Compute the circumradius of cyclic hexagon $ABCDEF$, which has side lengths $AB=BC=2$, $CD=DE=9$, and $EF=FA=12$.

Call admissible a set $A$ of integers that has the following property: If $x,y \in A$ (possibly $x=y$) then $x^2+kxy+y^2 \in A$ for every integer $k$. Determine all pairs $m,n$ of nonzero integers such that the only admissible set containing both $m$ and $n$ is the set of all integers. [i]Proposed by Warut Suksompong, Thailand[/i]

Seventeen football fans were planning to go to Korea to watch the World Cup football match. They selected 17 matches. The conditions of the admission tickets they booked were such that - One person should book at most one admission ticket for one match; - At most one match was same in the tickets booked by every two persons; - There was one person who booked six tickets. How many tickets did those football fans book at most?

Find the smallest positive integer that uses exactly two different digits when written in decimal notation and is divisible by all the numbers from $1$ to $9$.

Given an acute triangle $ABC$ such that $\angle C< \angle B< \angle A$. Let $I$ be the incenter of $ABC$. Let $M$ be the midpoint of the smaller arc $BC$, $N$ be the midpoint of the segment $BC$ and let $E$ be a point such that $NE=NI$. The line $ME$ intersects circumcircle of $ABC$ at $Q$ (different from $A, B$, and $C$). Prove that [b](i)[/b] The point $Q$ is on the smaller arc $AC$ of circumcircle of $ABC$. [b](ii)[/b] $BQ=AQ+CQ$

[b]a) [/b]Is the number $ 1111\cdots11$ (with $ 2010$ ones) a prime number? [b]b)[/b] Prove that every prime factor of $ 1111\cdots11$ (with $ 2011$ ones) is of the form $ 4022j\plus{}1$ where $ j$ is a natural number.

Physicists at Princeton are trying to analyze atom entanglement using the following experiment. Originally there is one atom in the space and it starts splitting according to the following procedure. If after $n$ minutes there are atoms $a_1, \dots, a_N$, in the following minute every atom $a_i$ splits into four new atoms, $a_i^{(1)},a_i^{(2)},a_i^{(3)},a_i^{(4)}$. Atoms $a_i^{(j)}$ and $a_k^{(j)}$ are entangled if and only the atoms $a_i$ and $a_k$ were entangled after $n$ minutes. Moreover, atoms $a_i^{(j)}$ and $a_k^{(j+1)}$ are entangled for all $1 \le i$, $k \le N$ and $j = 1$, $2$, $3$. Therefore, after one minute there is $4$ atoms, after two minutes there are $16$ atoms and so on. Physicists are now interested in the number of unordered quadruplets of atoms $\{b_1, b_2, b_3, b_4\}$ among which there is an odd number of entanglements. What is the number of such quadruplets after $3$ minutes? [i]Remark[/i]. Note that atom entanglement is not transitive. In other words, if atoms $a_i$, $a_j$ are entangled and if $a_j$, $a_k$ are entangled, this does not necessarily mean that $a_i$ and $a_k$ are entangled.

Let $n$ be a positive integer. Prove that $a(n) = n^5 +5^n$ is divisible by $11$ if and only if $b(n) = n^5 · 5^n +1$ is divisible by $11$. [i](Walther Janous)[/i]

$ \frac{3 \times 5}{9 \times 11} \times \frac{7 \times 9 \times 11}{3 \times 5 \times 7}\equal{}$ \[ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ \frac{1}{49} \qquad \textbf{(E)}\ 50 \]

Let $P$ be a cyclic polygon with circumcenter $O$ that does not lie on any diagonal, and let $S$ be the set of points on 2D plane containing $P$ and $O$. The $\textit{Matcha Sweep Game}$ is a game between two players $A$ and $B$, with $A$ going first, such that each choosing a nonempty subset $T$ of points in $S$ that has not been previously chosen, and such that if $T$ has at least $3$ vertices then $T$ forms a convex polygon. The game ends with all points have been chosen, with the player picking the last point wins. For which polygons $P$ can $A$ guarantee a win? [i]Proposed by Anzo Teh Zhao Yang[/i]