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.
On
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.
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 $ABC$.