42nd Putnam 1981 Problem B6 Let P be a convex polygon each of whose sides touches a circle C of radius 1. Let A be the set of points which are a distance 1 or less from P. If (x, y) is a point of A, let f(x, y) be the number of points in which a unit circle center (x, y) intersects P (so certainly f(x, y) ≥ 1). What is sup 1/|A| ∫ A f(x, y) dx dy, where the sup is taken over all possible polygons P? Solution
: 8/3. Construct a rectangle on each side of P, with the other dimension of each rectangle 1. Between each adjacent pair of rectangles construct a sector radius 1, center the common vertex. Then A comprises the interior of P, the rectangles and the sectors. By joining each vertex of P to the center of C we divide the interior of P into triangles, each with area 1/2 x length of side. So the area of P = k/2, where k is the perimeter of P. Obviously, the area of the rectangles is k and since the sectors form a circle their area is π. So |A| = π + 3k/2. The center of a unit circle which cuts a segment length δx must lie in one of two crescents centered on δx, which have total area 4δx. We may now integrate around P to get the area 4k. In this integration points are counted multiple times if the unit circles centered on them intersect P multiple times. In fact, the integral is exactly ∫ A f(x, y) dx dy. Note that most unit circles meeting P will meet it twice, but those with centres near a vertex may meet it 4 times. Thus 1/|A| ∫ A f(x, y) dx dy = 4k/(π + 3k/2) = 8/(3 + 2π/k). So clearly 8/3 is an upper bound. But we can make k arbitrarily large (by taking one vertex of P off to infinity), so 8/3 is also the least upper bound. 42nd Putnam 1981 © John Scholes jscholes@kalva.demon.co.uk 16 Jan 2001