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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.

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