Lost at Sea
Steven R. Finch and John A. Shonder
What is the path of minimum expected length for escaping a planar convex region $\Omega $? We rigorously obtain best $2$-segment and $3$-segment solutions when $\Omega $ is an infinite strip, and numerically examine $2$% -segment solutions when $\Omega $ is a disk.
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\begin{document}

\author{Steven R. Finch and John A. Shonder}
\title{Lost at Sea}
\date{November 23, 2004}
\maketitle

\begin{abstract}
What is the path of minimum expected length for escaping a planar convex
region $\Omega $? We rigorously obtain best $2$-segment and $3$-segment
solutions when $\Omega $ is an infinite strip, and numerically examine $2$%
-segment solutions when $\Omega $ is a disk.
\end{abstract}

\footnotetext{
Copyright \copyright\ 2004 by Steven R. Finch. All rights reserved.}A
swimmer is lost in a dense fog at sea. She knows that the sea is a planar
infinite strip 
\[
\left\{ (x,y)\in \Bbb{R}^2:0\leq x\leq 1\right\} 
\]
of unit width. Assume that the $x$-coordinate of her initial position is
uniformly distributed on the interval $[0,1]$; the distribution of the $y$%
-coordinate can be arbitrary. Assume as well that her initial orientation
(the angle between her initial velocity vector and the $x$-axis) is
uniformly distributed on $[-\pi ,\pi ]$ and that her speed is constant. What
escape trajectory should the swimmer follow that minimizes her expected time
to reach either shore?

The min-max analog of this problem was solved long ago (see \cite{FW} for a
survey). Zalgaller \cite{Za1, Za2} proposed a heuristic solution of the
above min-mean problem. Shonder \cite{Sh} independently determined the
precise $2$-segment solution of least expected escape time. (We agree that a 
$k$-segment path is a continuous, piecewise linear curve consisting of $\leq
k$ pieces.)\ \ The details underlying Shonder's computation appear here for
the first time. We find a $3$-segment path that improves slightly upon the $%
2 $-segment path and also demonstrate that Zalgaller's solution is far from
optimal.

Next, we examine a different sea:\ the disk 
\[
\left\{ (x,y)\in \Bbb{R}^2:x^2+y^2\leq 1\right\} 
\]
of unit radius. Assume that the starting point is uniformly distributed on
the disk and that we seek (as before) to minimize the expected time to reach
the boundary. Intuition suggests that the swimmer should follow a $1$%
-segment path. If this is true, then the mean length of an arbitrary escape
trajectory is $\geq 8/(3\pi )=0.8488263631...$ \cite{Co, Cu}. A
calculus-of-variations proof of this general inequality is not known. We can
confirm this only numerically for $2$-segment paths.

In the final section, there appears a first attempt at proving the above
inequality under special circumstances. The technique is due to Gevirtz \cite
{Ge} and yields results when the escape path curvature is sufficiently small.

\subsection{Infinite Strip:\ 2-Segment Scenario}

Without loss of generality, let the initial position be $(x,0)$ and the
initial orientation be $\theta $, where $0\leq x\leq 1$ and $0\leq \theta
\leq \pi $. Fix a distance $r\geq 0$ and an angle $0\leq \alpha \leq \pi $.
As shown in Figure 1, there are five distinct cases to consider.\\ \FRAME{%
ftbpFUX}{5.9689in}{2.4396in}{0pt}{\Qcb{Five cases for the infinite strip.}}{%
}{fig1.jpg}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width
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\textbf{Case 1}:\ the swimmer reaches the right-hand shore before traveling
a distance $r$, which happens when $x+r\geq 1$ and $\theta \leq \arccos
((1-x)/r)$. The path length is $(1-x)/\cos (\theta )$.

\textbf{Case 2}: the swimmer travels a distance $r$, does not reach the
right-hand shore, pivots an angle $\alpha $, and then travels until the
left-hand shore is reached. This happens when 
\[
\left\{ \left[ x+r\geq 1\text{ and }\arccos ((1-x)/r)<\theta \right] \text{
or }\left[ x+r<1\right] \right\} \,\text{ and }\left\{ \theta <\tfrac \pi
2-\alpha \right\} . 
\]
The path length is $r+\left( x+r\cos (\theta )\right) /\cos (\theta +\alpha
) $.

\textbf{Case 3}:\ the swimmer travels a distance $r$, does not reach a
shore, pivots an angle $\alpha $, and then travels until the right-hand
shore is reached. This happens when 
\begin{eqnarray*}
&&\left\{ \left[ x+r\geq 1\text{ and }x-r\leq 0\text{ and }\arccos
((1-x)/r)<\theta <\pi -\arccos (x/r)\right] \right. \\
&&\text{or }\left[ x+r<1\text{ and }x-r\leq 0\text{ and }\theta <\pi
-\arccos (x/r)\right] \text{ } \\
&&\text{or }\left[ x+r\geq 1\text{ and }x-r>0\text{ and }\arccos
((1-x)/r)<\theta \right] \text{ } \\
&&\left. \text{or }\left[ x+r<1\text{ and }x-r>0\right] \right\} \text{ and }%
\left\{ \tfrac \pi 2-\alpha \leq \theta \leq \tfrac{3\pi }2-\alpha \right\} .
\end{eqnarray*}
The path length is $r+\left( x-1+r\cos (\theta )\right) /\cos (\theta
+\alpha )$.

\textbf{Case 4}: the swimmer travels a distance $r$, does not reach the
left-hand shore, pivots an angle $\alpha $, and then travels until the
left-hand shore is reached. This happens when 
\[
\left\{ \left[ x-r\leq 0\text{ and }\theta <\pi -\arccos (x/r)\right] \text{
or }\left[ x-r>0\right] \right\} \,\text{ and }\left\{ \tfrac{3\pi }2-\alpha
<\theta \right\} . 
\]
The path length is $r+\left( x+r\cos (\theta )\right) /\cos (\theta +\alpha
) $.

\textbf{Case 5}:\ the swimmer reaches the left-hand shore before traveling a
distance $r$, which happens when $x-r\leq 0$ and $\pi -\arccos (x/r)\leq
\theta $. The path length is $-x/\cos (\theta )$.\\

We've restricted attention to $0\leq \theta \leq \pi $ for simplicity's
sake. The set of escape paths corresponding to $-\pi \leq \theta \leq 0$ is
obtained from the set of escape paths corresponding to $0\leq \theta \leq
\pi $ via reflection across the horizontal axis, followed by reflection
across the vertical axis. This composite transformation can be written as $%
(x,\theta )\mapsto (1-x,\theta -\pi )$.

If it is assumed that $r>1$ and $0<\frac \pi 2-\alpha <\arccos (1/r)$, then
only Cases 1, 3 and 5 enter into the calculations of the expected path
length: 
\[
\frac 1\pi \dint\limits_0^1\left[ \int\limits_0^{\arccos \left( \frac{1-x}%
r\right) }\frac{1-x}{\cos (\theta )}d\theta +\int\limits_{\arccos \left( 
\frac{1-x}r\right) }^{\pi -\arccos \left( \frac xr\right) }\left( r+\frac{%
x-1+r\cos (\theta )}{\cos (\theta +\alpha )}\right) d\theta
-\int\limits_{\pi -\arccos \left( \frac xr\right) }^\pi \frac x{\cos (\theta
)}d\theta \right] dx.
\]
All of these integrals can be evaluated in closed form. Hence the expected
path length, multiplied by $\pi $, becomes 
\begin{eqnarray*}
&&\ \ \ \ r\left[ \left( 1+\cos \alpha \right) \left( \frac \pi 2+\sqrt{r^2-1%
}-r\right) -\left( 2+\cos \alpha \right) \arccos \left( \frac 1r\right)
+\frac \pi 2\right] + \\
&&\ \ \ \ \ln \left( \sqrt{r^2-1}+r\right) +\frac 12r^2\ln (1-\cos \alpha
)\sin ^2\alpha +r\sin \alpha \ln \left( \frac{\sqrt{r^2-1}\sin \alpha -\cos
\alpha }{r\sin \alpha }\right) + \\
&&\ \ \ \ \frac 14\left( r^2\sin ^2\alpha +1\right) \ln \left( \sqrt{r^2-1}%
\cos \alpha +\sin \alpha +r\right) + \\
&&\ \ \ \ \frac 14\left( r^2\sin ^2\alpha -1\right) \ln \left( \sqrt{r^2-1}%
\cos \alpha -\sin \alpha +r\right) - \\
&&\ \ \ \ \frac 14\left( r^2\sin ^2\alpha +1\right) \ln \left[ -\sqrt{r^2-1}%
\cos ^2\alpha -\left( \sin \alpha +\sqrt{r^2-1}-r\right) \cos \alpha -\sin
\alpha +r\right] - \\
&&\ \ \ \ \frac 14\left( r^2\sin ^2\alpha -1\right) \ln \left[ -\sqrt{r^2-1}%
\cos ^2\alpha +\left( \sin \alpha -\sqrt{r^2-1}+r\right) \cos \alpha +\sin
\alpha +r\right] .
\end{eqnarray*}
Minimizing this expression gives $r=1.0432668686...$ and $\alpha
=1.3734935859...\approx 78.7^{\circ }$ as the optimal parameter values for
the $2$-segment scenario. Therefore the least expected path length is $%
0.8869669056...$. See Figure 2 for several sample realizations (with fixed $r
$ and $\alpha $).\FRAME{ftbpFUX}{2.4292in}{3.8787in}{0pt}{\Qcb{Four sample
realizations with $r=1.043$, $\alpha =78.7^{\circ }$, $x=0.1,0.3,0.5,0.9$
and $\theta =80^{\circ },160^{\circ },50^{\circ },140^{\circ }$ (involving
three Cases).}}{}{fig2.jpg}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width
2.4292in;height 3.8787in;depth 0pt;cropleft "0";croptop "1.0003";cropright
"0.9997";cropbottom "0";filename 'C:/FIG2.JPG';file-properties "XNPEU";}}

Here is a new problem. For each $(r,\alpha )$, there exists a probability
distribution of escape times obtained by sampling $(x,\theta )$ uniformly
from $[0,1]\times [0,\pi ]$. Compute the median of each distribution and
then determine $(r,\alpha )$ corresponding to the least median. Monte Carlo
simulation suggests that a solution is $r=\infty $ ($\alpha $ can be
arbitrary), consistent with the idea that medians are less sensitive to
outliers than means. Define a random variable $q$ to be $(1-x)/\cos (\theta
) $ if $\theta <\pi /2$ and $-x/\cos (\theta )$ if $\theta >\pi /2$. The
path length is the median of $q$, which is estimated to be $0.78...$.

\subsection{Infinite Strip:\ 3-Segment Scenario}

We assume that $r>1$ and $0<\frac \pi 2-\alpha <\arccos (1/r)$ as at the end
of the preceding section. Cases 1 and 5 thus remain unchanged and only Case
3 needs to be refined. Fix a second distance $s\geq 0$ and a second angle $%
0\leq \beta \leq \pi $. See Figure 3.\\ \FRAME{ftbpFUX}{4.1796in}{3.0995in}{%
0pt}{\Qcb{Two subcases of Case 3.}}{}{fig3.jpg}{\special{language
"Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display
"USEDEF";valid_file "F";width 4.1796in;height 3.0995in;depth 0pt;cropleft
"0";croptop "1.0009";cropright "0.9998";cropbottom "0";filename
'C:/FIG3.JPG';file-properties "XNPEU";}}

\textbf{Subcase 3.1}:\ the swimmer travels a distance $r$, does not reach a
shore, pivots an angle $\alpha $, and then reaches the right-hand shore
before traveling a distance $s$. This happens when 
\[
\arccos ((1-x)/r)<\theta \leq \theta _0(x,r,s,\alpha ) 
\]
where $\theta _0$ is a solution of the equation 
\[
1-x-r\cos (\theta )+s\cos (\theta +\alpha )=0; 
\]
an exact formula for $\theta _0$ is 
\[
\theta _0=\arcsin \left( \frac{s\sin (\alpha )}{\sqrt{r^2+s^2-2\,r\,s\cos
(\alpha )}}\right) +\arccos \left( \frac{1-x}{\sqrt{r^2+s^2-2\,r\,s\cos
(\alpha )}}\right) . 
\]
The path length is $r+\left( x-1+r\cos (\theta )\right) /\cos (\theta
+\alpha )$.

\textbf{Subcase 3.2}:\ the swimmer travels a distance $r$, does not reach a
shore, pivots an angle $\alpha $, travels a distance $s$, does not reach a
shore, pivots an angle $\beta $, and then travels until the right-hand shore
is reached. This happens when 
\[
\theta _0(x,r,s,\alpha )<\theta <\pi -\arccos (x/r) 
\]
and $\beta $ satisfies other constraints (for example, if $\beta $ is too
small, then the $3$-segment might cross itself and reach the left-hand shore
rather than the right-hand shore). We will not attempt to prescribe these
additional inequalities. The path length is $r+s+\left( 1-x+s\cos (\theta
+\alpha )-r\cos (\theta )\right) /\cos (\theta +\alpha +\beta ).$\\

Under the proper circumstances, therefore, the expected path length is 
\begin{eqnarray*}
&&\ \ \ \frac 1\pi \dint\limits_0^1\left[ \int\limits_0^{\arccos \left( 
\frac{1-x}r\right) }\frac{1-x}{\cos (\theta )}d\theta +\int\limits_{\arccos
\left( \frac{1-x}r\right) }^{\theta _0}\left( r+\frac{x-1+r\cos (\theta )}{%
\cos (\theta +\alpha )}\right) d\theta +\right. \\
&&\ \ \ \left. \int\limits_{\theta _0}^{\pi -\arccos \left( \frac xr\right)
}\left( r+s+\frac{1-x+s\cos (\theta +\alpha )-r\cos (\theta )}{\cos (\theta
+\alpha +\beta )}\right) d\theta -\int\limits_{\pi -\arccos \left( \frac
xr\right) }^\pi \frac x{\cos (\theta )}d\theta \right] dx.
\end{eqnarray*}
A closed-form expression is unlikely here because of the presence of $\theta
_0$ in the middle two integrals:\ only integration with respect to $\theta $
seems feasible. Let 
\[
\begin{array}{ccc}
\kappa =\sqrt{r^2+s^2-2\,r\,s\cos (\alpha )}, &  & \rho =\arcsin \left(
s\sin (\alpha )/\kappa \right) ,
\end{array}
\]
\[
\begin{array}{ccc}
u_1=r\sin (\alpha ), &  & u_2=s\sin (\beta )-r\sin (\alpha +\beta ),
\end{array}
\]
\[
\begin{array}{ccc}
v_1=r(1+\cos (\alpha )), &  & v_2=r+s+s\cos (\beta )-r\cos (\alpha +\beta ),
\end{array}
\]
\[
\begin{array}{ccc}
\xi _1=\alpha +\arccos \left( (1-x)/r\right) , &  & \xi _2=\alpha +\beta
+\rho +\arccos \left( (1-x)/\kappa \right) ,
\end{array}
\]
\[
\begin{array}{ccc}
\eta _1=\alpha +\rho +\arccos \left( (1-x)/\kappa \right) , &  & \eta
_2=\alpha +\beta +\pi -\arccos (x/r)
\end{array}
\]
and, for $j=1,2$, define 
\[
p_j=u_j\ln \left| \frac{\cos (\xi _j)}{\cos (\eta _j)}\right| +\frac{%
(-1)^j(1-x)}2\ln \left( \frac{\sin (\xi _j)\left( \sin (\eta _j)+1\right)
-\sin (\eta _j)-1}{\sin (\xi _j)\left( \sin (\eta _j)-1\right) +\sin (\eta
_j)-1}\right) +v_j(\eta _j-\xi _j). 
\]
Then the expected path length, multiplied by $\pi $, becomes 
\[
\ln \left( \sqrt{r^2-1}+r\right) +r\left( r-\sqrt{r^2-1}\right)
+\dint\limits_0^1\left( p_1+p_2\right) dx. 
\]
This can be minimized by use of Leibniz's rule for differentiation under the
integral sign. We obtain $r=1.0255050653...$, $\alpha
=1.4909825316...\approx 85.4^{\circ }$, $s=0.5306340577...$ and $\beta
=2.7495709960...\approx 157.5^{\circ }$ as the optimal parameter values for
the $3$-segment scenario, with least expected path length $0.8835534788...$.

A third subcase, for which $\pi \leq \beta \leq 2\pi $, awaits examination.
Either shore could be the final resting place.

\subsection{Infinite Strip:\ Zalgaller's Proposed Solution}

Consider the path in Figure 4, where the points $A=(0,0)$, $B=(0.814,0)$, $%
C=(0.8460,0.0005)$, $D=(1.3017,0.0151)$, $E=(0.814,1)$ are joined by line
segments, with the exception of points $B$, $C$ which are joined by a tiny
circular arc of radius $1$, center $E$ and angle $0.032.$ Zalgaller \cite
{Za1, Za2} claimed that, if the swimmer follows this path, then her expected
time to reach either shore is approximately minimal. We examined his claim
by computing best $2$-segment and $3$-segment fits to Zalgaller's path, and
then calculating the expected escape time via our formulas. For example, in
the $2$-segment fit, we obtained $r=1.3017$, $\alpha =64.3^{\circ }$ and
hence the expected escape time is $0.9188$. This is somewhat consistent with
Zalgaller's estimate $0.9523$. More importantly, however, Shonder's optimal $%
2$-segment path \cite{Sh} has expected escape time $0.8870$ which is
considerably less than $0.9523$. There must be an error somewhere in the
details of \cite{Za1}. We aim someday to better understand Zalgaller's
elaborate construction, in the hope that his procedure (once corrected) will
lead to escape trajectories that outperform even our optimal $3$-segment
path.\FRAME{ftbpFUX}{2.7994in}{2.7389in}{0pt}{\Qcb{Zalgaller's escape path.}%
}{}{fig4.jpg}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width
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"1.0017";cropbottom "0";filename 'C:/FIG4.JPG';file-properties "XNPEU";}}

\subsection{Disk:\ 2-Segment Scenario}

Without loss of generality, let the starting point be $(x,0)$ and the
initial orientation be $\theta $, where $0\leq x\leq 1$ and $-\pi \leq
\theta \leq \pi $. Fix a distance $0\leq r\leq 2$ and an angle $0\leq \alpha
\leq \pi $. As shown in Figure 5, there are two distinct cases to consider. 
\FRAME{ftbpFUX}{4.5792in}{2.1395in}{0pt}{\Qcb{Two cases for the disk.}}{}{%
fig5.jpg}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width
4.5792in;height 2.1395in;depth 0pt;cropleft "0";croptop "1.0006";cropright
"1.0007";cropbottom "0";filename 'C:/FIG5.JPG';file-properties "XNPEU";}}For
convenience, define 
\[
\begin{array}{ccc}
\varphi (x,r)=\arccos \left( \dfrac{1-x^2-r^2}{2xr}\right) , &  & \psi
(x,r)=\arccos \left( -\dfrac rx\right) .
\end{array}
\]

\textbf{Case 1}$^{\prime }$:\ the swimmer reaches the shore before traveling
a distance $r$, which happens when $x\geq |r-1|$ and $-\varphi \leq \theta
\leq \varphi $, or when $x<r-1$. The path length is $q=-x\cos (\theta )+%
\sqrt{1-x^2\sin ^2(\theta )}$.

\textbf{Case 2}$^{\prime }$: the swimmer travels a distance $r$, does not
reach the shore, pivots an angle $\alpha $, and then travels until the shore
is reached. The path length is $r+s$, where 
\[
s=y\cos (\alpha +\omega )+\sqrt{1-y^2\sin ^2(\alpha +\omega )}, 
\]
\[
y=\sqrt{x^2+r^2+2\,x\,r\cos (\theta )}, 
\]
\[
\omega =\left\{ 
\begin{array}{lll}
\arcsin \left( \dfrac{x\sin (\theta )}y\right) &  & \text{if }\left( x\geq r%
\text{ and }-\psi \leq \theta \leq \psi \right) \text{ or }x<r, \\ 
\pi -\arcsin \left( \dfrac{x\sin (\theta )}y\right) &  & \text{if }x\geq r%
\text{ and }\left( \theta \leq -\psi \text{ or }\psi \leq \theta \right) .
\end{array}
\right. 
\]
The proof of Case 1$^{\prime }$ makes use of the Law of Cosines and the fact
that $\varphi $ exists if and only if $x\geq |r-1|$. The Law of Sines
appears in the proof of Case 2$^{\prime }$ when defining the angle $\omega $
of an auxiliary triangle with side lengths $r$, $x$ and $y$; the Law of
Cosines is then applied to the triangle with side lengths $1$, $s$ and $y$
to obtain $s$.

The expected path length is therefore $I+J$, where 
\[
\begin{array}{ccc}
I=\dfrac 1\pi \dint\limits_{|r-1|}^1f\,(x,r,\alpha )\,x\,dx, &  & J=\left\{ 
\begin{array}{ccc}
\dfrac 1\pi \dint\limits_0^{r-1}g\,(x)\,x\,dx &  & \text{if }r\geq 1, \\ 
\dfrac 1\pi \dint\limits_0^{-r+1}h\,(x,r,\alpha )\,x\,dx &  & \text{if }r<1
\end{array}
\right. 
\end{array}
\]
and 
\[
\begin{array}{ccccc}
f=\dint\limits_{-\varphi }^\varphi q\,d\theta +\dint\limits_\varphi ^\pi
(r+s)\,d\theta +\dint\limits_{-\pi }^{-\varphi }(r+s)\,d\theta , &  & 
g=\dint\limits_{-\pi }^\pi q\,d\theta , &  & h=\dint\limits_{-\pi }^\pi
(r+s)\,d\theta .
\end{array}
\]
Closed-form expressions are not possible here. Numerical computations
confirm that $8/(3\pi )$ is the value of $I+J$ when $\alpha =\pi $, for any $%
r\geq 0$, and that otherwise $I+J$ is strictly larger than $8/(3\pi )$. See
Figure 6 for several sample realizations (with fixed $r$ but variable $%
\alpha $).\FRAME{ftbpFUX}{2.9992in}{3.0095in}{0pt}{\Qcb{Sample realizations
with $r=0.5$, $x=0.1,0.3,0.5$, $\alpha =110^{\circ },160^{\circ },20^{\circ }
$ and $\theta =200^{\circ },80^{\circ },45^{\circ }$(all from Case 2$%
^{\prime }$).}}{}{fig6.jpg}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width
2.9992in;height 3.0095in;depth 0pt;cropleft "0";croptop "0.9986";cropright
"0.9988";cropbottom "0";filename 'C:/FIG6.JPG';file-properties "XNPEU";}}

We mention a corresponding min-median result: Monte Carlo simulation
suggests that a solution is $r=2$ ($\alpha $ can be arbitrary). The path
length is the median of the random variable $-x\cos (\theta )+\sqrt{%
1-x^2\sin (\theta )^2}$, which is estimated to be $0.94...$.

These kinds of ``lost at sea'' problems deserve to be more thoroughly
studied! The only other published reference we know regarding the min-mean
problem is \cite{Gl}.

\subsection{Disk: Gevirtz's Technique}

Let $\gamma :[0,\infty )\rightarrow \Bbb{C}$ be a differentiable curve in
the complex plane satisfying $\gamma (0)=0$. Parametrize $\gamma $ by
arclength $s$ and let $\gamma ^{\prime }(s)=\exp (i\,\phi (s))$, where we
assume that $\phi (s)=\arg (\gamma ^{\prime }(s))$ is real, $\phi (0)=0$ and 
$|\phi (s)|\leq \phi _0<\frac \pi 2$. The function $s\mapsto \lambda
(s)=|\gamma (s)|$ is one-to-one and hence possesses an inverse $\lambda
\mapsto s(\lambda )$. Let $D$ denote the unit disk in $\Bbb{C}$ and define $%
C(z)$ to be the circle of radius $1$ centered at $z$; as an example, $C(0)$
is the boundary of $D$.

We wish to minimize the average 
\[
A(\gamma )=\frac 1\pi \dint\limits_D\sigma (z,\gamma )\,dx\,dy 
\]
where $\sigma (z,\gamma )$ denotes the first (and only) value of $s$ for
which $|\gamma (s)-z|=1$. (By symmetry of $D$, we have substituted $-z$ for $%
z$.) Now, 
\[
\dint\limits_D\sigma (z,\gamma )\,dx\,dy=\dint\limits_0^\infty s\,d\mu (s) 
\]
where $\mu (s)$ is the area of the portion of $D$ not enclosed by $C(\gamma
(s))$. This identity follows via a Riemann sum argument.

Consider the family of circular arcs $F=\left\{ D\cap C(\gamma
(s)):0<s<\infty \right\} $. If $\phi _0$ is small enough, then no two
distinct arcs in $F$ can intersect. As a consequence of the above identity,
if the curvature $\phi ^{\prime }(s)$ is suitably small, then 
\[
A(\gamma )=\frac 2\pi \dint\limits_0^{s^{*}}s\sqrt{1-\left( \frac{|\gamma
(s)|}2\right) ^2}\cos \left( \phi (s)-\arg (\gamma (s))\right) ds 
\]
where $s^{*}$ satisfies $|\gamma (s^{*})|=2$. Note that $s^{*}$ depends on $%
\gamma $, which complicates any variational approach to this problem.
Because $\lambda ^{\prime }(s)=\cos \left( \phi (s)-\arg (\gamma (s))\right) 
$ and $s(\lambda )\geq \lambda $, we have 
\[
A(\gamma )=\frac 2\pi \dint\limits_0^2s(\lambda )\sqrt{1-\left( \frac
\lambda 2\right) ^2}d\lambda \geq \frac 2\pi \dint\limits_0^2\lambda \sqrt{1-%
\frac{\lambda ^2}4}d\lambda =\frac 8{3\pi }, 
\]
as was to be shown. It is unclear what to do in the general situation for
which distinct arcs in $F$ are not necessarily disjoint. We also wonder if a
similar technique exists for the infinite strip sea.

\subsection{Acknowledgements}

We are grateful to Julian Gevirtz and Patrice Le Conte for their help.

\begin{thebibliography}{9}
\bibitem{FW}  S. R. Finch and J. E. Wetzel, Lost in a forest, \textit{Amer.
Math. Monthly} 11 (2004) 645-654.

\bibitem{Za1}  V. A. Zalgaller, A discussion about a question of Bellman (in
Russian), \textit{Zap. Nauchn. Sem. S.-Peterburg. Otdel. Mat. Inst. Steklov
(POMI)} 299 (2003) 54--86, 327--328; Engl. transl. available from S. Finch
upon request (\textit{J. Math. Sci.}, to appear?); MR2038255 (2004i:90174).

\bibitem{Za2}  V. A. Zalgaller, Extremal problems concerning the convex hull
of a space curve (in Russian), \textit{Algebra i Analiz}, v. 8 (1996) n. 3,
1--13; Engl. transl. in \textit{St. Petersburg Math.} J. 8 (1997) 369--379;
MR1402285 (97c:58028).

\bibitem{Sh}  J. A. Shonder, challenge and solution statement in F. Morgan,
Math Chat, http://www.maa.org/features/mathchat/mathchat\_8\_3\_00.html.

\bibitem{Co}  R. Coleman, Random paths through convex bodies, \textit{J.
Appl. Probab.} 6 (1969) 430--441; MR0254890 (40 \#8097).

\bibitem{Cu}  P. R. Coulton, Expectation relations for escape problems,
unpublished note (2003).

\bibitem{Gl}  B. Gluss, An alternative solution to the ``lost at sea''
problem, \textit{Naval Res. Logist. Quart.} 8 (1961) 117--121; MR0129005 (23
\#B2042).

\bibitem{Ge}  J. Gevirtz, private communication (1999).\\

\begin{tabular}{llll}
& Steven R. Finch &  & John A. Shonder \\ 
& Clay Mathematics Institute &  & Oak Ridge National Laboratory \\ 
& One Bow Street &  & P.O. Box 2008, Bldg 3147, MS-6070 \\ 
& Cambridge, MA 02138 &  & Oak Ridge, TN 37831 \\ 
& \textit{Steven.Finch@inria.fr} &  & \textit{shonderja@ornl.gov}
\end{tabular}
\end{thebibliography}

\end{document}

math0411518.extracted_bib
\begin{thebibliography}{9}
\bibitem{FW}  S. R. Finch and J. E. Wetzel, Lost in a forest, \textit{Amer.
Math. Monthly} 11 (2004) 645-654.

\bibitem{Za1}  V. A. Zalgaller, A discussion about a question of Bellman (in
Russian), \textit{Zap. Nauchn. Sem. S.-Peterburg. Otdel. Mat. Inst. Steklov
(POMI)} 299 (2003) 54--86, 327--328; Engl. transl. available from S. Finch
upon request (\textit{J. Math. Sci.}, to appear?); MR2038255 (2004i:90174).

\bibitem{Za2}  V. A. Zalgaller, Extremal problems concerning the convex hull
of a space curve (in Russian), \textit{Algebra i Analiz}, v. 8 (1996) n. 3,
1--13; Engl. transl. in \textit{St. Petersburg Math.} J. 8 (1997) 369--379;
MR1402285 (97c:58028).

\bibitem{Sh}  J. A. Shonder, challenge and solution statement in F. Morgan,
Math Chat, http://www.maa.org/features/mathchat/mathchat\_8\_3\_00.html.

\bibitem{Co}  R. Coleman, Random paths through convex bodies, \textit{J.
Appl. Probab.} 6 (1969) 430--441; MR0254890 (40 \#8097).

\bibitem{Cu}  P. R. Coulton, Expectation relations for escape problems,
unpublished note (2003).

\bibitem{Gl}  B. Gluss, An alternative solution to the ``lost at sea''
problem, \textit{Naval Res. Logist. Quart.} 8 (1961) 117--121; MR0129005 (23
\#B2042).

\bibitem{Ge}  J. Gevirtz, private communication (1999).\\

\begin{tabular}{llll}
& Steven R. Finch &  & John A. Shonder \\ 
& Clay Mathematics Institute &  & Oak Ridge National Laboratory \\ 
& One Bow Street &  & P.O. Box 2008, Bldg 3147, MS-6070 \\ 
& Cambridge, MA 02138 &  & Oak Ridge, TN 37831 \\ 
& \textit{Steven.Finch@inria.fr} &  & \textit{shonderja@ornl.gov}
\end{tabular}
\end{thebibliography}