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Difference between revisions of "2014 EGMO Problems"

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Revision as of 13:45, 24 December 2022

Day 1

Problem 1

Determine all real constants $t$ such that whenever $a$, $b$ and $c$ are the lengths of sides of a triangle, then so are $a^2+bct$, $b^2+cat$, $c^2+abt$.

Solution

Problem 2

Let $D$ and $E$ be points in the interiors of sides $AB$ and $AC$, respectively, of a triangle $ABC$, such that $DB = BC = CE$. Let the lines $CD$ and $BE$ meet at $F$. Prove that the incentre $I$ of triangle $ABC$, the orthocentre $H$ of triangle $DEF$ and the midpoint $M$ of the arc $BAC$ of the circumcircle of triangle $ABC$ are collinear.

Solution

Problem 3

We denote the number of positive divisors of a positive integer $m$ by $d(m)$ and the number of distinct prime divisors of $m$ by $\omega(m)$. Let $k$ be a positive integer. Prove that there exist infinitely many positive integers $n$ such that $\omega(n) = k$ and $d(n)$ does not divide $d(a^2+b^2)$ for any positive integers $a, b$ satisfying $a + b = n$.


Solution

Day 2

Problem 4

Determine all positive integers $n\geq 2$ for which there exist integers $x_1,x_2,\ldots ,x_{n-1}$ satisfying the condition that if $0<i<n,0<j<n, i\neq j$ and $n$ divides $2i+j$, then $x_i<x_j$.

Solution

Problem 5

Let $n$ be a positive integer. We have $n$ boxes where each box contains a non-negative number of pebbles. In each move we are allowed to take two pebbles from a box we choose, throw away one of the pebbles and put the other pebble in another box we choose. An initial configuration of pebbles is called solvable if it is possible to reach a configuration with no empty box, in a finite (possibly zero) number of moves. Determine all initial configurations of pebbles which are not solvable, but become solvable when an additional pebble is added to a box, no matter which box is chosen.

Prove that $\angle ACP = \angle QCB$.

Solution

Problem 6

Determine all functions $f:\mathbb R\rightarrow\mathbb R$ satisfying the condition \[f(y^2+2xf(y)+f(x)^2)=(y+f(x))(x+f(y))\] for all real numbers $x$ and $y$.

Solution

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