Compound of five octahedra

(Redirected from Small icosicosahedron)
Compound of five octahedra
Rank3
TypeWeakly regular compound
Notation
Bowers style acronymSe
Elements
Components5 octahedra
Faces40 triangles as 20 golden hexagrams
Edges60
Vertices30
Vertex figureSquare, edge length 1
Measures (edge length 1)
Circumradius${\displaystyle {\frac {\sqrt {2}}{2}}\approx 0.70711}$
Inradius${\displaystyle {\frac {\sqrt {6}}{6}}\approx 0.40825}$
Volume${\displaystyle {\frac {5{\sqrt {2}}}{3}}\approx 2.35702}$
Dihedral angle${\displaystyle \arccos \left(-{\frac {1}{3}}\right)\approx 109.47122^{\circ }}$
Central density5
Number of external pieces120
Level of complexity6
Related polytopes
ArmyId, edge length ${\displaystyle {\frac {{\sqrt {10}}-{\sqrt {2}}}{4}}}$
RegimentSe
DualCompound of five cubes
ConjugateCompound of five octahedra
Convex coreIcosahedron
Abstract & topological properties
Flag count240
Schläfli type{3,4}
OrientableYes
Properties
SymmetryH3, order 120
ConvexNo
NatureTame

The small icosicosahedron, se, or compound of five octahedra is a weakly-regular polyhedron compound. It consists of 40 triangles which form 20 coplanar pairs, combining into golden hexagrams. 4 triangles join at each vertex.

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.

It can be derived as a rectified chiricosahedron. It is also related to the icosicosahedron. If each stella octangula in the icosicosahedron is replaced with the intersection of the two tetrahedra (an octahedron), the result is a small icosicosahedron.

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

Vertex coordinates

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

• ${\displaystyle \left(\pm {\frac {\sqrt {2}}{2}},\,0,\,0\right)}$,

Plus all even permutations of:

• ${\displaystyle \left(\pm {\frac {\sqrt {2}}{4}},\,\pm {\frac {{\sqrt {2}}+{\sqrt {10}}}{8}},\,\pm {\frac {{\sqrt {10}}-{\sqrt {2}}}{8}}\right)}$.