\documentclass{amsart}

\usepackage{graphics}
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\title{Covering Regions with a Hexagon}

\author{Math 165: Calculus I}

\date{Fal, 1999}

\begin{document}

\maketitle

Any region of diameter 1 is contained in a region of constant width 1
(any two parallel lines tangent to the region must be a distance of 1
apart). In this note I'll sketch the proof of the following fact:

\begin{quote}
\textsl{Any region $R$ of constant width 1 can be covered by a regular hexagon
whose opposite sides are a distance of 1 apart.}
\end{quote}

Take a 60 degree angle, oriented in any direction, and move it towards the
region $R$ until both sides of the angle are tangent to~$R$.  Next  bring
a line parallel to $AB$ up towards the region until it is tangent to $R$,
and bring another line parallel to $AC$ towards $R$ until it is tangent to~$R$.
We get a figure like the following:

\[ \resizebox{4in}{!}{\includegraphics{hexagon1.eps}} \]

%A___________B____
% \    RRRRRR 
%  \  RRRRRRRR
%   \RRRRRRRR
%    \RRRRRR
%     \RRRR
%      \
%      C\

%A__________\B____
% \    RRRRRR\
%  \  RRRRRRRR\
%   \RRRRRRRR  \
%    \RRRRRR    \
%     \RRRR      \
%  ----\----------\----
%      C\          \
%

We now have $R$ inscribed in a parallelogram whose opposite sides must
be a distance of 1 apart since R has constant width~1. This means that
the parallelogram is actually a rhombus---the sides must have equal
length.  Since the triangles $ABC$ and $BDC$ are isosceles, and the angles
at $A$ and $D$ are 60 degrees, it follows that the angles at $B$ and $C$ must
be 120 degrees.

%A___________B
% \    RRRRRR\
%  \  RRRRRRRR\
%   \RRRRRRRR  \
%    \RRRRRR    \
%     \RRRR      \
%      \----------\ 
%      C          D

Now take two parallel lines that are perpendicular to the line $AD$ and
move them towards $R$ until they are both tangent to~$R$. These lines must
also be a distance of 1 apart since $R$ has constant width.
We now have $R$ inscribed in (a possibly non-regular) hexagon, $EBFGCH$, whose
opposite sides are a distance of 1 apart.

\[ \resizebox{4in}{!}{\includegraphics{hexagon2.eps}} \]

%      /
%A____/______B
% \  / RRRRRR\ /
%  \/ RRRRRRRR/
%  /\RRRRRRRR/ \
% /  \RRRRRR/   \
%     \RRRR/     \
%      \--/-------\ 
%      C /        D

%A____E______B
% \  / RRRRRR\
%  \/ RRRRRRRR\F
%  H\RRRRRRRR/ \
%    \RRRRRR/   \
%     \RRRR/     \
%      \--/-------\ 
%      C  G       D

Since the line $AD$ bisects the 60 degree angle at $A$ and is
perpendicular to $EH$, the angles $AEH$ and $HEA$ must also be 60
degrees. Therefore, the angles $CHE$ and $BEH$ are 120 degrees. A similar
argument implies that the angles $BFG$ and $CGF$ are 120 degrees. Thus,
the interior angles of the hexagon are all 120 degrees. The hexagon
need not be regular, since the sides do not necessarily have the same
length. However, since the interior angles are all the same, if $EH$
and $FG$ have the same length, then all the sides have the same length
and $EBFGCH$ is a regular hexagon.

Recall that at the start of the construction, we chose an arbitrary
orientation for the angle $A$. We can repeat the construction for any
starting orientation of $A$ and the lengths of $EH$ and $FG$ would change
accordingly. Suppose $EH < FG$, as in the above diagram. Then by rotating
the original orientation of the angle around 180 degrees we would end
up with a diagram like the one below where $EH > FG$. Since the lengths
$EH$ and $FG$ change continuously, there must be a starting orientation
for the angle $A$ for which the above construction gives $EH = FG$. The
resulting hexagon $EBFGCH$ is then the desired regular hexagon whose
opposite sides are a distance of 1 apart that contains~$R$.

\[ \resizebox{4in}{!}{\includegraphics{hexagon3.eps}} \]

%D____G______C
% \  / RRRRRR\
%  \/ RRRRRRRR\H
%  F\RRRRRRRR/ \
%    \RRRRRR/   \
%     \RRRR/     \
%      \--/-------\ 
%      B  E       A

\end{document}
