Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 4 / 11E (Geometry)

The Poincare disc model has metricRotations about the origin are hyperbolic isometries, so a Euclidean circle centred there is a hyperbolic circle. Its radius is the radial length

The hyperbolic law of sines isSet $\gamma =\pi /2$ in the cosine laws for sides and angles and eliminate $\cos \alpha$ and $\cos \beta$. Straight algebra gives the right-triangle identities

Gauss-Bonnet theorem gives the area of a curvature-$-1$ triangle as $\pi -(\alpha +\beta +\gamma )$. Split the regular polygon into $2n$ congruent right triangles. If $\theta$ is a polygon half-angle, the right-triangle identity and part (a) giveThe polygon area is $(n-2)\pi -2n\theta$. Rewriting $\theta =\pi /2-{\cot }^{-1}(\tan \theta )$ yields

A regular hyperbolic hexagon has angle tending to $2\pi /3$ as its radius tends to zero and to zero as its vertices approach the ideal boundary. Continuity therefore gives a radius at which every angle is $\pi /2$.

Two congruent right-angled hexagons glued along three alternating sides form a hyperbolic pair of pants with geodesic boundary. Glue $2g-2$ such pairs of pants along boundary curves in the usual pants decomposition. The right angles match smoothly and produce a closed hyperbolic surface of genus $g$ for every $g\ge 2$.