How big are regular (hyperbolic) polygons? Given a hyperbolic surface of constant curvature $K=-1/a^2$ embedded in $\mathbb{R}^3$, is there a known formula for the length of the edges of a  regular polygon?
I know that the Gauss–Bonnet theorem tells us that the area of an $n$-gon is given by:
$$A=(n-2)\pi-\sum_{i=1}^n \alpha_i = n(\pi-\alpha) - 2\pi$$
But i'm not sure how one would go about calculating the edge length without explicitly specifying the surface. Is an explicit formula even possible?
 A: Suppose $r$ is the distance one of the vertices of the polygon to its center. Draw the perpendicular bisector of the edge; by symmetry it will pass through the center of the polygon too. This creates a right triangle where


*

*The central angle is $\pi/n$.

*The hypotenuse is $r$.

*The opposite leg is half the edge length.


Applying the appropriate rule for a right hyperbolic triangle we get
$$ \sin \frac{\pi}{n} = \frac{\sinh(x/2a)}{\sinh(r/a)} $$
where $x$ is your sought edge length.
This is easily solved to get
$$ x = 2a \sinh^{-1} ( \sin(\pi/n) \sinh(r/a) ) $$

If instead of the radius you know the interior angle $\theta$ at each corner of the polygon, you would use a different rule and get
$$ \cosh(x/2a) = \frac{\cos(\pi/n)}{\sin(\theta/2)} $$
giving
$$ x = 2a \cosh^{-1}\left( \frac{\cos(\pi/n)}{\sin(\theta/2)} \right) $$
A: Law of cosines to the rescue:
If we know that the angle at each vertex is $\theta$, and the edge length to be found is $x$, then:
$$\cos(2\pi/n)=-\cos^2(\theta/2)+\sin^2(\theta/2)\cosh(x/a)$$
Or:
$$x = a\cosh^{-1}\left(\frac{\cos(2\pi/n)+\cos^2(\theta/2)}{\sin^2(\theta/2)}\right)$$
