# Pentagonal gyrocupolarotunda

Pentagonal gyrocupolarotunda
Rank3
TypeCRF
SpaceSpherical
Notation
Bowers style acronymPegycuro
Coxeter diagramxoxo5ofxx&#xt
Elements
Faces5+5+5 triangles, 5 squares, 1+1+5 pentagons
Edges5+5+5+5+10+10+10
Vertices5+5+5+10
Vertex figures5 isosceles trapezoids, edge lengths 1, 2, (1+5}0/2, 2
10 rectangles, edge lengths 1 and (1+5)/2
10 irregular tetragons, edge lengths 1, 1, 2, (1+5)/2
Measures (edge length 1)
Volume${\displaystyle 5\frac{11+5\sqrt5}{12} ≈ 9.24181}$
Dihedral angles3–4: ${\displaystyle \arccos\left(-\frac{\sqrt3+\sqrt{15}}{6}\right) ≈ 159.09484°}$
4–5 cupolaic: ${\displaystyle \arccos\left(-\sqrt{\frac{5+\sqrt5}{10}}\right) ≈ 148.28253°}$
3–5: ${\displaystyle \arccos\left(-\sqrt{\frac{5+2\sqrt5}{15}}\right) ≈ 142.62263°}$
3–3: ${\displaystyle \arccos\left(-\frac{\sqrt5}{5}\right) ≈ 116.56505°}$
4–5 join: ${\displaystyle \arccos\left(-\sqrt{\frac{25-11\sqrt5}{50}}\right) ≈ 95.15242°}$
Central density1
Related polytopes
ArmyPegycuro
RegimentPegycuro
Abstract & topological properties
Euler characteristic2
SurfaceSphere
OrientableYes
Genus0
Properties
SymmetryH2×I, order 10
ConvexYes
NatureTame

The pentagonal gyrocupolarotunda is one of the 92 Johnson solids (J33). It consists of 5+5+5 triangles, 5 squares, and 1+1+5 pentagons. It can be constructed by attaching a pentagonal cupola and a pentagonal rotunda at their decagonal bases, such that the two pentagonal bases are rotated 36° with respect to each other.

If the cupola and rotunda are joined such that the bases are in the same orientation, the result is the pentagonal orthocupolarotunda.

## Vertex coordinates

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

• ${\displaystyle \left(0,\,\sqrt{\frac{5+\sqrt5}{10}},\,\sqrt{\frac{5+2\sqrt5}{5}}\right),}$
• ${\displaystyle \left(±\frac12,\,-\sqrt{\frac{5+2\sqrt5}{20}},\,\sqrt{\frac{5+2\sqrt5}{5}}\right),}$
• ${\displaystyle \left(±\frac{1+\sqrt5}{4},\,\sqrt{\frac{5-\sqrt5}{40}},\,\sqrt{\frac{5+2\sqrt5}{5}}\right),}$
• ${\displaystyle \left(0,\,-\sqrt{\frac{5+2\sqrt5}{5}},\,\sqrt{\frac{5+\sqrt5}{10}}\right),}$
• ${\displaystyle \left(±\frac{1+\sqrt5}{4},\,\sqrt{\frac{25+11\sqrt5}{40}},\,\sqrt{\frac{5+\sqrt5}{10}}\right),}$
• ${\displaystyle \left(±\frac{3+\sqrt5}{4},\,-\sqrt{\frac{5+\sqrt5}{40}},\,\sqrt{\frac{5+\sqrt5}{10}}\right),}$
• ${\displaystyle \left(±\frac12,\,±\frac{\sqrt{5+2\sqrt5}}{2},\,0\right),}$
• ${\displaystyle \left(±\frac{3+\sqrt5}{4},\,±\sqrt{\frac{5+\sqrt5}{8}},\,0\right),}$
• ${\displaystyle \left(±\frac{1+\sqrt5}{2},\,0,\,0\right),}$
• ${\displaystyle \left(±\frac12,\,\sqrt{\frac{5+2\sqrt5}{20}},\,-\sqrt{\frac{5-\sqrt5}{10}}\right),}$
• ${\displaystyle \left(±\frac{1+\sqrt5}{4},\,-\sqrt{\frac{5+\sqrt5}{40}},\,-\sqrt{\frac{5-\sqrt5}{10}}\right),}$
• ${\displaystyle \left(0,\,-\sqrt{\frac{5+\sqrt5}{10}},\,-\sqrt{\frac{5-\sqrt5}{10}}\right).}$

## Related polyhedra

A decagonal prism can be inserted between the two halves of the pentagonal gyrocupolarotunda to produce the elongated pentagonal gyrocupolarotunda.