Compound of five cubes

(Redirected from Rhombihedron)
Compound of five cubes
Rank3
TypeWeakly regular compound
Notation
Bowers style acronymRhom
Elements
Components5 cubes
Faces30 squares
Edges60
Vertices20
Vertex figureGolden hexagram, edge length 2
Measures (edge length 1)
Circumradius${\displaystyle {\frac {\sqrt {3}}{2}}\approx 0.86603}$
Inradius${\displaystyle {\frac {1}{2}}=0.5}$
Volume5
Dihedral angle90°
Central density5
Number of external pieces360
Level of complexity18
Related polytopes
ArmyDoe, edge length ${\displaystyle {\frac {{\sqrt {5}}-1}{2}}}$
RegimentSidtid
DualCompound of five octahedra
ConjugateCompound of five cubes
Convex coreRhombic triacontahedron
Abstract & topological properties
Flag count240
Schläfli type{4,3}
OrientableYes
Properties
SymmetryH3, order 120
Flag orbits2
ConvexNo
NatureTame
History
Discovered byEdmond Hess
First discovered1876

The rhombihedron, rhom, or compound of five cubes is a weakly-regular polyhedron compound. It consists of 30 squares. The vertices coincide in pairs, leading to 20 vertices where 6 squares join.

It has the same edges as the small ditrigonary icosidodecahedron.

This compound is sometimes called regular, but it is not flag-transitive, despite the fact it is vertex-, edge-, and face-transitive. It is however regular if you consider conjugacies along with its other symmetries.

Its quotient prismatic equivalent is the cubic pentagyroprism, which is seven-dimensional.

Vertex coordinates

The vertices of a rhombihedron of edge length 1 are given by:

• ${\displaystyle \left(\pm {\frac {1}{2}},\,\pm {\frac {1}{2}},\,\pm {\frac {1}{2}}\right)}$,

along with all even permutations of:

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