5th Balkan 1988 Problem 1 ABC is a triangle area 1. AH is an altitude, M is the midpoint of BC and K is the point where the angle bisector at A meets the segment BC. The area of the triangle AHM is 1/4 and the area of AKM is 1 - (√3)/2. Find the angles of the triangle. Solution Solution by Nizameddin Ordulu and Ali Adalý:
: A = 90 o , B = 30 o , C = 60 o . Assume AB ≥ AC. Then BK/KC = AB/AC ≥ 1, so K lies between M and C. Put MH = x. Then since area AHM = 1/4 area ABC, we have BC = 4x, and hence BM = 2x, HC = x. Since area AKM = 1 - (√3)/2, we have MK = (4 - 2√3) x. Hence AB/AC = (BM + MK)/(MC - MK) = (6 - 2√3)/(2√3 - 2) = √3. We have AB 2 - BH 2 = AH 2 = AC 2 - CH 2 , so AB 2 - AC 2 = (3x) 2 - x 2 . Hence AC 2 = 4x 2 . So AC : AB : BC = 1 : √3 : 2. Hence A = 90 o , B = 30 o , C = 60 o . 5th Balkan 1988 © John Scholes jscholes@kalva.demon.co.uk 11 Apr 2002