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?

Take the case of a cube, which is the simplest six-sided figure. The length of a side will be 2R. This gives a surface area of 6*(4R^2) = 24*R^2. The volume of the cube will be (2R)^3, or rather 8R^3. The ratio of this cube's surface area to its volume is 3/R, or rather the volume is R/3 times its surface area.

A tetrahedron works similarly, and in fact the ratio for any figure should be 3/R.

Posted by Eric on 2004-06-23 14:32:21