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?
(In reply to
Cube by Bob)
Eric wrote: "A tetrahedron works similarly [to a cube], and in fact the ratio for any figure should be 3/R."
Prove it!
re: Cube
