lew@ihuxr.UUCP (08/18/83)
Perhaps not everyone will believe me when I say that my "proof" was given tongue-in-cheek. Anyway, many people easily saw the flaw in it. Takashi Iwasawa pointed out that every polyhedron circumscribed on a sphere (faces tangent to the sphere) has a surface/volume ratio of 3/r. You can see this by dividing the enclosed volume into pyramids with the faces as bases, and radii of the sphere as heights. the 3/r comes from the volume of a pyramid being B*h/3. Steve Sommars wondered about the generalization of the minimal surface property of a sphere to N dimensions. Actually, I don't know how you'd prove it in 3 dimensions, for that matter. There must be a clever way. The 3/r S/V ratio of the circumscribed polyhedra seems to provide a way, but this only shows that the sphere has the minimum surface area among a restricted class of shapes (of a given volume!) Lew Mammel, Jr. ihuxr!lew