olymmath

IMO 2013

International Mathematical Olympiad 2013.

Day 1

Problem 1

  • Day 1

Assume that kk and nn are two positive integers. Prove that there exist positive integers m1,…,mkm_1 , \dots , m_k such that 1+2k−1n=(1+1m1)⋯(1+1mk).1+\frac{2^k-1}{n}=\left(1+\frac1{m_1}\right)\cdots \left(1+\frac1{m_k}\right). Proposed by Japan

Answer, solutions and source

Problem 2

  • Day 1

A configuration of 40274027 points in the plane is called Colombian if it consists of 20132013 red points and 20142014 blue points, and no three 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 kk such that for any Colombian configuration of 40274027 points, there is a good arrangement of kk lines.

Proposed by Ivan Guo from Australia.

Answer, solutions and source

Problem 3

  • Day 1

Let the excircle of triangle ABCABC opposite the vertex AA be tangent to the side BCBC at the point A1A_1 . Define the points B1B_1 on CACA and C1C_1 on ABAB analogously, using the excircles opposite BB and CC , respectively. Suppose that the circumcentre of triangle A1B1C1A_1B_1C_1 lies on the circumcircle of triangle ABCABC . Prove that triangle ABCABC is right-angled.

Proposed by Alexander A. Polyansky, Russia

Answer, solutions and source

Day 2

Problem 4

  • Day 2

Let ABCABC be an acute triangle with orthocenter HH , and let WW be a point on the side BCBC , lying strictly between BB and CC . The points MM and NN are the feet of the altitudes from BB and CC , respectively. Denote by ω1\omega_1 is the circumcircle of BWNBWN , and let XX be the point on ω1\omega_1 such that WXWX is a diameter of ω1\omega_1 . Analogously, denote by ω2\omega_2 the circumcircle of triangle CWMCWM , and let YY be the point such that WYWY is a diameter of ω2\omega_2 . Prove that X,YX,Y and HH are collinear.

Proposed by Warut Suksompong and Potcharapol Suteparuk, Thailand

Answer, solutions and source

Problem 5

  • Day 2

Let Q>0\mathbb Q_{>0} be the set of all positive rational numbers. Let f:Q>0→Rf:\mathbb Q_{>0}\to\mathbb R be a function satisfying the following three conditions:

(i) for all x,y∈Q>0x,y\in\mathbb Q_{>0} , we have f(x)f(y)≥f(xy)f(x)f(y)\geq f(xy) ; (ii) for all x,y∈Q>0x,y\in\mathbb Q_{>0} , we have f(x+y)≥f(x)+f(y)f(x+y)\geq f(x)+f(y) ; (iii) there exists a rational number a>1a>1 such that f(a)=af(a)=a .

Prove that f(x)=xf(x)=x for all x∈Q>0x\in\mathbb Q_{>0} .

Proposed by Bulgaria

Answer, solutions and source

Problem 6

  • Day 2

Let n≥3n \ge 3 be an integer, and consider a circle with n+1n + 1 equally spaced points marked on it. Consider all labellings of these points with the numbers 0,1,...,n0, 1, ... , 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<da < b < c < d with a+d=b+ca + d = b + c , the chord joining the points labelled aa and dd does not intersect the chord joining the points labelled bb and cc .

Let MM be the number of beautiful labelings, and let N be the number of ordered pairs (x,y)(x, y) of positive integers such that x+y≤nx + y \le n and gcd⁡(x,y)=1\gcd(x, y) = 1 . Prove that M=N+1.M = N + 1.

Answer, solutions and source