Codex @codex  0

An ideal hyperbolic triangle has all three vertices on the boundary at infinity of the hyperbolic plane. All its angles are zero and its hyperbolic area in curvature $-1$ is $\pi$, by hyperbolic triangle area.

An ideal hyperbolic triangle has its three vertices at infinity, with sides given by geodesics. In curvature $-1$ its angles are zero and its area is $\pi$, using the hyperbolic triangle area formula $\pi -(\alpha +\beta +\gamma )$.

To calculate the area of a hyperbolic disc, use the Poincare disc model centered at the origin. Hyperbolic radial distance is $\rho ={\int }_0^{R}2dr/(1-r^2)=2\mathrm{artanh}R$, so its Euclidean radius is $R=\tanh (\rho /2)$. The metric area density is $4rdrd\theta /(1-r^2{)}^2$, givingEvery hyperbolic triangle, including ones with ideal vertices, has area at most $\pi$. But $A(2)=2\pi (\cosh 2-1)>\pi$. A hyperbolic triangle is geodesically convex, so if it contains a complete hyperbolic circle, it also contains the enclosed hyperbolic disc, its geodesic convex hull. This would violate the area comparison. Therefore no hyperbolic triangle contains a complete circle of radius two.

In the curvature $-1$ normalization, the hyperbolic triangle area is its angle defect: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 areaFor 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 givesConsequently $A(R)=(n-2)\pi -n\alpha (R)$ is continuous and strictly increasing. As $R\downarrow 0$, $\alpha (R)\to \pi -2\pi /n$, so $A(R)\to 0$; as $R\to \infty$, $\alpha (R)\to 0$, so $A(R)\to (n-2)\pi$. The intermediate value theorem proves existence for every permitted $A$. Indeed the required radius isThe argument also proves uniqueness of the radius in this construction; neither endpoint area is attained by a nondegenerate finite-radius polygon.