Truncated cube

The truncated cube, the truncated hexahedron, or tic, is one of the 13 Archimedean solids. It consists of 8 triangles and 6 octagons. Each vertex joins one triangle and two octagons. As the name suggests, it can be obtained by truncation of the cube.

Vertex coordinates
A truncated cube of edge length 1 has vertex coordinates given by all permutations of:
 * $$\left(±\frac{1+\sqrt2}{2},\,±\frac{1+\sqrt2}{2},\,±\frac12\right).$$

Representations
A truncated cube has the following Coxeter diagrams:


 * x4x3o (full symmetry)
 * xwwx4xoox&#xt (BC2 axial, octagon-first)
 * xwwxoo3ooxwwx&#xt (A2 axial, triangle-first)
 * wx3oo3xw&#zx (A3 subsymmetry, as hull of 2 small rhombitetratetrahedra)
 * wx xw4xo&#zx (BC2×A1 symmetry)
 * wwx wxw xww&#zx (A1×A1×A1 symmetry)
 * oxwUwxo xwwxwwx&#xt (A1×A1 axial)

Semi-uniform variant
The truncated cube has a semi-uniform variant of the form x4y3o that maintains its full symmetry. This variant has 8 triangles of size y and 6 ditetragons as faces.

With edges of length a (between two ditetragons) and b (between a ditetragon and a triangle), its circumradius is given by $$\frac{\sqrt{3a^2+4b^2+4ab\sqrt2}}{2}$$ and its volume is given by $$a^3+6ab^2+(9a^2b+5b^3)\frac{\sqrt2}{3}$$.

It has coordinates given by all permutations of:


 * $$\left(±\frac{a+b\sqrt2}{2},\,±\frac{a+b\sqrt2}{2},\,±\frac{a}{2}\right).$$

Related polyhedra
A truncated cube can be augmented by attaching a square cupola to one of its octagonal faces, forming the augmented truncated cube. If a second square cupola is attached to the opposite octagonal face, the result is the biaugmented truncated cube.

The truncated rhombihedron is a uniform polyhedron compound composed of 5 truncated cubes.