Let S be a sphere of unit radius in three dimensional Euclidean space, R^3. Given a positive real number e, does there always exist a convex polyhedron P in R^3 such that: (1) S is a subset of P (2) The boundary of P is homeomorphic to the boundary of S (3) The volume of P does not exceed the volume of S by more than e? It is not required that S be tangent to any of the faces of P.
$\begingroup$
$\endgroup$
2
asked Dec 11, 2010 at 15:58
-
$\begingroup$ (2) S has no boundary, so I assume you mean boundary of P is homeomorphic to S. Otherwise, the answer is YES, and this looks an awful lot like a homework problem. $\endgroup$– Igor RivinCommented Dec 11, 2010 at 16:11
-
$\begingroup$ I interpreted "sphere" to mean "ball". I would not have answered except the OP gives his name and has asked some reasonable questions in the past. $\endgroup$– Bill JohnsonCommented Dec 11, 2010 at 16:21
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
For small $\epsilon$ let $Q$ be the convex symmetric hull of a finite $\epsilon$ net for the boundary of $S$ and let $P=(1+\epsilon) Q$.
answered Dec 11, 2010 at 16:19
-
$\begingroup$ To Bill Johnson: I should have written "ball" instead oF "sphere". Your answer beautifully clarified for me what should be the right approach to the problem. If we begin with a ball B concentric to S whose volume exceeds that of S by exactly e, then by taking a small enough positive number v, it should be possible to prove the existence of a finite v-net for B whose convex hull, H, contains S as a subset. Then H is the sought for polyhedron. $\endgroup$ Commented Dec 11, 2010 at 20:25
$\begingroup$
$\endgroup$
2
This is proposition 17 of book 12 of Euclid's elements.
answered Dec 11, 2010 at 16:39
-
$\begingroup$ aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII17.html $\endgroup$– S. Carnahan ♦Commented Dec 11, 2010 at 17:07
-
1$\begingroup$ @Scott. Proof that you can make a trivial problem hard if you look at it the wrong way. :) $\endgroup$ Commented Dec 11, 2010 at 18:26