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Consider the (regular) dodecahedron $D\subset\Bbb R^3$. I want to continuously deform it so that throughout the deformation

  1. it stays a convex polytope,
  2. it stays a combinatorial dodecahedron (i.e. its edge-graph does not change), and
  3. all edge lengths stay the same.

Can I do this? If No, can I do it for some other realizations of the dodecahedron that is not necessarily regular? If Yes, is this possible for all realizations?

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    $\begingroup$ Rigidity requires the graph to have at least $3n-6$ edges. Here $n=20$ and the dodecahedron has $30$ edges. So I think there are not enough edges to make it rigid. $\endgroup$
    – Joseph O'Rourke
    Commented Nov 17, 2022 at 13:39
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    $\begingroup$ @Joseph Note that this is not rigidity of frameworks. You are forgetting to count the constraints that keep the faces flat. If my calculations are correct then you get exactly as many constraints as DOFs (for every 3-polytope). $\endgroup$
    – M. Winter
    Commented Nov 17, 2022 at 13:43
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    $\begingroup$ Well then Cauchy's rigidity theorem applies: all convex polyhedra are rigid. I must still be misunderstanding... $\endgroup$
    – Joseph O'Rourke
    Commented Nov 17, 2022 at 15:09
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    $\begingroup$ @JosephO'Rourke That theorem only applies if you require the faces to keep their shape. I don't think this is what OP wants. I think what they are asking about is whether there is something similar to how you can deform a cuboid into parallelepipeds. $\endgroup$
    – Wojowu
    Commented Nov 17, 2022 at 15:16
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    $\begingroup$ @RavenclawPrefect This seems reasonable. I have seen the tetartoid which looks like its edges are of unit length, but I don't know. There is also this deformation, which is suggestive but does not work. Maybe some symmetry can be retained, but I don't care too much. $\endgroup$
    – M. Winter
    Commented Nov 18, 2022 at 10:55

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