If P and Q are convex polyhedra (or nondegenerate convex objects described peicewise by a finite set of algebraic surfaces of bounded degree) in 3D, then I claim that either P intersect boundary(Q) or Q intersect boundary(P) must contain at least one component that is a topological disk. That solves Propp's problem -- showing "P pierces Q" and "Q pierces P" are not simultaneously possible, because the disk will cut either P or Q, as a consequence of whatever the Jordan-curve-theorem analogue is for well-behaved surfaces and sets in 3D. Proof of claim (modulo the fact, which is probably highly apparent, that I am not a topologist): Walk round the boundary of a component of P intersect boundary(Q) . This curve must close. So you get a circle-like curve on a sphere-like-surface. So each component of P intersect boundary(Q) must be a disc with holes in it. If it has no holes, done. If it has holes, then consider one of them as part of Q intersect boundary(P) and then redo the argument starting from there recursively. In this way we keep descending the "tree of recursion" to smaller & smaller topologically-circular loops. Since for polyhedra said tree must be finite, we eventually must stop, and which point we get a circle with no holes and proof is done. For nonpolyhedral convex objects, perhaps this tree could be infinite and the argument will never hit bottom. (Can anybody construct two convex surfaces with that behavior??) However, for convex bodies described by algebraic surfaces of bounded polynomial degree, with finite number of pieces, the argument always will hit bottom. (I have implicitly used well known facts about algebraic surfaces and curves...) -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)