Given the five platonic solids nested in the order from innermost to outermost of:
Icosahedron, octahedron, tetrahedron, cube, and dodecahedron --
the problem is to determine the shortest distance and also the longest possible distance between any vertex on the icosahedron, and any vertex on the dodecahedron.
The surface area of the icosahedron = 3.307 units. So since surface area of an icosahedron = 5√3 * s², I have the edge length of the icosahedron (s) = 0.618 units.
I've been given the hint that it is possible to inscribe the icosahedron within the octahedron two different ways which vary only in their rotation or "handedness.". The rotation pictured in the imagery will lead to 8 identical sets of 12 lengths per set, plus another 12 identical sets of 12 lengths each. Using the other rotation to inscribe the icosahedron, will result in an outcome containing 20 identical sets of 12 lengths each, and will not result in the longest/ shortest length possible.
So what are those distances?
