37th Putnam 1976 Problem A5 Let P be a convex polygon. Let Q be the interior of P and S = P ∪ Q. Let p be the perimeter of P and A its area. Given any point (x, y) let d(x, y) be the distance from (x, y) to the nearest point of S. Find constants α, β, γ such that ∫ U e -d(x,y) dx dy = α + βp + γA, where U is the whole plane. Solution
: A + p + 2π. For any point in S we have d(x, y) = 0. Hence the integral over S is just A. The locus of points a distance z from S is a set of segments parallel to the sides of P and displaced a distance z outwards, together with a set of arcs joining them. Each arc is centered on a vertex of P and has radius z. Together the arcs can be translated to form a complete circle radius z. Thus the set of points a distance z to z + δz from S is a strip of area p δz + 2πz δz. Thus the integral outside S is just ∫ 0 ∞ exp( - z) (p + 2πz) dz. This evaluates easily to p + 2π. 37th Putnam 1976 © John Scholes jscholes@kalva.demon.co.uk 23 Jan 2001