5

This is a follow-up to Do these four properties imply a polyhedron is a regular icosahedron?.

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

  1. Each vertex has exactly 5 edges coming out.
  2. Each face has exactly 3 edges.
  3. No self-intersections.
  4. 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.
  5. All faces are regular polygons congruent to each other.

Previously I conjectured that properties #1-#4 imply that the polyhedron is a regular icosahedron. This turned out to be false by a counterexample. I am wondering whether property #5 salvages the conjecture.

If the salvage is successful, what is the proof that properties #1-#5 imply the polyhedron is a regular icosahedron? If the conjecture is false, what is a suitable counterexample? Please do not assume convexity, symmetry properties, etc. unless they can be proven from #1-#5.

MaximusIdeal
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  • Without giving an answer, my feeling is the conditions determine that the polyhedron is a regular polyhedron, and the regular polyhedra are already classified, and only one of them has the vertex order you require. – Jeppe Stig Nielsen Dec 25 '24 at 16:19
  • Note that the accepted answer has $4$ faces meeting at an edge. This may disqualify it, depending on the definition of "polyhedron". – mr_e_man Jan 13 '25 at 03:22
  • @mr_e_man Thank you for the note. I see what you mean, and unfortunately, I'm confused by the ambiguity of the definition of polyhedron (this could be due to my lack of competence). If anyone can provide either (1) an improved counterexample in which each edge has $2$ faces or (2) a proof of the original request, then I would switch the accepted answer to the new one. – MaximusIdeal Jan 13 '25 at 03:39

1 Answers1

6

Consider a bunch of congruent regular tetrahedra. All faces have three edges, and all faces are regular polyhedra congruent to each other. Glue two such tetrahedra along an edge. Then the vertices of this edge have exactly 5 edges coming out.

We can again glue such tetrahedra at the edges opposite this glued edge. In this way we get a chain of congruent regular tetrahedra, satisfying properties 2 and 5, and satisfying property 1 except at the ends of the chain.

If we glue each pair of tetrahedra at the same angle then also property 4 is satisfied. If we choose our angle wisely then this chain of tetrahedra will form a loop, thus satisfying property 1. If we choose it to avoid self-intersections, then also property 3 is satisfied and we are done.

To see that such a wise choice exists, for each tetrahedron consider the line segment connecting the midpoints of its two glued edges, which are opposite. We can glue together six such tetrahedra, such that these segments form a hexagonal walk around a cube, like so: enter image description here

My drawing skills are unfortunately poor, but the result will look very roughly like this: enter image description here

enter image description here

Edit: Apparently this is known as a kaleidocycle.

Servaes
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