9th VMO 1970

ABC is a triangle. Show that sin A/2 sin B/2 sin C/2 < 1/4.

Solution

Put x = A/2, y = B/2. We have sin C/2 = sin(90^o-x-y) = cos(x+y). So we need to show that sin x sin y cos(x+y) < 1/4, or (cos(x-y) - cos(x+y) )cos(x+y) < 1/2, or 2 cos(x-y) cos(x+y) < 1 + 2 cos²(x+y). But 2 cos(x-y) cos(x+y) ≤ cos²(x+y) + cos²(x-y) ≤ 1 + cos²(x+y) < 1 + 2 cos²(x+y) (since 0 < x,y < 90^o.

Thanks to Suat Namli

9th VMO 1970

© John Scholes
jscholes@kalva.demon.co.uk
7 March 2004
Last corrected/updated 6 Mar 04