4th Balkan 1987

ABC is a triangle with BC = 1. Put sA = sin(A/2), cA = cos(A/2), sB = sin(B/2), cB = cos(B/2). We have sA²³cB⁴⁸ = sB²³cA⁴⁸. Find AC.

Solution

Answer: 1.

Since ABC is a triangle, both A/2 and B/2 lie strictly between 0 and π/2. Over this range sin x is strictly increasing and cos x is strictly decreasing, so if A/2 > B/2, then sin A/2 > sin B/2 and 1/cos A/2 > 1/cos B/2, so the equality cannot hold. Similarly if A/2 < B/2. Hence A/2 = B/2 and the triangle is isosceles.

Comment. This is too easy!

4th Balkan 1987

© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002