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 90o at P, (2) if Q is any point on the line CA, then BQ subtends an angle 90o at P, and (3) if Q is any point on the line AB, then CQ subtends an angle 90o at P.
Solution
It is sufficient to find a point O such that ∠AOB = ∠BOC = ∠COA = 90o. For then AO is normal to the plane BOC, so if P is any point in the plane BOC, we have ∠AOP = 90o, 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.
© John Scholes
jscholes@kalva.demon.co.uk
1 Nov 2003
Last corrected/updated 1 Nov 03