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?

OK, let's make it a regular tetrahedron. You then have four identical tetrahedra. As with most geometry problems, you want to subdivide the polyhedra into smaller polyhedra with more manageable statistics. So, if you take each face of the individual tetrahedron as the base of a new tetrahedron and the point at the center of the sphere, you have 4 new tetrahedra, all identical, with height R (the original radius).

OK, so we have 4 tetrahedra now. The area of each base is some variable X. I can't solve for it off the top of my head, but I don't need to. Since the volume of a tetrahedron is XR/3, using our variables, we have a total volume of the tetrahedron of 4XR/3. The total surface area is 4X. The ratio then of surface to volume is 3/R.

Edited on June 23, 2004, 2:44 pm

Posted by Eric on 2004-06-23 14:43:44