Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 4 / 12H (Geometry)

In the Poincare disc model, hyperbolic lines are diameters and circular arcs meeting the boundary circle orthogonally. In the upper half-plane model, they are vertical lines and semicircles with centres on the real axis. The metric is conformal, so hyperbolic perpendicularity agrees with Euclidean perpendicularity.

Use an isometry of the hyperbolic plane to send an arbitrary line to the positive imaginary axis. For $P=x+iy$ with $x\ne 0$, a point on that axis is $Q=iv$ with $v>0$. The hyperbolic distance formula givesPut $R=\sqrt{x^2+y^2}$. The last expression is $(v+R^2/v)/(2y)$, uniquely minimized at $v=R$, since $v+R^2/v\ge 2R$ with equality only there. Hence the unique closest point is $Q_0=iR$. The semicircle $|z|=R$ passes through $P$ and $Q_0$ and meets the imaginary axis at a right angle. Conversely a semicircle centred at a real number $b$ can meet that axis orthogonally only if $b=0$; a vertical line cannot furnish such a crossing. Thus the perpendicular is unique. Isometries transfer both uniqueness and minimization back to the original line.

For $P$ on the specified second semicircle, $a-r\le x\le a+r$ and $R=|P|\le a+r$. Its perpendicular to the axis is the semicircle just found, with radius $R$. MoreoverSince $t\le \mathrm{artanh}t$ for $0\le t<1$, we obtainThis is uniform over the whole semicircle, including points arbitrarily close to the ideal endpoints.

Finally send any first line of an ultraparallel pair to the axis by an isometry of the hyperbolic plane. The second cannot be a vertical line, since that would share the ideal point at infinity. Its two finite endpoints lie on the same side of zero; otherwise its semicircle would meet the axis, or share the endpoint zero. Reflect horizontally if necessary, so those endpoints are $a-r,a+r$ with $a>r>0$. The uniform estimate above then givesThus ultraparallel hyperbolic lines have strictly positive separation; disjointness in the interior alone would not suffice.