Does every unfolding of an icosahedron has the same number of edges to be glued to construct it back to the solid? If yes, what are those numbers for Platonic solids? If no, which unfoldings have the least number of edges to be glued for Platonic solids?
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
Every unfolding corresponds to a spanning tree on the faces of the solid, giving $F-1$ edges which are already joined. So, you still have to glue $E-F+1$ edges.
The question has been answered, but if you don't mind a bit of additional information which I find interesting: An icosahedron may be cut open and refolded to a flat, doubly covered parallelogram. I wrote a note on this, from which the figure below is copied.
Addendum. I was not addressing the issue, just showing a neat unfolding. :-) To address the issue more directly: for an icosahedron, $E=30$, $F=20$, and $E-F+1=11$. Note that in the illustrated unfolding, the cut tree is a Hamiltonian path of 11 edges. Of course, every spanning cut tree must have the same number of edges (as per Rahul's answer). So the unfolding is bounded by 22 half-edges, each corresponding to the cutting of one of the 11 edges of the path. But notice that many of the original polyhedron edges end up collinear in the unfolding. As a polygon, the unfolding has 16 edges, or 16 "half-edges" if you think of each as deriving from 8 straight slices on the icosahedron surface. So there is a sense in which one need only glue 8 segments to fold the unfolding back to the icosahedron.