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Imagine you are constructing regular polyhedra from rigid equilateral triangles, squares, and regular pentagons. Adjacent faces will be connected edge-to-edge by freely-swinging hinges. The minimum number of faces of a regular tetrahedron one needs to put in place to produce a rigid structure is three: two faces can clearly swing, but three faces make a triangular pyramid which is rigid (because the missing face is constrained to be an equilateral triangle).

For a cube or dodecahedron, again connecting all three faces that surround one vertex is rigid by similar reasoning, so the minimum is three for these polyhedra as well.

For an octahedron, however, it's not difficult to convince yourself by comparing the octahedron and the trigonal or pentagonal bipyramid (or by building a paper model) that whenever two adjacent faces are missing, the resulting model will not be rigid. Thus, the minimum number of faces required to create a rigid model is six (delete two opposite faces).

What's the corresponding minimum for a regular icosahedron?

EDIT

There are two cases, according to whether vertices are considered to be connected (with free pivots) as well.

In the case without vertex connections, the minimum is 16. To see this, note that the icosahedron with two edge-adjacent faces missing is not rigid (as again a paper model quickly verifies). On the other hand, if two faces sharing a vertex but not an edge are missing, then the face between them is only attached at one edge and can swing. Thus, for the configuration to be rigid, no two missing faces can share a vertex. Since there are twelve vertices, that means at most four faces can be missing for rigidity. On the other hand, deleting four faces whose centers make a regular tetrahedron produces a rigid structure. So the minimum is 16.

The middle step of the above argument fails if vertices are also considered to be attached (and pivot freely), suggesting there might be a rigid structure with fewer faces in that case.

In the case with vertex connections, the minimum for an octahedron is actually four, produced by using every other face.

Arctic Char
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Glen Whitney
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  • For the case with vertex connections, someone, I believe https://math.stackexchange.com/users/248217/oscar-lanzi, briefly posted but took down a solution showing the minimum is 12, based on inscribing the icosahedron in an octahedron. If that person can re-post that answer, I will accept it (for that case). – Glen Whitney Sep 08 '21 at 23:56
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    What was wrong with the "rigidity" tag? Yes, it was a new tag, but there is a large and well-developed mathematical literature focusing on rigidity, and I searched for a tag that would bring out that aspect, which is central to this question - in fact, I'd say it's more about rigidity than polyhedra. If I missed a tag that would capture this, please direct me to it and I'll be happy to add it; but I'd like to add some additional tag that better characterizes this question, and if there is no tag that relates to the mathematical study of rigidity, that's a gap that should be filled. Thanks. – Glen Whitney Sep 09 '21 at 17:10
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    I wish you had not taken out the rigidity tag. For one thing, there is afik no criterion of rigidity connecting $V,F,E$ of Euler relation for rigidity/stability definition. I think this had tacitly been recognized by those who up voted your question. – Narasimham Sep 10 '21 at 00:13

2 Answers2

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Suppose we've removed some faces such that no two of them share an edge. Then if vertices are considered to be connected, the resulting polyhedron is still rigid, because every edge length remains fixed so we know that each hole is bounded by three edges of equal length - so it still has to be an equilateral triangle. (In other words, because the icosahedron has triangular faces, it is determined by its skeleton, so if we continue to constrain all the edges nothing can change.)

On the other hand removing two faces that do share an edge violates rigidity already, so the question is just "how many faces can be chosen on an icosahedron with no two sharing an edge?"

Since every vertex can be missing at most two triangles in such an arrangement (and each missing triangle contributes to three vertices), the best we could hope for is $12\cdot 2/3 = 8$ missing triangles. But this is possible, as shown at this gif (still frame below):

enter image description here

So $12$ remaining triangles is the maximum.

RavenclawPrefect
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enter image description here

A double cover rough tessellation / development of an icosahedron at left is shown.

From it examined by symmetry which faces need to be deleted for the single cover... in order that all neighbor triangle cells are populated in the development .

That leaves the development with two cells/triangles deleted labeled $(1,2)$ leaving 18 faces intact as shown.

Hoping the procedure serves as aid to retain/delete cells (faces).

Narasimham
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