Stereographic projection of a regular tetrahedron inscribed in the Riemann sphere?

Question

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!

Answer 1

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 $(-0.485492, -0.840896, -0.239146)$.)

Answer 2

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.