# Great hexagonal hexecontahedron

Great hexagonal hexecontahedron Rank3
TypeUniform dual
SpaceSpherical
Notation
Coxeter diagramp5/3p5/2p3*a
Elements
Faces60 irregular hexagons
Edges60+60+60
Vertices20+60+24
Vertex figure20+60 triangles, 24 pentagrams
Measures (edge length 1)
Inradius$\frac{\sqrt2}{4} ≈ 0.35356$ Dihedral angle90°
Central density10
Number of pieces240
Related polytopes
DualGreat snub dodecicosidodecahedron
Abstract properties
Flag count720
Euler characteristic–16
Topological properties
OrientableYes
Properties
SymmetryH3+, order 60
ConvexNo
NatureTame

The great hexagonal hexecontahedron is a uniform dual polyhedron. It consists of 60 irregular hexagons, each with two short, two medium, and two long edges.

If its dual, the great snub dodecicosidodecahedron, has an edge length of 1, then the hexagon faces have short edge length $\frac{\sqrt{2\left(\sqrt5-1-2\sqrt{\sqrt5-2}\right)}}{4} ≈ 0.18177$ , medium edge length $\frac{\sqrt{2\left(\sqrt5-1+2\sqrt{\sqrt5-2}\right)}}{4} ≈ 0.52533$ , and long edge length $\frac{\sqrt2}{2} ≈ 0.70711$ . The hexagons have one interior angle of $\arccos\left(-\phi^{-1}\right) ≈ 128.17271°$ , one of $360°-\arccos\left(\phi^{-1}\right) ≈ 231.82792°$ , and four of 90°, where $\phi$ is the golden ratio.

The great hexagonal hexecontahedron and the cube are the only finite non-degenerate isohedral polyhedra with right dihedral angles.

## Vertex coordinates

A great hexagonal hexecontahedron with dual edge length 1 has vertex coordinates given by all even permutations of:

• $\left(±\frac{\sqrt{10}-\sqrt2}{8},\,±\frac{\sqrt{10}+\sqrt2}{8},\,0\right),$ • $\left(±\frac{1+\sqrt5}{4},\,±\frac{\sqrt{\sqrt5-1}}{4},\,0\right),$ • $\left(±\frac{\sqrt2}{4},\,±\frac{\sqrt2}{4},\,±\frac{\sqrt2}{4}\right),$ as well as all even permutations and even sign changes of:

• $\left(\frac{\sqrt{2\left(\sqrt5-2\right)}}{4},\,\frac{\sqrt{2\left(1-2\sqrt{\sqrt5-2}\right)}}{4},\,\frac{\sqrt{2\left(4-\sqrt5+2\sqrt{\sqrt5-2}\right)}}{4}\right),$ • $\left(\frac{\sqrt{10}-\sqrt2}{8},\,\frac{\sqrt{10}-3\sqrt2}{8},\,\frac{\sqrt{\sqrt5-1}}{2}\right),$ • $\left(\frac{\sqrt{2\left(4-\sqrt5-2\sqrt{\sqrt5-2}\right)}}{4},\,\frac{\sqrt{2\left(\sqrt5-2\right)}}{4},\,\frac{\sqrt{2\left(1+2\sqrt{\sqrt5-2}\right)}}{4}\right),$ • $\left(\frac{\sqrt{\sqrt5-1+2\sqrt{2\left(5\sqrt5-1\right)}}}{4},\,-\frac{\sqrt{2\left(3-\sqrt5-\sqrt{2\left(5\sqrt5-1\right)}\right)}}{4},\,\frac{1+\sqrt5}{4}\right),$ • $\left(\frac{\sqrt{\sqrt5-1-2\sqrt{2\left(5\sqrt5-1\right)}}}{4},\,\frac{\sqrt{2\left(3-\sqrt5+\sqrt{2\left(5\sqrt5-1\right)}\right)}}{4},\,\frac{1+\sqrt5}{4}\right).$ 