to draw a line which intersects on of the current drawn segments. If you reach an intersection , the algorithm is over. Is it true that for any $n$ and at any initial position , we can start from a point st in the algorithm , we reach all the points?

In the acute triangle $\triangle ABC$ , $H$ is the orthocenter. $S$ is a point on $(AHC)$ st $\angle ASB = 90$. $P$ is on $AC$ and not on the extention of $AC$ from $A$ , st $\angle APS=\angle BAS$.Prove that $CS$ , the circle $(BPC)$ and the circle with diameter $AC$ are concurrent.

Find all integers $n > 4$ st for every two subsets $A,B$ of $\{0,1,....,n-1\}$ , there exists a polynomial $f$ with integer coefficients st either $f(A) = B$ or $f(B) = A$ where the equations are considered mod n. We say two subsets are equal mod n if they produce the same set of reminders mod n. and the set $f(X)$ is the set of reminders of $f(x)$ where $x \in X$ mod n.

Let $N$ be the number of ordered pairs $(x,y)$ st $1 \leq x,y \leq p(p-1)$ and : $$x^{y} \equiv y^{x} \equiv 1 \pmod{p}$$where $p$ is a fixed prime number. Show that : $$(\phi {(p-1)}d(p-1))^2 \leq N \leq ((p-1)d(p-1))^2$$where $d(n)$ is the number of divisors of $n$

Let $K$ be an odd number st $S_2{(K)} = 2$ and let $ab=K$ where $a,b$ are positive integers. Show that if $a,b>1$ and $l,m >2$ are positive integers st:$S_2{(a)} < l$ and $S_2{(b)} < m$ then : $$K \leq 2^{lm-6} +1$$($S_2{(n)}$ is the sum of digits of $n$ written in base 2)

Let $n$ and $a \leq n$ be two positive integers. There's $2n$ people sitting around a circle reqularly. Two people are friend iff one of their distance in the circle is $a$(that is , $a-1$ people are between them). Find all integers $a$ in terms of $n$ st we can choose $n$ of these people , no two of them positioned in front of each other(means they're not antipodes of each other in the circle) and the total friendship between them is an odd number.

Find the number of permutations of $\{1,2,...,n\}$ like $\{a_1,...,a_n\}$ st for each $1 \leq i \leq n$: $$a_i | 2i$$

There's infinity of the following blocks on the table:$1*1 , 1*2 , 1*3 ,.., 1*n$. We have a $n*n$ table and Ali chooses some of these blocks so that the sum of their area is at least $n^2$. Then , Amir tries to cover the $n*n$ table so that none of blocks go out of the table and they don't overlap and he wanna maximize the covered area in the $n*n$ table with those blocks chosen by Ali. Let $k$ be the maximum coverable area independent of Ali's choice. Prove that: $$n^2 - \lceil \frac{n^2}{4} \rceil \leq k \leq n^2 - \lfloor \frac{n^2}{8} \rfloor$$ *Note : the blocks can be placed only vertically or horizontally.

Given $12$ complex numbers $z_1,...,z_{12}$ st for each $1 \leq i \leq 12$: $$|z_i|=2 , |z_i - z_{i+1}| \geq 1$$prove that : $$\sum_{1 \leq i \leq 12} \frac{1}{|z_i\overline{z_{i+1}}+1|^2} \geq \frac{1}{2}$$

find all $f : \mathbb{C} \to \mathbb{C}$ st: $$f(f(x)+yf(y))=x+|y|^2$$for all $x,y \in \mathbb{C}$

For numbers $a,b \in \mathbb{R}$ we consider the sets: $$A=\{a^n | n \in \mathbb{N}\} , B=\{b^n | n \in \mathbb{N}\}$$Find all $a,b > 1$ for which there exists two real , non-constant polynomials $P,Q$ with positive leading coefficients st for each $r \in \mathbb{R}$: $$ P(r) \in A \iff Q(r) \in B$$

In triangle $\triangle ABC$ , $I$ is the incenter and $M$ is the midpoint of arc $(BC)$ in the circumcircle of $(ABC)$not containing $A$. Let $X$ be an arbitrary point on the external angle bisector of $A$. Let $BX \cap (BIC) = T$. $Y$ lies on $(AXC)$ , different from $A$ , st $MA=MY$ . Prove that $TC || AY$ (Assume that $X$ is not on $(ABC)$ or $BC$)

In triangle $\triangle ABC$ , $M$ is the midpoint of arc $(BAC)$ and $N$ is the antipode of $A$ in $(ABC)$. The line through $B$ perpendicular to $AM$ , intersects $AM , (ABC)$ at $D,P$ respectively and a line through $D$ perpendicular to $AC$ , intersects $BC,AC$ at $F,E$ respectively. Prove that $PE,MF,ND$ are concurrent.

In triangle $\triangle ABC$ points $M,N$ lie on $BC$ st : $\angle BAM= \angle MAN= \angle NAC$ . Points $P,Q$ are on the angle bisector of $BAC$, on the same side of $BC$ as A , st : $$\frac{1}{3} \angle BAC = \frac{1}{2} \angle BPC = \angle BQC$$Let $E = AM \cap CQ$ and $F = AN \cap BQ$ . Prove that the common tangents to $(EPF), (EQF)$ and the circumcircle of $\triangle ABC$ , are concurrent.