Trigyrate rhombicosidodecahedron

The trigyrate rhombicosidodecahedron is one of the 92 Johnson solids (J75). It consists of 1+1+3+3+6+6 triangles, 3+3+3+3+6+6+6 squares, and 3+3+3+3 pentagons. It can be constructed by rotating three mutually non-adjacent pentagonal cupolaic caps of the small rhombicosidodecahedron by 36°.

Vertex coordinates
A trigyrate rhombicosidodecahedron of edge length 1 has vertices given by:
 * $$\left(±\frac{5+\sqrt5}{4},\,0,\,\frac{3+\sqrt5}{4}\right),$$
 * $$\left(\frac{5+\sqrt5}{4},\,0,\,-\frac{3+\sqrt5}{4}\right),$$
 * $$\left(0,\,±\frac{3+\sqrt5}{4},\,-\frac{5+\sqrt5}{4}\right),$$
 * $$\left(±\frac{3+\sqrt5}{4},\,±\frac{5+\sqrt5}{4},\,0\right),$$
 * $$\left(±\frac12,\,±\frac12,\,\frac{2+\sqrt5}{2}\right),$$
 * $$\left(\frac12,\,±\frac12,\,-\frac{2+\sqrt5}{2}\right),$$
 * $$\left(±\frac{2+\sqrt5}{2},\,±\frac12,\,±\frac12\right),$$
 * $$\left(±\frac12,\,±\frac{2+\sqrt5}{2},\,-\frac12\right),$$
 * $$\left(±\frac{3+\sqrt5}{4},\,±\frac{1+\sqrt5}{4},\,\frac{1+\sqrt5}{2}\right),$$
 * $$\left(\frac{3+\sqrt5}{4},\,±\frac{1+\sqrt5}{4},\,-\frac{1+\sqrt5}{2}\right),$$
 * $$\left(±\frac{1+\sqrt5}{2},\,±\frac{3+\sqrt5}{4},\,±\frac{1+\sqrt5}{4}\right),$$
 * $$\left(±\frac{1+\sqrt5}{4},\,±\frac{1+\sqrt5}{2},\,-\frac{3+\sqrt5}{4}\right),$$
 * $$\left(±\frac12,\,±\frac{5+4\sqrt5}{10},\,\frac{10+3\sqrt5}{10}\right),$$
 * $$\left(±\frac{1+\sqrt5}{4},\,±\frac{5+2\sqrt5}{5},\,\frac{15+\sqrt5}{20}\right),$$
 * $$\left(0,\,±\frac{15+13\sqrt5}{20},\,\frac{5+\sqrt5}{20}\right),$$
 * $$\left(-\frac{10+3\sqrt5}{10},\,±\frac12,\,-\frac{5+4\sqrt5}{10}\right),$$
 * $$\left(-\frac{15+\sqrt5}{20},\,±\frac{1+\sqrt5}{4},\,-\frac{5+2\sqrt5}{5}\right),$$
 * $$\left(-\frac{5+\sqrt5}{20},\,0,\,-\frac{15+13\sqrt5}{20}\right).$$