30th IMO 1989 shortlist Problem 25 ABC is an acute-angled triangle with circumcenter O and orthocenter H. AO = AH. Find all possible values for the angle A.
Let the altitude from C be CF and the circumradius be R. ∠HAF = 90 o - B, so AF = AH sin B = R sin B. Hence CF = R sin B tan A. But ∠BOC = 2A, so BC = 2 R sin A, and CF = BC sin B = 2R sin A sin B. Hence cos A = ½. Hence A = 60 o . 30th IMO shortlist 1989 © John Scholes jscholes@kalva.demon.co.uk 28 Dec 2002 Last corrected/updated 28 Dec 02