Constructing a spherical triangle of a given surface area

Question

Given three points $p_1$, $p_2$ and $p_3$ on the unit sphere $S^2$, we can construct a spherical triangle. The angles associated with the points are $A$, $B$ and $C$.

Keeping the points $p_1$ and $p_2$, the question is how to find a point $q$ on the original spherical triangle such that the surface area of this new spherical triangle is a given fraction $f \in [0,1]$ of the surface area $\Delta$ of the original spherical triangle. That is, the area of the new spherical triangle should be $f\Delta = f(A+B+C-\pi)$.

One approach might be to slide point $q$ along the arc from $p_1$ to $p_3$ (or likewise slide it along the arc from $p_2$ to $p_3$) until the area matches $f\Delta$. Using this method we already know one angle, that is, A (or B if we slide $q$ along the other arc). The other two angles (let's call those $D$ and $E$) are unknown. I don't seem to have enough information to compute $q$ this way (e.g. using the spherical sine/cosine rules) using a direct approach — I could use an iterative approach, but that is not what I'm after.

Another approach might be to somehow use a spherical isosceles triangle, but I'm not familiar enough with spherical geometry to know whether — given an arc $p_1p_2$ — there exists a spherical triangle where the other two arcs $p_2q$ and $qp_1$ are of equal length and where their angles with the given arc are the same such that its area is $f\Delta$.

Answer 1

I think you have enough information to compute $q$. Let $b'$ be the angle between $p_1$ and $q$, and $a'$ between $p_2$ and $q$. Then by using the spherical law of cosines, you can get three equations $\cos a' = \ldots$, $\cos B' = \ldots$, and $\cos C' = \ldots$, with four unknowns $a', b', B', C'$, where $B'$ is the angle opposite $b'$ and $C'$ is the angle opposite $c$ (which is opposite $C$), both in the new triangle. The fourth equation is then simply $B'+C' = f\Delta - A$. Solving this system of four equations in four unknowns should give you $b'$, and thus $q$.