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:
Answer: A = 90o, B = 30o, C = 60o.
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 AB2 - BH2 = AH2 = AC2 - CH2, so AB2 - AC2 = (3x)2 - x2. Hence AC2 = 4x2. So AC : AB : BC = 1 : √3 : 2. Hence A = 90o, B = 30o, C = 60o.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002