Double analysis of a 300-hedron

Let as start with a hexecontahedron. Put 12 triacontahedra on it, and two prolate rhombohedra at each 2-pol. But we could think of the solid as consisting of 12 icosahedra and 30 dodecahedra on its surface. Let us remove the icosahedra first and then the dodecahedra. We are left with a solid with 60 prolate rhombohedra on its surface. Remove these rhombohedra and we are back at the hexecontahedron.







Converted by Mathematica      May 14, 2001