Gyroelongated pentagonal birotunda

The gyroelongated pentagonal birotunda is one of the 92 Johnson solids (J48). It consists of 10+10+10+10 triangles and 2+10 pentagons. It can be constructed by attaching pentagonal rotundas to the bases of the decagonal antiprism.

It is one of five Johnson solids to be chiral.

Vertex coordinates
A gyroelongated pentagonal birotunda of edge length 1 has the following vertices:
 * $$\left(±\frac12,\,-\sqrt{\frac{5+2\sqrt5}{20}},\,\sqrt{\frac{5+2\sqrt5}{5}}+H\right),$$
 * $$\left(±\frac{1+\sqrt5}{4},\,\sqrt{\frac{5-\sqrt5}{40}},\,\sqrt{\frac{5+2\sqrt5}{5}}+H\right),$$
 * $$\left(0,\,\sqrt{\frac{5+\sqrt5}{10}},\,\sqrt{\frac{5+2\sqrt5}{5}}+H\right),$$
 * $$\left(±\frac{1+\sqrt5}{4},\,\sqrt{\frac{25+11\sqrt5}{40}},\,\sqrt{\frac{5+\sqrt5}{10}}+H\right),$$
 * $$\left(±\frac{3+\sqrt5}{4},\,-\sqrt{\frac{5+\sqrt5}{40}},\,\sqrt{\frac{5+\sqrt5}{10}}+H\right),$$
 * $$\left(0,\,-\sqrt{\frac{5+2\sqrt5}{5}},\,\sqrt{\frac{5+\sqrt5}{10}}+H\right),$$
 * $$\left(±\frac12,\,±\frac{\sqrt{5+2\sqrt5}}2,\,H\right),$$
 * $$\left(±\frac{3+\sqrt5}4,\,±\sqrt{\frac{5+\sqrt5}8},\,H\right),$$
 * $$\left(±\frac{1+\sqrt5}2,\,0,\,H\right),$$
 * $$\left(±\frac{\sqrt{5+2\sqrt5}}2,\,±\frac12,\,-H\right),$$
 * $$\left(±\sqrt{\frac{5+\sqrt5}8},\,±\frac{3+\sqrt5}4,\,-H\right),$$
 * $$\left(0,\,±\frac{1+\sqrt5}2,\,-H\right),$$
 * $$\left(-\sqrt{\frac{5+2\sqrt5}{20}},\,±\frac12,\,-\sqrt{\frac{5+2\sqrt5}{5}}+H\right),$$
 * $$\left(\sqrt{\frac{5-\sqrt5}{40}},\,±\frac{1+\sqrt5}{4},\,-\sqrt{\frac{5+2\sqrt5}{5}}+H\right),$$
 * $$\left(\sqrt{\frac{5+\sqrt5}{10}},\,0,\,-\sqrt{\frac{5+2\sqrt5}{5}}+H\right),$$
 * $$\left(\sqrt{\frac{25+11\sqrt5}{40}},\,±\frac{1+\sqrt5}{4},\,-\sqrt{\frac{5+\sqrt5}{10}}+H\right),$$
 * $$\left(-\sqrt{\frac{5+\sqrt5}{40}},\,±\frac{3+\sqrt5}{4},\,-\sqrt{\frac{5+\sqrt5}{10}}+H\right),$$
 * $$\left(-\sqrt{\frac{5+2\sqrt5}{5}},\,0,\,-\sqrt{\frac{5+\sqrt5}{10}}+H\right).$$

where H = $\sqrt{5}$/2 is the distance between the decagonal antiprism's center and the center of one of its bases.