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{\bf Question}

Let $\ell$ be the closed hyperbolic line segment in the upper
half-plane \textbf{H} joining
$\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$ and
$-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$. Determine the two
points which break $\ell$ into three segments of equal hyperbolic
length. [Note that you won't be able to find these points
explicitly, but you will be able to derive conditions determining
them.]
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{\bf Answer}

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By symmetry, the two points to be calculated are $e^{i\theta}$ and
$e^{i(\pi-\theta)}$ where
$\ds\frac{\pi}{4}<\theta<\ds\frac{\pi}{2}$ (since
$\ds\frac{1}{\sqrt 2}+\ds\frac{1}{\sqrt 2}i=e^{i\frac{\pi}{4}}$.

The hyperbolic length of the line segment joining
$e^{i\frac{\pi}{4}}$ and $e^{i\theta}$

is \begin{eqnarray*} \ds\int_{\frac{\pi}{4}}^{\theta}
\ds\frac{1}{\sin(t)} \,dt & = &
\ln|\csc(t)-\cot(t)|_{\frac{\pi}{4}}^{\theta}\\ & = &
\ln\ds\frac{|\csc(\theta)-\cot(\theta)|}{|\csc(\frac{\pi}{4})-\cot(\frac{\pi}{4})|}\\
& = & \ln\ds\frac{|\csc(\theta)-\cot(\theta)|}{\sqrt{2}-1}
\end{eqnarray*}

(Parametrizing the line segment by $z(t)=e^{it}$ and using that
the element of arc length is $\ds\frac{1}{{\rm{Im}}(z)}|\,dz|$)

\newpage
The hyperbolic length of the line segment joining $e^{i\theta}$
and $e^{i(\pi-\theta)}$ is twice the length of the line segment
joining $e^{i\theta}$ and $e^{i\frac{\pi}{2}}$,

which is then

\begin{eqnarray*} 2 \ds\int_{\theta}^{\frac{\pi}{2}}
\ds\frac{1}{\sin(t)} \,dt & = & 2\ln
|\csc(t)-\cot(\theta)|_{\theta}^{\frac{\pi}{2}}\\ & = &
2\ln\ds\frac{|\csc(\frac{\pi}{2})-\cot(\frac{\pi}{2})|}{|\csc(\theta)-\cot(\theta)|}\\
& = & 2\ln\ds\frac{1}{|\csc(\theta)-\cot(\theta)|} \end{eqnarray*}

Hence, $\theta$ satisfies

$$\ln\ds\frac{(\csc(\theta)-\cot(\theta))}{\sqrt{2}-1}=
\ln\ds\frac{1}{(\csc(\theta)-\cot(\theta))^2}$$

and so $(\csc(\theta)-\cot(\theta))^3 = \sqrt{2}-1$

\un{$(1-\cos(\theta)^3=(\sqrt{2}-1)\sin^3(\theta)$}

(far enough since they don't have calculations.)
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