Point within a spherical triangle given areas Consider a spherical triangle like this:

where $A_1, A_2, A_3,$ and $P$ are points on the sphere and $t_1, t_2, t_3$ are the proportion of the area of the large triangle contained within the small triangle: that is, if $\Omega$ is the area of $A_1 A_2 A_3$, the area of $A_1 A_2 P$ is $t_3 \Omega$.
Given $A_1 A_2 A_3$ and $t_1, t_2, t_3$, what is $P$? Is there an analytic expression for $P$, or can it only be found numerically? Is there an analytic expression if $A_1 A_2 A_3$ is an equilateral spherical triangle (i.e. $A_1, A_2,$ and $A_3$ are equidistant)?
(If this was a planar triangle, this would reduce to simple barycentric coordinates in the plane. In fact, that's where I stole the image from...)
 A: Here's an answer that is correct but probably not as simplified as it could be; feel free to improve on it.
Start with the formula for solid angle of a spherical triangle given on  wikipedia, and assume all vectors are of unit length, then for a triangle defined by vertices $\vec a\vec b\vec c$:
$$\tan \left( \frac{\Omega_{abc}}{2}  \right) =
  \frac{\vec a\cdot \vec b\times \vec c}{1 + \left(\vec a \cdot \vec b\right) + \left(\vec a \cdot \vec c\right) + \left(\vec b \cdot \vec c\right)}$$
Let $\vec x$ be the point inside the spherical triangle. The area of the triangle $\vec a \vec b \vec x$ can be grouped and rearranged as so:
$$x \cdot \left(\cot \frac{\Omega_{abx}}{2}\vec a \times \vec b - \vec a - \vec b\right) = 1 + \vec a \cdot \vec b$$
Define $\vec L_{abx} = \cot (\Omega_{abx}/2) \vec a \times \vec b - \vec a - \vec b$ and $h_{abx} = 1 + \vec a \cdot \vec b$. 
The other triangles $\vec b\vec c\vec x$ and $\vec c \vec a \vec x$ are similar, resulting in $\vec L_{bcx}, \vec L_{cax}, h_{bcx}$, and $h_{cax}$.(Be careful with the order of points here: they all have to be in counterclockwise order, or else your signs get confused.) These quantities are related to $\vec x$ by the linear system
$$ \begin{bmatrix} \vec L_{abx} & \vec L_{bcx} & \vec L_{cax} \end{bmatrix} \vec x = \begin{bmatrix} h_{abx}  \\ h_{bcx} \\ h_{cax}  \end{bmatrix}$$
where $\begin{bmatrix} \vec L_{abx} & \vec L_{bcx} & \vec L_{cax} \end{bmatrix}$ is the matrix where the first column is the vector $\vec L_{abx}$, etc. Define $\vec h = [h_{abx}, h_{bcx}, h_{cax}] $. Cramer's rule can be used to get the components of $\vec x = [x_1, x_2, x_3]$ :
$$x_1 = \frac{\begin{vmatrix}\vec h & \vec L_{bcx} & \vec L_{cax}\end{vmatrix}}{\begin{vmatrix}\vec L_{abx} & \vec L_{bcx} & \vec L_{cax}\end{vmatrix}}, \, etc.$$
This is an explicit, analytic expression, but is complicated. My attempts to simplify this form have only resulted in frustration; maybe someone with access to a better computer algebra system can whittle it down?
A: For any point $X$ on the unit sphere, let $\vec{X}$ be corresponding unit vector.
We will assume

*

*The spherical triangle $A_1A_2A_3$ is small in the sense that it falls within a hemisphere.

*Vertices are ordered in counterclockwise manner. i.e. $\Delta \stackrel{def}{=} \vec{A}_1 \cdot ( \vec{A}_2 \times \vec{A}_3 ) > 0$.

*The desired point $P$ lies inside $A_1A_2A_3$. i.e. $t_1,t_2,t_3 \ge 0, t_1 + t_2 + t_3 = 1$.

Express $P$ as a "linear combination" of $A_1,A_2,A_3$. i.e.
$$\vec{P} = u_1 \vec{A}_1 + u_2 \vec{A}_2 + u_3\vec{A}_3$$
for parameter $u_1,u_2,u_3$ to be determined.
Let $(i,j,k)$ be any cyclic permutation of $(1,2,3)$. Define

*

*$v_k = \frac{u_k}{1 + u_1+u_2+u_3} \iff u_k = \frac{v_k}{1 - v_1 - v_2 - v_3}$

*$\alpha_k = \vec{A}_i \cdot \vec{A}_j$ and $\beta_k = \frac{\alpha_i + \alpha_j}{1 + \alpha_k}$

*$\tau = \tan\left(\frac{\Omega}{2}\right)$, $\tau_k = \tan\left(\frac{t_k\Omega}{2}\right)$ and $\lambda_k = \frac{\tau_k}{\tau}$.

Apply the formula of solid angle in OP's own answer to triangle $A_1A_2A_3$, we
obtain
$$\tau = \frac{\Delta}{1 + \alpha_1 + \alpha_2 + \alpha_3} = \frac{\Delta}{(1+\alpha_k)(1+\beta_k)}$$
Apply same formula to triangle $PA_iA_j$, we obtain
$$\begin{align}
\tau_k  
&= \frac{\vec{A}_i \cdot (\vec{A}_j \times \vec{P})}{
1+ \vec{A}_i \cdot \vec{A}_j + \vec{A}_i\cdot\vec{P} + \vec{A}_j\cdot\vec{P}}\\
&=  \frac{u_k \Delta}{
1 + \alpha_k + (u_i + u_j \alpha_k + u_k \alpha_j) + (u_i \alpha_k + u_j + u_k \alpha_i)}\\
&= \frac{u_k \Delta}{
(1 + \alpha_k)(1 + u_i + u_j) + (\alpha_j + \alpha_k)u_k}\\
&= \frac{u_k \Delta}{
(1 + \alpha_k)(1 + u_i + u_j + \beta_k u_k)}\\
&= \frac{v_k \Delta}{
(1 + \alpha_k)(1  - (1 -\beta_k) v_k)}\\
\end{align}
$$
This leads to
$$v_k = \frac{1}{\frac{\Delta}{\tau_k(1+\alpha_k)} + 1 - \beta_k}
= \frac{1}{\frac{(1+\beta_k)\tau}{\tau_k} + (1-\beta_k)}
= \frac{\lambda_k}{(1+\beta_k) + (1-\beta_k)\lambda_k}$$
In the special case $A_1A_2A_3$ is equilateral with side length $\theta$, we have
$$\alpha_1 = \alpha_2 = \alpha_3 = \cos\theta
\implies  \beta_1 = \beta_2 = \beta_3 = \beta \stackrel{def}{=}\frac{2\cos\theta}{1 + \cos\theta}$$
The formula for $P$ simplifies to
$$\vec{P} = \frac{v_1 \vec{A}_1 + v_2 \vec{A}_2 + v_3 \vec{A}_3}{1-v_1-v_2-v_3}$$
where
$$v_k = \frac{\lambda_k}{(1+\beta) + (1-\beta)\lambda_k},\quad
\lambda_k = \frac{\tan\left(\frac{t_k \Omega}{2}\right)}{\tan\left(\frac{\Omega}{2}\right)}
$$
In particular, when $\theta = \frac{\pi}{2}$, $A_1A_2A_3$ is one-eight of the unit sphere. $\vec{A}_1,\vec{A}_2, \vec{A}_3$ are orthonormal and $\beta = 0$. The coefficients $v_k$ become something very simple.
$$v_k = \frac{1}{\cot\left(\frac{t_k\pi}{4}\right) + 1}$$
