Gyroelongated pentagonal cupola

Gyroelongated pentagonal cupola
(3D model) (OFF file)
Rank	3
Type	CRF
Space	Spherical
Notation
Bowers style acronym	Gyepcu
Coxeter diagram	oxo10sox&#xt
Elements
Faces	5+5+5+10 triangles, 5 squares, 1 pentagon, 1 decagon
Edges	5+5+5+10+10+10+10
Vertices	5+5+5+10
Vertex figures	5 isosceles trapezoids, edge lengths 1, √2, (1+√5)/2, √2
	10 irregular pentagons, edge lengths 1, 1, 1, 1, √2
	10 isosceles trapezoids, edge lengths 1, 1, 1, √(5+√5)/2
Measures (edge length 1)
Volume	$\frac{5+4\sqrt5+5\sqrt{2-2\sqrt5+2\sqrt{650+290\sqrt5}}}{6} ≈ 9.07333$
Dihedral angles	3–3 antiprismatic: $\arccos\left(\frac{1-\sqrt{10+2\sqrt5}}{3}\right) ≈ 159.18651°$
	3–4 cupolaic: $\arccos\left(-\frac{\sqrt3+\sqrt{15}}{6}\right) ≈ 159.09484°$
	4–5: $\arccos\left(-\sqrt{\frac{5+\sqrt5}{10}}\right) ≈ 148.28253°$
	3–3 join: $\arccos\left(\sqrt{\frac{5+2\sqrt5}{15}}\right)+\arccos\left(-\sqrt{\frac{11+4\sqrt5-2\sqrt{50+22\sqrt5}}{3}}\right) ≈ 132.62401°$
	3–4 join: $\arccos\left(-\sqrt{\frac{5+\sqrt5}{10}}\right)+\arccos\left(-\sqrt{\frac{11+4\sqrt5-2\sqrt{50+22\sqrt5}}{3}}\right) ≈ 126.96412°$
	3–10: $\arccos\left(-\sqrt{\frac{11+4\sqrt5-2\sqrt{50+22\sqrt5}}{3}}\right) ≈ 95.24664°$
Central density	1
Related polytopes
Army	Gyepcu
Regiment	Gyepcu
Dual	Pentadeltadecapentagonal hemitrapezohedron
Conjugate	Gyroelongated retrograde pentagrammic cupola
Abstract & topological properties
Euler characteristic	2
Surface	Sphere
Orientable	Yes
Genus	0
Properties
Symmetry	H2×I, order 10
Convex	Yes
Nature	Tame

The gyroelongated pentagonal cupola is one of the 92 Johnson solids (J₂₄). It consists of 5+5+5+10 triangles, 5 squares, 1 pentagon, and 1 decagon. It can be constructed by attaching a decagonal antiprism to the decagonal base of the pentagonal cupola.

If a second cupola is attached to the other decagonal base of the antiprism, the result is the gyroelongated pentagonal bicupola.

Vertex coordinates

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

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

where H = $\sqrt{\frac{-4-2\sqrt5+\sqrt{50+22\sqrt{5}}}8}$is the distance between the decagonal antiprism's center and the center of one of its bases.

External links

• Klitzing, Richard. "gyepcu".

• Quickfur. "The Gyroelongated Pentagonal Cupola".

• Wikipedia Contributors. "Gyroelongated pentagonal cupola".
• McCooey, David. "Gyroelongated Pentagonal Cupola"