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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).2012-01-18
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    @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?2012-01-19
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    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...2012-01-19
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    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.2012-01-22

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