\section{Polya's Orchard Problem}

You are standing in the center of a circular orchard of integer radius $R$. A tree of raduis $r$ has been planted at every integer point in the circle. If $r$ is small, you will have a clear line of sight through the orchard. If $r$ is large, there will be no clear line of sight through in any direction:

\begin{center}
	\hfill
	\begin{tikzpicture}[scale=0.4]
		\draw[color=black, very thick] (0,0) circle (5);

		\draw[fill=black] (0,0) circle (0.2);
		\draw [->, very thick](0,0) -- (0.7, 0.7);
		\draw [->, very thick](0,0) -- (-0.7, 0.7);

		\foreach \x in {-4, ..., 4} {
		\foreach \y in {-4, ..., 4} {
		\ifthenelse{
			% Remove center
			\( \x=0 \AND \y=0 \) \OR
			%
			% Remove corners
			\( \x=4 \AND \y=-4 \) \OR
			\( \x=4 \AND \y=4 \) \OR
			\( \x=-4 \AND \y=4 \) \OR
			\( \x=-4 \AND \y=-4 \) \OR
			%
			% Remove corner corners
			\( \x=4 \AND \y=-3 \) \OR
			\( \x=4 \AND \y=3 \) \OR
			\( \x=-4 \AND \y=3 \) \OR
			\( \x=-4 \AND \y=-3 \) \OR
			\( \x=3 \AND \y=-4 \) \OR
			\( \x=3 \AND \y=4 \) \OR
			\( \x=-3 \AND \y=4 \) \OR
			\( \x=-3 \AND \y=-4 \)
		}{}{
			\draw[fill=teal] (\x,\y) circle (0.4);
		}}}

		\draw[fill=teal] (0, 5) circle (0.4);
		\draw[fill=teal] (5, 0) circle (0.4);
		\draw[fill=teal] (0, -5) circle (0.4);
		\draw[fill=teal] (-5, 0) circle (0.4);
	\end{tikzpicture}
	\hfill
	\begin{tikzpicture}[scale=0.4]
		\draw[color=black, very thick] (0,0) circle (5);

		\draw[fill=black] (0,0) circle (0.2);

		\draw [->, very thick](0,0) -- (-5,3.7);
		\draw [->, very thick](0,0) -- (5, 3.7);

		\foreach \x in {-4, ..., 4} {
		\foreach \y in {-4, ..., 4} {
		\ifthenelse{
			% Remove center
			\( \x=0 \AND \y=0 \) \OR
			%
			% Remove corners
			\( \x=4 \AND \y=-4 \) \OR
			\( \x=4 \AND \y=4 \) \OR
			\( \x=-4 \AND \y=4 \) \OR
			\( \x=-4 \AND \y=-4 \) \OR
			%
			% Remove corner corners
			\( \x=4 \AND \y=-3 \) \OR
			\( \x=4 \AND \y=3 \) \OR
			\( \x=-4 \AND \y=3 \) \OR
			\( \x=-4 \AND \y=-3 \) \OR
			\( \x=3 \AND \y=-4 \) \OR
			\( \x=3 \AND \y=4 \) \OR
			\( \x=-3 \AND \y=4 \) \OR
			\( \x=-3 \AND \y=-4 \)
		}{}{
			\draw[fill=teal] (\x,\y) circle (0.2);
		}}}

		\draw[fill=teal] (0, 5) circle (0.2);
		\draw[fill=teal] (5, 0) circle (0.2);
		\draw[fill=teal] (0, -5) circle (0.2);
		\draw[fill=teal] (-5, 0) circle (0.2);
	\end{tikzpicture}
	\hfill ~
\end{center}

\problem{}
Show that you will have at least one clear line of sight if $r < \frac{1}{\sqrt{R^2 + 1}}$. \par
\hint{Consider the line segment from $(0, 0)$ to $(R, 1)$. Calculate the distance from the closest integer points to the ray.}

\begin{solution}
	Consider the ray from the origin to the point $(R, 1)$.

	The two lattice points closest to this ray are $(1, 0)$ and $(R - 1, 1)$. Say the distance from each of these points to the ray is $\delta$. \par

	Now, consider the triangle with vertices $(0, 0)$, $(1, 0)$, and $(R, 1)$. The area of this triangle is $\frac{1}{2}$.

	The area of this triangle is also equal to $\frac{1}{2} \delta \sqrt{R^2 + 1}$. By algebra, $\delta = \frac{1}{\sqrt{R^2+1}}$. \par

	Therefore, if $r < \frac{1}{\sqrt{R^2+1}}$, we will have a clear line of sight given by this ray.
\end{solution}


\vfill
\pagebreak

\problem{}
Show that there is no line of sight through the orchard if $r > \frac{1}{R}$. You may want to use the following steps:
\begin{itemize}
	\item Show that there is no line of sight if $r \geq 1$.
	\item Suppose $r < 1$ and $r > \frac{1}{R}$. Then, $R \geq 2$. Choose a potential line of sight passing through an arbitrary point $P$ on the circle. Thicken this line of sight equally on both sides into a rectangle of width $2r$ tangent to $P$ and $-P$. From here, use Minkowski's theorem to get a contradiction. Don't forget
	to rule out any lattice points that sit outside the orchard but inside the rectangle.
\end{itemize}

% Any interest in counting the number of trees in the orchard? If so, google the Gauss circle problem. If orchards are not your slice of fruit, maybe rational approximations are your cup of tea

\begin{solution}
	Suppose $r < 1$ and let $L$ be a potential line of sight. Consider the rectangle of width $2r$ tangent to $P$ and $-P$. Then this is convex and symmetric with respect to the origin. Its area is $(2R)(2r) > 4 \frac{R}{R} = 4$. By Minkowski, we have a nonzero integral point in this rectangle. Suppose first that the integer point is within the orchard. Then this means that there is a tree whose distance to the line is at most $r$. Therefore, this tree blocks the line of sight. Now notice that there is a part of this rectangle that sits outside the orchard. Can the integer point be in this region? This would mean its distance to the origin, $D$, would satisfy $D > R$. Now since this point is within a distance of $r$ of our line $L$, we have that $D < \sqrt{R^2 + r^2} < \sqrt{R^2 + 1}$. So we have that $R < D < \sqrt{R^2 + 1}$. Then $R^2 < D^2 < R^2 + 1$, but $D^2$ is an integer so this is impossible.
\end{solution}

\vfill

\problem{Challenge}
Prove that there exists a rational approximation of $\sqrt{3}$ within $10^{-3}$ with denominator at most $501$. Come up with an upper bound for the smallest denominator of a $\epsilon$-close rational approximation of any irrational number $\alpha > 0$. Your bound can have some dependence on $\alpha$ and should get smaller as $\alpha$ gets larger. \\
\hint{Use the orchard.}

\begin{solution}
	Take the line through the origin of slope $\sqrt{3}$. We would like an orchard for which $r = 10^{-3}$ gives no line of sight, since this will guarantee an integer point within a distance of $10^{-3}$. Then by our previous problem, we can take $10^{-3} > \frac{1}{R}$, so take $R > 1000$. Now since this line intersects the boundary of the orchard at $( \frac{R}{2}, \frac{\sqrt{3R}}{2} )$, we have that the $x$-coordinate is at most $\frac{R}{2} = 501$. Then we have that our lattice point $(x, y)$ satisfies $\sqrt{3x} - 10^{-3} < y < \sqrt{3x} + 10^{-3}$, so $\sqrt{3} - 10^{-3} < \frac{x}{y} < \sqrt{3} + \frac{10^{-3}}{x}$. Therefore, $\frac{y}{x}$ is a rational approximation that is $10^{-3}$-close to $\sqrt{3}$ and has denominator at most $501$. Notice that we got closer than we need to.

	Repeating this same process, our upper bound for the denominator of an $\epsilon$-close approximation of $\alpha$ is $\frac{\text{cos}({\text{atan}({\alpha})})}{\epsilon} = \frac{1}{\sqrt{\alpha^2 + 1 \epsilon}}$.
\end{solution}

\vfill
\pagebreak