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?

"For each face of the tetrahedron, construct a new tetrahedron with that face as the base and the center of the sphere as the fourth vertex. Now the original tetrahedron is divided into four smaller ones, each of height R. The volume of a tetrahedron is Ah/3 where A is the area of the base and h the height; in this case h=R.  Combine the four tetrahedra algebraically to find that the volume of the original tetrahedron is R/3 times its surface area." http://rec-puzzles.org/new/sol.pl/geometry/tetrahedron 

I suppose that the similar result holds for the n-dimensional simplex: The "internal content" is one nth of the "boundary content" times the inradius.  This result is very clear for a triangle where the area is the semiperimeter times the inradius (a well-known formula). For the ratio of boundary content to internal content to depend on the inradius seems really remarkable to me.

Posted by Richard on 2004-06-23 23:23:19