Bijection between spherical and planar triangle surfaces I subdivide a unit sphere, centered at origin, onto 20 spherical triangles. For the sake of argument let's take one such triangle $Ts$, in $\mathbb{R}^3$, that has vertices $Normalize(-1,0,g), Normalize(0,g,1)$, and $Normalize(1,0,g)$, where $g = \frac{1.0+\sqrt{5}}{2}$. In the same time there is another equilateral planar triangle $Tp$, in $\mathbb{R}^2$, that has the following vertices $(0,0), (1,0),(0.5, \frac{\sqrt{3}}{2})$. I'm looking for a projection (exact formula) that would create bijection of surface points of $Ts$ to surface points of $Tp$. It's crucial that such mapping maximize uniformity of distortions (not sure how to express this in strict math terms).

Suppose then you applied a suitable transformation, so that the vertices of the spherical triangle are the same as the vertices of the plane triangle, sitting then on plane $z=0$. Let $O=(x_0,y_0,z_0)$ be the center of the transformed sphere and $P=(x,y,z)$ any point on the spherical triangle. You can then:

1. Apply a translation carrying $O$ at the origin: $P\to P-O=(x-x_0,y-y_0,z-z_0)$;

2. Slide $P$ along $PO$ until its $z$ coordinate becomes $-z_0$: $$ P\to {-z_0\over z-z_0}P=\left(-{x-x_0\over z-z_0}z_0,-{y-y_0\over z-z_0}z_0,-z_0 \right); $$

3. Translate back: $$ P\to P+(x_0,y_0,z_0)=\left({x_0z-xz_0\over z-z_0},{y_0z-yz_0\over z-z_0},0 \right). $$ Please ask if anything is not clear.