Triangular hebesphenorotunda

Triangular hebesphenorotunda
(3D model) (OFF file)
Rank	3
Type	CRF
Notation
Bowers style acronym	Thawro
Coxeter diagram	xfox3oxFx&#xt
Elements
Faces	1+3+3+6 triangles, 3 squares, 3 pentagons, 1 hexagon
Edges	3+3+3+3+6+6+6+6
Vertices	3+3+6+6
Vertex figures	3 rectangles, edge lengths 1 and (1+√5)/2
	6 tetragons, edge lengths 1, √2, 1, (1+√5)/2
	3 trapezoids, edge lengths 1, 1, 1, (1+√5)/2
	6 tetragons, edge lengths 1, 1, √2, √3
Measures (edge length 1)
Volume	${\frac {15+7{\sqrt {5}}}{6}}\approx 5.10875$
Dihedral angles	3–4 lunaic: $\arccos \left(-{\frac {{\sqrt {3}}+{\sqrt {15}}}{6}}\right)\approx 159.09484^{\circ }$
	3–5 rotundaic: $\arccos \left(-{\sqrt {\frac {5+2{\sqrt {5}}}{15}}}\right)\approx 142.62263^{\circ }$
	3–3: $\arccos \left(-{\frac {\sqrt {5}}{3}}\right)\approx 138.18969^{\circ }$
	3–6: $\arccos \left(-{\frac {\sqrt {5}}{3}}\right)\approx 138.18969^{\circ }$
	3–4 medial: $\arccos \left(-{\frac {{\sqrt {15}}-{\sqrt {3}}}{6}}\right)\approx 110.95106^{\circ }$
	4-6 medial: $\arccos \left(-{\frac {{\sqrt {15}}-{\sqrt {3}}}{6}}\right)\approx 110.95106^{\circ }$
	3–5 join: $\arccos \left(-{\sqrt {\frac {5-2{\sqrt {5}}}{15}}}\right)\approx 100.81237^{\circ }$
Central density	1
Number of external pieces	20
Level of complexity	24
Related polytopes
Army	Thawro
Regiment	Thawro
Dual	Hexatetragotrideltohexatetragotrideltoidal octadecahedron
Conjugate	Great triangular hebesphenorotunda
Abstract & topological properties
Flag count	144
Euler characteristic	2
Surface	Sphere
Orientable	Yes
Genus	0
Properties
Symmetry	A2×I, order 6
Convex	Yes
Nature	Tame

The triangular hebesphenorotunda is one of the 92 Johnson solids (J₉₂). It consists of 1+3+3+6 triangles, 3 squares, 3 pentagons, and 1 hexagon.

It is one of several polyhedra near the end of the list of Johnson solids with no obvious relation to the uniform polyhedra. However, there are hidden connections to the icosidodecahedron. The part around the top triangle is congruent with the corresponding part of an icosidodecahedron. In fact, it can be thought of as an icosidodecahedron cut in half in triangular symmetry, with the bottom face (a hexagon of edge length ${\frac {1+{\sqrt {5}}}{2}}$) shrunken down to unit edge length.

The triangular hebesphenorotunda also has a connection with the regular icosahedron, being a partial Stott expansion of a triangular-symmetric faceting of the icosahedron. Because of this, the clusters of triangles at the 3.3.3.5 vertices are congruent with the corresponding triangles of an icosahedron.

The triangular hebesphenorotunda also has a minor connection with the small rhombicosidodecahedron. The triangles and squares forming the "lune" portion of the shape are congruent with corresponding triangles and squares of the small rhombicosidodecahedron.

Vertex coordinates

A triangular hebesphenorotunda of edge length 1 has vertices given by the following coordinates:

• $\left(\pm {\frac {1}{2}},\,-{\frac {\sqrt {3}}{6}},\,{\frac {3{\sqrt {3}}+{\sqrt {15}}}{6}}\right)$,
• $\left(0,\,{\frac {\sqrt {3}}{3}},\,{\frac {3{\sqrt {3}}+{\sqrt {15}}}{6}}\right)$,
• $\left(\pm {\frac {1}{2}},\,{\frac {2{\sqrt {3}}+{\sqrt {15}}}{6}},\,{\frac {{\sqrt {3}}+{\sqrt {15}}}{6}}\right)$,
• $\left(\pm {\frac {3+{\sqrt {5}}}{4}},\,-{\frac {{\sqrt {15}}-{\sqrt {3}}}{12}},\,{\frac {{\sqrt {3}}+{\sqrt {15}}}{6}}\right)$,
• $\left(\pm {\frac {1+{\sqrt {5}}}{4}},\,-{\frac {5{\sqrt {3}}+{\sqrt {15}}}{12}},\,{\frac {{\sqrt {3}}+{\sqrt {15}}}{6}}\right)$,
• $\left(\pm {\frac {3+{\sqrt {5}}}{4}},\,{\frac {3{\sqrt {3}}+{\sqrt {15}}}{12}},\,{\frac {\sqrt {3}}{3}}\right)$,
• $\left(0,\,-{\frac {3{\sqrt {3}}+{\sqrt {15}}}{6}},\,{\frac {\sqrt {3}}{3}}\right)$,
• $\left(\pm {\frac {1}{2}},\,\pm {\frac {\sqrt {3}}{2}},\,0\right)$,
• $\left(\pm 1,\,0,\,0\right)$.

External links

• Klitzing, Richard. "thawro".

• Quickfur. "The Triangular Hebesphenorotunda".

• Wikipedia Contributors. "Triangular hebesphenorotunda".
• McCooey, David. "Triangular Hebesphenorotunda"