TITLE: Smallest tile to tessellate the hyperbolic plane
QUESTION [20 upvotes]: Is it known what the smallest tile (in terms of area) that can tessellate the hyperbolic plane is? In particular, it should tessellate the plane by itself.
I think it will be a Triangle group, but I'm not sure.
(In spherical geometry, the answer is that there is no smallest tile, because you can make bipyramids with arbitrarily small faces. I don't think this will be the case with the hyperbolic plane though.)

REPLY [32 votes]: Binary Tiling
In fact, one can tile the hyperbolic plane with arbitrarily small tiles. There is a tiling of the hyperbolic plane (apparently due to Boroczky) by pentagons. 

The horizontal edges are horocycles in the upper half-space model of the hyperbolic plane, and the vertical lines geodesics. The edge at the top of each tile is half the length of the one at the bottom. One can make these arbitrarily thin, and hence have arbitrarily small area.

REPLY [19 votes]: The tilings mentioned by Ian Agol are related to an action of a Baumslag-Solitar group $\{ a,b \bigg| b^{-1}a^2b=a  \}$ on the hyperbolic plane. They have arbitrarily small area, but diameter uniformly bounded away from $0$. It is possible to tesselate the hyperbolic plane with a single tile with arbitrarily small diameter, too. Let there be $n$ arcs on top and $n+1$ arcs on the bottom. As $n \to \infty$ the distance between the top and bottom goes to $0$. The region is naturally related to a (non-faithful) action of a Baumslag-Solitar group $\{a,b \bigg|b^{-1}a^{n+1}b=a^n\}$ on the hyperbolic plane.