# Triangular hebesphenorotunda

Triangular hebesphenorotunda
Rank3
TypeCRF
Notation
Bowers style acronymThawro
Coxeter diagramxfox3oxFx&#xt
Elements
Faces
Edges4×3+4×6
Vertices3+3+6+6
Vertex figures3 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${\displaystyle {\frac {15+7{\sqrt {5}}}{6}}\approx 5.10875}$
Dihedral angles3–4 lunaic: ${\displaystyle \arccos \left(-{\frac {{\sqrt {3}}+{\sqrt {15}}}{6}}\right)\approx 159.09484^{\circ }}$
3–5 rotundaic: ${\displaystyle \arccos \left(-{\sqrt {\frac {5+2{\sqrt {5}}}{15}}}\right)\approx 142.62263^{\circ }}$
3–3: ${\displaystyle \arccos \left(-{\frac {\sqrt {5}}{3}}\right)\approx 138.18969^{\circ }}$
3–6: ${\displaystyle \arccos \left(-{\frac {\sqrt {5}}{3}}\right)\approx 138.18969^{\circ }}$
3–4 medial: ${\displaystyle \arccos \left(-{\frac {{\sqrt {15}}-{\sqrt {3}}}{6}}\right)\approx 110.95106^{\circ }}$
4-6 medial: ${\displaystyle \arccos \left(-{\frac {{\sqrt {15}}-{\sqrt {3}}}{6}}\right)\approx 110.95106^{\circ }}$
3–5 join: ${\displaystyle \arccos \left(-{\sqrt {\frac {5-2{\sqrt {5}}}{15}}}\right)\approx 100.81237^{\circ }}$
Central density1
Number of external pieces20
Level of complexity24
Related polytopes
ArmyThawro
RegimentThawro
ConjugateGreat triangular hebesphenorotunda
Abstract & topological properties
Flag count144
Euler characteristic2
SurfaceSphere
OrientableYes
Genus0
Properties
SymmetryA2×I, order 6
Flag orbits24
ConvexYes
NatureTame

The triangular hebesphenorotunda (OBSA: thawro) 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 ${\displaystyle {\frac {1+{\sqrt {5}}}{2}}}$) 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:

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