The bi-icositetradiminished hexacosichoron or bidex , also known as the tetragonal-antiwedge intersected pyritooctaswirlic tetracontoctachoron , is a convex noble scaliform polychoron that consists of 48 tridiminished icosahedra . Six cells join at each vertex.
It is a diminishing of the regular hexacosichoron , where 48 vertices corresponding to the vertices of two inscribed icositetrachora are removed. It can also be thought of as a snub disicositetrachoron with the vertices of an inscribed icositetrachoron removed.
Diminishing one further set of vertices corresponding to an icositetrachoron yields a tri-icositetradiminished hexacosichoron (tridex), the dual of this polychoron. Diminishing two such sets yields a quatro-icositetradiminished hexacosichoron , and diminishing all three of those here remaining subsets results in the hecatonicosachoron . In fact, all of these various steps of zero to five diminishings result in pairs of dual polychora, and do not necessarily reflect the removal of vertices. Instead, the diminished vertices create intersections that yield the aforementioned polychora.
A bi-icositetradiminished hexacosichoron of edge length 1 has vertex coordinates given by:
(
0
,
1
+
5
4
,
±
3
+
5
4
,
1
2
)
{\displaystyle \left(0,\,{\frac {1+{\sqrt {5}}}{4}},\,\pm {\frac {3+{\sqrt {5}}}{4}},\,{\frac {1}{2}}\right)}
(
0
,
−
1
+
5
4
,
3
+
5
4
,
±
1
2
)
{\displaystyle \left(0,\,-{\frac {1+{\sqrt {5}}}{4}},\,{\frac {3+{\sqrt {5}}}{4}},\,\pm {\frac {1}{2}}\right)}
(
0
,
±
1
+
5
4
,
−
3
+
5
4
,
−
1
2
)
{\displaystyle \left(0,\,\pm {\frac {1+{\sqrt {5}}}{4}},\,-{\frac {3+{\sqrt {5}}}{4}},\,-{\frac {1}{2}}\right)}
(
0
,
1
2
,
1
+
5
4
,
±
3
+
5
4
)
{\displaystyle \left(0,\,{\frac {1}{2}},\,{\frac {1+{\sqrt {5}}}{4}},\,\pm {\frac {3+{\sqrt {5}}}{4}}\right)}
(
0
,
−
1
2
,
±
1
+
5
4
,
3
+
5
4
)
{\displaystyle \left(0,\,-{\frac {1}{2}},\,\pm {\frac {1+{\sqrt {5}}}{4}},\,{\frac {3+{\sqrt {5}}}{4}}\right)}
(
0
,
±
1
2
,
−
1
+
5
4
,
−
3
+
5
4
)
{\displaystyle \left(0,\,\pm {\frac {1}{2}},\,-{\frac {1+{\sqrt {5}}}{4}},\,-{\frac {3+{\sqrt {5}}}{4}}\right)}
(
0
,
3
+
5
4
,
±
1
2
,
−
1
+
5
4
)
{\displaystyle \left(0,\,{\frac {3+{\sqrt {5}}}{4}},\,\pm {\frac {1}{2}},\,-{\frac {1+{\sqrt {5}}}{4}}\right)}
(
0
,
−
3
+
5
4
,
1
2
,
±
1
+
5
4
)
{\displaystyle \left(0,\,-{\frac {3+{\sqrt {5}}}{4}},\,{\frac {1}{2}},\,\pm {\frac {1+{\sqrt {5}}}{4}}\right)}
(
0
,
±
3
+
5
4
,
−
1
2
,
1
+
5
4
)
{\displaystyle \left(0,\,\pm {\frac {3+{\sqrt {5}}}{4}},\,-{\frac {1}{2}},\,{\frac {1+{\sqrt {5}}}{4}}\right)}
(
1
+
5
4
,
0
,
−
1
2
,
±
3
+
5
4
)
{\displaystyle \left({\frac {1+{\sqrt {5}}}{4}},\,0,\,-{\frac {1}{2}},\,\pm {\frac {3+{\sqrt {5}}}{4}}\right)}
(
−
1
+
5
4
,
0
,
±
1
2
,
3
+
5
4
)
{\displaystyle \left(-{\frac {1+{\sqrt {5}}}{4}},\,0,\,\pm {\frac {1}{2}},\,{\frac {3+{\sqrt {5}}}{4}}\right)}
(
±
1
+
5
4
,
0
,
1
2
,
−
3
+
5
4
)
{\displaystyle \left(\pm {\frac {1+{\sqrt {5}}}{4}},\,0,\,{\frac {1}{2}},\,-{\frac {3+{\sqrt {5}}}{4}}\right)}
(
1
+
5
4
,
1
2
,
±
3
+
5
4
,
0
)
{\displaystyle \left({\frac {1+{\sqrt {5}}}{4}},\,{\frac {1}{2}},\,\pm {\frac {3+{\sqrt {5}}}{4}},\,0\right)}
(
−
1
+
5
4
,
±
1
2
,
3
+
5
4
,
0
)
{\displaystyle \left(-{\frac {1+{\sqrt {5}}}{4}},\,\pm {\frac {1}{2}},\,{\frac {3+{\sqrt {5}}}{4}},\,0\right)}
(
±
1
+
5
4
,
−
1
2
,
−
3
+
5
4
,
0
)
{\displaystyle \left(\pm {\frac {1+{\sqrt {5}}}{4}},\,-{\frac {1}{2}},\,-{\frac {3+{\sqrt {5}}}{4}},\,0\right)}
(
1
+
5
4
,
3
+
5
4
,
0
,
±
1
2
)
{\displaystyle \left({\frac {1+{\sqrt {5}}}{4}},\,{\frac {3+{\sqrt {5}}}{4}},\,0,\,\pm {\frac {1}{2}}\right)}
(
−
1
+
5
4
,
±
3
+
5
4
,
0
,
1
2
)
{\displaystyle \left(-{\frac {1+{\sqrt {5}}}{4}},\,\pm {\frac {3+{\sqrt {5}}}{4}},\,0,\,{\frac {1}{2}}\right)}
(
±
1
+
5
4
,
−
3
+
5
4
,
0
,
−
1
2
)
{\displaystyle \left(\pm {\frac {1+{\sqrt {5}}}{4}},\,-{\frac {3+{\sqrt {5}}}{4}},\,0,\,-{\frac {1}{2}}\right)}
(
1
2
,
0
,
3
+
5
4
,
±
1
+
5
4
)
{\displaystyle \left({\frac {1}{2}},\,0,\,{\frac {3+{\sqrt {5}}}{4}},\,\pm {\frac {1+{\sqrt {5}}}{4}}\right)}
(
−
1
2
,
0
,
±
3
+
5
4
,
−
1
+
5
4
)
{\displaystyle \left(-{\frac {1}{2}},\,0,\,\pm {\frac {3+{\sqrt {5}}}{4}},\,-{\frac {1+{\sqrt {5}}}{4}}\right)}
(
±
1
2
,
0
,
−
3
+
5
4
,
1
+
5
4
)
{\displaystyle \left(\pm {\frac {1}{2}},\,0,\,-{\frac {3+{\sqrt {5}}}{4}},\,{\frac {1+{\sqrt {5}}}{4}}\right)}
(
1
2
,
±
1
+
5
4
,
0
,
3
+
5
4
)
{\displaystyle \left({\frac {1}{2}},\,\pm {\frac {1+{\sqrt {5}}}{4}},\,0,\,{\frac {3+{\sqrt {5}}}{4}}\right)}
(
−
1
2
,
1
+
5
4
,
0
,
±
3
+
5
4
)
{\displaystyle \left(-{\frac {1}{2}},\,{\frac {1+{\sqrt {5}}}{4}},\,0,\,\pm {\frac {3+{\sqrt {5}}}{4}}\right)}
(
±
1
2
,
−
1
+
5
4
,
0
,
−
3
+
5
4
)
{\displaystyle \left(\pm {\frac {1}{2}},\,-{\frac {1+{\sqrt {5}}}{4}},\,0,\,-{\frac {3+{\sqrt {5}}}{4}}\right)}
(
1
2
,
±
3
+
5
4
,
1
+
5
4
,
0
)
{\displaystyle \left({\frac {1}{2}},\,\pm {\frac {3+{\sqrt {5}}}{4}},\,{\frac {1+{\sqrt {5}}}{4}},\,0\right)}
(
−
1
2
,
3
+
5
4
,
±
1
+
5
4
,
0
)
{\displaystyle \left(-{\frac {1}{2}},\,{\frac {3+{\sqrt {5}}}{4}},\,\pm {\frac {1+{\sqrt {5}}}{4}},\,0\right)}
(
±
1
2
,
−
3
+
5
4
,
−
1
+
5
4
,
0
)
{\displaystyle \left(\pm {\frac {1}{2}},\,-{\frac {3+{\sqrt {5}}}{4}},\,-{\frac {1+{\sqrt {5}}}{4}},\,0\right)}
(
3
+
5
4
,
0
,
1
+
5
4
,
±
1
2
)
{\displaystyle \left({\frac {3+{\sqrt {5}}}{4}},\,0,\,{\frac {1+{\sqrt {5}}}{4}},\,\pm {\frac {1}{2}}\right)}
(
−
3
+
5
4
,
0
,
±
1
+
5
4
,
1
2
)
{\displaystyle \left(-{\frac {3+{\sqrt {5}}}{4}},\,0,\,\pm {\frac {1+{\sqrt {5}}}{4}},\,{\frac {1}{2}}\right)}
(
±
3
+
5
4
,
0
,
−
1
+
5
4
,
−
1
2
)
{\displaystyle \left(\pm {\frac {3+{\sqrt {5}}}{4}},\,0,\,-{\frac {1+{\sqrt {5}}}{4}},\,-{\frac {1}{2}}\right)}
(
3
+
5
4
,
±
1
+
5
4
,
−
1
2
,
0
)
{\displaystyle \left({\frac {3+{\sqrt {5}}}{4}},\,\pm {\frac {1+{\sqrt {5}}}{4}},\,-{\frac {1}{2}},\,0\right)}
(
−
3
+
5
4
,
1
+
5
4
,
±
1
2
,
0
)
{\displaystyle \left(-{\frac {3+{\sqrt {5}}}{4}},\,{\frac {1+{\sqrt {5}}}{4}},\,\pm {\frac {1}{2}},\,0\right)}
(
±
3
+
5
4
,
−
1
+
5
4
,
1
2
,
0
)
{\displaystyle \left(\pm {\frac {3+{\sqrt {5}}}{4}},\,-{\frac {1+{\sqrt {5}}}{4}},\,{\frac {1}{2}},\,0\right)}
(
3
+
5
4
,
1
2
,
0
,
±
1
+
5
4
)
{\displaystyle \left({\frac {3+{\sqrt {5}}}{4}},\,{\frac {1}{2}},\,0,\,\pm {\frac {1+{\sqrt {5}}}{4}}\right)}
(
−
3
+
5
4
,
±
1
2
,
0
,
−
1
+
5
4
)
{\displaystyle \left(-{\frac {3+{\sqrt {5}}}{4}},\,\pm {\frac {1}{2}},\,0,\,-{\frac {1+{\sqrt {5}}}{4}}\right)}
(
±
3
+
5
4
,
−
1
2
,
0
,
1
+
5
4
)
{\displaystyle \left(\pm {\frac {3+{\sqrt {5}}}{4}},\,-{\frac {1}{2}},\,0,\,{\frac {1+{\sqrt {5}}}{4}}\right)}
These can be obtained from the vertices of a snub disicositetrachoron by removing 24 vertices.
