1st Austrian-Polish Mathematics Competition 1978 Problems



1st Austrian-Polish Mathematics Competition 1978 Problems

1.  Find all real-valued functions f on the positive reals which satisfy f(x + y) = f(x2 + y2) for all positive x, y.
2.  A parallelogram has its vertices on the boundary of a regular hexagon and its center at the center of the hexagon. Show that its area is at most 2/3 the area of the hexagon.


3.  Let x = 1o. Show that (tan x tan 2x ... tan 44x)1/44 < √2 - 1 < (tan x + tan 2x + ... + tan 44x)/44.
4.  Given a positive rational k not equal to 1, show that we can partition the positive integers into sets Ak and Bk, so that if m and n are both in Ak or both in Bk then m/n does not equal k.
5.  The sets A1, A2, ... , A1978 each have 40 elements and the intersection of any two distinct sets has just one element. Show that the intersection of all the sets has one element.
6.  S is a set of disks in the plane. No point belongs to the interior of more than one disk. Each disk has a point in common with at least 6 other disks. Show that S is infinite.
7.  S is a finite set of lattice points in the plane such that we can find a bijection f: S → S satisfying |P - f(P)| = 1 for all P in S. Show that we can find a bijection g: S → S such that |P - g(P)| = 1 for all P in S and g(g(P) ) = P for all P in S.
8.  k is a positive integer. Define a1 = √k, an+1 = √(k + an). Show that the sequence an converges. Find all k such that the limit is an integer. Show that if k is odd, then the limit is irrational.
9.  P is a convex polygon. Some of the diagonals are drawn, so that no interior point of P lies on more than one diagonal. Show that at least two vertices of P do not lie on any diagonals.

Solutions

1st APMC 1978 Problem 1

Find all real-valued functions f on the positive reals which satisfy f(x + y) = f(x2 + y2) for all positive x, y.
Solution
If x + y = k and x2 + y2 = h then xy = (k2 - h)/2, so x, y are the roots of the polynomial z2 - kz + (k2 - h)/2 = 0. This polynomial has positive real roots provided that (1) k > 0, (2) k2 ≥ 2(k2 - h) or h ≥ k2/2, and (3) k2 > k2 - 2(k2-h) or h < k2. So if k > 0 and k2/2 ≤ h < k2, then f(k) = f(h). In other words, f must be constant on the interval [k2/2, k2). Take An to be the interval with k = (1.2)n. The the intervals An and An+1 overlap and ... A-2, A-1, A0, A1, A2, ... cover the positive reals, so f is constant on the positive reals.
Thanks to Suat Namli

1st APMC 1978 Problem 3

Let x = 1o. Show that (tan x tan 2x ... tan 44x)1/44 < √2 - 1 < (tan x + tan 2x + ... + tan 44x)/44.
Solution
Note that (3-√8) + (√8-2) = 1 and (3-√8) = (√8-2)2/4, so if the positive integers A, B, satisfy A + B = 1, A < B2/4, then A < 3-√8.
Now for 0 < y < 45o, we have 1 = tan 45o = (tan y + tan(45o-y) )/(1 - tan y tan(45o-y)), so (tan y + tan(45o-y)) + tan y tan(45o-y) = 1. But by AM/GM (tan y + tan(45o-y)) < (tan y tan(45o-y))2/4, provided y ≠ 45o/2. So by the result above we have (tan y + tan(45o-y)) < 3 - √8, and hence (tan x tan 2x ... tan 44x)1/44 < (3 - √8)22/44 = √2 - 1. Also √8 - 2 < (tan y + tan(45o-y)) and hence √2 - 1 = (22/44)(√8 - 2) < (tan x + tan 2x + ... + tan 44x)/44.
Thanks to Suat Namli
 
 


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