# Bidodecateron

Bidodecateron
File:Bidodecateron.png
Rank5
TypeNoble
Notation
Bowers style acronymBidot
Coxeter diagramo3o3m3o3o
Elements
Tera20 triangular duotegums
Cells90 tetragonal disphenoids
Faces120 isosceles triangles
Edges30+30
Vertices12
Vertex figureJoined pentachoron
Measures (based on 2 hexatera of edge length 1)
Edge lengthsLacing edges (30): ${\frac {\sqrt {6}}{3}}\approx 0.81650$ Base edges (30): 1
Circumradius${\frac {\sqrt {15}}{6}}\approx 0.64550$ Inradius${\frac {\sqrt {6}}{6}}\approx 0.40825$ Diteral angle$\arccos \left(-{\frac {1}{3}}\right)\approx 109.47122^{\circ }$ Height${\frac {\sqrt {6}}{3}}\approx 0.81650$ Central density1
Related polytopes
ArmyBidot
RegimentBidot
DualDodecateron
Abstract & topological properties
Euler characteristic2
OrientableYes
Properties
SymmetryA5×2, order 1440
ConvexYes
NatureTame

The bidodecateron or bidot, also known as the triangular-duotegmatic icosateron or triangular duotegmatic alterprism, is a convex noble polyteron with 20 identical triangular duotegums as facets. 10 facets join at each vertex, with the vertex figure being a joined pentachoron. It can be obtained as the convex hull of a hexateron and its central inversion (or, equivalently, its dual). It is also the triangular member of an infinite series of isogonal duotegmatic alterprisms.

The ratio between the longest and shortest edges is $1:{\frac {\sqrt {6}}{2}}\approx 1:1.22474$ .

## Vertex coordinates

The vertices of a bidodecateron, based on two hexatera of edge length 1, centered at the origin, are given by:

• $\pm \left(\pm {\frac {1}{2}},\,-{\frac {\sqrt {3}}{6}},\,-{\frac {\sqrt {6}}{12}},\,-{\frac {\sqrt {10}}{20}},\,-{\frac {\sqrt {15}}{30}}\right),$ • $\pm \left(0,\,{\frac {\sqrt {3}}{3}},\,-{\frac {\sqrt {6}}{12}},\,-{\frac {\sqrt {10}}{20}},\,-{\frac {\sqrt {15}}{30}}\right),$ • $\pm \left(0,\,0,\,{\frac {\sqrt {6}}{4}},\,-{\frac {\sqrt {10}}{20}},\,-{\frac {\sqrt {15}}{30}}\right),$ • $\pm \left(0,\,0,\,0,\,{\frac {\sqrt {10}}{5}},\,-{\frac {\sqrt {15}}{30}}\right),$ • $\left(0,\,0,\,0,\,0,\,\pm {\frac {\sqrt {15}}{6}}\right).$ 