Area of polygon of hyperbolic disc

Let us consider hyperbolic disc. I use uniform tessellation {5,4}.Here 5 stands for pentagon, 4 for number of polygons sharing the same vertex.

{hyperbolic disc} There exists formula which defines area of convex hyperbolic polygon: $$A_{m} = \{ \pi(m-2) - (a(1)+...+a(m))\} \frac{1}{-K}$$ where $a(i)$ is interior angle. K is Gaussian curvature, which I define: $$K=-\frac{1}{l^2}$$

If I apply this formula for pentagon polygon I obtain that area of each polygon is given by: $$A_{5} = \frac{\pi}{2} l^2$$

So it follows that type of uniform tessellation together with Gaussian curvature define area associated with polygon. Is it correct?

Is it possible to have tessellations with {5,n} where n is not 4?

The computation looks fine.

And yes, there is a tesselation of type $\{5,n\}$ for every $n \ge 4$.

To see why, you need the fact that if $\alpha_0$ is the interior angle of a regular Euclidean pentagon then for any $\alpha < \alpha_0$ one can construct a regular hyperbolic pentagon $P$ having interior angles equal to $\alpha$.

One then computes $\alpha_0 = \frac{3\pi}{5}$, and if $n \ge 4$ then the desired interior angle of $P$ is $$\alpha = \frac{2\pi}{n} \le \frac{2\pi}{4} < \frac{3\pi}{5} = \alpha_0$$

• (+1): OP may also be interested in Don Hatch's hyperbolic tessellations page. Jun 25 '16 at 14:14

First I have to point you at a misconception you have, and I fear it will be confusing. (I only understand just about half of it myself)

The disk model you use is the Poincare disk model (see https://en.wikipedia.org/wiki/Poincar%C3%A9_disk_model ) and has a fixed absolute distance and a fixed curvature of -1 (this curvature is just build in in the model)

off course you are free to use another distance scale with some formula $1a = i^2 c$ with $1a$ meaning the length of a segment of one absolute distance and $i^2$ your distance measurement of which $c$ is an unit length.

And this will result in a change of curvature. (but I do hear different opinions on what the curvature then or even what the unit of curvature is. so hopefully somebody else will help you here.

In absolute measure the pentagon you describe has a fixed area of $$A_{5} = \{ 3\pi - (5 \frac{\pi}{2}) \} = \frac{1}{2}\pi$$

And using the formulas at https://en.wikipedia.org/wiki/Hyperbolic_triangle#Trigonometry you can calculate the side lenghts

There is a hyperbolic tesselation {5,n} for every $n > 3$ see for example https://en.wikipedia.org/wiki/Order-5_pentagonal_tiling and https://en.wikipedia.org/wiki/Order-6_pentagonal_tiling