To double triacontahedron

From the table of space angles we read that a space angle of the prolate rhombohedron is of the full space angle, so 20 rhombohedra form a hexecontahedron. We can then put 12 icosahedra to get a 120-hedron. Filing the niches with 20 oblate rhombohedra we get doube triacontahedron.

From double triacontahedron to fourfolded triacontahedron

But we could put 12 triacontaherda on the hexecontahedron. Then we fill niches at points of 2-fold symmetry with 2 prolate rhombohedra and then put one prolate rhombohedron at each point of 3-fold symmetry.

At this stage we put 3 dodecahedra around each point of 3-fold symmetry and then fill niches at points of 5-fold symmetry with two layers of prolate rhombohedra.

We then put two dodecahedra at points of 2-fold symmetry, a oblate rhombohedron at each point of 3-fold symmetry and an icosahedron at each point of 5-fold symmetry.

Then we fill niches with layers of oblate rhombohedra and get a fourfold triacontahedron.

Converted by