olymmath

IMO 2014

International Mathematical Olympiad 2014.

Day 1

Problem 1

  • Day 1

Let a0<a1<a2<…a_0 < a_1 < a_2 < \dots be an infinite sequence of positive integers. Prove that there exists a unique integer n≥1n\geq 1 such that an<a0+a1+a2+⋯+ann≤an+1.a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.

Proposed by Gerhard Wöginger, Austria.

Answer, solutions and source

Problem 2

  • Day 1

Let n≥2n \ge 2 be an integer. Consider an n×nn \times n chessboard consisting of n2n^2 unit squares. A configuration of nn rooks on this board is peaceful if every row and every column contains exactly one rook. Find the greatest positive integer kk such that, for each peaceful configuration of nn rooks, there is a k×kk \times k square which does not contain a rook on any of its k2k^2 unit squares.

Answer, solutions and source

Problem 3

  • Day 1

Convex quadrilateral ABCDA B C D has ∠ABC=∠CDA=90∘\angle A B C = \angle C D A = 90^{\circ} . Point HH is the foot of the perpendicular from AA to BD‾\overline{{B D}} . Points SS and TT lie on sides ABA B and ADA D , respectively, such that HH lies inside triangle SCTS C T and

∠CHS−∠CSB=90∘,∠THC−∠DTC=90∘.\angle C H S - \angle C S B = 90^{\circ}, \angle T H C - \angle D T C = 90^{\circ}.

Prove that line BDB D is tangent to the circumcircle of triangle TSHT S H

Answer, solutions and source

Day 2

Problem 4

  • Day 2

Let PP and QQ be on segment BCBC of an acute triangle ABCABC such that ∠PAB=∠BCA\angle PAB=\angle BCA and ∠CAQ=∠ABC\angle CAQ=\angle ABC . Let MM and NN be the points on APAP and AQAQ , respectively, such that PP is the midpoint of AMAM and QQ is the midpoint of ANAN . Prove that the intersection of BMBM and CNCN is on the circumference of triangle ABCABC .

Proposed by Giorgi Arabidze, Georgia.

Answer, solutions and source

Problem 5

  • Day 2

For each positive integer nn , the Bank of Cape Town issues coins of denomination 1n\frac1n . Given a finite collection of such coins (of not necessarily different denominations) with total value at most most 99+1299+\frac12 , prove that it is possible to split this collection into 100100 or fewer groups, such that each group has total value at most 11 .

Answer, solutions and source

Problem 6

  • Day 2

A set of lines in the plane is in general position if no two are parallel and no three pass through the same point. A set of lines in general position cuts the plane into regions, some of which have finite area; we call these its finite regions. Prove that for all sufficiently large nn , in any set of nn lines in general position it is possible to colour at least n\sqrt{n} lines blue in such a way that none of its finite regions has a completely blue boundary.

Note: Results with n\sqrt{n} replaced by cnc\sqrt{n} will be awarded points depending on the value of the constant cc .

Answer, solutions and source