Mercurial, the Spectre wrote:In 4D the examples Klitzing provided contain cells that are rectified prisms, which have isosceles triangles and hence not a solution.
Mercurial, the Spectre wrote:If you mean isogonal faces, then you need to make sure that the triangles and pentagons have to be regular (since heptagons and above only occur in prisms) as per symmetry group. Therefore the set of such polyhedra is equal to the isogonal polyhedra because any meetup of three rectangles implies a triangular leftover (four rectangles will give a flat tiling) and the rectangles have to be identical for the triangle to be equilateral. The same applies to hexagons, with the limit being two, and that there must be no other configuration. The only solutions are those whose symmetries are polyhedral (the Archimedean solids) and dihedral (prisms and antiprisms).
Mercurial, the Spectre wrote:(since heptagons and above only occur in prisms)
mr_e_man wrote:Here's an example of a polyhedron that's vertex-transitive but not semi-uniform. It's "between" a prism and an antiprism, with trapezoids instead of triangles. The polyhedron is vertex-transitive, but the trapezoid faces are not. That's the opposite of what I want to consider here.
Here's an example of a vertex-transitive-faced polyhedron that's not vertex-transitive. Take an octahedron, and expand it by different amounts along the three axes, placing rectangles between the triangles. The result is combinatorially the same as the small rhombicuboctahedron, and it has the same dihedral angles (135° and 144.74°), but it has less symmetry.
This illustrates something interesting. Conjecture: Any convex equiangular-faced polyhedron can be continuously deformed into a convex regular-faced polyhedron (of the same abstract combinatorial type), while keeping all face angles, dihedral angles, and solid angles constant. And a similar statement for higher-dimensional polytopes. What do you think? This would imply that CRF polytopes are equivalent to CEF polytopes, and that edge lengths are completely irrelevant.
Note, no such thing happens with the "dual" concept, "convex edge-transitive-faced" or "convex equilateral-faced" polyhedra, as shown by the rhombic dodecahedron. (A rhombus is edge-transitive.) It has four 4-gons around a vertex, but no convex regular-faced or equiangular-faced polyhedron can have four 4-gons around a vertex.
Mercurial, the Spectre wrote:Expanding an octahedron gives a rhombicuboctahedron, which has a pyritohedral subsymmetry. Of course it has half the symmetry, and can be formed from a great rhombicuboctahedron by the process of a bialternatosnub (aka. snubbing the octagons into rectangles in a way that the hexagons alternate into triangles). You're correct on this. PS: I made mistakes earlier (for heptagons and higher, I mean odd-sided polygons, and for the three rectangles, it's a cuboid if the three rectangles form no gap, otherwise you get a rhombicuboctahedron with three squares and one equilateral triangle per vertex.)
Mercurial, the Spectre wrote:For this conjecture, it's true since the set equals to a convex uniform polyhedron (Archimedeans + prisms/antiprisms).
mr_e_man wrote:Mercurial, the Spectre wrote:Expanding an octahedron gives a rhombicuboctahedron, which has a pyritohedral subsymmetry. Of course it has half the symmetry, and can be formed from a great rhombicuboctahedron by the process of a bialternatosnub (aka. snubbing the octagons into rectangles in a way that the hexagons alternate into triangles). You're correct on this. PS: I made mistakes earlier (for heptagons and higher, I mean odd-sided polygons, and for the three rectangles, it's a cuboid if the three rectangles form no gap, otherwise you get a rhombicuboctahedron with three squares and one equilateral triangle per vertex.)
I said "expand by different amounts". If one or two of these amounts is 0, we get a square orthobicupola or an elongated square bipyramid. The expanded polyhedron doesn't necessarily have cubic or pyritohedral symmetry.Mercurial, the Spectre wrote:For this conjecture, it's true since the set equals to a convex uniform polyhedron (Archimedeans + prisms/antiprisms).
No. I've already said that the polyhedron doesn't need to be vertex-transitive, and that this set contains the Johnson solids as a subset.
Klitzing wrote:But the use of Johnson solids wouldn't be allowed for 3D faces within 4D vertex transitive faced polytopes, just because those aren't vertex transitive themselves.
--- rk
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