Tetrahedron
Cartesian coordinates

One way to construct a regular tetrahedron is by using the following Cartesian coordinates, defining the four vertices of a tetrahedron with edge length 2, centered at the origin, and two-level edges: (± 1, 0, − 1 2) and (0, ± 1, 1 2) {\displaystyle \left(\pm 1,0,-{\frac {1}{\sqrt {2}}}\right)\quad {\mbox{and}}\quad \left(0,\pm 1,{\frac {1}{\sqrt {2}}}\right)}

Expressed symmetrically as 4 points on the unit sphere, centroid at the origin, with lower face parallel to the x y {\displaystyle xy} plane, the vertices are: (8 9, 0, − 1 3), (− 2 9, 2 3, − 1 3), (− 2 9, − 2 3, − 1 3), (0, 0, 1) {\displaystyle {\begin{aligned}\left({\sqrt {\frac {8}{9}}},0,-{\frac {1}{3}}\right),&\quad \left(-{\sqrt {\frac {2}{9}}},{\sqrt {\frac {2}{3}}},-{\frac {1}{3}}\right),\\\left(-{\sqrt {\frac {2}{9}}},-{\sqrt {\frac {2}{3}}},-{\frac {1}{3}}\right),&\quad (0,0,1)\end{aligned}}} with the edge length of 2 6 3 {\textstyle {\frac {2{\sqrt {6}}}{3}}}.

A regular tetrahedron can be embedded inside a cube in two ways such that each vertex is a vertex of the cube, and each edge is a diagonal of one of the cube's faces. For one such embedding, the Cartesian coordinates of the vertices are (1, 1, 1), (1, − 1, − 1), (− 1, 1, − 1), (− 1, − 1, 1). {\displaystyle {\begin{aligned}(1,1,1),&\quad (1,-1,-1),\\(-1,1,-1),&\quad (-1,-1,1).\end{aligned}}} This yields a tetrahedron with edge-length 2 2 {\displaystyle 2{\sqrt {2}}}, centered at the origin. For the other tetrahedron (which is dual to the first), reverse all the signs. These two tetrahedra's vertices combined are the vertices of a cube, demonstrating that the regular tetrahedron is the 3- demicube, a polyhedron that is by alternating a cube. This form has Coxeter diagram and Schläfli symbol h { 4, 3 } {\displaystyle \mathrm {h} \{4,3\}}.
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