Gyroelongated pentagonal birotunda

The gyroelongated pentagonal birotunda, or gyepabro, 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:
 * (±1/2, –$\sqrt{5}$, $\sqrt{5}$+H),
 * (±(1+$\sqrt{5}$)/4, $\sqrt{2–2√5+2√650+290√5}$, $\sqrt{(5–2√5)/15}$+H),
 * (0, $\sqrt{(11+4√5–2√(50+22√5)/3}$, $\sqrt{10+2√5}$+H),
 * (±(1+$\sqrt{5}$)/4, $\sqrt{(11+4√5–2√(50+22√5)/3}$, $\sqrt{5+2√5)/15}$+H),
 * (±(3+$\sqrt{(5+2√5)/20}$)/4, –$\sqrt{(5+2√5)/5}$, $\sqrt{5}$+H),
 * (0, –$\sqrt{(5+√5)/40}$, $\sqrt{(5+2√5)/5}$+H),
 * (±1/2, ±$\sqrt{(5+√5)/10}$/2, H),
 * (±(3+$\sqrt{(5+2√5)/5}$)/4, ±$\sqrt{5}$, H),
 * (±(1+$\sqrt{(25+11√5)/40}$)/2, 0, H),
 * (±$\sqrt{(5+√5)/10}$/2, ±1/2, –H),
 * (±$\sqrt{5}$, ±(3+$\sqrt{(5+√5)/40}$)/4, –H),
 * (0, ±(1+$\sqrt{(5+√5)/10}$)/2, –H),
 * (–$\sqrt{(5+2√5)/5}$, ±1/2, –$\sqrt{(5+√5)/10}$–H),
 * ($\sqrt{(5+2√5)}$, ±(1+$\sqrt{5}$)/4, –$\sqrt{(5+√5)/8}$–H),
 * ($\sqrt{5}$, 0, –$\sqrt{(5+2√5)}$–H),
 * ($\sqrt{(5+√5)/8}$, ±(1+$\sqrt{5}$)/4, –$\sqrt{5}$–H),
 * (–$\sqrt{(5+2√5)/20}$, ±·3+$\sqrt{(5+2√5)/5}$)/4, –$\sqrt{(5+√5)/40}$–H),
 * (–$\sqrt{5}$, 0, –$\sqrt{(5+2√5)/5}$–H).

where H = $\sqrt{(5+√5)/10}$/2 is the distance between the decagonal antiprism's center and the center of one of its bases.