Re: [math-fun] poly (regular) polyhedra
Rich writes: << I've been working out counts of polypolyhedra by hand. (regular polyhedra, left-right mirror pairs count as only 1) volume 1 2 3 4 5 polytetrahedra 1 1 1 3 7 polycubes 1 1 2 7 ... polyoctahedra 1 1 3 11 polydodecahedra 1 1 3 16 polyicosahedra 1 1 4 33 I find the visualization is hard, and diagrams only help a little. Notes: 3 dodecahedra will fit around an edge, and 4 will fit around a vertex. 3 icosahedra will not fit around an edge, nor 4 around a vertex. 3 icosahedra will fit around a vertex.
This might also be fun to try in the "other two geometries", S^3 and H^3, where a regular polyhedron's max # per vertex (or per edge) will depend on its size (and the question would presumably be for joining congruent regular polylhedra). H^3 as you may know is especially flexible, in that a polyhedron's size can be chosen so that *any* preassigned number (greater than the Euclidean case) will fit around an edge -- and any preassigned number at all will fit about a vertex. --Dan
On 8/3/06, Daniel Asimov <dasimov@earthlink.net> wrote: ...
H^3 as you may know is especially flexible, in that a polyhedron's size can be chosen so that *any* preassigned number (greater than the Euclidean case) will fit around an edge -- and any preassigned number at all will fit about a vertex.
5 cubes around a vertex? I don't think so! WFL
On 8/3/06, Fred lunnon <fred.lunnon@gmail.com> wrote:
On 8/3/06, Daniel Asimov <dasimov@earthlink.net> wrote: ...
H^3 as you may know is especially flexible, in that a polyhedron's size can be chosen so that *any* preassigned number (greater than the Euclidean case) will fit around an edge -- and any preassigned number at all will fit about a vertex.
5 cubes around a vertex? I don't think so! WFL
Uh-huh --- apologies, I was getting confused with tilings. WFL
participants (2)
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Daniel Asimov -
Fred lunnon