13th Putnam 1953 Problem A7 p(x) ≡ x 3 + ax 2 + bx + c has three positive real roots. Find a necessary and sufficient condition on a, b, c for the roots to be cos A, cos B, cos C for some triangle ABC.
If A + B + C = 180 o , then cos A = - cos(B + C) = sin B sin C - cos B cos C. Squaring, we get cos 2 A + cos 2 B + cos 2 C + 2 cos A cos B cos C = 1. So a necessary condition is a 2 - 2b - 2c = 1. Conversely, suppose that this condition holds. Then if the roots are p, q, r, we have p 2 + q 2 + r 2 + 2pqr = 1 (*). We are given that the roots are all positive, so 2pqr > 0, hence p 2 < 1 and so p < 1. Similarly for q and r. So we can find angles A, B, C greater than 0 and less than 90 o such that p = cos A, q = cos B, r = cos C. Now we can rewrite (*) as (1 - cos 2 B)(1 - cos 2 C) = cos 2 A + 2 cos A cos B cos C + cos 2 B cos 2 C = (cos A + cos B cos C) 2 . But (1 - cos 2 B) = sin 2 B, (1 - cos C) = sin 2 C, so we have sin B sin C = ±(cos A + cos B cos C). But we know that A, B, C are between 0 and 90 o , so cos A, cos B, cos C, sin B, sin C are all positive. Hence we must use the + sign and we have cos A = sin B sin C - cos B cos C, so A + B + C = 180 o . Hence the condition is also sufficient. 13th Putnam 1953 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002