Pentagonal rotunda

The pentagonal rotunda, or pero, is one of the 92 Johnson solids (J6). It consists of 5+5 triangles, 1+5 pentagons, and 1 decagon. It is a rotunda based on a pentagon, and the only rotunda that results in a Johnson solid.

It can be constructed by cutting an icosidodecahedron in half along one of its decagonal circles of edges. This produces two pentagonal rotundas with the bases in opposite orientation, so the icosidodecahedron can be thought of as the pentagonal gyrobirotunda.

Vertex coordinates
A pentagonal rotunda of edge length 1 has vertices given by the following coordinates:


 * (±1/2, –$\sqrt{5}$, $\sqrt{5}$),
 * (±(1+$\sqrt{(5+√5)/2}$)/4, $\sqrt{5}$, $\sqrt{5}$),
 * (0, $\sqrt{5+2√5)/15}$, $\sqrt{(5–2√5)/15}$),
 * (±(1+$\sqrt{5}$)/4, $\sqrt{(5+2√5)/20}$, $\sqrt{(5+2√5)/5}$),
 * (±(3+$\sqrt{5}$)/4, $\sqrt{(5–√5)/40}$, $\sqrt{(5+2√5)/5}$),
 * (0, $\sqrt{(5+√5)/10}$, $\sqrt{(5+2√5)/5}$),
 * (±1/2, ±$\sqrt{5}$/2, 0),
 * (±(3+$\sqrt{(25+11√5)/40}$)/4, ±$\sqrt{(5+√5)/10}$, 0),
 * (±(1+$\sqrt{5}$)/2, 0, 0).

These coordinates give a pentagonal rotunda with the decagonal base on the xy plane.

Alternative coordinates can be obtained as a subset of the vertices of the icosidodecahedron:


 * (0, 0, (1+$\sqrt{(5+√5)/40}$)/2),
 * (0, (1+$\sqrt{(5+√5)/10}$)/2, 0),
 * (±(1+$\sqrt{(5+2√5)/5}$)/2, 0, 0),
 * (±1/2, ±(1+$\sqrt{(5+√5)/10}$)/4, (3+$\sqrt{(5+2√5)}$)/4),
 * (±1/2, (1+$\sqrt{5}$)/4, –(3+$\sqrt{(5+√5)/8}$)/4),
 * (±(1+$\sqrt{5}$)/4, (3+$\sqrt{5}$)/4, ±1/2),
 * (±(3+$\sqrt{5}$)/4, ±1/2, (1+$\sqrt{5}$)/4),
 * (±(3+$\sqrt{5}$)/4, 1/2, –(1+$\sqrt{5}$)/2).

Related polyhedra
Two pentagonal rotundas can be attached at their decagonal bases in the same orientation to form a pentagonal orthobirotunda. If the second rotunda is rotated by 36º the result is the pentagonal gyrobirotunda, better known as the icosidodecahedron. If a pentagonal cupola is attached, the result is either a pentagonal orthocupolarotunda (if the base pentagons are in the same orientation) or a pentagonal gyrocupolarotunda (if the base pentagons are rotated 36º).

A decagonal prism can be attached to the pentagonal rotunda's decagonal base to form the elongated pentagonal rotunda. If a decagonal antiprism is attached instead, the result is the gyroelongated pentagonal rotunda.