My extremely fuzzy 4D visualization skills suggest that there are no good answers for the 5-cell. The best I've been able to do is set it on a triangular face, with a horizontal edge forming the peak. The horizontal cross-sections have order-12 symmetry; they are triangular prisms with various aspect ratios, and the one halfway up is a semiregular triangular prism (with square sides). Notice that the equator has no more symmetry than any of its parallel cross-sections. On Thu, May 14, 2020 at 1:24 AM Scott Kim <scott@scottkim.com> wrote:
I'm going to have fun working out equators of regular 4d polytopes. Probably some of you already know the answer. I know from Banchoff's hypercube movie that the balancing-on-a-vertex equator of a hypercube is an octahedron, and I can see that the balancing-on-a-facet equator of the 16-cell (cross-polytope) is a cuboctahedron. It looks to me like the equator of a a balancing-on-a-facet equator of the 24-cell is also a cuboctahedron, but a very special one — the edges of the equatorial cuboctahedron are edges of the 24-cell itself. So...what's the most symmetrical way to balance a 5-cell? It's certainly not on a vertex!
On Wed, May 13, 2020 at 7:38 PM Hans Havermann <gladhobo@bell.net> wrote:
JP: "Why isn’t this better known? Has anyone built models?"
There's a 2009 paper in the Moscow Mathematical Journal (Interlocking of convex polyhedra: towards a geometric theory of fragmented solids) by Kanel-Belov, Dyskin, Estrin, Pasternak, & Ivanov-Pogodaev that appears to be what we are discussing here.
https://www.researchgate.net/publication/23709852_Interlocking_of_Convex_Pol...
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