# Find points of a regular tetrahedron

I'm given one of the vertices of a regular tetrahedron and the radius of the circumsphere. I also know the center point of the circumsphere. How can I find the remaining three vertices? (It was pointed out in the comments that "given any solution, you can rotate it around the line through the given vertex and the circumcentre." I think this doesn't matter since any possible solution will work for my application).

So far I've mostly just tried guessing. I'm working on an OpenGL program and I need to draw a regular tetrahedron algorithmically. I don't have much of a math background so I'm a bit lost on even where to begin on this.

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Mostly just guessing. I'm working on an OpenGL program and I need to draw a tetrahedron algorithmically. I don't have much of a math background so I'm a bit lost on even where to begin on this. – gregghz Jan 7 '12 at 3:57
This is not homework. It's for a programming project (this project is also not homework) I'm working on. I'm just under-mathed is all. :( – gregghz Jan 7 '12 at 3:58
OK, thanks for the clarifications. It might be a good idea to edit the question and state these explicitly, for the benefit of others. – Srivatsan Jan 7 '12 at 3:59
This is strongly underspecified. You can take any point whose distance from the given vertex is the given radius, and then take any three vertices (not coplanar with the given vertex) on the sphere around that point with that radius. – joriki Jan 7 '12 at 4:07
Are you sure you mean an isosceles tetrahedron and not a regular one? I was half suspecting that you might have these additional constraints, but even then it's underspecified -- given any solution, you can rotate it around the line through the given vertex and the circumcentre. – joriki Jan 7 '12 at 4:19

All you really need to know is that, when taking vectors from the center of the sphere to the vertices of the regular tetrahedron, the angle between any two such radii is $$\arccos \frac{-1}{3}$$ or about 109 degrees. The fundamental relation for dot product, $$u \cdot v = |u| |v| \cos \theta$$ where $\theta$ is the angle between them. So, if you have the center of the sphere, one vertex, and a plane containing the sphere center and the first vertex and another vertex, you can finish this.

Now, once you have the sphere center $C$ and two correct vertices $v_1,v_2$ let $w$ be the radius that gives the angle bisector of $v_1,v_2.$ The plane that contains $C, v_3, v_4$ also contains the line $w$ but is perpendicular to the plane with $C, v_1, v_2.$

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Let $\vec r$ be the vector from the circumcentre to the given vertex. You can construct a solution for the remaining vertices if you have two vectors perpendicular to $\vec r$ and each other. If you don't have such vectors, one way to obtain them would be to take any vector $\vec x$ not parallel to $\vec r$ and form $\vec b_1=\vec r\times\vec x$ and $\vec b_2=\vec r\times(\vec r\times x)$. Let $\vec u_1$ and $\vec u_2$ be the unit vectors corresponding to $\vec b_1$ and $\vec b_2$. Then the vectors from the circumcentre to the three remaining vertices of the regular tetrahedron are given by

$$\begin{eqnarray} \vec v_1 &=& -\frac13\vec r+\frac{\sqrt8}3r\vec u_1\;, \\ \vec v_2 &=& -\frac13\vec r+\frac{\sqrt8}3r\left(-\frac12\vec u_1+\frac{\sqrt3}2\vec u_2\right)\;, \\ \vec v_3 &=& -\frac13\vec r+\frac{\sqrt8}3r\left(-\frac12\vec u_1-\frac{\sqrt3}2\vec u_2\right)\;, \end{eqnarray}$$

where $r=|\vec r|$ is the circumradius. (You can cancel a factor of $2$ in $\vec v_2$ and $\vec v_3$; I wrote it like this to make it easier to see that these are vectors of length $r$.)

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