# Compound of ten triangular prisms

(Redirected from Chirorhombicosahedron)
Compound of ten triangular prisms
Rank3
TypeUniform
Notation
Bowers style acronymKri
Elements
Components10 triangular prisms
Faces20 triangles, 30 squares
Edges30+60
Vertices60
Vertex figureIsosceles triangle, edge lengths 1, 2, 2
Measures (edge length 1)
Circumradius${\displaystyle {\frac {\sqrt {21}}{6}}\approx 0.76376}$
Volume${\displaystyle {\frac {5{\sqrt {3}}}{2}}\approx 4.33013}$
Dihedral angles4–3: 90°
4–4: 60°
Central density10
Number of external pieces180
Level of complexity30
Related polytopes
ArmySemi-uniform Ti, edge lengths ${\displaystyle {\frac {{\sqrt {15}}-{\sqrt {3}}}{6}}}$ (pentagons), ${\displaystyle {\frac {3{\sqrt {3}}-{\sqrt {15}}}{6}}}$ (between ditrigons)
RegimentKri
DualCompound of ten triangular tegums
ConjugateCompound of ten triangular prisms
Convex coreRhombic triacontahedron
Abstract & topological properties
Flag count360
OrientableYes
Properties
SymmetryH3+, order 60
ConvexNo
NatureTame

The chirorhombicosahedron, kri, or compound of ten triangular prisms is a uniform polyhedron compound. It consists of 30 squares and 20 triangles, with one triangle and two squares joining at a vertex.

Its quotient prismatic equivalent is the triangular prismatic decayottoorthowedge, which is twelve-dimensional.

## Vertex coordinates

The vertices of a chirorhombicosahedron of edge length 1 are given by all permutations of:

• ${\displaystyle \left(\pm {\frac {\sqrt {15}}{6}},\,\pm {\frac {\sqrt {3}}{6}},\,\pm {\frac {\sqrt {3}}{6}}\right),}$

plus all even permutations of:

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

## Related polyhedra

This compound is chiral. The compound of the two enantiomorphs is the disrhombicosahedron.