Subcontests
(24)IMO Shortlist 2014 A2
Define the function f:(0,1)→(0,1) by f(x)={x+21x2if x<21if x≥21 Let a and b be two real numbers such that 0<a<b<1. We define the sequences an and bn by a0=a,b0=b, and an=f(an−1), bn=f(bn−1) for n>0. Show that there exists a positive integer n such that (an−an−1)(bn−bn−1)<0.Proposed by Denmark IMO Shortlist 2014 N6
Let a1<a2<⋯<an be pairwise coprime positive integers with a1 being prime and a1≥n+2. On the segment I=[0,a1a2⋯an] of the real line, mark all integers that are divisible by at least one of the numbers a1,…,an . These points split I into a number of smaller segments. Prove that the sum of the squares of the lengths of these segments is divisible by a1.Proposed by Serbia IMO Shortlist 2014 G7
Let ABC be a triangle with circumcircle Ω and incentre I. Let the line passing through I and perpendicular to CI intersect the segment BC and the arc BC (not containing A) of Ω at points U and V , respectively. Let the line passing through U and parallel to AI intersect AV at X, and let the line passing through V and parallel to AI intersect AB at Y . Let W and Z be the midpoints of AX and BC, respectively. Prove that if the points I,X, and Y are collinear, then the points I,W, and Z are also collinear.Proposed by David B. Rush, USA IMO Shortlist 2014 G6
Let ABC be a fixed acute-angled triangle. Consider some points E and F lying on the sides AC and AB, respectively, and let M be the midpoint of EF . Let the perpendicular bisector of EF intersect the line BC at K, and let the perpendicular bisector of MK intersect the lines AC and AB at S and T , respectively. We call the pair (E,F) <spanclass=′latex−italic′>interesting</span>, if the quadrilateral KSAT is cyclic.
Suppose that the pairs (E1,F1) and (E2,F2) are interesting. Prove that ABE1E2=ACF1F2
Proposed by Ali Zamani, Iran IMO Shortlist 2014 G4
Consider a fixed circle Γ with three fixed points A,B, and C on it. Also, let us fix a real number λ∈(0,1). For a variable point P∈{A,B,C} on Γ, let M be the point on the segment CP such that CM=λ⋅CP . Let Q be the second point of intersection of the circumcircles of the triangles AMP and BMC. Prove that as P varies, the point Q lies on a fixed circle.Proposed by Jack Edward Smith, UK IMO Shortlist 2014 G2
Let ABC be a triangle. The points K,L, and M lie on the segments BC,CA, and AB, respectively, such that the lines AK,BL, and CM intersect in a common point. Prove that it is possible to choose two of the triangles ALM,BMK, and CKL whose inradii sum up to at least the inradius of the triangle ABC.Proposed by Estonia IMO Shortlist 2014 C6
We are given an infinite deck of cards, each with a real number on it. For every real number x, there is exactly one card in the deck that has x written on it. Now two players draw disjoint sets A and B of 100 cards each from this deck. We would like to define a rule that declares one of them a winner. This rule should satisfy the following conditions:
1. The winner only depends on the relative order of the 200 cards: if the cards are laid down in increasing order face down and we are told which card belongs to which player, but not what numbers are written on them, we can still decide the winner.
2. If we write the elements of both sets in increasing order as A={a1,a2,…,a100} and B={b1,b2,…,b100}, and ai>bi for all i, then A beats B.
3. If three players draw three disjoint sets A,B,C from the deck, A beats B and B beats C then A also beats C.
How many ways are there to define such a rule? Here, we consider two rules as different if there exist two sets A and B such that A beats B according to one rule, but B beats A according to the other.Proposed by Ilya Bogdanov, Russia IMO Shortlist 2014 A3
For a sequence x1,x2,…,xn of real numbers, we define its <spanclass=′latex−italic′>price</span> as 1≤i≤nmax∣x1+⋯+xi∣. Given n real numbers, Dave and George want to arrange them into a sequence with a low price. Diligent Dave checks all possible ways and finds the minimum possible price D. Greedy George, on the other hand, chooses x1 such that ∣x1∣ is as small as possible; among the remaining numbers, he chooses x2 such that ∣x1+x2∣ is as small as possible, and so on. Thus, in the i-th step he chooses xi among the remaining numbers so as to minimise the value of ∣x1+x2+⋯xi∣. In each step, if several numbers provide the same value, George chooses one at random. Finally he gets a sequence with price G. Find the least possible constant c such that for every positive integer n, for every collection of n real numbers, and for every possible sequence that George might obtain, the resulting values satisfy the inequality G≤cD.Proposed by Georgia