= Solution
In the curvature $-1$ normalization, the <hyperbolic triangle area> is its angle defect:
$$
\boxed{\operatorname{area}(T)=\pi-\alpha-\beta-\gamma}.
$$
Join one vertex of a convex geodesic polygon to its nonadjacent vertices. This gives $n-2$ <hyperbolic triangles> with disjoint interiors. Their angles at every polygon vertex sum to that vertex's interior angle. Adding the <hyperbolic triangle areas> therefore gives the <hyperbolic polygon area>
$$
\boxed{A=(n-2)\pi-\sum_{j=1}^n\alpha_j}.
$$
For a <regular hyperbolic polygon with prescribed area>, place $n$ equally spaced vertices on a <hyperbolic circle> of radius $R>0$ and join consecutive vertices by <geodesics>. Rotations through $2\pi/n$ and reflections in radial lines show that this is a convex regular polygon. The triangle from its centre to a vertex and the midpoint of an adjacent side is right angled, with angles $\pi/n$, $\alpha(R)/2$, and $\pi/2$, and hypotenuse $R$. The angle form of the <hyperbolic law of cosines> gives
$$
\cosh R=\cot(\pi/n)\cot(\alpha(R)/2),\qquad \alpha(R)=2\arctan\!\left(\frac{\cot(\pi/n)}{\cosh R}\right).
$$
Consequently $A(R)=(n-2)\pi-n\alpha(R)$ is continuous and strictly increasing. As $R\downarrow0$, $\alpha(R)\to\pi-2\pi/n$, so $A(R)\to0$; as $R\to\infty$, $\alpha(R)\to0$, so $A(R)\to(n-2)\pi$. The <intermediate value theorem> proves existence for every permitted $A$. Indeed the required radius is
$$
\boxed{R=\operatorname{arcosh}\!\left[\frac{\cot(\pi/n)}{\tan(((n-2)\pi-A)/(2n))}\right]}.
$$
The argument also proves uniqueness of the radius in this construction; neither endpoint area is attained by a nondegenerate finite-radius polygon.
Back to article page
