randy@umcp-cs.UUCP (07/02/84)
The recent mail on geometric duality brings to mind a question I've had for some time. Maybe someone out there can help me. I'm wondering what the list of all semi-regular polyhedra is. (Recall that a semi- regular polyhedron has all faces regular but not necessarily congruent, but every vertex is congruent. Thus the cuboctahedron is composed of triangles and squares with exactly two triangles and two squares meeting at every vertex.) I tried to make a list and found that there seem to be *lots* more than I expected. For example, there is an infinite set of "cylinders" built as follows: For base and top choose a regular K sided polygon. Then for the walls of the cylinder use K squares. At every vertex, you'll have one K-gon and 2 squares meeting. (You can also get a similar infinite set using triangles as wall constructors providing K > 3. Then you need 2*K triangles.) Anyway, I've found all sorts of unexpected (and rather pleasing) polygons with the help of a lisp program to generate legal possibilities. I can send details if anyone is interested. But this *must* have been done somewhere before. Anybody got pointers? Thanks. - Randy -- Randy Trigg ...!seismo!umcp-cs!randy (Usenet) randy%umcp-cs@CSNet-Relay (Arpanet)
ellis@flairvax.UUCP (Michael Ellis) (07/09/84)
The Semi-Regular polyhedra, for those who are unfamiliar with the term,
have faces that are all regular polygons, and identical vertices. From
this definition, one may infer that all edges are likewise identical,
and that the vertices all lie on a sphere.
I believe that all possible 3-D semiregulars are listed in the table below,
divided into three categories, depending on whether the symmetry groups
(not counting reflections) are A4, S4 or A5:
SYMBOL NAME F E V
3-3-3 tetrahedron 4 6 4
3-6-6 truncated tetrahedron 8 18 12
4-4-4 cube (hexahedron) 6 12 8
3-8-8 truncated cube 14 36 24
3-3-3-3 octahedron 8 12 6
4-6-6 truncated octahedron 14 36 24
3-4-3-4 cuboctahedron 14 24 12
3-4-4-4 ?? 26 48 24
4-6-8 ?? 26 72 48
3-3-3-3-4 snub cube 38 60 24
5-5-5 dodecahedron 12 30 20
3-10-10 truncated dodecahedron 32 90 60
3-3-3-3-3 icosahedron 20 30 12
5-6-6 truncated icosahedron 32 90 60
3-5-3-5 icosidodecahedron 32 60 30
3-4-5-4 ?? 62 120 60
4-6-10 ?? 62 180 120
3-3-3-3-5 snub dodecahedron 92 150 60
Besides these 18 polyhedra there are also two infinite classes, with
symmetry groups Dn (ignoring reflections):
4-4-n prisms 2+n 3n 2n
3-3-3-n antiprisms 2+2n 4n 2n
I believe these solids were all investigated by the ancient greek geometers.
Notes:
1. The symbol `3-4-5-4' means a polyhedron with a regular triangle, square,
regular pentagon, and square at every vertex. This is NOT the Schlaefli
symbolism, by which, for instance, a cube is represented {3,4}.
2. The terms An and Sn mean the Alternating and Symmetric groups of n
elements, respectively. Dn is the Dihedral group of the regular n-gon.
3. The snub polyhedra 3-3-3-3-4 and 3-3-3-3-5 have left and right handed
variants.
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One can generate all the `interesting' polyhedra from the tetrahedron
with six `operators', crudely described below:
V - swap faces and vertices (duality)
T - each edge generates a pair vertices (truncation)
A - each edge generates one new vertex
B - each n-gonal face generates n new vertices; each original edge
resulting in a square face.
C - each edge generates four new vertices
D - each n-gonal face generates n new vertices, each original edge
resulting in a pair of triangular faces.
V T T*V A=A*V B=B*V C=C*V D=D*V
3-3-3 3-3-3 3-6-6 3-6-6 3-3-3-3 3-4-3-4 4-6-6 3-3-3-3-3
3-3-3-3 4-4-4 4-6-6 3-8-8 3-4-3-4 3-4-4-4 4-6-8 3-3-3-3-4
3-3-3-3-3 5-5-5 5-6-6 3-10-10 3-5-3-5 3-4-5-4 4-6-10 3-3-3-3-5
These also operate on the infinite plane tessellations, which can be regarded
as `degenerate' polyhedra:
4-4-4-4 4-4-4-4 4-8-8 4-8-8 4-4-4-4 4-4-4-4 4-8-8 4-3-4-3-3
3-3-3-3-3-3 6-6-6 6-6-6 3-12-12 3-6-3-6 3-4-6-4 4-6-12 3-3-3-3-6
There is at least one tessellation I've omitted, namely the ugly 4-4-3-3-3.
-michael