Prove that if BC + AC = tan C/2 (BC tan A + AC tan B), then the triangle ABC is isosceles.
Solution
A straight slog works. Multiply up to get (a + b) cos A cos B cos C/2 = a sin A cos B sin C/2 + b cos A sin B sin C/2 (where a = BC, b = AC, as usual). Now use cos(A + C/2) = cos A cos C/2 - sin A sin C/2 and similar relation for cos (B + C/2) to get: a cos B cos(A + C/2) + b cos A cos (B + C/2) = 0. Using C/2 = 90o - A/2 - B/2, we find that cos(A + C/2) = - cos(B + C/2) (and = 0 only if A = B). Result follows.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
29 Sep 1998
Last corrected/updated 26 Sep 2003