Consider the hyperbolic 3-space $(\mathbb{H}^3,ds^2)$ with
$$\mathbb{H}^3:=\{(x,y,z)\in\mathbb{R}^3|z>0\}, \quad ds^2=\frac{dx^2+dy^2+dz^2}{z^2}$$
Geodesics for this space are circular arcs normal to $\{z=0\}$ and vertical rays normal to $\{z=0\}$.
Given any two points, for example $p1=(2,-1,3),p_2=(1,2,4)$ (I've chosen them so they don't lie on a vertical ray) is it possible to obtain a closed formula for the coordinates of the midpoint $m$ of $p_1$ and $p_2$? (i.e. the point $m$ is such that $d(m,p_1)=d(m,p_2)=d(p_1,p_2)/2$ where $d$ is the metric induced by $ds^2$)
I just can think of this method: given $p_1$ and $p_2$ find the plane $H$ orthogonal to $\{z=0\}$ and which contains both $p_1$ and $p_2$. Then in $H$ find the circular arc $g:[0,d(p_1,p_2)]\rightarrow H$ orthogonal to $\{z=0\}$ with $g(0)=p_1$, $g(1)=p_2$ and $|\dot g(t)|_{ds^2}=1$. Finally $m=g(d(p_1,p_2)/2)$.
But my method is rather long and complicated. Do you know of a better one which could be written directly as a formula? I'm interested in the case of $(\mathbb{H}^3,ds^2)$ to find a method which could be generalized for other riemannian and metric spaces.
A clean way to do this is to work with the hyperboloid model of the hyperbolic space where $H^n$. Then, given two points $p, q$ in the (upper) hyperboloid $H^n\subset R^{1,n}$, their midpoint is $$\frac{p+q}{\sqrt{<p+q, p+q>}}.$$ If you know how to go back and forth between the upper half space and the hyperboloid model, you can translate this formula into the upper half space model. (I think it will not be pretty.) See also this question.