### 13th Putnam 1953

Problem A7

p(x) ≡ x3 + ax2 + 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.

Solution

If A + B + C = 180o, then cos A = - cos(B + C) = sin B sin C - cos B cos C. Squaring, we get cos2A + cos2B + cos2C + 2 cos A cos B cos C = 1. So a necessary condition is a2 - 2b - 2c = 1.

Conversely, suppose that this condition holds. Then if the roots are p, q, r, we have p2 + q2 + r2 + 2pqr = 1 (*). We are given that the roots are all positive, so 2pqr > 0, hence p2 < 1 and so p < 1. Similarly for q and r. So we can find angles A, B, C greater than 0 and less than 90o such that p = cos A, q = cos B, r = cos C. Now we can rewrite (*) as (1 - cos2B)(1 - cos2C) = cos2A + 2 cos A cos B cos C + cos2B cos2C = (cos A + cos B cos C)2. But (1 - cos2B) = sin2B, (1 - cosC) = sin2C, so we have sin B sin C = ±(cos A + cos B cos C). But we know that A, B, C are between 0 and 90o, 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 = 180o. Hence the condition is also sufficient.