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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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I don't think 'proof assistant' means what you think it means. – HJRW Nov 18 '10 at 17:32
Having just spent an hour lecturing about the $p$-adic numbers, this question instantly makes me wonder about the classification of non-Archimedean solids :-/ – Kevin Buzzard Nov 18 '10 at 18:00
up vote 11 down vote accepted

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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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. – Tyler Clark Nov 18 '10 at 20:42
I'm glad you found it helpful. – Andrey Rekalo Nov 18 '10 at 21:05

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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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. – Tyler Clark Nov 18 '10 at 20:40
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. – Nathan Reading Feb 10 '15 at 14:10

Incidentally, you may be interested in the article by Joseph Malkevitch, "Milestones in the history of polyhedra," which appeared in Shaping Space: A Polyhedral Approach, Marjorie Senechal and George Fleck, editors, pages 80-92. 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.
alt text

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I will definitely try to find this article and read through it. Thanks for your feedback! – Tyler Clark Nov 18 '10 at 20:42

I use a slightly different approach than Cromwell. Please see the Exercises at the end of Chapter 5 here:

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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My proof can be found here:

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