# Minimize $\sum_i \arccos(v_i\cdot x)^2$ subject to the constraints $\|v_i\|=1$ and $\|x\|=1$?

Some background: (skip to the end for the actual question) Recently I have been trying to define some notion of an average of points on the surface of a sphere.

My original idea was to ignore the fact that the points are on a sphere and just find the vector $x$ such that the distance to all the other points squared $D=\sum_i\|v_i-x\|^2$ is minimized. Assuming $x$ must also lie on the sphere I expanded and simplified to get $$x=\frac{\sum_iv_i}{\|\sum_iv_i\|}$$ Doing some simulations though, I found that this definition was unsatisfactory.

Instead I found the following alternate definition worked better: Let us denote the average to be the point $x$. Then $x$ is the point such that the sum of the distances along the surface to the other points squared is minimized i.e. $$D=\min_{\|x\|=1}\sum_idist_{S}(v_i,x)^2$$ where $dist_S(v_i,x)$ is the length of the geodesic between $v_i$ and $x$ on the surface $S$.

Since we are working on the sphere however, the distance function is a rather nice and simple: $$dist_S(v_i,x)=\arccos(v_i\cdot x)$$ It follows from the fact that a geodesic on the sphere is an arc and that $$v_i\cdot x=\|v_i\| \|x\|\cos\theta=\cos\theta$$ My question: does there exist a closed form for the vector $x$ such that $\displaystyle\sum_i\arccos(v_i^T x)^2$ is minimized? If not, is there some other way of determining what $x$ is?

I would not expect a closed form in a minimization problem with non-unique solution. E.g., if $v_i$ are the vertices of inscribed tetrahedron, then (on the grounds of symmetry) there are multiple minimizers. If $v_i$ are perturbed slightly, any of them can become the unique minimizer. Hence, $x$ is not a continuous function of $v_i$.
• What if you restrict it such that all the $v_i$'s lie on the upper hemisphere?