In four-dimensional geometry, the rectified 5-cell is a uniform 4-polytope composed of 5 regular tetrahedral and 5 regular octahedral cells. Each edge has one tetrahedron and two octahedra. Each vertex has two tetrahedra and three octahedra. In total it has 30 triangle faces, 30 edges, and 10 vertices. Each vertex is surrounded by 3 octahedra and 2 tetrahedra; the vertex figure is a triangular prism.

Rectified 5-cell
Schlegel diagram with the 5 tetrahedral cells shown.
Type	Uniform 4-polytope
Schläfli symbol	t1{3,3,3} or r{3,3,3} {32,1} = ${\displaystyle \left\{{\begin{array}{l}3\\3,3\end{array}}\right\}}$
Coxeter-Dynkin diagram
Cells	10	5 {3,3} 5 3.3.3.3
Faces	30 {3}
Edges	30
Vertices	10
Vertex figure	Triangular prism
Symmetry group	A4, [3,3,3], order 120
Petrie polygon	Pentagon
Properties	convex, isogonal, isotoxal
Uniform index	1 2 3

Topologically, under its highest symmetry, [3,3,3], there is only one geometrical form, containing 5 regular tetrahedra and 5 rectified tetrahedra (which is geometrically the same as a regular octahedron). It is also topologically identical to a tetrahedron-octahedron segmentochoron.^{[clarification needed]}

The vertex figure of the rectified 5-cell is a uniform triangular prism, formed by three octahedra around the sides, and two tetrahedra on the opposite ends.^[1]

Despite having the same number of vertices as cells (10) and the same number of edges as faces (30), the rectified 5-cell is not self-dual because the vertex figure (a uniform triangular prism) is not a dual of the polychoron's cells.

Seen in a configuration matrix, all incidence counts between elements are shown. The diagonal f-vector numbers are derived through the Wythoff construction, dividing the full group order of a subgroup order by removing one mirror at a time.^[2]

More information A4, k-face ...

A4		k-face	fk	f0	f1	f2		f3		k-figure	Notes
A2A1		( )	f0	10	6	3	6	3	2	{3}x{ }	A4/A2A1 = 5!/3!/2 = 10
A1A1		{ }	f1	2	30	1	2	2	1	{ }v( )	A4/A1A1 = 5!/2/2 = 30
A2A1		{3}	f2	3	3	10	*	2	0	{ }	A4/A2A1 = 5!/3!/2 = 10
A2				3	3	*	20	1	1		A4/A2 = 5!/3! = 20
A3		r{3,3}	f3	6	12	4	4	5	*	( )	A4/A3 = 5!/4! = 5
A3		{3,3}		4	6	0	4	*	5

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Together with the simplex and 24-cell, this shape and its dual (a polytope with ten vertices and ten triangular bipyramid facets) was one of the first 2-simple 2-simplicial 4-polytopes known. This means that all of its two-dimensional faces, and all of the two-dimensional faces of its dual, are triangles. In 1997, Tom Braden found another dual pair of examples, by gluing two rectified 5-cells together; since then, infinitely many 2-simple 2-simplicial polytopes have been constructed.^[3][4]

It is one of three semiregular 4-polytopes made of two or more cells which are Platonic solids, discovered by Thorold Gosset in his 1900 paper. He called it a tetroctahedric for being made of tetrahedron and octahedron cells.^[5]

E. L. Elte identified it in 1912 as a semiregular polytope, labeling it as tC₅.

• Tetroctahedric (Thorold Gosset)
• Dispentachoron
• Rectified 5-cell (Norman W. Johnson)
• Rectified 4-simplex
• Fully truncated 4-simplex
• Rectified pentachoron (Acronym: rap) (Jonathan Bowers)
• Ambopentachoron (Neil Sloane & John Horton Conway)
• (5,2)-hypersimplex (the convex hull of five-dimensional (0,1)-vectors with exactly two ones)

The Cartesian coordinates of the vertices of an origin-centered rectified 5-cell having edge length 2 are:

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Coordinates

${\displaystyle \left({\sqrt {\frac {2}{5}}},\ {\frac {2}{\sqrt {6}}},\ {\frac {2}{\sqrt {3}}},\ 0\right)}$ ${\displaystyle \left({\sqrt {\frac {2}{5}}},\ {\frac {2}{\sqrt {6}}},\ {\frac {-1}{\sqrt {3}}},\ \pm 1\right)}$ ${\displaystyle \left({\sqrt {\frac {2}{5}}},\ {\frac {-2}{\sqrt {6}}},\ {\frac {1}{\sqrt {3}}},\ \pm 1\right)}$ ${\displaystyle \left({\sqrt {\frac {2}{5}}},\ {\frac {-2}{\sqrt {6}}},\ {\frac {-2}{\sqrt {3}}},\ 0\right)}$

${\displaystyle \left({\frac {-3}{\sqrt {10}}},\ {\frac {1}{\sqrt {6}}},\ {\frac {1}{\sqrt {3}}},\ \pm 1\right)}$ ${\displaystyle \left({\frac {-3}{\sqrt {10}}},\ {\frac {1}{\sqrt {6}}},\ {\frac {-2}{\sqrt {3}}},\ 0\right)}$ ${\displaystyle \left({\frac {-3}{\sqrt {10}}},\ -{\sqrt {\frac {3}{2}}},\ 0,\ 0\right)}$

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More simply, the vertices of the rectified 5-cell can be positioned on a hyperplane in 5-space as permutations of (0,0,0,1,1) or (0,0,1,1,1). These construction can be seen as positive orthant facets of the rectified pentacross or birectified penteract respectively.

The rectified 5-cell is the vertex figure of the 5-demicube, and the edge figure of the uniform 2₂₁ polytope.

Pentachoron polytopes

The rectified 5-cell is one of 9 Uniform 4-polytopes constructed from the [3,3,3] Coxeter group.

More information Name, 5-cell ...

Name	5-cell	truncated 5-cell	rectified 5-cell	cantellated 5-cell	bitruncated 5-cell	cantitruncated 5-cell	runcinated 5-cell	runcitruncated 5-cell	omnitruncated 5-cell
Schläfli symbol	{3,3,3} 3r{3,3,3}	t{3,3,3} 2t{3,3,3}	r{3,3,3} 2r{3,3,3}	rr{3,3,3} r2r{3,3,3}	2t{3,3,3}	tr{3,3,3} t2r{3,3,3}	t0,3{3,3,3}	t0,1,3{3,3,3} t0,2,3{3,3,3}	t0,1,2,3{3,3,3}
Coxeter diagram
Schlegel diagram
A4 Coxeter plane Graph
A3 Coxeter plane Graph
A2 Coxeter plane Graph

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Semiregular polytopes

The rectified 5-cell is second in a dimensional series of semiregular polytopes. Each progressive uniform polytope is constructed as the vertex figure of the previous polytope. Thorold Gosset identified this series in 1900 as containing all regular polytope facets, containing all simplexes and orthoplexes (tetrahedrons and octahedrons in the case of the rectified 5-cell). The Coxeter symbol for the rectified 5-cell is 0₂₁.

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k21 figures in n dimensions
Space	Finite						Euclidean	Hyperbolic
En	3	4	5	6	7	8	9	10
Coxeter group	E3=A2A1	E4=A4	E5=D5	E6	E7	E8	E9 = ${\displaystyle {\tilde {E}}_{8}}$ = E8+	E10 = ${\displaystyle {\bar {T}}_{8}}$ = E8++
Coxeter diagram
Symmetry	[3−1,2,1]	[30,2,1]	[31,2,1]	[32,2,1]	[33,2,1]	[34,2,1]	[35,2,1]	[36,2,1]
Order	12	120	1,920	51,840	2,903,040	696,729,600	∞
Graph							-	-
Name	−121	021	121	221	321	421	521	621

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Isotopic polytopes

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Isotopic uniform truncated simplices

Dim.	2	3	4	5	6	7	8
Name Coxeter	Hexagon = t{3} = {6}	Octahedron = r{3,3} = {31,1} = {3,4} ${\displaystyle \left\{{\begin{array}{l}3\\3\end{array}}\right\}}$	Decachoron 2t{33}	Dodecateron 2r{34} = {32,2} ${\displaystyle \left\{{\begin{array}{l}3,3\\3,3\end{array}}\right\}}$	Tetradecapeton 3t{35}	Hexadecaexon 3r{36} = {33,3} ${\displaystyle \left\{{\begin{array}{l}3,3,3\\3,3,3\end{array}}\right\}}$	Octadecazetton 4t{37}
Images
Vertex figure	( )∨( )	{ }×{ }	{ }∨{ }	{3}×{3}	{3}∨{3}	{3,3}×{3,3}	{3,3}∨{3,3}
Facets		{3}	t{3,3}	r{3,3,3}	2t{3,3,3,3}	2r{3,3,3,3,3}	3t{3,3,3,3,3,3}
As intersecting dual simplexes	∩	∩	∩	∩	∩	∩	∩

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