olymmath

IMO 2004

International Mathematical Olympiad 2004.

Day 1

Problem 1

  • Day 1

Let ABCABC be an acute-angled triangle with AB≠ACAB\neq AC . The circle with diameter BCBC intersects the sides ABAB and ACAC at MM and NN respectively. Denote by OO the midpoint of the side BCBC . The bisectors of the angles ∠BAC\angle BAC and ∠MON\angle MON intersect at RR . Prove that the circumcircles of the triangles BMRBMR and CNRCNR have a common point lying on the side BCBC .

Attachments:

logo2004.jpg

Answer, solutions and source

Problem 2

  • Day 1

Find all polynomials ff with real coefficients such that for all reals a,b,ca,b,c such that ab+bc+ca=0ab+bc+ca = 0 we have the following relations

f(a−b)+f(b−c)+f(c−a)=2f(a+b+c).f(a-b) + f(b-c) + f(c-a) = 2f(a+b+c).

Attachments:

logo2004.jpg

Answer, solutions and source

Problem 3

  • Day 1

Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure.

[asy] unitsize(0.5 cm);

draw((0,0)--(1,0)); draw((0,1)--(1,1)); draw((2,1)--(3,1)); draw((0,2)--(3,2)); draw((0,3)--(3,3)); draw((0,0)--(0,3)); draw((1,0)--(1,3)); draw((2,1)--(2,3)); draw((3,1)--(3,3)); [/asy] Determine all m×nm\times n rectangles that can be covered without gaps and without overlaps with hooks such that

  • the rectangle is covered without gaps and without overlaps
  • no part of a hook covers area outside the rectangle.

Answer, solutions and source

Day 2

Problem 4

  • Day 2

Let n≥3n \geq 3 be an integer. Let t1t_1 , t2t_2 , ..., tnt_n be positive real numbers such that n2+1>(t1+t2+⋯+tn)(1t1+1t2+⋯+1tn).n^2 + 1 > \left( t_1 + t_2 + \cdots + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \cdots + \frac{1}{t_n} \right). Show that tit_i , tjt_j , tkt_k are side lengths of a triangle for all ii , jj , kk with 1≤i<j<k≤n1 \leq i < j < k \leq n .

Attachments:

logo2004.jpg

Answer, solutions and source

Problem 5

  • Day 2

In a convex quadrilateral ABCDABCD , the diagonal BDBD bisects neither the angle ABCABC nor the angle CDACDA . The point PP lies inside ABCDABCD and satisfies ∠PBC=∠DBAand∠PDC=∠BDA.\angle PBC=\angle DBA \text{and} \angle PDC=\angle BDA. Prove that ABCDABCD is a cyclic quadrilateral if and only if AP=CPAP=CP .

Attachments:

logo2004.jpg

Answer, solutions and source

Problem 6

  • Day 2

We call a positive integer alternating if every two consecutive digits in its decimal representation are of different parity.

Find all positive integers nn such that nn has a multiple which is alternating.

Attachments:

logo2004.jpg

Answer, solutions and source