Can a dodecahedron be deformed into a great stellated dodecahedron?

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?

• 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 Aug 29, 2021 at 20:23
• 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.) Aug 30, 2021 at 1:51
• The great stellated dodecahedron’s faces are pentagrammic, not triangular. Aug 30, 2021 at 3:19
• @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 $\mathbb{R}^{60}$. Aug 30, 2021 at 11:51
• 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)). Aug 30, 2021 at 12:00

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.