Putnam 1998 Problem B3 Let S be the sphere center the origin and radius 1. Let P be a regular pentagon in the plane z = 0 with vertices on S. Find the surface area of the part of the sphere which lies above (z > 0) P or its interior. Solution
: π(5 cos π/5 - 3) = 3.2832. Straightforward. Call the area on a sphere bounded by a circle a cap . The depth of the cap is the distance from its central point to the plane through its boundary. A simple integration gives that the area of a cap depth d is 2πrd, where r is the radius of the sphere. In this case the vertical plane through a side of the pentagon defines a cap. The five caps do not overlap and so the area required is (area S - 5 area cap)/2. We have that d = 1 - cos π/5. Hence area required = π(5 cos π/5 - 3). Putnam 1998 © John Scholes jscholes@kalva.demon.co.uk 12 Dec 1998