# Projection from triangle to spherical triangle

Consider a triangle, $T$, in $\mathbb{R}^3$ with vertices $(0,0,1), (0,1,0)$, and $(1,0,0)$. Let $S$ denote the sphere centered at the origin with radius 1 and let $S_1$ denote the portion of the sphere in the same quadrant at the triangle. We can define a map $f: T \rightarrow S_1$ by $$f(x) = \frac{x}{|x|}$$ which is one-to-one and onto.

The map is a bijection, so the inverse should exist. Maybe I am losing my mind but, what does the explicit formula for $f^{-1}: S_1 \rightarrow T$ look like?

• By "look like", do you mean that you want an explicit formula for $f^{-1}$? – Travis Willse Sep 4 '15 at 20:31
• $(x,y,z) \mapsto \left(\frac{x}{x+y+z},\frac{y}{x+y+z},\frac{z}{x+y+z}\right)$. – achille hui Sep 4 '15 at 20:31
• Yep, that's what I meant. Thanks for your answer. I can mark it as an answer if you post it. – user1515302 Sep 4 '15 at 20:35

Notice under the direct mapping, any point $p$ on $T$ get mapped to a point $f(p)$ on $S$ which is a scalar multiple of $p$. So for any point $q = (x,y,z) \in S$, the inverse mapping $f^{-1}$ will send $q$ to a point which is a scalar multiple of $q$. This means there exists a function $\lambda : S \to \mathbb{R}$ such that
$$q = (x,y,z)\quad\mapsto\quad f^{-1}(q) = ( \lambda(q)x, \lambda(q)y, \lambda(q)z)$$
It is clear the triangle $T$ lies on the plane $x + y + z = 1$, this means $\lambda(q)$ satisfy the constraint:
$$\lambda(q)(x+y+z) = 1 \quad\iff\quad \lambda(q) = \frac{1}{x+y+z}$$
$$f^{-1}(x,y,z) = \left(\frac{x}{x+y+z},\frac{y}{x+y+z},\frac{z}{x+y+z}\right)$$