Perimeter of $(p,q)$ tiling of the hyperbolic plane Consider a $(p,q)$ regular tiling of the hyperbolic plane projected on the Poincare disc (that is, a tiling of q p-gons joining at each vertex).
Obviously the area of all tilings converge to $\pi$, but what about the total perimeter? that is, the sum of all the lengths of the tiling edges?
Is the total perimeter infinite? and if it is finite, is there a known formula for the total length?
 A: The total perimeter will be infinite.
You can see that like this: start with one polygon in the center of the Poincaré disk. Now add to that every adjacent polygon, i.e. every polygon which has at least one vertex in common with the central one. The outer boundary will be a sequence of hyperbolic line segments which goes once around your central polygon. So the length of all together will be greater than the perimeter of the Euclidean convex hull of the central polygon. Furthermore, none of these line segments is an edge of the central polygon, so the two sets of edges are disjoint.
Repeat this, adding all the polygons which have at least one vertex in common with your union from the previous step. At each step you'll be considering a new set of edges, which sum up to more than the central convex hull. You can repeat this process an infinite number of times. And the sum of an infinite number of lengths with a common positive lower bound must be infinite. This isn't even accounting for the edges between polygons of the same step.
