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Suppose we have a 3D polyhedron with the following properties:

  • Each vertex has exactly 5 edges coming out.
  • Each face has exactly 3 edges.
  • No self-intersections.
  • Each vertex is locally congruent in the following sense: If you take an open ball centered at a vertex, intersect it with the polyhedron, the results will be identical up to isometries.

I have a conjecture that these properties imply that the polyhedron is a regular icosahedron.

The complication to proving it is that we are working with very little assumptions about the polyhedron here! So I emphasize: we cannot use convexity (unless we prove it applies), we cannot invoke global symmetries (unless we prove they apply), etc., etc. The Cauchy rigidity theorem seems apt here, but it relies on the shape being convex, so it cannot be invoked. For some reason, this problem has resisted a lot of attempts to tackle it.

The obstacles I am facing is that (1) I am not assuming the shape is convex, and (2) a vertex with 5 faces is not "rigid" in the sense that there are many ways to attach faces to a vertex (in fact it is possible for the vertex to be "locally non-convex").

How can one prove or disprove that the polyhedron is necessarily a regular icosahedron?

Servaes
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MaximusIdeal
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2 Answers2

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The cuboctahedron is another icosahedron that satisfies all the given requirements. One just has to divide each square face into two triangles. Here’s an image comparing it with regular icosahedron. Image source: https://en.m.wikipedia.org/wiki/Icosahedron.

cuboctahedron

Pranay
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    Thank you, I am impressed with this example! This is a bit embarrassing on my part, but I made a variant of this question that you might be interested in here: https://math.stackexchange.com/q/5015546 I suspect adding one additional property might change the scenario, but of course, you are not obligated to answer it. – MaximusIdeal Dec 25 '24 at 08:56
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In fact, it is not only the (regular) icosahedron and the cuboctahedron (with dissected squares), it rather is a full continuous gyrohedron: consider the icosahedron put into pyritohedral subsymmetry, as given in the other answer's picture by the red triangles and the respectively opposite yellow ones, use an axis of rotation through their centers, and let the remaining 6 edges (between the adjacent blue triangles) change their size continuously, while the inradius towards the red/yellow triangles adapts synchroniously so that their vertices still connect.

This shows, that the outcome does not even need to remain convex, while still being non-intersecting. However, the gyration well could be continued beyond that intersection point, then continuously continuing up to the great icosahedron (before returning again).

Cf. pyritosnub tetrahedron or pyritohedral icosahedron or (2,3,3)-twister

Also: Dr Hugo Verheyen: "The complete set of jitterbug transformers and the analysis of their motion" a. Comp. Math. Appl., 17, Pergamon, 203-250. Oxford, 1989. b. Symmetry 2: Unifying Human Understanding, ed. I. Hargittai. Pergamon, Oxford, 1989

--- rk

Dr. Richard Klitzing
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