36th Eötvös 1932

An acute-angled triangle has angles A < B < C. Show that sin 2A > sin 2B > sin 2C.

Solution

If 2B ≤ 90^o, then A+B < 2B ≤ 90^o, so C > 90^o. Contradiction. Also 2C < 180^o. Hence 2C > 2B > 90^o. But sin 2x is decreasing over this range, so sin 2B > sin 2C.

If 2A ≥ 90^o, then we have sin 2A > sin 2B for the same reason. So we need to consider the case 2A < 90^o < 2B. Since C < 90^o, we have A + B > 90^o and hence 2B > 180^o - 2A. Hence sin 2B < sin(180^o - 2A) = sin 2A.

36th Eötvös 1932

© John Scholes
jscholes@kalva.demon.co.uk
1 Nov 2003
Last corrected/updated 1 Nov 03