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There are 11 non-congruent nets of a cube as well as 11 distinct nets of an octahedron. Both a dodecahedron and an icosahedron have 43380 distinct nets.

Is it true that any pair of dual convex polyhedra always have the same number of distinct nets (let's forget about self-intersections for simplicity)? I believe, there should be some simple proof.

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I am not sure how to define a net in general, but if we disregard self-intersection, a net should correspond to a choice of edges that are cut, so that the faces are still connected, but there are no circles (which would prohibit the flattening of the net).

So if I regard the polyhedron as a graph, a net is a choice of edges=dual edges to remove, so that the dual graph becomes a tree. So I think that you are looking at spanning trees of graphs and their duals.

And the complement of the edges of a spanning trees viewed as dual edges form a spanning tree of the dual graph, so their number is equal.

(In particular, if you regard the corresponding edges of the cube and the octahedron, each net of the cube corresponds to a net of the octahedron where one has cut open exactly those edges that have not been cut in the cube.)

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Look at the case of two regular tetrahedra pasted together along an equilateral triangle face (bi-pyramid) vs. a triangular prism, with squares and equilateral triangles, and do the cuts for each of these convex polyhedra along edges to get their "nets."

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