In message <ken.pledger-1E8759.firstname.lastname@example.org>, Ken Pledger <email@example.com> writes >Here's an intuitive line of thought, not a complete proof. > > Starting from a face of the cube, the four adjacent edges are >parallel. That seems to be what permits the deformation. None of the >other regular polyhedra has parallel edges adjacent to a face, so I >suspect none can be deformed.
and here's an intuitive line of thought leading to the opposite conclusion.
Among the polygons, only the triangles cannot be deformed. So any polyhedron with all triangular sides is likely to also be non-deformable.
Polyhedrons with no triangular sides are probably deformable.
Consider a dodecahedral frame standing on one face. Push down on the top face. Each of the five surrounding faces pivots around the edge in common with the top face, widening the angle between the faces. The far ends of the edges adjoining the common edge move apart, the angle between the two further edges increases to allow that. The bottom half mirrors this. The deformation can continue until the angles between the further edges becomes 180 degrees. The "equator" of the dodecahedron, which was a non-planar decahedron has become a planar pentagon.
The two further sides "lock straight" when the other angles in the face become 60 and 120 degrees (two of each). If the deformation could continue until squashed flat the larger angles would be 144 degrees, so that is not possible. However, you could now twist the top and bottom faces which would lower the overall height further. I think that allows the whole thing to be squashed flat. It will look like a pentagon with sides of length 2 containing two concentric pentagons of with sides of length 1 each rotated wrt to the outer pentagon so that its vertices are each at length one from a linked vertex of the outer pentagon. Rotating by 36 degrees appears to put that length at a little over 1, so a slightly smaller rotation should do it. -- David Hartley