Find all positive integers $n$ such that the equation $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ has exactly $2011$ positive integer solutions $(x,y)$ where $x \leq y$.

The diagonals $AC,BD$ of the quadrilateral $ABCD$ intersect at $E$. Let $M,N$ be the midpoints of $AB,CD$ respectively. Let the perpendicular bisectors of the segments $AB,CD$ meet at $F$. Suppose that $EF$ meets $BC,AD$ at $P,Q$ respectively. If $MF\cdot CD=NF\cdot AB$ and $DQ\cdot BP=AQ\cdot CP$, prove that $PQ\perp BC$.

The positive reals $a,b,c,d$ satisfy $abcd=1$. Prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{9}{{a + b + c + d}} \geqslant \frac{{25}}{4}$.

A tennis tournament has $n>2$ players and any two players play one game against each other (ties are not allowed). After the game these players can be arranged in a circle, such that for any three players $A,B,C$, if $A,B$ are adjacent on the circle, then at least one of $A,B$ won against $C$. Find all possible values for $n$.

A real number $\alpha \geq 0$ is given. Find the smallest $\lambda = \lambda (\alpha ) > 0$, such that for any complex numbers ${z_1},{z_2}$ and $0 \leq x \leq 1$, if $\left| {{z_1}} \right| \leq \alpha \left| {{z_1} - {z_2}} \right|$, then $\left| {{z_1} - x{z_2}} \right| \leq \lambda \left| {{z_1} - {z_2}} \right|$.

Do there exist positive integers $m,n$, such that $m^{20}+11^n$ is a square number?

There are $n$ boxes ${B_1},{B_2},\ldots,{B_n}$ from left to right, and there are $n$ balls in these boxes. If there is at least $1$ ball in ${B_1}$, we can move one to ${B_2}$. If there is at least $1$ ball in ${B_n}$, we can move one to ${B_{n - 1}}$. If there are at least $2$ balls in ${B_k}$, $2 \leq k \leq n - 1$ we can move one to ${B_{k - 1}}$, and one to ${B_{k + 1}}$. Prove that, for any arrangement of the $n$ balls, we can achieve that each box has one ball in it.

The $A$-excircle $(O)$ of $\triangle ABC$ touches $BC$ at $M$. The points $D,E$ lie on the sides $AB,AC$ respectively such that $DE\parallel BC$. The incircle $(O_1)$ of $\triangle ADE$ touches $DE$ at $N$. If $BO_1\cap DO=F$ and $CO_1\cap EO=G$, prove that the midpoint of $FG$ lies on $MN$.