Triaugmented truncated dodecahedron

The triaugmented truncated dodecahedron is one of the 92 Johnson solids (J71). It consists of 1+1+3+3+3+6+6+6+6 triangles, 3+6+6 squares, 3 pentagons, and 3+3+3 decagons. It can be constructed by attaching pentagonal cupolas to three mutually non-adjacent decagonal faces of the truncated dodecahedron.

Vertex coordinates
A triaugmented truncated dodecahedron of edge length 1 has vertices given by all even permutations of: plus the following additional vertices:
 * $$\left(0,\,±\frac12,\,±\frac{5+3\sqrt5}{4}\right),$$
 * $$\left(±\frac12,\,±\frac{3+\sqrt5}{4},\,±\frac{3+\sqrt5}{2}\right),$$
 * $$\left(±\frac{3+\sqrt5}{4},\,±\frac{1+\sqrt5}{2},\,±\frac{2+\sqrt5}{2}\right),$$
 * $$\left(±\frac12,\,±\frac{15+13\sqrt5}{20},\,3\frac{5+\sqrt5}{10}\right),$$
 * $$\left(±\frac{1+\sqrt5}{4},\,±\frac{25+13\sqrt5}{20},\,\frac{25+\sqrt5}{20}\right),$$
 * $$\left(0,\,±\frac{10+9\sqrt5}{10},\,\frac{15+\sqrt5}{20}\right),$$
 * $$\left(-3\frac{5+\sqrt5}{10},\,±\frac12,\,-\frac{15+13\sqrt5}{20}\right),$$
 * $$\left(-\frac{25+\sqrt5}{20},\,±\frac{1+\sqrt5}{4},\,-\frac{25+13\sqrt5}{20}\right),$$
 * $$\left(-\frac{15+\sqrt5}{20},\,0,\,-\frac{10+9\sqrt5}{10}\right).$$