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{\bf Question}

Let $\ell_0$ and $\ell_1$ be ultraparallel hyperbolic lines in
${\bf H}$.  Label the endpoints at infinity of $\ell_0$ as $z_0$
and $z_1$, and the endpoints at infinity of $\ell_1$ as $w_0$ and
$w_1$, so that they occur in the order $z_0$, $w_0$, $w_1$, and
$z_1$ moving counter-clockwise around ${\bf R}$.  Prove that
\[ {\rm tanh}^2\left(\frac{1}{2} {\rm d}_{\bf H}(\ell_0, \ell_1)
\right) =\frac{1}{1 -[z_0, w_0; w_1, z_1]}. \]

\medskip

{\bf Answer}

By the ordering of the parts around ${\bf{R}}$, there exists an
element of M\"{o}b(${\bf{H}}$) taking $z_0$ to 0, $z_1$ to
$\infty$, $w_0$ to 1, and $w_1$ to $x>1$, so that

$$[z_0w_0;w_1z_1]=[0,1;x,\infty]=\ds\frac{x-1}{0-1}=1-x$$

and so $1-[z_0w_0;w_1z_1]=x$.

\newpage
\un{$d_{\bf{H}}(\ell_0\ell_1)$}:

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we used to determine the perpendicular bisector of $\ell_0\ell_1$:

By the euclidean pythagorean theorem:

$\begin{array} {l}
\left(\ds\frac{1+x}{2}\right)^2=r^2+\left(\ds\frac{-1+x}{2}\right)^2\\
(1+x)^2=4r^2+(x-1)^2\\1+2x+x^2=4r^2+x^2-2x+1\\4r^2=4x\\r=\sqrt x
\end{array}$

$\cos(\alpha)=\ds\frac{2r}{1+x}=\ds\frac{2\sqrt x}{1+x}$

$\sin(\alpha)=\ds\frac{x-1}{2}\cdot\ds\frac{2}{x+1}=\ds\frac{x-1}{x+1}$

\begin{eqnarray*} d_{\bf{H}}(\ell_0\ell_1) & = &
\ds\int_{\alpha}^{\frac{\pi}{2}} \ds\frac{1}{\sin(t)} \,dt\\ & = &
-\ln|\csc(\alpha)-\cot(\alpha)|\\ & = &
\ln\left|\ds\frac{\sin(\alpha)}{1-\cos(\alpha)}\right| =
\ln\left|\ds\frac{x-1}{1+x-2\sqrt x}\right| \end{eqnarray*}

\begin{eqnarray*} d_{\bf{H}}(\ell_0\ell_1) & = &
\ln\left(\ds\frac{x-1}{(\sqrt {x}-1)^2}\right)\\ & = &
\ln\left(\ds\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}-1)^2}\right)\\
& = & \ln\left(\ds\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)
\end{eqnarray*}

\newpage
\begin{eqnarray*} \tanh^2(a) & = &
\ds\frac{\sinh^2(a)}{\cosh^2(a)}\\ & = &
\ds\frac{(e^a-e^{-a})^2}{(e^a+e^{-a})^2}\\ & = &
\ds\frac{e^{2a}+e^{-2a}-2}{e^{2a}+e^{-2a}+2} \end{eqnarray*}

\begin{eqnarray*} \tanh^2\left(\ds\frac{1}{2}d_{\bf{H}}(\ell_0\ell_1\right) & =
&
\tanh^2\left(\ds\frac{1}{2}\ln\left(\ds\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\right)\\
& = &
\ds\frac{\ds\frac{\sqrt{x}+1}{\sqrt{x}-1}+\ds\frac{\sqrt{x}-1}{\sqrt{x}+1}-2}
{\ds\frac{\sqrt{x}+1}{\sqrt{x}-1}+\ds\frac{\sqrt{x}-1}{\sqrt{x}+1}+2}\\
& = &
\ds\frac{(\sqrt{x}-1)^2+(\sqrt{x}-1)^2-2(x-1)}{(\sqrt{x}+1)^2+(\sqrt{x}-1)^2+2(x-1)}\\
& = & \ds\frac{4}{4x} = \ds\frac{1}{x} =
\ds\frac{1}{1-[z_0,w_0;w_1,z_1]} \end{eqnarray*}


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