The distance between two farthest points on the Bolza surface?

The Bolza surface $M$ is the closed hyperbolic surface of genus $2$ that can be obtained by identifying the opposite sides of the regular octagon in $\mathbb{H}^2$.

What two points on $M$ are farthest away from each other?

Some thoughts:

It is known that $M = \mathbb{H}^2/G$, where $G \cong \left< a, b, c, d \mid abcda^{-1}b^{-1}c^{-1}d^{-1} \right>$. Let us fix a point $p \in M$ and let $\tilde{p}$ be a lift of $p$ in $\mathbb{H}^2$. Consider the Dirichlet region $D_{\tilde{p}}$ of $G$ centered at the point $\tilde{p}$. Clearly, $D_{\tilde{p}}$ is a convex polygon with finite number of sides. Let $\tilde{v}$ be one of the farthest vertices of $D_{\tilde{p}}$ from $\tilde{p}$. Then the projection $v$ of $\tilde{v}$ on $M$ is furthest away from $p$.

If $D_{\tilde{p}}$ is a regular octagon, then the maximal distance from $p$ to a point on $M$ can be computed explicitly (the radius of the circle circumscribing the octagon). However in general $D_{\tilde{p}}$ can be a convex polygon with number of sides from $8$ to $18$, and the question about the distance between two farthest points on $M$ seems not obvious.

One conjecture is that the maximal distance between two farthest points $p, v$ is attained when $D_{\tilde{p}}$ is a regular octagon.

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As you noticed, the answer to your question in general is that diameter of any hyperbolic surface is the maximal radius of Dirichlet domains $D_x$. This is not a smooth function of $x$ but it is smooth generically, so you can set up a variational problem and see what differential equations on $x$ you get. – Misha Aug 1 '13 at 12:05
Thank you for the comment! I have a question, might be trivial. Is it clear that the maximal radius of $D_x$ for a closed hyperbolic surface is smooth generically. Is there a reference to read in this direction? – Mikhail Aug 2 '13 at 9:42
First, let $\Gamma$ be a Fuchsian group. There is a nowhere dense closed subset $E\subset H^2$, whose complement is union of components $U_i$ so that for each $U_i$, the domain $D_x$ is bounded by bisectors $B_{ij}=B_{x,\gamma_{ij}(x)}$ of a certain finite set of elements $\gamma_{ij}\in \Gamma$. Now, fix $i$ and drop the corresponding subscript. Then for $x\in U$, each vertex $v_j=v_j(x)$ of $D_x$ is the intersection of bisectors $B_{j}, B_{j+1}$... – Misha Aug 2 '13 at 14:04
...This intersection points varies real-analytically with $x$. The diameter $d(x)=diam(D_x)$ is $max$ of the functions $f_j(x)=d(x, v_j(x))$. Since these are real-analytic functions, there is an open and dense subset $U'$ in $U$, so that on each component of $U$ exactly one of the functions $f_j(x)$ gives maximum. – Misha Aug 2 '13 at 14:05
Thanks for the detailed answer! – Mikhail Aug 3 '13 at 11:32