I like the idea of a concave polyhedron that cannot rest on any of its faces (stably or unstably), i.e., a polyhedron none of whose faces belong to supporting planes. (Sort of like a holeyhedron, but a whole lot easier to construct.) How few faces can such a polyhedron have? Take a solid tetrahedron and excavate triangular-pyramidal holes on each of its four faces; we get a polyhedron with twelve faces that can't rest (stably or unstably) on any of them. Or: stick two tetrahedra together along a face, but give one of the tetrahedra a 180-degree twist, so that the two abutting triangular faces form a star of David. We again get a polyhedron with twelve faces that can't rest on any of them. Can anyone do better? (I can prove by brute force that the pentagram is the optimal solution to the 2D version of the question. Is there a slick way to see that a tetragram can't have the property? This last question smells like an Olympiad problem to me.) Jim Propp On Monday, October 12, 2015, Bill Gosper <billgosper@gmail.com> wrote:
Odd barrel dice have the problem that two faces are always uppermost, so how do you label them? Alternative to face-transitivity, there are fair dice which cannot rest on any face, and yet have one face uppermost. Spoiler: gosper.org/5&7dice.png --rwg I assume these are old ideas. Check out Wikipedia if you haven't seen spherical, 6 outcome dice. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun