Relative side lengths of dual dodecahedron and icosahedron

Question

If the side length of a dodecahedron is $1$, then what is the side length of its dual icosahedron whose vertices occupy the same space as the mid-points of the faces of the dodecahedron. I've read that the answer to this is $1.618$ (golden ratio). But my calculations make it $1.1708$ ($\tan 54^\circ\times\sin 58.2825^\circ$ [half the dihedral angle]). I've even made paper models and my version looks OK. If I make the icosahedron with side lengths of $1.618$ then there's no way it will fit inside the dodecahedron. What am I not understanding.

Answer 1

What they are doing is having the edges of the dual figures bisect each other. As a result, the vertices of the icosahedron and the dodecahedron together make up the vertices of a rhombic triacontahedron, I will get a picture: alright, see HHEERRE where, in the lower right of the page, they show:

A rhombic triacontahedron with an inscribed dodecahedron (blue) and icosahedron (purple). (Click here for rotating model)

You can do this yourself, make 30 rhombususeses with diagonal ratio = $\frac{1 + \sqrt 5}{2}$ out of cardboard, put it together, then you can draw the figures in one color each.

Two inexpensive books I have that I can recommend, Daud Sutton and Alan Holden