Let $a$ and $b$ be positive integers such that $a! + b!$ divides $a!b!$. Prove that $3a \ge 2b + 2$.

Find all function $f:\mathbb{Z}\rightarrow\mathbb{Z}$ such that $f(f(x)+f(y))+f(x)f(y)=f(x+y)f(x-y)$ for all integer $x,y$

Let $O$ be the circumcenter of triangle $ABC$, and $\omega$ be the circumcircle of triangle $BOC$. Line $AO$ intersects with circle $\omega$ again at the point $G$. Let $M$ be the midpoint of side $BC$, and the perpendicular bisector of $BC$ meets circle $\omega$ at the points $O$ and $N$. Prove that the midpoint of the segment $AN$ lies on the radical axis of the circumcircle of triangle $OMG$, and the circle whose diameter is $AO$.

Let $\left< F_n\right>$ be the Fibonacci sequence, that is, $F_0=0$, $F_1=1$, and $F_{n+2}=F_{n+1}+F_{n}$ holds for all nonnegative integers $n$. Find all pairs $(a,b)$ of positive integers with $a < b$ such that $F_n-2na^n$ is divisible by $b$ for all positive integers $n$.

In Lineland there are $n\geq1$ towns, arranged along a road running from left to right. Each town has a left bulldozer (put to the left of the town and facing left) and a right bulldozer (put to the right of the town and facing right). The sizes of the $2n$ bulldozers are distinct. Every time when a left and right bulldozer confront each other, the larger bulldozer pushes the smaller one off the road. On the other hand, bulldozers are quite unprotected at their rears; so, if a bulldozer reaches the rear-end of another one, the first one pushes the second one off the road, regardless of their sizes. Let $A$ and $B$ be two towns, with $B$ to the right of $A$. We say that town $A$ can sweep town $B$ away if the right bulldozer of $A$ can move over to $B$ pushing off all bulldozers it meets. Similarly town $B$ can sweep town $A$ away if the left bulldozer of $B$ can move over to $A$ pushing off all bulldozers of all towns on its way. Prove that there is exactly one town that cannot be swept away by any other one.

Let $x,y$ be positive real numbers such that $x+y=1$. Prove that$\frac{x}{x^2+y^3}+\frac{y}{x^3+y^2}\leq2(\frac{x}{x+y^2}+\frac{y}{x^2+y})$.

Let $ABC$ be an acute triangle with orthocenter $H$. Let $G$ be the point such that the quadrilateral $ABGH$ is a parallelogram. Let $I$ be the point on the line $GH$ such that $AC$ bisects $HI$. Suppose that the line $AC$ intersects the circumcircle of the triangle $GCI$ at $C$ and $J$. Prove that $IJ = AH$.

Let $m$ and $n$ be positive integers such that $m>n$. Define $x_k=\frac{m+k}{n+k}$ for $k=1,2,\ldots,n+1$. Prove that if all the numbers $x_1,x_2,\ldots,x_{n+1}$ are integers, then $x_1x_2\ldots x_{n+1}-1$ is divisible by an odd prime.

There is a grid of equilateral triangles with a distance 1 between any two neighboring grid points. An equilateral triangle with side length $n$ lies on the grid so that all of its vertices are grid points, and all of its sides match the grid. Now, let us decompose this equilateral triangle into $n^2$ smaller triangles (not necessarily equilateral triangles) so that the vertices of all these smaller triangles are all grid points, and all these small triangles have equal areas. Prove that there are at least $n$ equilateral triangles among these smaller triangles.

Suppose that a sequence $a_1,a_2,\ldots$ of positive real numbers satisfies \[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]for every positive integer $k$. Prove that $a_1+a_2+\ldots+a_n\geq n$ for every $n\geq2$.

Let $n$ be a positive integer. Two players $A$ and $B$ play a game in which they take turns choosing positive integers $k \le n$. The rules of the game are: (i) A player cannot choose a number that has been chosen by either player on any previous turn. (ii) A player cannot choose a number consecutive to any of those the player has already chosen on any previous turn. (iii) The game is a draw if all numbers have been chosen; otherwise the player who cannot choose a number anymore loses the game. The player $A$ takes the first turn. Determine the outcome of the game, assuming that both players use optimal strategies. Proposed by Finland

Let $AXYZB$ be a convex pentagon inscribed in a semicircle with diameter $AB$, and let $K$ be the foot of the altitude from $Y$ to $AB$. Let $O$ denote the midpoint of $AB$ and $L$ be the intersection of $XZ$ with $YO$. Select a point $M$ on line $KL$ with $MA=MB$ , and finally, let $I$ be the reflection of $O$ across $XZ$. Prove that if quadrilateral $XKOZ$ is cyclic then so is quadrilateral $YOMI$. Proposed by Evan Chen