Prismatorhombated hecatonicosachoron Rank 4 Type Uniform Space Spherical Notation Bowers style acronym Prahi Coxeter diagram x5o3x3x ( Elements Cells 720 pentagonal prisms , 1200 hexagonal prisms , 600 truncated tetrahedra , 120 small rhombicosidodecahedra Faces 2400 triangles , 3600+3600 squares , 1440 pentagons , 2400 hexagons Edges 3600+7200+7200 Vertices 7200 Vertex figure Isosceles trapezoidal pyramid , base edge lengths 1, √2 , (1+√5 )/2, √2 ; lateral edge lengths √2 , √2 , √3 , √3 Measures (edge length 1) Circumradius
5
2
+
3
10
2
≈
8.27895
{\displaystyle \frac{5\sqrt2+3\sqrt{10}}{2} ≈ 8.27895}
Hypervolume
25
1814
+
779
5
4
≈
22224.3560
{\displaystyle 25\frac{1814+779\sqrt5}{4} ≈ 22224.3560}
Dichoral angles Tut–6–hip:
arccos
(
−
6
+
30
8
)
≈
172.23876
∘
{\displaystyle \arccos\left(-\frac{\sqrt6+\sqrt{30}}{8}\right) ≈ 172.23876^\circ}
Pip–4–hip:
arccos
(
−
10
+
2
5
15
)
≈
169.18768
∘
{\displaystyle \arccos\left(-\sqrt{\frac{10+2\sqrt5}{15}}\right) ≈ 169.18768^\circ}
Srid–5–pip: 162° Srid–4–hip:
arccos
(
−
3
+
15
6
)
≈
159.09484
∘
{\displaystyle \arccos\left(-\frac{\sqrt3+\sqrt{15}}{6}\right)≈ 159.09484^\circ}
Srid–3–tut:
arccos
(
−
7
+
3
5
4
)
≈
157.76124
∘
{\displaystyle \arccos\left(-\frac{\sqrt{7+3\sqrt5}}{4}\right) ≈ 157.76124^\circ}
Central density 1 Number of external pieces 2640 Level of complexity 16 Related polytopes Army Prahi Regiment Prahi Dual Deltopyramidal heptachiliadiacosichoron Conjugate Prismatoquasirhombated great grand stellated hecatonicosachoron Abstract & topological properties Flag count230400 Euler characteristic 0 Orientable Yes Properties Symmetry H4 , order 14400Convex Yes Nature Tame
The prismatorhombated hecatonicosachoron , or prahi , also commonly called the runcitruncated 600-cell , is a convex uniform polychoron that consists of 720 pentagonal prisms , 1200 hexagonal prisms , 600 truncated tetrahedra , and 120 small rhombicosidodecahedra . 1 pentagonal prism, 2 hexagonal prisms, 1 truncated tetrahedron, and 1 small rhombicosidodecahedron join at each vertex. As one of its names suggests, it can be obtained by runcitruncating the hexacosichoron .
The vertices of a prismatorhombated hecatonicosachoron of edge length 1 are given by all even permutations of:
(
±
1
2
,
±
1
2
,
±
2
+
5
2
,
±
7
+
4
5
2
)
,
{\displaystyle \left(±\frac12,\,±\frac12,\,±\frac{2+\sqrt5}{2},\,±\frac{7+4\sqrt5}{2}\right),}
(
±
1
2
,
±
1
2
,
±
3
+
2
5
2
,
±
8
+
3
5
2
)
,
{\displaystyle \left(±\frac12,\,±\frac12,\,±\frac{3+2\sqrt5}{2},\,±\frac{8+3\sqrt5}{2}\right),}
(
±
3
+
5
4
,
±
7
+
5
4
,
±
11
+
5
5
4
,
±
11
+
5
5
4
)
,
{\displaystyle \left(±\frac{3+\sqrt5}{4},\,±\frac{7+\sqrt5}{4},\,±\frac{11+5\sqrt5}{4},\,±\frac{11+5\sqrt5}{4}\right),}
(
±
5
+
5
4
,
±
5
+
5
4
,
±
9
+
5
5
4
,
±
13
+
5
5
4
)
,
{\displaystyle \left(±\frac{5+\sqrt5}{4},\,±\frac{5+\sqrt5}{4},\,±\frac{9+5\sqrt5}{4},\,±\frac{13+5\sqrt5}{4}\right),}
(
±
3
1
+
5
4
,
±
7
+
3
5
4
,
±
9
+
5
5
4
,
±
9
+
5
5
4
)
,
{\displaystyle \left(±3\frac{1+\sqrt5}{4},\,±\frac{7+3\sqrt5}{4},\,±\frac{9+5\sqrt5}{4},\,±\frac{9+5\sqrt5}{4}\right),}
(
±
5
+
3
5
4
,
±
5
+
3
5
4
,
±
7
+
5
5
4
,
±
11
+
5
5
4
)
,
{\displaystyle \left(±\frac{5+3\sqrt5}{4},\,±\frac{5+3\sqrt5}{4},\,±\frac{7+5\sqrt5}{4},\,±\frac{11+5\sqrt5}{4}\right),}
(
0
,
±
1
2
,
±
9
+
5
5
4
,
±
5
3
+
5
4
)
,
{\displaystyle \left(0,\,±\frac12,\,±\frac{9+5\sqrt5}{4},\,±5\frac{3+\sqrt5}{4}\right),}
(
0
,
±
3
+
5
4
,
±
7
+
4
5
2
,
±
5
+
5
4
)
,
{\displaystyle \left(0,\,±\frac{3+\sqrt5}{4},\,±\frac{7+4\sqrt5}{2},\,±\frac{5+\sqrt5}{4}\right),}
(
0
,
±
2
+
5
2
,
±
15
+
7
5
4
,
±
7
+
5
4
)
,
{\displaystyle \left(0,\,±\frac{2+\sqrt5}{2},\,±\frac{15+7\sqrt5}{4},\,±\frac{7+\sqrt5}{4}\right),}
(
0
,
±
3
+
5
2
,
±
5
+
3
5
2
,
±
(
3
+
5
)
)
,
{\displaystyle \left(0,\,±\frac{3+\sqrt5}{2},\,±\frac{5+3\sqrt5}{2},\,±(3+\sqrt5)\right),}
(
0
,
±
5
+
3
5
4
,
±
3
2
+
5
2
,
±
11
+
3
5
4
)
,
{\displaystyle \left(0,\,±\frac{5+3\sqrt5}{4},\,±3\frac{2+\sqrt5}{2},\,±\frac{11+3\sqrt5}{4}\right),}
(
±
1
2
,
±
1
+
5
4
,
±
7
+
3
5
2
,
±
7
+
5
5
4
)
,
{\displaystyle \left(±\frac12,\,±\frac{1+\sqrt5}{4},\,±\frac{7+3\sqrt5}{2},\,±\frac{7+5\sqrt5}{4}\right),}
(
±
1
2
,
±
1
,
±
11
+
5
5
4
,
±
13
+
5
5
4
)
,
{\displaystyle \left(±\frac12,\,±1,\,±\frac{11+5\sqrt5}{4},\,±\frac{13+5\sqrt5}{4}\right),}
(
±
1
2
,
±
1
,
±
5
+
3
5
4
,
±
15
+
7
5
4
)
,
{\displaystyle \left(±\frac12,\,±1,\,±\frac{5+3\sqrt5}{4},\,±\frac{15+7\sqrt5}{4}\right),}
(
±
1
2
,
±
5
+
5
4
,
±
5
+
3
5
2
,
±
11
+
5
5
4
)
,
{\displaystyle \left(±\frac12,\,±\frac{5+\sqrt5}{4},\,±\frac{5+3\sqrt5}{2},\,±\frac{11+5\sqrt5}{4}\right),}
(
±
1
2
,
±
2
+
5
2
,
±
8
+
3
5
2
,
±
4
+
5
2
)
,
{\displaystyle \left(±\frac12,\,±\frac{2+\sqrt5}{2},\,±\frac{8+3\sqrt5}{2},\,±\frac{4+\sqrt5}{2}\right),}
(
±
1
2
,
±
3
1
+
5
4
,
±
7
+
3
5
2
,
±
3
3
+
5
4
)
,
{\displaystyle \left(±\frac12,\,±3\frac{1+\sqrt5}{4},\,±\frac{7+3\sqrt5}{2},\,±3\frac{3+\sqrt5}{4}\right),}
(
±
1
2
,
±
3
3
+
5
4
,
±
9
+
5
5
4
,
±
(
3
+
5
)
)
,
{\displaystyle \left(±\frac12,\,±3\frac{3+\sqrt5}{4},\,±\frac{9+5\sqrt5}{4},\,±(3+\sqrt5)\right),}
(
±
1
2
,
±
(
2
+
5
)
,
±
11
+
5
5
4
,
±
11
+
3
5
4
)
,
{\displaystyle \left(±\frac12,\,±(2+\sqrt5),\,±\frac{11+5\sqrt5}{4},\,±\frac{11+3\sqrt5}{4}\right),}
(
±
1
+
5
4
,
±
3
+
5
4
,
±
1
+
5
2
,
±
7
+
4
5
2
)
,
{\displaystyle \left(±\frac{1+\sqrt5}{4},\,±\frac{3+\sqrt5}{4},\,±\frac{1+\sqrt5}{2},\,±\frac{7+4\sqrt5}{2}\right),}
(
±
1
+
5
4
,
±
5
+
5
4
,
±
3
1
+
5
4
,
±
15
+
7
5
4
)
,
{\displaystyle \left(±\frac{1+\sqrt5}{4},\,±\frac{5+\sqrt5}{4},\,±3\frac{1+\sqrt5}{4},\,±\frac{15+7\sqrt5}{4}\right),}
(
±
1
+
5
4
,
±
4
+
5
2
,
±
11
+
5
5
4
,
±
(
3
+
5
)
)
,
{\displaystyle \left(±\frac{1+\sqrt5}{4},\,±\frac{4+\sqrt5}{2},\,±\frac{11+5\sqrt5}{4},\,±(3+\sqrt5)\right),}
(
±
1
+
5
4
,
±
7
+
3
5
4
,
±
13
+
5
5
4
,
±
11
+
3
5
4
)
,
{\displaystyle \left(±\frac{1+\sqrt5}{4},\,±\frac{7+3\sqrt5}{4},\,±\frac{13+5\sqrt5}{4},\,±\frac{11+3\sqrt5}{4}\right),}
(
±
1
,
±
3
+
5
4
,
±
3
2
+
5
2
,
±
9
+
5
5
4
)
,
{\displaystyle \left(±1,\,±\frac{3+\sqrt5}{4},\,±3\frac{2+\sqrt5}{2},\,±\frac{9+5\sqrt5}{4}\right),}
(
±
1
,
±
3
+
5
4
,
±
8
+
3
5
2
,
±
7
+
3
5
4
)
,
{\displaystyle \left(±1,\,±\frac{3+\sqrt5}{4},\,±\frac{8+3\sqrt5}{2},\,±\frac{7+3\sqrt5}{4}\right),}
(
±
1
,
±
1
+
5
2
,
±
7
+
3
5
2
,
±
(
2
+
5
)
)
,
{\displaystyle \left(±1,\,±\frac{1+\sqrt5}{2},\,±\frac{7+3\sqrt5}{2},\,±(2+\sqrt5)\right),}
(
±
3
+
5
4
,
±
3
1
+
5
4
,
±
8
+
3
5
2
,
±
3
+
5
2
)
,
{\displaystyle \left(±\frac{3+\sqrt5}{4},\,±3\frac{1+\sqrt5}{4},\,±\frac{8+3\sqrt5}{2},\,±\frac{3+\sqrt5}{2}\right),}
(
±
3
+
5
4
,
±
5
+
3
5
4
,
±
5
3
+
5
4
,
±
3
3
+
5
4
)
,
{\displaystyle \left(±\frac{3+\sqrt5}{4},\,±\frac{5+3\sqrt5}{4},\,±5\frac{3+\sqrt5}{4},\,±3\frac{3+\sqrt5}{4}\right),}
(
±
3
+
5
4
,
±
3
3
+
5
4
,
±
7
+
5
5
4
,
±
11
+
5
5
4
)
,
{\displaystyle \left(±\frac{3+\sqrt5}{4},\,±3\frac{3+\sqrt5}{4},\,±\frac{7+5\sqrt5}{4},\,±\frac{11+5\sqrt5}{4}\right),}
(
±
1
+
5
2
,
±
5
+
5
4
,
±
5
+
3
5
4
,
±
8
+
3
5
2
)
,
{\displaystyle \left(±\frac{1+\sqrt5}{2},\,±\frac{5+\sqrt5}{4},\,±\frac{5+3\sqrt5}{4},\,±\frac{8+3\sqrt5}{2}\right),}
(
±
1
+
5
2
,
±
4
+
5
2
,
±
9
+
5
5
4
,
±
11
+
5
5
4
)
,
{\displaystyle \left(±\frac{1+\sqrt5}{2},\,±\frac{4+\sqrt5}{2},\,±\frac{9+5\sqrt5}{4},\,±\frac{11+5\sqrt5}{4}\right),}
(
±
1
+
5
2
,
±
3
+
2
5
2
,
±
13
+
5
5
4
,
±
3
3
+
5
4
)
,
{\displaystyle \left(±\frac{1+\sqrt5}{2},\,±\frac{3+2\sqrt5}{2},\,±\frac{13+5\sqrt5}{4},\,±3\frac{3+\sqrt5}{4}\right),}
(
±
5
+
5
4
,
±
2
+
5
2
,
±
5
3
+
5
4
,
±
(
2
+
5
)
)
,
{\displaystyle \left(±\frac{5+\sqrt5}{4},\,±\frac{2+\sqrt5}{2},\,±5\frac{3+\sqrt5}{4},\,±(2+\sqrt5)\right),}
(
±
5
+
5
4
,
±
5
+
3
5
4
,
±
7
+
3
5
2
,
±
4
+
5
2
)
,
{\displaystyle \left(±\frac{5+\sqrt5}{4},\,±\frac{5+3\sqrt5}{4},\,±\frac{7+3\sqrt5}{2},\,±\frac{4+\sqrt5}{2}\right),}
(
±
2
+
5
2
,
±
7
+
5
4
,
±
7
+
3
5
4
,
±
7
+
3
5
2
)
,
{\displaystyle \left(±\frac{2+\sqrt5}{2},\,±\frac{7+\sqrt5}{4},\,±\frac{7+3\sqrt5}{4},\,±\frac{7+3\sqrt5}{2}\right),}
(
±
2
+
5
2
,
±
3
+
5
2
,
±
7
+
5
5
4
,
±
13
+
5
5
4
)
,
{\displaystyle \left(±\frac{2+\sqrt5}{2},\,±\frac{3+\sqrt5}{2},\,±\frac{7+5\sqrt5}{4},\,±\frac{13+5\sqrt5}{4}\right),}
(
±
2
+
5
2
,
±
4
+
5
2
,
±
3
+
2
5
2
,
±
3
2
+
5
2
)
,
{\displaystyle \left(±\frac{2+\sqrt5}{2},\,±\frac{4+\sqrt5}{2},\,±\frac{3+2\sqrt5}{2},\,±3\frac{2+\sqrt5}{2}\right),}
(
±
2
+
5
2
,
±
(
2
+
5
)
,
±
7
+
5
5
4
,
±
9
+
5
5
4
)
,
{\displaystyle \left(±\frac{2+\sqrt5}{2},\,±(2+\sqrt5),\,±\frac{7+5\sqrt5}{4},\,±\frac{9+5\sqrt5}{4}\right),}
(
±
3
1
+
5
4
,
±
3
+
2
5
2
,
±
11
+
5
5
4
,
±
(
2
+
5
)
)
,
{\displaystyle \left(±3\frac{1+\sqrt5}{4},\,±\frac{3+2\sqrt5}{2},\,±\frac{11+5\sqrt5}{4},\,±(2+\sqrt5)\right),}
(
±
5
+
3
5
4
,
±
7
+
3
5
4
,
±
3
+
2
5
2
,
±
5
+
3
5
2
)
.
{\displaystyle \left(±\frac{5+3\sqrt5}{4},\,±\frac{7+3\sqrt5}{4},\,±\frac{3+2\sqrt5}{2},\,±\frac{5+3\sqrt5}{2}\right).}
The prismatorhombated hecatonicosachoron has a semi-uniform variant of the form x5o3y3z that maintains its full symmetry. This variant uses 120 small rhombicosidodecahedra of form x5o3y, 600 truncated tetrahedra of form z3y3o, 720 pentagonal prisms of form z x5o, and 1200 ditrigonal prisms of form x y3z as cells, with 3 edge lengths.
With edges of length a, b, and c (such that it forms a5o3b3c), its circumradius is given by
14
a
2
+
10
b
2
+
3
c
2
+
22
a
b
+
11
a
c
+
10
b
c
+
(
6
a
2
+
4
b
2
+
c
2
+
10
a
b
+
5
a
c
+
4
b
c
)
5
2
{\displaystyle \sqrt{\frac{14a^2+10b^2+3c^2+22ab+11ac+10bc+(6a^2+4b^2+c^2+10ab+5ac+4bc)\sqrt5}{2}
}}
The prismatorhombated hecatonicosachoron is the colonel of a 3-member regiment that also includes the small prismatohexacosidishecatonicosachoron and the small rhombiprismic dishecatonicosachoron .
