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Can a convex regular dodecahedron be deformed into a great stellated dodecahedron while keeping all pentagons planar and all edges of nonzero length the whole time?

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    $\begingroup$ If you're like me and don't know what great stellated dodecahedron means, this will save you half a second en.m.wikipedia.org/wiki/Great_stellated_dodecahedron $\endgroup$ Aug 29, 2021 at 20:23
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    $\begingroup$ I assume (but perhaps it is better if I state and see if the original poster confirms) that the 5 points which start on a face of the dodecahedron must remain coplanar at all times, but move through non-convex positions until they finally are arranged at the vertices of a 5 pointed star. (This means that, at some times, one of these points must be on the line segment through some of the others.) $\endgroup$ Aug 30, 2021 at 1:51
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    $\begingroup$ The great stellated dodecahedron’s faces are pentagrammic, not triangular. $\endgroup$ Aug 30, 2021 at 3:19
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    $\begingroup$ @RyanBudney - here are the rules, as I understand them. Let $D$ be the usual dodecahedron, embedded in space. Consider the set of maps of the one-skeleton of $D$ into three-space so that (i) edges are sent to line segments (of positive length!) and (ii) five-cycles lie in a plane. (That is, for each map and for each five-cycle, there is some plane containing the image of the five-cycle.). We topologise the set as a subspace of $\mathbb{R}^{60}$. $\endgroup$
    – Sam Nead
    Aug 30, 2021 at 11:51
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    $\begingroup$ Another (equivalent) way to topologise the space is to use the face normals. This gives us a dimension count of 36; there are 12 faces and each requires three dimensions (unit normal (2) plus offset (1)). $\endgroup$
    – Sam Nead
    Aug 30, 2021 at 12:00

1 Answer 1

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Talking with Saul Schleimer, we came up with the following:

Orthogonally project the great stellated dodecahedron into the $z=0$ plane, choosing a direction that does not result in any zero length edges. Do the same for the dodecahedron. We can realise each of these projections as the endpoint of a homotopy through affine maps $(x,y,z) \to (x,y,(1-t)z)$, so planarity is preserved.

Now that everything is in the plane, the planarity of faces is easy to maintain. Note that the condition that an edge has length zero is codimension two. Therefore we can move the vertices of the flat dodecahedron to the vertices of the flat great stellated dodecahedron in a generic way.

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  • $\begingroup$ Oops. This shouldn’t be allowed, but I didn’t consider this particular case. I’ve edited the question. $\endgroup$ Aug 30, 2021 at 23:25
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    $\begingroup$ @DanielSebald, editing the question in a way that invalidates existing answers is not generally approved. Asking another question would probably be better. $\endgroup$
    – LSpice
    Aug 31, 2021 at 0:03
  • $\begingroup$ Okay. Good to know. $\endgroup$ Aug 31, 2021 at 0:31
  • $\begingroup$ Strong agree with LSpice here. It is not so fun to play a game where the rules change at the whim of just one of the players. :( $\endgroup$
    – Sam Nead
    Aug 31, 2021 at 8:50

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