olymmath

EGMO 2015

European Girls' Mathematical Olympiad 2015.

Day 1

Problem 1

  • Day 1

Let △ABC\triangle A B C be an acute-angled triangle, and let DD be the foot of the altitude from CC. The angle bisector of ∠ABC\angle A B C intersects CDC D at EE and meets the circumcircle ω\omega of triangle △ADE\triangle A D E again at FF. If ∠ADF=45∘\angle A D F=45^{\circ}, show that CFC F is tangent to ω\omega. (Luxembourg)

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Problem 2

  • Day 1
  • Answer

A domino is a 2×12 \times 1 or 1×21 \times 2 tile. Determine in how many ways exactly n2n^{2} dominoes can be placed without overlapping on a 2n×2n2 n \times 2 n chessboard so that every 2×22 \times 2 square contains at least two uncovered unit squares which lie in the same row or column. (Turkey)

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Problem 3

  • Day 1

Let n,mn, m be integers greater than 1 , and let a1,a2,…,ama_{1}, a_{2}, \ldots, a_{m} be positive integers not greater than nmn^{m}. Prove that there exist positive integers b1,b2,…,bmb_{1}, b_{2}, \ldots, b_{m} not greater than nn, such that

gcd⁡(a1+b1,a2+b2,…,am+bm)<n\operatorname{gcd}\left(a_{1}+b_{1}, a_{2}+b_{2}, \ldots, a_{m}+b_{m}\right)<n

where gcd⁡(x1,x2,…,xm)\operatorname{gcd}\left(x_{1}, x_{2}, \ldots, x_{m}\right) denotes the greatest common divisor of x1,x2,…,xmx_{1}, x_{2}, \ldots, x_{m}.

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Day 2

Problem 4

  • Day 2

Determine whether there exists an infinite sequence a1,a2,a3,…a_{1}, a_{2}, a_{3}, \ldots of positive integers which satisfies the equality

an+2=an+1+an+1+ana_{n+2}=a_{n+1}+\sqrt{a_{n+1}+a_{n}}

for every positive integer n. (Japan)

Answer, solutions and source

Problem 5

  • Day 2

Let m,nm, n be positive integers with m>1m>1. Anastasia partitions the integers 1,2,…,2m1,2, \ldots, 2 m into mm pairs. Boris then chooses one integer from each pair and finds the sum of these chosen integers. Prove that Anastasia can select the pairs so that Boris cannot make his sum equal to nn. (Netherlands)

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Problem 6

  • Day 2

Let HH be the orthocenter and GG be the centroid of acute-angled triangle △ABC\triangle A B C with AB≠ACA B \neq A C. The line AGA G intersects the circumcircle of △ABC\triangle A B C at AA and PP. Let P′P^{\prime} be the reflection of PP in the line BCB C. Prove that ∠CAB=60∘\angle C A B=60^{\circ} if and only if HG=GP′H G=G P^{\prime}. (Ukraine)

Answer, solutions and source