olymmath

IMO 1975

International Mathematical Olympiad 1975.

Official paper

The official paper in 1 language

English

Day 1

Problem 1

  • Day 1

We consider two sequences of real numbers x1≥x2≥…≥xnx_{1} \geq x_{2} \geq \ldots \geq x_{n} and  y1≥y2≥…≥yn.\ y_{1} \geq y_{2} \geq \ldots \geq y_{n}. Let z1,z2,.…,znz_{1}, z_{2}, .\ldots, z_{n} be a permutation of the numbers y1,y2,…,yn.y_{1}, y_{2}, \ldots, y_{n}. Prove that ∑i=1n(xi− yi)2≤∑i=1n\sum \limits_{i=1}^{n} ( x_{i} -\ y_{i} )^{2} \leq \sum \limits_{i=1}^{n} (xi−zi)2.( x_{i} - z_{i})^{2}.

Answer, solutions and source

Problem 2

  • Day 1

Let a1,…,ana_{1}, \ldots, a_{n} be an infinite sequence of strictly positive integers, so that ak<ak+1a_{k} < a_{k+1} for any k.k. Prove that there exists an infinity of terms am,a_{m}, which can be written like am=x⋅ap+y⋅aqa_m = x \cdot a_p + y \cdot a_q with x,yx,y strictly positive integers and p≠q.p \neq q.

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Problem 3

  • Day 1

In the plane of a triangle ABC,ABC, in its exterior ,, we draw the triangles ABR,BCP,CAQABR, BCP, CAQ so that ∠PBC=∠CAQ=45∘\angle PBC = \angle CAQ = 45^{\circ} , ∠BCP=∠QCA=30∘\angle BCP = \angle QCA = 30^{\circ} , ∠ABR=∠RAB=15∘\angle ABR = \angle RAB = 15^{\circ} .

Prove that a.) ∠QRP=90 ∘,\angle QRP = 90\,^{\circ}, and b.) QR=RP.QR = RP.

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Day 2

Problem 4

  • Day 2

When 444444444444^{4444} is written in decimal notation, the sum of its digits is A.A. Let BB be the sum of the digits of A.A. Find the sum of the digits of B.B. ( AA and BB are written in decimal notation.)

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Problem 5

  • Day 2

Can there be drawn on a circle of radius 11 a number of 19751975 distinct points, so that the distance (measured on the chord) between any two points (from the considered points) is a rational number?

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Problem 6

  • Day 2

Determine the polynomials P of two variables so that:

a.) for any real numbers t,x,yt,x,y we have P(tx,ty)=tnP(x,y)P(tx,ty) = t^n P(x,y) where nn is a positive integer, the same for all t,x,y;t,x,y;

b.) for any real numbers a,b,ca,b,c we have P(a+b,c)+P(b+c,a)+P(c+a,b)=0;P(a + b,c) + P(b + c,a) + P(c + a,b) = 0;

c.) P(1,0)=1.P(1,0) =1.

Answer, solutions and source