Finding centroid of spherical triangle Given three vectors $u,v,w\in S^2$, and a spherical triangle $[u,v,w]$, find its centroid, i.e. the point of intersection of the three medians $\left[u, \frac{v+w}{|v+w|}\right]$, $\left[v, \frac{u+w}{|u+w|}\right]$ and $\left[w, \frac{u+v}{|u+v|}\right]$.
The problem is that I don't even know how to approach this problem at all, the reason being that I'm not sure how to derive equations for the three medians. Your help would be much appreciated.
 A: The spherical centroid exists by the spherical version of Ceva's theorem.
Assuming that $A,B,C$ are three points on a unit sphere centered at $O$, we may join $A$ with the midpoint $M_A$ of the $BC$ side in the spherical triangle $ABC$. The plane through $A,M_A,O$ meets the $ABC$ plane at a line $\ell_A$, the plane through $B,M_B,O$ meets the $ABC$ plane at a line $\ell_B$. Assuming that $\ell_A$ and $\ell_B$ meet at $G$ in the $ABC$ plane, the spherical centroid of the spherical triangle $ABC$ is just the intersection between the $OG$ ray and the original sphere, i.e. $\frac{G}{\left\|G\right\|}$.
It follows that you just need to compute the (planar) centroid of the euclidean triangle $ABC$, since the $OM_A$ ray meets the $BC$ segment at its midpoint on so on.
Long story short, the answer is just given by $\color{red}{\frac{u+v+w}{\left\|u+v+w\right\|}}$, since the spherical medians are given by the central projections of the medians of the planar triangle $ABC$.
A: The way this question is posed is contradictory because the centroid on a sphere wouldn't normally be defined in this way.  As normally understood the centroid is the center of gravity and in this case we would treat a triangular piece of a massive spherical shell.  The centroid is then given by J. E. Brock, The Inertia Tensor for a Spherical Triangle, J. Applied Mechanics 42, 239 (1975).  The result is in a form that can be easily generalized to arbitrary polygons.
