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Does anyone know of any good resources for the proof of the number of Archimedean solids (also known as semiregular polyhedra)?

I have seen a couple of algebraic discussions but no true proof. Also, I am looking more at trying to prove it topologically, but for now, any resource will help.*

*I worked on this project a bit as an undergraduate and am just now getting back into it.

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    $\begingroup$ I don't think 'proof assistant' means what you think it means. $\endgroup$
    – HJRW
    Commented Nov 18, 2010 at 17:32
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    $\begingroup$ Having just spent an hour lecturing about the $p$-adic numbers, this question instantly makes me wonder about the classification of non-Archimedean solids :-/ $\endgroup$
    – Kevin Buzzard
    Commented Nov 18, 2010 at 18:00
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    $\begingroup$ By the way the computer programs result shows that you could get the output by hand with out the program. Just using the fact that there are between 3 and 5 polygons at a vertex and you are going thru the possibilities like a speedometer. e.g. 3,3,3 followed by 3,3,4...3,4,4...3,4,5... It would take a cople of hours to get all of the results. Gerson Sparer PhD $\endgroup$
    – user88011
    Commented Feb 22, 2016 at 17:01

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A proof of the enumeration theorem for the Archimedean solids (which basically dates back to Kepler) can be found in the beautiful book "Polyhedra" by P.R. Cromwell (Cambridge University Press 1997, pp. 162-167).

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  • $\begingroup$ This is indeed a very good proof resource. I just glanced briefly at it, but will explore it further when I get a chance. I really appreciate your suggestion - it is actually the closest to the algorithm I am using than any I have found so far. Thanks again. $\endgroup$
    – Tyler Clark
    Commented Nov 18, 2010 at 20:42
  • $\begingroup$ I'm glad you found it helpful. $\endgroup$
    – Andrey Rekalo
    Commented Nov 18, 2010 at 21:05
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Following up on Joseph's comment: Branko Grünbaum and others have pointed out that besides the 13 or 14, there are also two infinite families of polyhedra meeting the definition of Archimedean, although generally not considered to be Archimedean. Why prisms and antiprisms are excluded from the list has never been clear to me.

In any case, this is not just a historical curiosity --- in any attempt you make to classify them, you should run into these two infinite families.

If you use a modern definition, i.e. vertex-transitive, then you will also get 13 others. And a little group theory can help in the classification. If you use a more classical definition, i.e. "locally vertex-regular," you will indeed find a 14th.

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  • $\begingroup$ You are indeed correct. The algorithm I worked on with another classmate did in fact give us the prisms and antiprisms. It has been a while since I have worked on this and I am just now getting back into it. I cannot remember why we excluded the prisms and antiprisms - I will have to take a closer look at that issue. $\endgroup$
    – Tyler Clark
    Commented Nov 18, 2010 at 20:40
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    $\begingroup$ I know this is an old question, but one rationale for excluding the prisms/antiprisms (possibly even on Kepler's mind, although he could not have said it this way) is a notion of "irreducible" versus "reducible" that comes up, for example, in the classification of finite Coxeter groups or root systems. The polytopes we call Archimedean are irreducibly 3-dimensional, whereas the others admit a decomposition related to the direct-sum decomposition of R^3 into R^2 + R. $\endgroup$
    – Nathan Reading
    Commented Feb 10, 2015 at 14:10
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Incidentally, you may be interested in the article by Joseph Malkevitch, "Milestones in the history of polyhedra," which appeared in

Shaping Space: Exploring Polyhedra in Nature, Art, and the Geometrical Imagination, Marjorie Senechal, editor, pages 53-63. Springer, 2013. (Earlier edition: Birkhauser, Boston, 1988).

There he makes the case (following Grünbaum) that there should be 14 Archimedean solids rather than 13, including the pseudorhombicuboctahedron as the 14th.
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    $\begingroup$ I will definitely try to find this article and read through it. Thanks for your feedback! $\endgroup$
    – Tyler Clark
    Commented Nov 18, 2010 at 20:42
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I use a slightly different approach than Cromwell. Please see the Exercises at the end of Chapter 5 here: http://staff.imsa.edu/~vmatsko/pgsCh1-5.pdf.

This is a draft of a textbook I am writing, and currently using to teach a course on polyhedra. The level of the text is mid-level undergraduate, so strictly speaking, the Exercises are really an outline of a rigorous enumeration. Symmetry considerations are glossed over.

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    $\begingroup$ The link is dead. $\endgroup$
    – Gerry Myerson
    Commented Feb 22, 2016 at 22:13
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My proof can be found here: http://ywhmaths.webs.com/Geometry/ArchimedeanSolids.pdf $ $

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  • $\begingroup$ It appears you don't count the pseudorhombicuboctahedron (as in the answer posted by @Joseph). $\endgroup$
    – Gerry Myerson
    Commented Feb 22, 2016 at 22:08
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I write a program to find all of the sets of regular polygons whose angles sum to less than 360 degrees. I use the fact that there are between 3 and 5 such polygons. The program has nested loops and they run like a speedometer. This spits out all such sets including the antiprisms and prisms. Then using the local conditions on these results. e.g. 3.m.n is only possible at a vertex if m equals n. This knocks out all but the constructable ones. But existence proofs are necessary for these. One can use truncations and snubbing arguments to finish the proofs. I have some nice proofs for some of these using the principle of continuity. This is a rough outline. If you would like to see the output of the program, I would be happy to send it. Also a copy of the program written in qb64 can be made available. Gerson Sparer PhD

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