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I've been reading about stereographic projections. I did a problem about finding the stereographic projection of a cube inscribed inside the Riemann sphere with edges parallel to the coordinate axes. This was simple since the 8 vertices have coordinates $(\pm a,\pm a,\pm a)$, with $3a^2=1$.

Trying it with a regular tetrahedron is a little tougher for me. If a regular tetrahedron is inscribed in the Riemann sphere in general position, with two vertices $(x_1,x_2,x_3)$ and $(x'_1,x'_2,x'_3)$, is there some way to compute the coordinates of the other 2 vertices in terms of $x_1,x_2,x_3,x'_1,x'_2,x'_3$ in order to compute the stereographic projection of the vertices? How could this be done otherwise, if not?

Thanks!

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One vertex won't quite determine the coordinates of the other vertices—the tetrahedron can be rotated about the altitude from the known vertex to the opposite face. The coordinates will all lie on the circle that is the intersection of the sphere with a plane orthogonal to the diameter of the sphere through the known vertex, some specific distance from that vertex (which can be found using right triangles, but I don't have scratch paper in front of me at the moment). –  Isaac Jan 18 '12 at 23:40
    
@Isaac, Thanks, I see that now, darn it. If you know the coordinates of two points are then, is it possible to find the other two in relation to them? –  JPan Jan 19 '12 at 0:05
    
Yes, and probably a good bit easier, though I'm guessing a little off the top of my head. Any two vertices are the endpoints of a single edge. The other two vertices are the endpoints of an edge that lies in the plane that perpendicularly bisects the first edge, so that puts the points on the circle that is the intersection of that plane with the sphere. The two unknown vertices are the same distance apart as the two known vertices and together with the midpoint of the known-vertices-edge form an isosceles triangle. I think that's sufficient to find them... –  Isaac Jan 19 '12 at 0:10
    
Note that the vertices of a regular tetrahedron form a subset of the vertices of a cube. So, if you just want some tetrahedron, then take the coordinates $(\pm a, \pm a, \pm a)$ of your cube such that an even number of "$\pm$"s are "$-$". Granted, the tetrahedron isn't general position; but, then, neither was the cube in the first problem. –  Blue Jan 22 '12 at 6:11

2 Answers 2

Given your two vertices of the tetrahedron, $A=(x_1,x_2,x_3)$ and $B=(x_1',x_2',x_3')$, the plane that perpendicularly bisects the edge determined by $A$ and $B$, which contains the other two vertices, has equation $$X\cdot(A-B)=\left(\frac{A+B}{2}\right)\cdot(A-B)$$ (where $\cdot$ is the vector dot product and $X$ is a general point in 3-space). The other two vertices must also lie on the unit sphere, $$|X|=1.$$ And, the other two vertices must be the same distance from both $A$ and $B$ as $A$ and $B$ are from one another: $$|A-X|=|A-B|$$ $$|B-X|=|A-B|.$$ Solving all four (or perhaps any three) of these equations simultaneously should give the coordinates of the other two vertices. I strongly doubt that this yields anything nice in the fully general case. Even putting one vertex at $(0,0,1)$ and letting a second vertex have $y$-coordinate $0$, the expressions for the coordinates of the vertices weren't pretty. (Numerical approximations: $(0.970984, 0, -0.239146)$, $(-0.485492, 0.840896, -0.239146)$, and $(-0.485492, -0.840896, -0.239146)$.)

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Given tetrahedral vertices $P$ and $Q$ on a unit sphere centered at the origin $O$, let $M$ be the midpoint of edge $PQ$, and let $N$ be the reflection of $M$ in the origin. (That is, $N := -\frac{1}{2}(P+Q)$.) Then, the tetrahedron's other vertices, $R$ and $S$, determine vectors $NR$ and $NS$ that are (1) perpendicular to plane $OPQ$ (and, hence, are parallel to the vector $P\times Q$), and (2) of length $\frac{1}{2}|PQ|$.

Noting that $\cos\angle POQ =-\frac{1}{3}$ and $|PQ| = \frac{2\sqrt{2}}{\sqrt{3}}$ [*], we deduce that vertices $R$ and $S$ have the form

$$\begin{eqnarray*} N \pm \frac{1}{2}|PQ| \frac{ P \times Q }{|P\times Q|} &=& -\frac{1}{2}(P+Q)\pm \frac{\sqrt{2}}{\sqrt{3}}\frac{P\times Q}{\sin\angle POQ}\\ &=& -\frac{1}{2}(P+Q)\pm \frac{\sqrt{3}}{2}\;( P \times Q ) \end{eqnarray*}$$

You can sanity-check this using your cube coordinates, as in my comment on your question. If $P := (a,a,a)$ and $Q := (a,-a,-a)$ with $3a^2=1$, then the formula gives vertices $(-a,a,-a)$ and $(-a,-a,a)$.

[*] You can derive the values of $\cos\angle POQ$ and $|PQ|$ using the cube coordinates.

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