olymmath

EGMO 2014

European Girls' Mathematical Olympiad 2014.

Day 1

Problem 1

  • Day 1
  • Answer

Determine all real constants tt such that whenever a,b,ca, b, c are the lengths of the sides of a triangle, then so are a2+bct,b2+cat,c2+abta^{2}+b c t, b^{2}+c a t, c^{2}+a b t.

Proposed by S. Khan, UNK The answer is the interval [2/3,2][2 / 3,2].

Answer, solutions and source

Problem 2

  • Day 1

Let DD and EE be two points on the sides ABA B and ACA C, respectively, of a triangle ABCA B C, such that DB=BC=CED B=B C=C E, and let FF be the point of intersection of the lines CDC D and BEB E. Prove that the incenter II of the triangle ABCA B C, the orthocenter HH of the triangle DEFD E F and the midpoint MM of the arc⁡BAC\operatorname{arc} B A C of the circumcircle of the triangle ABCA B C are collinear.

Proposed by Danylo Khilko, UKR

Answer, solutions and source

Problem 3

  • Day 1

We denote the number of positive divisors of a positive integer mm by d(m)d(m) and the number of distinct prime divisors of mm by ω(m)\omega(m). Let kk be a positive integer. Prove that there exist infinitely many positive integers nn such that ω(n)=k\omega(n)=k and d(n)d(n) does not divide d(a2+b2)d\left(a^{2}+b^{2}\right) for any positive integers a,ba, b satisfying a+b=na+b=n.

Proposed by JPN

Answer, solutions and source

Day 2

Problem 4

  • Day 2
  • Answer

Determine all integers n≥2n \geq 2 for which there exist integers x1,x2,…,xn−1x_{1}, x_{2}, \ldots, x_{n-1} satisfying the condition that if 0<i<n,0<j<n,i≠j0<i<n, 0<j<n, i \neq j and nn divides 2i+j2 i+j, then xi<xjx_{i}<x_{j}.

Proposed by Merlijn Staps, NLD The answer is that n=2kn=2^{k} with k≥1k \geq 1 or n=3⋅2kn=3 \cdot 2^{k} where k≥0k \geq 0.

Answer, solutions and source

Problem 5

  • Day 2

Let nn be a positive integer. We have nn boxes where each box contains a nonnegative 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.

Proposed by Dan Schwarz, ROU

The answer is any configuration with 2n−22 n-2 pebbles which has even numbers of pebbles in each box.

Answer, solutions and source

Problem 6

  • Day 2

Determine all functions f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} satisfying the condition

f(y2+2xf(y)+f(x)2)=(y+f(x))(x+f(y))f\left(y^{2}+2 x f(y)+f(x)^{2}\right)=(y+f(x))(x+f(y))

for all real numbers xx and yy. Proposed by Daniël Kroes, NLD The answer is the functions f(x)=x,f(x)=−xf(x)=x, f(x)=-x and f(x)=12−xf(x)=\frac{1}{2}-x.

Answer, solutions and source