# What are the vertices of a regular tetrahedron embeded in a sphere of radius R

Imagine you had a sphere of radius R centered at the origin. What are the coordinates of the vertices of the regular tetrahedron which is circumscribed by the sphere? One of the vertices of the tetrahedron is (0,0,R) and one of the vertices lies in the z,x plane.

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A hint rather than a proper answer: alternating vertices of a cube (e.g. the vertices $(-1, -1, -1), (-1, 1, 1), (1, -1, 1),$ and $(1, 1, -1)$ of the cube $[-1..1]^3$) form the vertices of a regular tetrahedron. This allows you to easily calculate the internal angle of the tetrahedron (i.e., the angle between the lines from the center to any two vertices), and that internal angle provides the position of the vertex in the $xz$ plane. Once you have that vertex, you can find the others by rotating its position $\pm120$ degrees about the $z$ axis.

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Good one. Things are apparently easier for a lot of people when they are shown the embedding of a tetrahedron within a cube. –  Guess who it is. May 1 '12 at 4:50
@J.M. Absolutely so; it's by far the best way I know of calculating most of the angles and distances involved on the tetrahedron. –  Steven Stadnicki May 1 '12 at 7:18

I found at this site

The tetrahedron coordinates:

Vertex coordinate 0, x= 0.000, y= 0.000, z= 1.000 1, x= 0.943, y= 0.000, z=-0.333 2, x=-0.471, y= 0.816, z=-0.333 3, x=-0.471, y=-0.816, z=-0.333

Length of every edge 1.6329932

Of course, the real answers have square roots in them. You can scale by the radius

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