37th Swedish 1997 Problem 2 ABC is a triangle. BD is the angle bisector. The point E on AB is such that ∠ACE = (2/5) ∠ ACB. BD and CE meet at P. ED = DC = CP. Find the angles of ABC.
∠A = 45 o , ∠B = 60 o , ∠C = 75 o . Solution Let ∠C = 5x, so ∠ACE = 2x. Since DE = DC, ∠DEC = 2x. Since CD = CP, ∠CDP = ∠CPD = 90 o -x. So ∠B/2 = 90 o -4x, so ∠A = 3x. Also ∠ADE = ∠DEC + ∠DCE = 4x. Now DE/AD = DC/AD = BC/AB (BD angle bisector). But BC/AB = sin A/sin C, and DE/AD = sin A/sin AED. Hence ∠AED = 5x. So considering triangle ADE, 3x + 4x + 5x = 180 o , so x = 15 o . 37th Swedish 1997 © John Scholes jscholes@kalva.demon.co.uk 19 January 2004 Last corrected/updated 19 Jan 04