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{\bf Question}

 Given any angle $\theta$ in the interval $(0, \frac{\pi}{2})$
and any point $c$ in ${\bf R}$, consider the two Euclidean rays
$R_1$ and $R_2$ in ${\bf H}$, originating at $c$ and making angles
$\theta$ and $\pi -\theta$ with the positive real axis,
respectively.  Show that the hyperbolic distance between the two
points $R_1\cap L$ and $R_2\cap L$ is independent of $L$, where
$L$ is any horizontal Euclidean line in ${\bf H}$.
\medskip

{\bf Answer}


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\begin{picture}(8,4)
\put(0,0){\line(1,0){6}} \put(0,2){\line(1,0){6}}
\put(3,0){\line(1,1){3}} \put(3,0){\line(-1,1){3}}
\put(6.2,1.9){$L$} \put(7,1){\bf H} \put(5.9,3.1){$R_1$}
\put(-0.1,3.1){$R_2$} \put(2.9,-0.3){$c$} \put(3.4,0.1){$\theta$}
\put(2.4,0.1){$\theta$}

\end{picture}
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First involve symmetry; since the picture is invariant under
reflection through the vertical line through $c$, which is a
consequence of the choice of angles of $R_1R_2$ with ${\bf{R}}$,
not only are the imaginary points of the two points $R_1 \cap L$
and $R_2 \cap L$ equal, but both lie on a circle centred at $c$.
So, the hyperbolic line segment from $p_1=R_1 \cap L$ to $p_2=R_2
\cap L$ is parametrized by

$$f:[\theta,\pi-\theta] \longrightarrow {\bf{H}},\ \
f(t)=c+pe^{it}\ \ p=|c-p_1|=|c-p_2|$$

Then, Im$(f(t))=p\sin (t)$ (since $c \in {\bf{R}})$ and
$|f'(t0|=p$, and so

length$_{\bf{H}}(f)=d_{\bf{H}}(p_1p_2)=\ds\int_{\theta}^{\pi-\theta}
\ds\frac{1}{\sin(t)} \,dt$, which is independent of $L$, and in
fact depends only on the angles.

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