EGMO 2012

Let $A B C$ be a triangle with circumcentre $O$. The points $D, E$ and $F$ lie in the interiors of the sides $B C, C A$ and $A B$ respectively, such that $D E$ is perpendicular to $C O$ and $D F$ is perpendicular to $B O$. (By interior we mean, for example, that the point $D$ lies on the line $B C$ and $D$ is between $B$ and $C$ on that line.)

Let $K$ be the circumcentre of triangle $A F E$. Prove that the lines $D K$ and $B C$ are perpendicular. Origin. Netherlands (Merlijn Staps).

Let $n$ be a positive integer. Find the greatest possible integer $m$, in terms of $n$, with the following property: a table with $m$ rows and $n$ columns can be filled with real numbers in such a manner that for any two different rows $\left[a_{1}, a_{2}, \ldots, a_{n}\right]$ and $\left[b_{1}, b_{2}, \ldots, b_{n}\right]$ the following holds:

$$\max \left(\left|a_{1}-b_{1}\right|,\left|a_{2}-b_{2}\right|, \ldots,\left|a_{n}-b_{n}\right|\right)=1$$

Origin. Poland (Tomasz Kobos).

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$f(y f(x+y)+f(x))=4 x+2 y f(x+y)$$

for all $x, y \in \mathbb{R}$.

Origin. Netherlands (Birgit van Dalen).

A set $A$ of integers is called sum-full if $A \subseteq A+A$, i.e. each element $a \in A$ is the sum of some pair of (not necessarily different) elements $b, c \in A$. A set $A$ of integers is said to be zero-sum-free if 0 is the only integer that cannot be expressed as the sum of the elements of a finite nonempty subset of $A$.

Does there exist a sum-full zero-sum-free set of integers?

Origin. Romania (Dan Schwarz). Remark. The original formulation of this problem had a weaker definition of zero-sum-free that did not require all nonzero integers to be sums of finite nonempty subsets of $A$.

The numbers $p$ and $q$ are prime and satisfy

$$\frac{p}{p+1}+\frac{q+1}{q}=\frac{2 n}{n+2}$$

for some positive integer $n$. Find all possible values of $q-p$. Origin. Luxembourg (Pierre Haas).

There are infinitely many people registered on the social network Mugbook. Some pairs of (different) users are registered as friends, but each person has only finitely many friends. Every user has at least one friend. (Friendship is symmetric; that is, if $A$ is a friend of $B$, then $B$ is a friend of $A$.)

Each person is required to designate one of their friends as their best friend. If $A$ designates $B$ as her best friend, then (unfortunately) it does not follow that $B$ necessarily designates $A$ as her best friend. Someone designated as a best friend is called a 1-best friend. More generally, if $n>1$ is a positive integer, then a user is an $n$-best friend provided that they have been designated the best friend of someone who is an $(n-1)$-best friend. Someone who is a $k$-best friend for every positive integer $k$ is called popular. (a) Prove that every popular person is the best friend of a popular person. (b) Show that if people can have infinitely many friends, then it is possible that a popular person is not the best friend of a popular person.

Origin. Romania (Dan Schwarz) (rephrasing by Geoff Smith). Remark. The original formulation of this problem was: Given a function $f: X \rightarrow X$, let us use the notations $f^{0}(X):=X, f^{n+1}(X):=f\left(f^{n}(X)\right)$ for $n \geq 0$, and also $f^{\omega}(X):=\bigcap_{n \geq 0} f^{n}(X)$. Let us now impose on $f$ that all its fibres $f^{-1}(y):=\{x \in X \mid f(x)=y\}$, for $y \in f(X)$, are finite. Prove that $f\left(f^{\omega}(X)\right)=f^{\omega}(X)$.