36th Putnam 1975 Problem A2 For what region of the real (a, b) plane, do both (possibly complex) roots of the polynomial z 2 + az + b = 0 satisfy |z| < 1? Solution
: the interior of the triangle with vertices (±2, 1), (0, -1). Clearly we require |a| ≤ 2, |b| ≤ 1. But that is not sufficient. For example, if b = 0, then we require |a| ≤ 1. We consider first the region where a ≥ 0. If b > a 2 /4, then the roots are z = a/2 ± i √(b - a 2 /4), so these satisfy |z| < 1 iff (a/2) 2 + (b - a 2 /4) < 1 or b < 1. This gives the quasi-triangular region bounded by the two lines a = 0 and b = 1 on two sides and the curve 4b = a 2 on the third side. If b < a 2 /4, then the roots are z = a/2 ± √(a 2 /4 - b). These satisfy |z| < 1 iff a/2 + √(a 2 /4 - b) < 1 which gives a - b < 1. This gives the quasi-triangular region bounded by the lines a = 0 and a - b = 1 on two sides and the curve 4b = a 2 on the third. But since the line a - b = 1 touches the curve b = a 2 /4 at a = 2, the two regions fit together to give the interior of triangle with vertices (a, b) = (0, 1), (0, -1) and (2, 1). The region for a ≤ 0 is evidently the mirror image in the b-axis, so we get in total the (interior of the) triangular region with vertices (2, 1), (-2, 1) and (0, -1). 36th Putnam 1975 © John Scholes jscholes@kalva.demon.co.uk 27 Jan 2002