# Stella octangula

Stella octangula
Rank3
TypeRegular
SpaceSpherical
Notation
Bowers style acronymSo
Coxeter diagramβ4o3o ()
Elements
Components2 tetrahedra
Faces8 triangles
Edges12
Vertices8
Vertex figureEquilateral triangle, edge length 1
Measures (edge length 1)
Circumradius${\displaystyle \frac{\sqrt6}{4} \approx 0.61237}$
Inradius${\displaystyle \frac{\sqrt6}{12} \approx 0.20412}$
Volume${\displaystyle \frac{\sqrt2}{6} \approx 0.23570}$
Dihedral angle${\displaystyle \arccos\left(\frac13\right) \approx 70.52877^\circ}$
Central density2
Number of external pieces24
Level of complexity3
Related polytopes
ArmyCube, edge length ${\displaystyle \frac{\sqrt2}{2}}$
RegimentSo
DualStella octangula
ConjugateNone
Convex coreOctahedron
Abstract & topological properties
Flag count48
Schläfli type{3,3}
OrientableYes
Properties
SymmetryB3, order 48
ConvexNo
NatureTame

The stella octangula, stellated octahedron, so, or compound of two tetrahedra is a regular polyhedron compound. It's made out of 8 triangles, 3 joining at each vertex. It can be constructed by taking a tetrahedron and overlaying it with its central inversion. Alternatively, it can be created by stellating the octahedron.

It can also be considered an antiprism based on the compound of two digons {4/2}.

Its quotient prismatic equivalent is the hexadecachoron, which is four-dimensional.

## Representations

• (full symmetry)
• (β2β4o)
• (β2β2β)
• xo3oo3ox

## Vertex coordinates

The vertices of a stella octangula of edge length 1 can be given by:

• ${\displaystyle \left(\pm\frac{\sqrt2}{4},\,\pm\frac{\sqrt2}{4},\,\pm\frac{\sqrt2}{4}\right).}$

These arise from the fact that a tetrahedron can be constructed as the alternation of the cube. Taking even changes of sign and odd changes of sign reconstructs the two component tetrahedra.

Alternate coordinates can be derived from those of the triangle, by considering the tetrahedron as a triangular pyramid:

• ${\displaystyle \pm\left(\pm\frac12,\,-\frac{\sqrt3}{6},\,-\frac{\sqrt6}{12}\right)}$,
• ${\displaystyle \pm\left(0,\,\frac{\sqrt3}{3},\,-\frac{\sqrt6}{12}\right)}$,
• ${\displaystyle \left(0,\,0,\,\pm\frac{\sqrt6}{4}\right)}$.

These are more complicated, but generalize to two-simplex compounds of any dimension.