olymmath

IMO 1997

International Mathematical Olympiad 1997.

Day 1

Problem 1

  • Day 1

In the plane the points with integer coordinates are the vertices of unit squares. The squares are coloured alternately black and white (as on a chessboard). For any pair of positive integers mm and nn , consider a right-angled triangle whose vertices have integer coordinates and whose legs, of lengths mm and nn , lie along edges of the squares. Let S1S_1 be the total area of the black part of the triangle and S2S_2 be the total area of the white part. Let f(m,n)=∣S1−S2∣f(m,n) = | S_1 - S_2 | .

a) Calculate f(m,n)f(m,n) for all positive integers mm and nn which are either both even or both odd.

b) Prove that f(m,n)≤12max⁡{m,n}f(m,n) \leq \frac 12 \max \{m,n \} for all mm and nn .

c) Show that there is no constant C∈RC\in\mathbb{R} such that f(m,n)<Cf(m,n) < C for all mm and nn .

Answer, solutions and source

Problem 2

  • Day 1

It is known that ∠BAC\angle BAC is the smallest angle in the triangle ABCABC . The points BB and CC divide the circumcircle of the triangle into two arcs. Let UU be an interior point of the arc between BB and CC which does not contain AA . The perpendicular bisectors of ABAB and ACAC meet the line AUAU at VV and WW , respectively. The lines BVBV and CWCW meet at TT .

Show that AU=TB+TCAU = TB + TC .

Alternative formulation:

Four different points A,B,C,DA,B,C,D are chosen on a circle Γ\Gamma such that the triangle BCDBCD is not right-angled. Prove that:

(a) The perpendicular bisectors of ABAB and ACAC meet the line ADAD at certain points WW and V,V, respectively, and that the lines CVCV and BWBW meet at a certain point T.T.

(b) The length of one of the line segments AD,BT,AD, BT, and CTCT is the sum of the lengths of the other two.

Answer, solutions and source

Problem 3

  • Day 1

Let x1x_1 , x2x_2 , …\ldots , xnx_n be real numbers satisfying the conditions: {∣x1+x2+⋯+xn∣=1 ∣xi∣≤n+12  for i=1,2,…,n.\left\{\begin{array}{cccc} |x_1 + x_2 + \cdots + x_n | & = & 1 & \ |x_i| & \leq & \frac {n + 1}{2} & \ \textrm{ for }i = 1, 2, \ldots , n. \end{array} \right. Show that there exists a permutation y1y_1 , y2y_2 , …\ldots , yny_n of x1x_1 , x2x_2 , …\ldots , xnx_n such that ∣y1+2y2+⋯+nyn∣≤n+12.| y_1 + 2 y_2 + \cdots + n y_n | \leq \frac {n + 1}{2}.

Answer, solutions and source

Day 2

Problem 4

  • Day 2

An n×nn \times n matrix whose entries come from the set S={1,2,…,2n−1}S = \{1, 2, \ldots , 2n - 1\} is called a silver matrix if, for each i=1,2,…,ni = 1, 2, \ldots , n , the ii -th row and the ii -th column together contain all elements of SS . Show that:

(a) there is no silver matrix for n=1997n = 1997 ;

(b) silver matrices exist for infinitely many values of nn .

Answer, solutions and source

Problem 6

  • Day 2

For each positive integer nn , let f(n)f(n) denote the number of ways of representing nn as a sum of powers of 2 with nonnegative integer exponents. Representations which differ only in the ordering of their summands are considered to be the same. For instance, f(4)=4f(4) = 4 , because the number 4 can be represented in the following four ways: 4; 2+2; 2+1+1; 1+1+1+1.

Prove that, for any integer n≥3n \geq 3 we have 2n24<f(2n)<2n222^{\frac {n^2}{4}} < f(2^n) < 2^{\frac {n^2}2} .

Answer, solutions and source