# Gyroelongated pentagonal cupolarotunda

Gyroelongated pentagonal cupolarotunda
Rank3
TypeCRF
Notation
Bowers style acronymGyepcuro
Elements
Faces
Edges16×5
Vertices7×5
Vertex figures10 rectangles, edge lengths 1 and (1+5)/2
20 irregular pentagons, edge lengths 1, 1, 1, 1, (1+5)/2
5 isosceles trapezoids, edge lengths 1, 2, (1+5)/2, 2
Measures (edge length 1)
Volume${\displaystyle 5{\frac {11+5{\sqrt {5}}+2{\sqrt {2-2{\sqrt {5}}+2{\sqrt {650+290{\sqrt {5}}}}}}}{12}}\approx 15.99110}$
Dihedral angles3–3 rotundaic join: ${\displaystyle \arccos \left({\sqrt {\frac {5-2{\sqrt {5}}}{15}}}\right)+\arccos \left(-{\sqrt {\frac {11+4{\sqrt {5}}-2{\sqrt {50+22{\sqrt {5}}}}}{3}}}\right)\approx 174.43432^{\circ }}$
3–3 antiprismatic: ${\displaystyle \arccos \left({\frac {1-{\sqrt {10+2{\sqrt {5}}}}}{3}}\right)\approx 159.18651^{\circ }}$
3–4 cupolaic: ${\displaystyle \arccos \left(-{\frac {{\sqrt {3}}+{\sqrt {15}}}{6}}\right)\approx 159.09484^{\circ }}$
3–5 join: ${\displaystyle \arccos \left({\frac {\sqrt {5}}{5}}\right)+\arccos \left(-{\sqrt {\frac {11+4{\sqrt {5}}-2{\sqrt {50+22{\sqrt {5}}}}}{3}}}\right)\approx 158.68159^{\circ }}$
4–5: ${\displaystyle \arccos \left(-{\sqrt {\frac {5+{\sqrt {5}}}{10}}}\right)\approx 148.28253^{\circ }}$
3–5 rotundaic: ${\displaystyle \arccos \left(-{\sqrt {\frac {5+2{\sqrt {5}}}{15}}}\right)\approx 142.62263^{\circ }}$
3–3 cupolaic join: ${\displaystyle \arccos \left({\sqrt {\frac {5+2{\sqrt {5}}}{15}}}\right)+\arccos \left(-{\sqrt {\frac {11+4{\sqrt {5}}-2{\sqrt {50+22{\sqrt {5}}}}}{3}}}\right)\approx 132.62401^{\circ }}$
3–4 join: ${\displaystyle \arccos \left(-{\sqrt {\frac {5+{\sqrt {5}}}{10}}}\right)+\arccos \left(-{\sqrt {\frac {11+4{\sqrt {5}}-2{\sqrt {50+22{\sqrt {5}}}}}{3}}}\right)\approx 126.96412^{\circ }}$
Central density1
Number of external pieces47
Level of complexity64
Related polytopes
ArmyGyepcuro
RegimentGyepcuro
Abstract & topological properties
Flag count320
Euler characteristic2
SurfaceSphere
OrientableYes
Genus0
Properties
SymmetryH2+×I, order 5
Flag orbits64
ConvexYes
NatureTame

The gyroelongated pentagonal cupolarotunda (OBSA: gyepcuro) is one of the 92 Johnson solids (J47). It consists of 7×5 triangles, 5 squares, and 1+1+5 pentagons. It can be constructed by attaching a pentagonal cupola and a pentagonal rotunda to opposite bases of the decagonal antiprism.

It is one of five Johnson solids to be chiral.

## Vertex coordinates

A gyroelongated pentagonal cupolarotunda of edge length 1 has the following vertices:

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

where ${\displaystyle H={\sqrt {\frac {-4-2{\sqrt {5}}+{\sqrt {50+22{\sqrt {5}}}}}{8}}}}$is the distance between the decagonal antiprism's center and the center of one of its bases.