Suppose you have a sphere of radius R and you have four planes that are all tangent to the sphere such that they form an arbitrary tetrahedron (it can be irregular). What is the ratio of the surface area of the tetrahedron to its volume?

The tetrahedron encompassing the sphere can be looked at as the sum of four tetrahedra, each with its apex at the center of the sphere and its base being one of the four sides of the original tetrahedron.  The volume of a pyramid with triangular base is 1/3*base*height. Since of the smaller tetrahedra have a height equal to the radius R of the sphere, the sum of their volumes is

Ó(R/3)*base

Since the sum of the four bases equals the surface are of the original tetrahedron, and the sum of volumes of the smaller tetrahedra equals the volume of the original tetrahedron, the ratio of surface area to volume is 3/R.

Posted by Bryan on 2004-06-23 14:51:45