Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer
Good one. Things are apparently easier for a lot of people when they are shown the embedding of a tetrahedron within a cube. – J. M. 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

The tetrahedron coordinates:

  • (0.000, 0.000, 1.000)
  • (0.943, 0.000, -0.333)
  • (-0.471, 0.816, -0.333)
  • (-0.471, -0.816, -0.333)

Length of every edge: 1.6329932

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.