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

I was wondering if anyone could provide a 'clear' way of establishing the vertices of a tetrahedron. This is a regular/standard/all faces are congruent tetrahedron.

To make it more generalized, let's say that the center of the object sits at $(0,0,0)$ and the first known vertex sits at $(0,-1, 0)$ giving the sphere that contains the tetrahedron a radius of $1$ unit. A second vertex must sit at $(0, y > 0, z < 0)$.

Furthermore, $(x < 0, y > 0, z > 0)$ and $(x > 0, y > 0, z > 0)$.

The desired end result is that I could provide any initial coordinate for the vertex and receive the other three vertices in relation to the first.

I figured this was the best route get a solution... --

Correction, in my grasping for straws here, I mentioned 3-Simplex which appears to not be conducive to what I'm trying to learn here. I simply want a clear cut way to represent a REGULAR tetrahedron when one of the vertices is very specific to the sphere that encompasses said tetrahedron. By clear cut I mean... break the tetrahedron into it's separate triangles and multiply by the sin of 180/pi.

This is a personal problem that I'm sure is beneath the initial design of this Stack but it is keeping me up at night and scouring wikipedia and an old Trig book is not helping


marked as duplicate by Steven Stadnicki, Jyrki Lahtonen, t.b., J. M. is a poor mathematician, Henning Makholm Jul 20 '12 at 12:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ You write as if there is only one tetrahedron. 000, 100, 010, and 001 are the vertices of a tetrahedron, and indeed when I see the phrase "3-simplex" that's the tetrahedron I think of. But maybe you are only interested in a regular tetrahedron. If so, you might want to edit your question accordingly. $\endgroup$ – Gerry Myerson Jul 19 '12 at 8:53
  • $\begingroup$ Sorry, @Steven. I should have guessed that this question has been answered here already. $\endgroup$ – Jyrki Lahtonen Jul 19 '12 at 17:25
  • $\begingroup$ @JyrkiLahtonen No apology needed! I doubt I would have spotted it if I hadn't remembered answering the previous one. $\endgroup$ – Steven Stadnicki Jul 19 '12 at 17:26
  • $\begingroup$ I am quite aware of that answer, @Steve, as I did not just out of the blue come up with this...confounding question and have look through this site and other google-able sites. I was hoping for some math that I might already understand. Hence the 'clear' part. $\endgroup$ – Dialock Jul 20 '12 at 2:52

IMHO the simplest description of a regular tetrahedron has its vertices at four corners of a cube. So start with the vertices $(0,0,0),(1,1,0),(1,0,1),(0,1,1)$. If you want the center of the tetrahedron to be at the origin, translate by $(-1/2,-1/2,-1/2)$. If you want a specific vertex at some other point, do another translation. If you want the tetrahedron in some other orientation, apply a 3x3 orthogonal matrix. If you want a tetrahedron of a different size, apply a scaling factor.

  • $\begingroup$ The cube...yes I am aware of it. However, I need that cube rotated...So what's easier to rotate in space? a cube or a tetrahedron? The key words here are translation and orthogonal matrix. Both of which I know VERY little about. But, I thank you for giving me something to focus on that might actually answer this problem. $\endgroup$ – Dialock Jul 20 '12 at 2:57
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    $\begingroup$ @Dialock: an orthogonal matrix is just a way of representing rotations and/or reflections; if you for instance need to rotate some object whose coordinates you have by a certain angle and along a certain axis, you can always assemble an orthogonal matrix that you apply to your coordinates to yield a rotated version of your original object. $\endgroup$ – J. M. is a poor mathematician Jul 20 '12 at 7:40

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