Apologies -- I lost track of this conversation. What's our current understanding of the possible numbers of congruent faces on a convex polyhedron? Certainly all the "fair dice" qualify -- those are the polyhedra whose symmetry groups act transitively on their faces, so of course the faces are all congruent. That gives us the "2n for all n" families (which are actually a 2-parameter family whose generic members have faces with no symmetry, but which also includes the duals of the prisms and antiprisms). It also covers the 120-hedraon, which is a "DisdyakisTriacontahedron" if you want to cut rhombuses into quarters, but you can also visualize it as the barycentric subdivision of the icosahedron or dodecahedron, which is probably more accessible ("poke up" the center of each face & edge, so that each n-gon becomes 2n scalene triangles). But if you want all faces to be congruent *by symmetry*, then those are the best you can do: we know all the finite subgroups of SO(3), and those are the biggest ones available. So what was the 180-hedron that sparked this conversation? If it really exists, then it's going to have to have faces which are congruent to each other despite being in different symmetry classes, and I'd very much like to understand what that looks like -- but right now I don't. --Michael Kleber On Wed, Nov 26, 2008 at 7:01 AM, <rwg@sdf.lonestar.org> wrote:
rwg>there's "DisdyakisTriacontahedron", which looks to me like
it 120-sects its circumsphere.
Application: A puzzle box requiring an explosive decompression chamber. See http://gosper.org/slab.htm --rwg PHENOCRYSTIC PYROTECHNICS
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