Let $n$ be a positive integer. Each point $(x,y)$ in the plane, where $x$ and $y$ are non-negative integers with $x+y<n$, is coloured red or blue, subject to the following condition: if a point $(x,y)$ is red, then so are all points $(x',y')$ with $x'\leq x$ and $y'\leq y$. Let $A$ be the number of ways to choose $n$ blue points with distinct $x$-coordinates, and let $B$ be the number of ways to choose $n$ blue points with distinct $y$-coordinates. Prove that $A=B$.

Click for solution This was actually in the contest, wasn't it? It's kind of hard to put into words. The red points consists of a series of $n$ columns of decreasing height (I mean decreasing starting from left to right, and I don't mean strictly decreasing). The blue points consist of the completion of the red columns up to the border of the triangle having vertices $(0,0),(0,n-1),(n-1,0)$. At the same time, the blue and red sets can also be regarded as rows instead of columns. $A$ is simply the product of the lengths of all blue columns, while $B$ is the product of the lengths of all blue rows. If we show that the same numbers, with the same multiplicities, appear in the collections of a) blue column lengths and b) blue row lengths, then we are clearly done. Call this assertion $(*)$. The blue set can be regarded as the union of some maximal right triangles with the legs parallel to the axes of the plane (maximal in the sense that they aren't part of any other similar right triangles). The triangles are obtained as follows: from a point on the diagonal $x+y=n-1$ go down as far as you can without taking any red points. Now go to the right until you meet the diagonal again. The maximal triangles of this set are the ones we're looking for. We can now prove the assertion $(*)$ by induction on $n$. If there are points on the diagonal $x+y=n-1$ which are red, then it's easy to see that the number of blue columns of length $0$ is equal to the number of blue rows of length $0$, and we can look at the picture as the union of some smaller pictures (for smaller $n$'s), by breaking the diagonal where it has red points. We use the induction hypothesis on these smaller pictures. On the other hand, if no points on $x+y=n-1$ are red, then we can look at the points strictly below the diagonal $x+y=n-1$, and use the induction hypothesis on these. After we append the diagonal $x+y=n-1$ all lengths (of blue columns and rows) increase by $1$, so we have shown what we wanted: the number of blue columns of length $t$ is the same as the number of blue rows of length $t$ for all $0\le t\le n$. Watch out: I might have reversed "red" and "blue" in some places.

The circle $S$ has centre $O$, and $BC$ is a diameter of $S$. Let $A$ be a point of $S$ such that $\angle AOB<120{{}^\circ}$. Let $D$ be the midpoint of the arc $AB$ which does not contain $C$. The line through $O$ parallel to $DA$ meets the line $AC$ at $I$. The perpendicular bisector of $OA$ meets $S$ at $E$ and at $F$. Prove that $I$ is the incentre of the triangle $CEF.$

Click for solution Triangles $AOE,AOF$ are equilateral because $AE=OE=OA,\ AF=OF=OA$. The condition $\angle AOB<120^{\circ}\iff \angle AOC>60^{\circ}$ simply means that $F$ (assume $F$ is on the same side of $OA$ as $C$) lies on the arc $AC$ which doesn't contain $E$, so $CA$ is the internal bisector of $\angle ECF$ because $A$ is the midpoint of the arc $EF$ (without the condition, $AC$ would be the external bisector). This means that $J$ lies on the bisector of $\angle ECF$. On the other hand, we have $AD\|OJ$ and $OD\|AJ$ because they're both $\perp AB$, so $J$ is the reflection of $D$ in the midpoint of $OA$. This means $\angle EJF=\angle EDF=120^{\circ}$, and since $\angle ECF=\frac{\angle EOF}2=60^{\circ}$ and $J$ lies on the internal bisector of $\angle ECF$, it means that $J$ is the incenter (the only other position on $AC$ from which $EF$ is seen under an angle of $120^{\circ}$ is $A$, and it's clear that $J\ne A$).

Find all pairs of positive integers $m,n\geq3$ for which there exist infinitely many positive integers $a$ such that \[ \frac{a^m+a-1}{a^n+a^2-1} \] is itself an integer. Laurentiu Panaitopol, Romania

Let $n\geq2$ be a positive integer, with divisors $1=d_1<d_2<\,\ldots<d_k=n$. Prove that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ is always less than $n^2$, and determine when it is a divisor of $n^2$.

Click for solution $d_k = {n\over d_1}$ $d_{k-1} = {n\over d_2}$ $d_1 = {n\over d_k}$ So we can write it like this: $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}$ The Maximum that can be is: ${{n^2}\over 1*2}+{{n^2}\over 2*3}+\,\ldots\,+{{n^2}\over (n-1)n}$ We can see that the sum starting with the form: ${m\over {m+1}}{n^2}$ We check what happens if we add the next term: ${m\over {m+1}}{n^2} + {1\over {(m+1)(m+2)}}{n^2} = {{(m+1)^2}\over {(m+1)(m+2)}}{n^2} = {{m+1}\over {m+2}}{n^2}$ From here is easy to define that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}<{n^2}$ $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}$ is a divisor of ${n^2}$ when $n$ is a prime number: $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k} = n$ if $n$ isn't a prime we have at least 3 divisors of $n$ We proved that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}<{n^2}$ So if it's a divisor of ${n^2}$ the maximum that can be: ${{n^2}\over d_2}$ $d_1 = 1$ $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k} = {{n^2}\over d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k} > {{n^2}\over d_2}$ We see from here that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ is a divisor of ${n^2}$ only if $n$ is a prime!

Find all functions $f$ from the reals to the reals such that \[ \left(f(x)+f(z)\right)\left(f(y)+f(t)\right)=f(xy-zt)+f(xt+yz) \] for all real $x,y,z,t$.

Let $n\geq3$ be a positive integer. Let $C_1,C_2,C_3,\ldots,C_n$ be unit circles in the plane, with centres $O_1,O_2,O_3,\ldots,O_n$ respectively. If no line meets more than two of the circles, prove that \[ \sum\limits^{}_{1\leq i<j\leq n}{1\over O_iO_j}\leq{(n-1)\pi\over 4}. \]