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...)
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?