For what $n$ does a hyperbolic regular $n$-gon exist around a circle?

Solution 1:

In the picture (Klein model) below it is clearly shown that the half of the central angle $\Phi(=\frac{2\pi}{n})$ equals the the angle of parallelism belonging to $\rho$, the radius of the circle. That is, $$\Pi(\rho)\ge \frac{\pi}{n}$$ or equivalently $$n\ge \frac{\pi}{\Pi(\rho)}$$is the condition that the tangent lines meet within the plane or exactly on the absolute.

One of the known formulas for the angle of parallelism is $$\tan(\Pi(\rho))=\frac{1}{\sinh(\rho)}.$$ Whit this, our condition is $$\frac{1}{\sinh(\rho)} \ge \tan \frac{\pi}{n} >\frac{\pi}{n}, $$ since $n\ge 3.$ From here $$\pi \sinh(\rho)< n.$$ Or

$$\frac{\circ \rho}{2} < n,$$

where $\circ \rho$ is the circumference of the hyperbolic circle of radius $\rho$. Note that the last three inequalities give sufficient conditions only. But if $n$ is large enough they are sharp lower bounds for $n$. Think of ($\tan(x)>x$).

Solution 2:

Different aproach:

we can just assume that the circle you want to circumscribe is centererd at the centre of the disk of the Beltrami Klein Model. and that the segments of the n-gon are just lines in this model.

Then from Euclidean geometry we now from Regular n-gon between two concentric circles

we learn $n \geq \pi/\cos^{-1} r/R $

with $ r$ and $R$ being the euclidean measures

Taking $R=1$ we get $n \geq \pi/\arccos( r) $

Then in the Beltrami klein model we have for the length

$ h = 1/2 \ln(\frac{ 1+r}{1-r}) $

or $r = \frac{e^{2h}-1}{e^{2h}+1} $

where h = the hyperbolic radius of the circle

so putting it all together you get:

$$n \geq \frac{\pi}{\arccos( \frac{e^{2h}-1}{e^{2h}+1})} $$