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!
Best Answer
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 $(-0.485492, -0.840896, -0.239146)$.)