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A deltahedron is a polyhedron whose faces are equilateral triangles. It is well-known that there are exactly eight convex deltahedra, and it is easy to find out that this was first proved by Freudenthal and van der Waerden in 1947.

Unfortunately, the paper is in a rather obscure journal , and also is written in Dutch. (Freudenthal, H; van der Waerden, B. L. (1947), "Over een bewering van Euclides ("On an Assertion of Euclid")", Simon Stevin 25: 115–128). I was not able to obtain this article. I have spent a lot of time searching elsewhere for proofs. Most books and papers that I looked at that discussed the matter just referred back to the Freudenthal-van der Waerden paper. The only proof I found was quite ad-hoc and also unpersuasive: it depended on a lot of rather handwavy assertions about the geometric form of a deltahedron that I found not at all obvious.

If you have seen the Freudenthal-van der Waerden proof, how does it go? If you have not, but you have an idea for how to prove this, I would be glad to see that too.

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The Euler characteristic gives some clues here, but this has got me playing with my daughters' "Geomag" set. This demonstrates that whilst the conclusion is "obvious" in the "traditional" way (of waving hands), it requires the geometric and metric properties of the underlying space and is not a topological theorem. For example, there is a configuration with 2 vertices of order 3 (3 edges) 2 of order 4 and 2 of order 5 - but it is not convex. It could be made convex if the faces were not equilateral. So the proof is not going to be without its technical points. – Mark Bennet Mar 18 '12 at 21:55
@MarkB Yes, quite so. Of course the first thing I looked at myself was the Euler characteristic, but it does not get you very close. There are 19 configurations of vertex degrees that satisfy the Euler criterion. Some of these can be ruled out for topological reasons, but others require specific geometric properties of $R^3$, and of course the Euler characteristic tells you nothing about whether the polyhedron will be convex. I wrote a blog article about this particular approach a few years ago. Thanks very much for your comment. – MJD Mar 18 '12 at 22:54
MarkD - I was making some notes and used the same notation as your blog post. You mention there configuration (2,1,4) (two vertices of order 3, one of order 4 and four of order 5). This reveals a different block to a proof - it is possible to make a convex solid by adding a vertex of order 6 - which in the "natural" accounting system has cost zero - but the order 6 vertex is in the middle of one edge of the convex hull, so is a vertex of the graph but not of the convex solid. – Mark Bennet Mar 19 '12 at 6:55

Our library had a copy of the Freudenthal/van der Waerden paper. The proof is straightforward and uses only elementary geometric arguments. My native tongue is German, but I could easily make out what they are saying. I have put up a pdf-version of the paper here, so you can see for yourself.

[Addendum: The paper is no longer on Blatter's site, so I have placed a copy on my own server, where it is likely to remain for a long time. —MJD]

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Thanks. I would be very grateful for that. – MJD Mar 21 '12 at 13:20
I have downloaded it, but not looked at it closely yet. It is only about six pages long. Page 120 has an excellent diagram of the rather surprising symmetry of the snub disphenoid. Thanks very much. – MJD Mar 22 '12 at 2:14
Reading mathematics in Dutch is harder than I had hoped. – MJD Apr 14 '12 at 2:12
@MJD: I don't know why I had retrieved it. At any rate, the paper is again there now. Thank you for the trouble you took. – Christian Blatter Jan 3 '14 at 10:05

According to Math Reviews, the deltahedra are discussed in Chapter 8 of A R Rajwade, Convex polyhedra with regularity conditions and Hilbert's third problem, Texts and Readings in Mathematics, 21, Hindustan Book Agency, New Delhi, 2001. viii+120 pp. ISBN: 81-85931-28-3, MR1891668 (2003b:52007).

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