\definition{} A curve has {\em constant diameter} if every diameter is the same, no matter which direction the parallel support lines point. For curves with constant diameter, we can then talk about {\em the diameter} without ambiguity.

\begin{problem}{}<non-examples>
    Draw an example of a simple closed curve with constant diameter.

    Draw an example of a simple closed curve with non-constant diameter.
\end{problem}
\vspace{4cm}
\vfill
\pagebreak


\begin{problem}{(Open-ended)}
    More than likely, in \ref{non-examples}, your example of a curve with constant diameter was a circle. Can you come up with any non-circle examples of a curve with constant diameter?\\
    {\em Hint: begin with a simple polygon. See if you can "fix" it to have constant diameter.}
\end{problem}
\pagebreak



\begin{problem}{Reuleaux triangle}<reuleaux triangle>
    We'll now take you through a construction of some non-trivial shapes with constant diameter.

    Draw an equilateral triangle with side lengths $d$. \\
    Verify that this is a curve with non-constant diameter. Where is the diameter not equal to $d$? \\
    These are the problem regions that we need to address.

    Label the corners of your triangle $X,Y,Z$. Connect $Y$ to $Z$ with a circular arc of radius $d$ and center $X$. Similarly, connect $Z$ to $X$ with a circular arc centered at $Y$ and connect $X$ to $Y$ with a circular arc centered at $Z$.

    How did this process address our problem regions from before?\\
    {\em Note: for any pair of parallel support lines, one of the lines is touching a corner of the triangle.}
\end{problem}
\pagebreak

\begin{figure}
    \centering
    \includegraphics[width=0.5\linewidth]{img/reuleaux.png}
    \caption{Reuleaux triangle: Curve of constant diameter constructed from an equilateral triangle.}
\end{figure}
\begin{problem}{Generalization}
    Draw a regular pentagon. Repeat the above process to create another (non-circular) curve of constant diameter. \\
    Repeat this process for a heptagon.
\end{problem}
\vfill 

\begin{problem}{Generalization}
    Does this process work for all polygons? \\
    What happens if you try to do this process with a square? What fails?
\end{problem}
\begin{solution}
    As stated, this only works for regular, odd-sided polygons. This is because any pair of parallel support lines of an odd-sided polygon will intersect a corner. For even-sided shapes, this is not the case.
\end{solution}
\vfill 
\pagebreak


\begin{problem}{Circumference}
    For your curves of constant diameter, calculate their circumference. How does this compare to the circumference of a circle? 

    Conjecture what the circumference of a curve of constant diameter $d$ should be.\\
    {\em We will not prove this here.}
\end{problem}
\begin{solution}
    The triangle, pentagon, and heptagon we constructed all have circumference $2\pi d$, where $d$ is the diameter. We conjecture that this holds for all curves of constant diameter $d$.
\end{solution}

\vfill

\begin{problem}{Area}
    For the Reuleaux triangle, \ref{reuleaux triangle}, calculate its area. How does this compare to the area of a circle?\\
    {\em If you would like a} big {\em challenge, you can try to calculate the area of the constant diameter pentagon.}

    Conjecture what the area of a curve of constant diameter $d$ should be. \\
    {\em Hint: Your conjecture should be a bound on the area, not a specific area.}
\end{problem}
\begin{solution}
    The triangle has area 
    \begin{equation*}
        3*\tfrac{\pi d^2}{6} - 2*\tfrac{\sqrt{3}d^2}{4} = \tfrac{\pi - \sqrt{3}}{2}d^2.
    \end{equation*}
    The $\pi d^2/6$ comes from a $60^\circ$ arc of a circle with radius $d$. This overcounts the inner equilateral triangle twice, so we need to subtract that overcounting.

    Comparing this to a circle of diameter $d$, we find that the circle has larger area. We conjecture that the circle should have the largest area for a perscribed diameter.
\end{solution}

\vfill\pagebreak

\begin{problem}{Non-equilateral triangles}<non-equilateral>
    Draw a non-equilateral triangle with vertices $X,Y,Z$. See the figure for a visualization of the following process.
    \begin{enumerate}[label={(\alph*)}]
        \item 
            Start with point $X$. Fix a large length $\ell$. Draw a circular arc, centered at $X$, with radius $\ell$, that sweeps from direction $XY$ to $XZ$.
        \item 
            Extend our curve with a circular arc, centered at $Z$, that extends the curve to direction $YZ$.\\
            Repeat this with another arc centered at $Y$.
        \item 
            Extend our curve with a circular arc, centered at $Y$, that sweeps from direction $YZ$ to $YX$.\\
            Repeat this with another arc centered at $Z$.
        \item 
            Complete our closed curve by drawing a circular arc centered at $X$.
        \item 
            The resulting curve has constant diameter. Why? Provide a brief explanation and convince yourself that this is true. \\
        \item
            Show that the diameter of this curve is $\ell + \overline{YZ} - \overline{XZ} - \overline{XY}$.
    \end{enumerate}
\end{problem}
\begin{figure}[h]
    \label{non-equil}
    \centering
    \includegraphics[width=0.5\linewidth]{img/non-equilateral.jpg}
    \caption{A curve of constant diameter constructed from a non-equilateral triangle.}
\end{figure}
\begin{solution}
    The diameter of this curve follows easily from filling in the figure and comparing radii.
\end{solution}
\pagebreak

\begin{problem}{(Bonus) Area and circumference}
    Draw a triangle with nice angles and side lengths. (For instance, a 45-45-90 triangle.) \\
    Repeat the above process to create a curve with constant diameter. Verify your conjectures about the circumference and area of a curve with constant diameter.\\
\end{problem}
\vfill
\begin{problem}{(Bonus, challenge) Non-regular polygons}
    Draw a non-regular (convex) polygon with $5$ sides. Can you generalize the process from \ref{non-equilateral} to draw a curve of constant diameter?
\end{problem}
\begin{solution}
    The only modification needed for 5 points is to draw the rays from your starting point along the diagonals of the pentagon. Not along the sides.
\end{solution}
\vfill
\pagebreak