42nd Eötvös 1938 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.
Problem 1101
Official 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 x 2 + y 2 = c 2 , y 2 + z 2 = a 2 , z 2 + x 2 = b 2 , so x 2 = (b 2 + c 2 - a 2 )/2, y 2 = (a 2 - b 2 + c 2 )/2, z 2 = (a 2 + b 2 - c 2 )/2. Since the triangle is acute-angled, each of the brackets is positive, so we can choose such x, y, z. 42nd Eötvös 1938 © John Scholes [email protected] 1 Nov 2003 Last corrected/updated 1 Nov 03