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\begin{document}
\centerline{MATH 392 -- Geometry Through History}
\centerline{Finishing Bolyai-Lobachevsky Theorem}
\centerline{February 29}
\vskip 20pt
\noindent
To finish up our discussion of the Bolyai-Lobachevsky theorem
and hyperbolic geometry, we need to prove the following lemma
that we deferred last time.  (Note: some names of points changed
to agree with a diagram I'm copying!)


\vskip 10pt
\noindent
{\it Reminder about notation}:  $\overline{AB}$ is a segment of a (hyperbolic)
line; $AB$ is the length of the line segment; $\widehat{AB}$ is the 
\emph{horocycle arc}, or its arclength.

\vskip 10pt
\setcounter{lem}{5}
\begin{lem}
\label{BLLemma}
Let $\overrightarrow{OX} \perp \overrightarrow{OY}$, let $A$ be along the 
ray $\overrightarrow{OY}$ with $OA = u$, and $\overrightarrow{AA'}$ be parallel
to $\overrightarrow{OX}$ in the direction of $X$ and $A'$.
Let $\widehat{OB}$ be the horocycle arc through $O$ (for the pencil 
of parallel lines containing $\overleftrightarrow{OX}$ and $\overleftrightarrow{AA'}$) 
intersecting $\overrightarrow{AA'}$ at $B$.  Let $v = AB$ and $s = \widehat{OB}$.
Then the following relations hold:
$$e^{v/k} = \cosh(u/k),$$
and
$$s = \sigma \tanh(u/k)$$
where $\sigma$ is a constant that will be determined in the course of the proof.
\end{lem}

\vfill\eject

\noindent
\begin{figure}
\begin{centering}
\includegraphics[trim=2in 8in 0in 0in,clip,width=7truein,height=5truein]{BLLProof.pdf}
\caption{Figure for proof of Lemma~\ref{BLLemma}}
\end{centering}
\end{figure}

\begin{itemize}
\item Let $\overleftrightarrow{TT'}$
be parallel with $\overrightarrow{OX}$ in the direction of $X$ and $T$ \emph{and} with 
$\overrightarrow{OY}$ in the direction of $Y$ and $T'$.  (You showed how to get such a 
line in Problem Set 3.)
\item Construct $N$ as in diagram so $AN = u$ and let $\overrightarrow{NN'}$ be
perpendicular to $\overleftrightarrow{AA'}$ and
parallel to $\overrightarrow{OY}$ in the direction of $Y$.
\item Construct $U$ as in the diagram so $OU = u$, let $\overrightarrow{UU'}$
be parallel to $\overrightarrow{OX}$ in the direction of $X$, and extend the horocycle through
$O,B$ so that it meets $\overrightarrow{UU'}$ at point $V$.
\item Construct $M$ so $UM = u$ again, and let $\overrightarrow{MM'}$ be perpendicular
to $\overleftrightarrow{UU'}$ parallel
to $\overrightarrow{OY}$ in the direction of $Y$.
\item Construct the perpendiculars $\overline{ND'}, \overline{OC'}, \overline{ME'}$
to $\overleftrightarrow{TT'}$ as in the diagram.
\end{itemize}

\vfill\eject
\noindent
\begin{figure}
\begin{centering}
\includegraphics[trim=2in 8in 0in 0in,clip,width=7truein,height=5truein]{BLLProof.pdf}
\caption{Figure for proof of Lemma~\ref{BLLemma}}
\end{centering}
\end{figure}

\begin{itemize}
\item By construction, it follows that $ND' = OC' = ME'$ (since all equal the $x$
such that $\Pi(x) = 45^\circ$)
\item Consider the concentric horocycle arcs $\widehat{ND}, \widehat{OC}, \widehat{ME}$.
By the previous bullet, these are also all equal in length:
 $\widehat{ND} = \widehat{OC} = \widehat{ME} = \sigma$.  This defines the constant
in the second formula we are trying to prove.
\item By our previous result about ratios of horocycle arcs, since $BN = u + v$, 
and $\widehat{BC} = \sigma - s$, we have
\begin{equation}
\label{one}
\frac{\sigma - s}{\sigma} = e^{-(u+v)/k}
\end{equation}
\item Similarly, $VM = u - v$, and $\widehat{CV} = \sigma + s$, so
\begin{equation}
\label{two}
\frac{\sigma}{\sigma + s} = e^{-(u-v)/k} \Leftrightarrow \frac{\sigma + s}{\sigma} = e^{(u-v)/k}
\end{equation}
\end{itemize}
Adding (\ref{one}) and (\ref{two}) gives
$$2 = \frac{\sigma - s}{\sigma} + \frac{\sigma + s}{\sigma} = e^{-v/k}(e^{u/k} + e^{-u/k}),$$
which shows
$$e^{v/k} = \cosh(u/k)$$
This is the first part of Lemma~\ref{BLLemma}.
\vskip 10pt
The second part also comes from (\ref{one}) and (\ref{two}), which we copy for your
convenience:
$$\frac{\sigma - s}{\sigma} = e^{-(u+v)/k}\eqno(1)$$
and
$$
\frac{\sigma}{\sigma + s} = e^{-(u-v)/k} \Leftrightarrow \frac{\sigma + s}{\sigma} = e^{(u-v)/k}
\eqno(2)$$
If we subtract these instead of adding we get:
$$\frac{2s}{\sigma} = e^{(u-v)/k} - e^{-(u+v)/k} = e^{-v/k}(e^{u/k} - e^{-u/k})$$
so by the first part: 
$$s = \sigma e^{-v/k} \sinh(u/k) = \sigma \frac{\sinh(u/k)}{\cosh(u/k)} = \sigma \tanh(u/k)$$
and the proof is complete.  
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