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I wanted to know if I could start at one point on an icosahedron and traverse to all the others sequentially without visiting any one twice, which I assume I could model as a Hamiltonian path in a graph if I correctly mapped the polyhedron to a graph.

I didn't try because I was unclear on the graph. Each face is connected to three adjacent faces, and I found keeping track of the intersections to be confusing so instead I just built an icosahedron and numbered the sides and could see that it worked: I can start at one face and end at another face on the opposite side, having only visited adjacent faces and numbered them all from 1 to 20.

What I would like to know is if I did this to an arbitrary polyhedron where all faces are triangles, Is there always a path that starts at one face, and continues using only adjacent faces, convering every one?

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  • $\begingroup$ Labeling the faces and tracing out a Hamiltonian path along the edges are very different things. The Hamiltonian Path wiki page mentions that all platonic solids have a Hamiltonian path. $\endgroup$ – John Habert Jan 31 '14 at 15:07
  • $\begingroup$ Since the faces of an icosahedron are connected in the same way as the vertices of a dodecahedron, finding a face-to-face Hamiltonian path on the icosahedron is the same as finding a vertex-to-vertex Hamiltonian path on the dodecahedron. As the Wikipedia page says, both are possible. $\endgroup$ – MJD Jan 31 '14 at 15:18
  • $\begingroup$ What if split each face of an icosahedron into 4 triangles by dividing each line at the midpoint? That results in a non-platonic solid with 80 triangles. Is there a path for that one? $\endgroup$ – Dov Feb 1 '14 at 13:44

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