I was thinking sloppily when I wrote (as part of the math-fun thread "5 and 7 outcome fair dice") that the pentagram is optimal. For purposes of this problem, the pentagram should be viewed as a non-convex decagon. And 10 sides isn't optimal (see below). So here are my questions: 1) Call a side of a polygon "supporting" if it's part of a support line of the polygon (that is, the polygon lies in one of the closed half-planes determined by the line) and "unsupporting" otherwise, and call a polygon "totally unsupported" if it has no supporting sides. What is the smallest number of sides a totally unsupported polygon can have? My record is 6 (a hexagon whose sides alternate between concave angles and reflex angles). 2) Call a face of a polyhedron "supporting" if it's part of a support plane of the polyhedron (that is, the polyhedron lies in one of the closed half-spaces determined by the plane) and "unsupporting" otherwise, and call a polyhedron "totally unsupported" if it has no supporting faces. What is the smallest number of faces a totally unsupported polyhedron can have? My record is 12 (see below). Also: 3) Although this is more relevant to my mistake from last night than to the questions I'm raising here, I'd still like to know the right way to see that, given points A, B, C, and D in the plane, it isn't possible for line segment AB to intersect line segment CD *and* for line segment BC to intersect line segment AD unless two or more of the points coincide. Jim On Tue, Oct 13, 2015 at 12:07 AM, James Propp <jamespropp@gmail.com> wrote:
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 https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun