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The Johnson solids are the strictly convex polyhedra made out of regular faces, excluding the vertex-transitive ones (which instead enter into the uniform category). Victor Zalgaller proved in 1969, in his paper Convex Polyhedra With Regular Faces, that only $92$ such polyhedra existed. I've read that he did this by a tedious computer search. I’m not aware of any other document that covers this proof.

The problem is, even after scouring the internet (and even some not fully legitimate sites) for about an hour, I've been unable to find a copy of his paper anywhere (except for this one, in Russian). And I have no idea of how I would replicate the argument myself. It doesn't even seem trivial that there is a way to reduce the problem to finitely many calculations. In fact, the existence of prisms and antiprisms seems to effectively contradict that intuition: what's stopping us from taping a bunch of equilateral triangles to a $50$-gonal prism and creating a valid solid, for example? [EDIT: Ted's answer already redirects to another source explaining this.]

I'm not asking for a full rundown of the proof, that would be too much to ask. My question is:

What are the main ideas of Zalgaller's (or any other's) enumeration of the Johnson solids?

Any accessible reference is welcome.

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  • $\begingroup$ I couldn't find the English translation online either, but my university's library has a copy of the physical text. Do you have a library nearby you could reach out to? $\endgroup$
    – Santana Afton
    Commented Jan 6, 2020 at 4:15
  • $\begingroup$ @SantanaAfton I live near a university, which has a library I can access. But they unfortunately didn’t have the text either. $\endgroup$
    – ViHdzP
    Commented Jan 6, 2020 at 4:17

2 Answers 2

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A good start is this paper of Johnson, which establishes that the number of such solids is finite, apart from prisms and antiprisms. In particular, the theorem 2 should answer your question "what's stopping us from taping a bunch of equilateral triangles to a 50-gonal prism and creating a valid solid, for example?"

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  • $\begingroup$ I'm working through this paper; I've made it to the end of Lemma 1, and haven't yet verified the impossibility of $n=12$ or $14$ (a gyroelongated hexagonal or heptagonal rotunda). It would have to have $6$- or $7$-fold rotation symmetry, but not reflection symmetry since $\alpha_1\neq\alpha_2$.... Aha! Given the top $6$- or $7$-gon and its incident $3$-gons, which do have reflection symmetry, adding the $5$-gons preserves this symmetry, and there are no degrees of freedom left to break symmetry. By comparing with the pentagonal rotunda, we see that more $3$-gons don't fit between the $5$-gons. $\endgroup$
    – mr_e_man
    Commented Feb 18, 2020 at 7:21
  • $\begingroup$ The next part, "But even this is impossible since no Archimedean solid $3.11^2$ exists", doesn't make sense; a $3$-gon surrounded by three $11$-gons is valid locally, and is invalid globally only when the same configuration is repeated (because $11$ is odd). But this "trivalent" part of the argument in (i) does make sense if it's after the "tetravalent" part, where all other configurations are ruled out. Also, for Figure 8, my calculator says the angle at $D$ is $149.79^\circ$, not the $11$-gon's $147.27^\circ$ as claimed. $\endgroup$
    – mr_e_man
    Commented Mar 4, 2020 at 2:43
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    $\begingroup$ Here's a more serious problem: In Figure 14, when both vertices $V,W$ have type $(3^2.4.11)$, I find that the free angle at $Z$ (where three triangles meet) can be anything between $146.31^\circ$ and $180^\circ$, so it's not necessarily greater than $168^\circ$. In particular, another $11$-gon could fit there (though its adjacent vertices couldn't be closed up), or a triangle and a square or pentagon. $\endgroup$
    – mr_e_man
    Commented Mar 5, 2020 at 3:23
  • $\begingroup$ And in Lemma 3, vertices of type $(4.7.9)$ are not considered. This paper is flawed. But it is indeed a good start. $\endgroup$
    – mr_e_man
    Commented Mar 6, 2020 at 1:14
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I quote Robert R. Tupelo-Schneck:

The complete list [of Johnson polyhedra] was given by Norman Johnson in 1966 and proven complete by Victor Zalgaller in 1969. The proof proceeds by an exhaustive case-analysis over all ways to attach regular faces together which preserve convexity as well as the property of being noncomposite. A regular-faced polyhedron is noncomposite if there is no plane which divides the polyhedron into two regular-faced polyhedra. Zalgaller established that the noncomposite strictly convex regular-faced polyhedra were the prisms and antiprisms (excepting A4, the octahedron, which is composite) and a further 28 polyhedra. Zalgaller concluded by stating the theorem that all junctions of those polyhedra along entire faces which are strictly convex are in the list of 110 [Platonic, Archimedean, and Johnson solids], establishing the classification.

Actually, it doesn't seem like Zalgaller actually completed the details of the argument. Quoting A. V. Timofeenko:

The completeness of that list was stated in [Zalgaller 1967] on page 18 as a theorem; however, it was supplied with only a brief indication to a method of proof, namely, the listing of all possible adjoins of noncomposite polyhedra without conditional edges along entire faces.

Here, a noncomposite polyhedron is a Johnson solid that cannot divided into two of them by a plane. It only remains to see how the noncomposite polyhedra were found.

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