Consider the order-4 pentagonal tiling of the hyperbolic plane (shown in the figure Hyperbolic plane tiling with pentagons). Pick a vertex $s$ (in white), label it with $0$ and then label all the other vertices with their minimum distance from $s$ (some labels in black).
I would like to prove that:
For each pentagon, there is exactly $1$ edge $(a,b)$ such that $a$ and $b$ have the same label (light blue in the picture). Moreover, this label is either the smallest or the biggest among the ones of that pentagon.
I tried using the exponential divergence of geodesics (which I think holds in this right-angled Coxeter group), but without success. Any suggestions? Thanks
Image credits: Wikipedia
If you show that no three vertices of a pentagon have the same label, your claim follows immediately.
If you connect all points with the same label together in loops, you get concentric polygons. You can prove by induction that these polygons are convex.
Now for any given pentagon (marked in green below), let its smallest label be n. There are two possible cases for intersection of this pentagon with the n-th convex polygon (marked in yellow): either they touch in a point or they touch in an edge. In both cases, use the fact that the n-th polygon is convex and thus cannot extend past the cyan boundaries to prove that the vertices of the given pentagon are labelled as depicted.