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.

**Solution**

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.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002