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Question about Archimedean solids

repoman

repoman

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The definition I am using from Wikipedia is:

an Archimedean solid is one of the 13 solids first enumerated by Archimedes. They are the semi-regular convex polyhedra composed of regular polygons meeting in identical vertices, excluding the 5 Platonic solids (which are composed of only one type of polygon) and excluding the prisms and antiprisms. They differ from the Johnson solids, whose regular polygonal faces do not meet in identical vertices
Click to expand...

800px-Polyhedron_great_rhombi_12-20_max.png

But looking at the picture of a truncated icosidodecahedron there are light and dark gray vertices that if I use a right hand rule are (4,6,10) for dark and (4,10,6) for light.

Same for the truncated cuboctahedron, which is listed as have (4,6,8) vertex figure, but seems to have that split between (4,6,8) and (4,8,6).

800px-Polyhedron_great_rhombi_6-8_max (1).png

Does this silly nitpick I have mean anything?
 
Those different orderings are for the different asymmetries of each of those kinds of vertices -- each kind is a mirror image of the other.

 Archimedean solid with duals  Catalan solid

Dual: vertex <-> face and edge <-> edge.
Dualities of Platonic solids: Tetrahedron: self-dual, cube <-> octahedron, dodecahedron <-> icosahedron.

Symmetries:
  • Isohedral - face-transitive - faces alike to within reflection
  • Isotoxal - edge-transitive - edges alike
  • Isogonal - vertex-transitive - vertices alike to within reflection
The Archimedean solids are isogonal, while the Catalan solids are isohedral. The Platonic solids are both.

Their rotation/reflection symmetry groups are the quasi-spherical ones: tetrahedral, octahedral, icosahedral.

There are some additional isogonal and isohedral polyhedra, the axially-symmetric ones.

Prism (barrel with quad sides) - Bipyramid (triangles connected to poles and equator)
Antiprism (barrel with alternating-direction triangle sides) - Trapezohedron (quads connected to poles and making a zigzag equator)

I got interested in this issue out of interest in the fair-dice problem: what kinds of dice are there were all the faces behave the same? This includes all the isohedral polyhedra (Platonic, Catalan, bipyramids, trapezohedra), and with ignoring the end-cap faces, the prisms and antiprisms. There are some degenerate cases:
  • Tetrahedron: 2-antiprism
  • Cube: 4-prism, 3-trapezohedron
  • Octahedron: 3-antiprism
Endcaps are counted with the other faces, except for the 2-antiprism, where they degenerate into lines.
 
Next is the question of why there are only 13 Archimedean solids. There are three subsets of them: tetrahedral, octahedral, and icosahedral. The octahedral and icosahedral ones have 6 members, while the tetrahedral one has only 1 member.


ReferenceTetrahedronOctahedronIcosahedron
Dual(same)
CubeDodecahedron
Vertex truncatedTruncated tetraTruncated octaTruncated icosa
Vtx trunc of dual(same)
Truncated cubeTruncated dodeca
Full vtx truncOctahedronCuboctahedronIcosadodecahedron
Edge truncatedCuboctahedronRhombicuboctaRhombiicosadodeca
Vtx edge truncatedTruncated octaTruncated cuboctaTruncated icisadodeca
SnubIcosahedronSnub cubeSnub dodeca
Pyramid dualTriakis tetraTetrakis hexaPentakis dodeca
Pyramid(same)
Triakis octaTriakis icosa
Rhombic quadCubeRhombic dodecaRhombic triaconta
Deltoidal quadRhombic dodecaDeltoidal icosatetraDeltoidal hexeconta
Split PyramidTetrakis hexaDisdyakis dodecaDisdyakis triaconta
PentagonalDodecahedronPentagonal icosatetraPentagonal hexeconta
The colored ones are the Archimedean ones, and then the Catalan ones, in corresponding order.
 
 List of Euclidean uniform tilings has plane-tiling versions of the Platonic, Archimedean, and Catalan solids, and  Uniform tilings in hyperbolic plane has hyperbolic-plane versions of them.

Which one's which can be distinguished with the help of Euler's theorem. One calculates the Euler characteristic X from number of vertices V, edges E, and faces F:

X = V - E + F

The Euler characteristic is a topological invariant, independent of splitting edges with new vertices, splitting faces with edges that connect vertices, and those operations' inverses. For spherical topology, it is 2, for genus g (how many "holes"), it is 2(1-g), for the flat plane, it is 0, and for the hyperbolic plane, it is negative.
 
I will now find what partially transitive tilings there are.

Edge-transitive means all the edges look alike. There may be one or two kinds of vertices, and likewise for faces. In general,

X = V - E + F = E * ( (sum of 1/rk over k) - 1 + (sum of 1/nk over k) )

where each kind k of face has nk vertices and edges, and each kind k of vertex has rk[/sup] faces and edges. If there are two kinds of vertices, then every face has even n, while if there are two kinds of faces, then every vertex has even r.

For all different, X = E * ( 1/r1 + 1/r2 - 1 + 1/n1 + 1/n2 )

Here, r1, r2, n1, and n2 must be at least 4. For all equal to 4, then X = 0. If any of them are greater than 4, then X < 0 -- hyperbolic.

For face-transitive (Catalan) or vertex-transitive (Archimedean), one of the pairs is equal, and for convenience, I will consider the Catalan case. The Archimedean case is its dual. r1 and r2 are different, while n1 = n2 = even.

X = E * ( 1/r1 + 1/r2 - 1 + 2/n )

Here are all the non-hyperbolic ones:
r1 = r2 = 3, n = 4 ... o polyhedron
r1 = r2 = 3, n = 6 ... - planar
r1 = 3, r2 = 4, n = 4 ... o
r1 = 3, r2 = 5, n = 4 ... o
r1 = 3, r2 = 6, n = 4 ... -
r1 = r2 = 4, n = 4 ... -

The Platonic solids and their planar and hyperbolic relatives have n1 = n2 and r1 = r2.

X = E * ( 2/r - 1 + 2/n )

Non-hyperbolic:
r = 3, n = 3 ... o ... tetrahedron
r = 3, n = 4 ... o ... cube
r = 3, n = 5 ... o ... dodecahedron
r = 3, n = 6 ... - ... hexagon tiling
r = 4, n = 3 ... o ... octahedron
r = 4, n = 4 ... - ... square tiling
r = 5, n = 3 ... o ... icosahedron
r = 6, n = 3 ... - ... triangle tiling
 
The general face-transitive case (Catalan) satisfies

X = F * ( (sum of rk over k = 1 to n) - n/2 + 1)

The vertex-transitive case (Archimedean) is its dual, so solutions automatically carry over.

I will consider triangles here, n = 3, the r's being r1, r2, r3.

For all the vertices different, then the r's are even, since for each one vertex, the other two kinds alternate around it. That makes the r values even.

X = F * ( 1/r1 + 1/r2 + 1/r3 - 1/2 )

Non-hyperbolic:
r1 = r2 = 4, r3 = 2m ... o
r1 = 4, r2 = 6, r3 = 6 ... o
r1 = 4, r2 = 6, r3 = 8 ... o
r1 = 4, r2 = 6, r3 = 10 ... o
r1 = 4, r2 = 6, r3 = 12 ... -
r1 = 4, r2 = 8, r3 = 8 ... -
r1 = 6, r2 = 6, r3 = 6 ... -

Make the second and third vertices alike: r2 = r3. That makes r2 even.

X = F * ( 1/r1 + 2/r2 - 1/2 )

Non-hyperbolic:
r1 = m, r2 = 4 ... o
r1 = 3, r2 = 6 ... o
r1 = 3, r2 = 8 ... o
r1 = 3, r2 = 10 ... o
r1 = 3, r2 = 12 ... -
r1 = 4, r2 = 6 ... o
r1 = 4, r2 = 8 ... -
r1 = 5, r2 = 6 ... o
r1 = 6, r2 = 6 ... -

For all vertices alike, r1 = r2 = r3 = r, and we get the triangle-face Platonic solids and their relatives.

X = F * ( 3/r - 1/2 )

Non-hyperbolic:
r = 3 ... o
r = 4 ... o
r = 5 ... o
r = 6 ... -
 
For quadrangles, it's much more difficult. In general, the vertices have r = r1, r2, r3, r4, and

X = F * ( 1/r1 + 1/r2 + 1/r3 + 1/r4 - 1 )

If all the vertices are different, then all the r's are even, and the only non-hyperbolic case is for all the r's equaling 4.

That is also true if two neighbors are the same, and also is the remaining neighbors are the same, making two pairs of neighboring same vertices.

So I turn to two diagonal vertices being the same: r2 = r4. That makes r2 even. Deltoids or kites.

X = F * ( 1/r1 + 2/r2 + 1/r3 - 1 )

Non-hyperbolic:
r1 = 3, r2 = 4, r3 = 3 ... o
r1 = 3, r2 = 4, r3 = 4 ... o
r1 = 3, r2 = 4, r3 = 5 ... o
r1 = 3, r2 = 4, r3 = 6 ... -
r1 = 3, r2 = 6, r3 = 3 ... -
r1 = 4, r2 = 4, r3 = 4 ... -

Now for each pair of diagonal vertices being the same: r1 = r3, r2 = r4. Rhombi (rhombuses).

X = F * ( 2/r1 + 2/r2 - 1 )

Non-hyperbolic:
r1 = 3, r2 = 3 ... o
r1 = 3, r2 = 4 ... o
r1 = 3, r2 = 5 ... o
r1 = 3, r2 = 6 ... -
r1 = 4, r2 = 4 ... -

For three of the four vertices the same: r2 = r3 = r4. Kites again.

X = F * ( 1/r1 + 3/r2 - 1 )

Non-hyperbolic:
r1 = m, r2 = 3 ... o
r1 = 3, r2 = 4 ... o
r1 = 4, r2 = 4 ... -

For all four of the vertices the same:

X = F * ( 4/r - 1 )

Non-hyperbolic:
r = 3 ... o
r = 4 ... -
 
Pentagons are even more difficult, and the vertices have r = r1, r2, r3, r4, r5 and

X = F * ( 1/r1 + 1/r2 + 1/r3 + 1/r4 + 1/r5 - 3/2 )

In the general case, all the r's are even, making all solutions hyperbolic.

For two alike (12344), r4 = r5, and at least r1, r2, and r3 are even, making all solutions hyperbolic.

X = F * ( 1/r1 + 1/r2 + 1/r3 + 2/r4 - 3/2 )

An alternate for two alike (12343), r3 = r5, and at least r1 and r2 are even.

X = F * (1/r1 + r/r2 + 2/r3 + 1/r4 - 3/2)

Non-hyperbolic:
r1 = r2 = 4, r3 = r4 = 3 ... -

For three alike (12333), r3 = r4 = r5, and at least r1 and r2 are even.

X = F * ( 1/r1 + 1/r2 + 3/r3 - 3/2 )

Non-hyperbolic:
r1 = r2 = 4, r3 = 3 ... -

An alternate for three alike (12322), r2 = r4 = 45

X = F * ( 1/r1 + 3/r2 + 1/r3 - 3/2 )

Non-hyperbolic:
r1 = 3, r2 = 3, r3 = 3 ... o
r1 = 3, r2 = 3, r3 = 4 ... o
r1 = 3, r2 = 3, r3 = 5 ... o
r1 = 3, r2 = 3, r3 = 6 ... -
r4 = 3, r2 = 3, r3 = 4 ... -

For four alike (12222): r2 = r3 = r4 = r5

X = F * ( 1/r1 + 4/r2 - 3/2 )

Non-hyperbolic:
r1 = 3, r2 = 3 ... o
r1 = 4, r2 = 3 ... o
r1 = 5, r2 = 3 ... o
r1 = 6, r2 = 3 ... -

For two pairs (12233): r2 = r3, r4 = r5, r1 even

X = F * ( 1/r1 + 2/r2 + 2/r4 - 3/2 )

Non-hyperbolic:
r1 = 4, r2 = r3 = 3 ... o

For two pairs (12323): r2 = r4, r3 = r5, r1 even

X = F * ( 1/r1 + 2/r2 + 2/r3 - 3/2 )

Non-hyperbolic:
r1 = 4, r2 = r3 = 3 ... o

For two pairs (12332): r2 = r5, r3 = r4, r2, r3 even

X = F * ( 1/r1 + 2/r2 + 2/r3 - 3/2 )

All hyperbolic.

For three and two (11222): r1 = r2, r3 = r4 = r5, r1 even

Non-hyperbolic:
r1 = 4, r3 = 3 ... -

For three and two (12122): r1 = r3, r2 = r4 = r5

Non-hyperbolic:
r1 = 4, r2 = 3 ... -

For all five equal: r1 = r2 = r3 = r4 = r5

Non-hyperbolic:
r = 3 ... -
 
For hexagons, the only non-hyperbolic solutions have r = 3 for all vertices, making a planar shape: -

For different kinds of vertices, one has 121212, alternation between the two kinds.

For higher n-gons, all solutions are hyperbolic.

I have thus derived all 5 of the Platonic solids, all 13 of the Archimedean and Catalan solids, the 4 infinite families of axially-symmetric relatives, and also their plane-tiling counterparts.
 
Are there 4-D analogues to Archimedean solids?

Can't find any, but I may not be looking with the correct terminology.
 
repoman said:
Are there 4-D analogues to Archimedean solids?

Can't find any, but I may not be looking with the correct terminology.
Click to expand...

"Archimedean Polychora" are a subcategory of "Uniform Polychora" aka "Uniform 4-Polytopes". I believe there are 47 of them in 4-D, depending on how you count.
 
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