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 = 1^o. 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 A_k and B_k, so that if m and n are both in A_k or both in B_k 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, a_n+1 = √(k + a_n). Show that the sequence a_n 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 A_n to be the interval with k = (1.2)ⁿ. The the intervals A_n and A_n+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 = 1^o. 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 < 45^o, we have 1 = tan 45^o = (tan y + tan(45^o-y) )/(1 - tan y tan(45^o-y)), so (tan y + tan(45^o-y)) + tan y tan(45^o-y) = 1. But by AM/GM (tan y + tan(45^o-y)) < (tan y tan(45^o-y))²/4, provided y ≠ 45^o/2. So by the result above we have (tan y + tan(45^o-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(45^o-y)) and hence √2 - 1 = (22/44)(√8 - 2) < (tan x + tan 2x + ... + tan 44x)/44.
Thanks to Suat Namli