IMO Problems 2014

IMO Problems 2014

Day 1

IMO 2014 Problem 1

Let a₁, a₂, … be an infinite sequence of positive integers. Prove that there exist a unique integer n ≥ 1 such that:

aₙ ≤ (a₁ + a₂ + … +aₙ)/n ≤ aₙ₊₁.

IMO 2014 Problem 2

let n ≥ 2 be an integer. Consider an n×n chess board consisting of n² unit square. A configuration of n rooks on this board is peaceful if every row and every column contains exactly one rook. Find the greatest positive integer k such that, for each peaceful configuration of n rooks, there is a k×k square which does not contain a rook on any of its k² squares.

IMO 2014 Problem 3

Convex quadrilateral ABCD has angle(ABC) = angle(CDA) = 90ᵒ. Point H is the foot of the perpendicular from A to BD. Points S and T lie on sides AB and AD respectively, such that H lies inside the triangle SCT, angle(CHS) – angle(CSB) = 90ᵒ and angle(THC) – angle(DTC) = 90ᵒ.

Prove that line BD is tangent to the circumcircle of triangle TSH.

Day 2

IMO 2014 Problem 4

Let P and Q be on segment BC of an acute triangle ABC such that angle(PAB) = angle(BCA) and angle(CAQ) = angle(ABC). Let M and N be the points on AP and AQ respectively, such that P is the midpoint of AM and Q is the midpoint of AN. Prove that the intersection of BM and CN is on the circumference of triangle ABC.

IMO 2014 Problem 5

For each positive integer n, the bank of Cape Town issues coins of denomination 1/n. Given a finite collections of such coins (of not to necessarily different denominations) with total value at most
199/2. Prove that it is possible to split this collection into 100 or fewer groups, such that each group has total value at most one.

IMO 2014 Problem 6

A set of lines in the plane is in general position if no to a 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 n, in any set of n lines in general position. It is possible to colour at least √n lines blue in such a way that none of its finite regions has a completely blue boundary.

Note: Results with √n replaced by c√n will be awarded points depending on the value of the constant c.