42nd Eötvös Competition Problems 1938

1.  Show that a positive integer n is the sum of two squares iff 2n is the sum of two squares.

2.  Show that 1/n + 1/(n+1) + 1/(n+2) + ... + 1/n² > 1 for integers n > 1.

3.  Show that for every acute-angled triangle ABC there is a point in space P such that (1) if Q is any point on the line BC, then AQ subtends an angle 90^o at P, (2) if Q is any point on the line CA, then BQ subtends an angle 90^o at P, and (3) if Q is any point on the line AB, then CQ subtends an angle 90^o at P. 

Solutions

Problem 1

Show that a positive integer n is the sum of two squares iff 2n is the sum of two squares.

Solution

If n = a² + b², then 2n = (a+b)² + (a-b)². Conversely if 2n = a² + b², then n = ( (a+b)/2)² + ( (a-b)/2)², and since a²+b² is even, so are a±b, and hence (a±b)/2 are integers. 

Problem 2

Show that 1/n + 1/(n+1) + 1/(n+2) + ... + 1/n² > 1 for integers n > 1.

Solution

There are n²-n terms after the first. These terms are all at least 1/n² and at least one of them is > 1/n², so their sum > (n²-n)/n² = 1 - 1/n. Hence the sum of all the terms > 1. 

Problem 3

Show that for every acute-angled triangle ABC there is a point in space P such that (1) if Q is any point on the line BC, then AQ subtends an angle 90^o at P, (2) if Q is any point on the line CA, then BQ subtends an angle 90^o at P, and (3) if Q is any point on the line AB, then CQ subtends an angle 90^o at P.

Solution

It is sufficient to find a point O such that ∠AOB = ∠BOC = ∠COA = 90^o. For then AO is normal to the plane BOC, so if P is any point in the plane BOC, we have ∠AOP = 90^o, In particular, it is true for any point P on the line BC. Similarly for the other vertices.

Take AB = c, CA = b, BC = a, as usual. Suppose OA = x, OB = y, OC = z. Then x² + y² = c², y² + z² = a², z² + x² = b², so x² = (b² + c² - a²)/2, y² = (a² - b² + c²)/2, z² = (a² + b² - c²)/2. Since the triangle is acute-angled, each of the brackets is positive, so we can choose such x, y, z.