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{\bf Question}

Let $\ell_1$ and $\ell_2$ be parallel hyperbolic lines in ${\bf
H}$, where $\ell_1$ is contained in a vertical Euclidean line.
Prove that $\ell_1$ and $\ell_2$ are ultraparallel if and only if
there is a hyperbolic line $\ell$ perpendicular to both $\ell_1$
and $\ell_2$.
\medskip

{\bf Answer}

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\epsfig{file=407-2-1.eps, width=80mm}
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One way to proceed is by cases.  Suppose that $\ell_2$ is
ultraparallel to $\ell_1$ and has endpoints $a,\ b$ ($a&gt;0, b&gt;a$ as
drawn.  The case that $b&lt;a&lt;0$ is similar).

Any line perpendicular to $\ell_1$ is contained in a euclidean
circle centred at $\xi$ (where $\ell_1$ \lq intersects ' $\bf
{R}$).

Such a line is perpendicular to $\ell_2$ if and only if

$$r^2+r_2^2=\left(\ds\frac{1}{2}(b+a)-\xi\right)^2$$

where $r_2$ is the radius of the circle containing $\ell_2$ and
$r$ is the radius of the circle containing $\ell$ (and hence the
only variable in the equation). That is

$$r=\sqrt{\left(\ds\frac{1}{2}(b+a)-\xi\right)^2-r_2^2}$$

$\left(\rm{and\ since}\ \ell_1 \cap \ell_2=\emptyset
(\ell_1,\ell_2\ \rm{are\ disjoint})\
\ds\frac{1}{2}(b+a)-\xi&gt;r_2\right)$

So, such a circle $\ell$ exists, center $\xi$, radius $r$ as
above.

If $\ell_1\ell_2$ are parallel but not ultraparallel, then either
$\ell_2$ is a vertical euclidean line or is a euclidean circle
passing through $\xi$.

In the former case, no circle perpendicular to $\ell_1$ can also
be perpendicular to $\ell_2$, since the angle between $\ell$ and
$\ell_2$ is equal to the argument of the point of intersection of
$\ell$ and $\ell_2$ (as shown in the picture).

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\epsfig{file=407-2-3.eps, width=50mm}
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We may in the latter case use the law of cosines to calculate the
angle between $\ell$ (a circle perpendicular to $\ell_1$ with
radius $r$) and $\ell_2$ (with fixed center $c$ and fixed radius
$p$ to see that

$(c-\xi)^2=r^2+p^2-2rp\cos\theta$

$(c-\xi)^2-p^2=r^2-2p\cos\theta \cdot r$

The only way that $\theta=\ds\frac{\pi}{2}$ is that

$$(c-\xi)^2=r^2+p^2$$

But note that $c-\xi=p$ (since $\ell_1\ell_2$ are parallel) and so
$r=0$ which is not a circle.$\otimes$

So if $\ell_1\ell_2$ are ultraparallel there is a (unique) circle
(containing a hyperbolic line) perpendicular to both. If
$\ell_1\ell_2$ are parallel but not ultraparallel, no such circle
exists and so we are done.
\end{document}
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