IMO Problems 2013

IMO Problems 2013

Day 1

IMO 2013 Problem 1

Assume that n and k are two positive integers. Prove that there exist positive integers m₁, m₂, …, mₖ such that

1 + (2ᵏ – 1)/n = (1 + 1/m₁) + (1 + 1/m₂) + … + (1 + 1/mₖ).

IMO 2013 Problem 2

A configuration of 4027 points in the plane is called Colombian, if it consists of 2013 red points and 2013 blue points, and no 3 of the points of the configuration are collinear. By drawing some lines, the plane is divided into several regions. An arrangement of lines is good for a Colombian configuration if the following two conditions are satisfied:

(i) No line passes through any point of the configuration.
(ii) No region contains points of both colors.

Find the least value of k such that for any Colombian configuration of 4027 points, there is a good arrangement of k lines.

IMO 2013 Problem 3

Let the excircle of triangle ABC opposite to the vertex A be tangent to the side BC at the point A₁. Define the points B₁ on CA and C₁ on AB analogously, using the excircles opposite B and C respectively. Suppose that the circumcentre of triangle A₁B₁C₁ lies on the circumcircle of triangle ABC.

Prove that triangle ABC is right angled.

Day 2

IMO 2013 Problem 4

Let ABC be an acute triangle with orthocentre H and let W be a point on the side BC, lying strictly between B and C. The points M and N are the feet of the altitudes from B and C respectively. Let ω₁ is the circumcircle of BWN and let X be the point on ω₁ such that WX is the diameter of ω₁ analogously. Let ω₂ be the circumcircle of triangle CWM ,and let Y with the points such that WY is a diameter of ω₂.

Prove that X, Y and H are collinear.

IMO 2013 Problem 5

Let Q⁺ be the set of all positive rational numbers. Let f: Q⁺ → R be a function satisfying the following conditions:

(i) For all x and y ∈ Q⁺, we have f(x)f(y) ≥ f(xy).
(ii) For all x and y ∈ Q⁺, we have f(x + y) ≥ f(x) + f(y).
(iii) There exist a rational number a > 1 such that f(a) = a.

Prove that f(x) = x for all x ∈ Q⁺.

IMO 2013 Problem 6

Let n ≥ 3 be an integer and consider a circle with n + 1 equally spaced points marked on it. Consider all labellings of these points with the numbers 0, 1, 2, …, n such that each label is used exactly once. Two such labellings are considered to be the same if one can be obtained from the other by a rotation of the circle. A labelling is called beautiful, if for any four labels a < b < c < d with
a + d = b + c. The chord joining the points labelled a and d does not intersect the chord joining the points labelled b and c. Let M be the number of beautiful labelling and let N be the number of ordered pairs (x, y) of positive integers such that x + y ≤ n and gcd(x, y) = 1.

Prove that M = N + 1.