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Advanced/Lattices/parts/2 orchard.tex
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\section{Polya's Orchard Problem}
You are standing in the center of a circular orchard of integer radius R. A tree was planted each integer lattice point, and each has grown to the same radius $r$. If the radius is small enough, you will have a clear line of sight through the orchard in certain directions. If the radius is too large, there is no line of sight through the orchard in any direction. See the figure below:
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
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\end{center}
\problem{}
Show that if $r < \frac{1}{\sqrt{R^2 + 1}}$, you have at least one directon with a clear line of sight. \\
\hint{Take a look at the ray through the point $(R, 1)$ and calculate the distance from the closest integer points to the ray.}
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 through the point $(R, 1)$. Clearly, the two closest lattice points are $(1, 0)$ and $(R - 1, 1)$. They are equally far from the ray so let's calculate the distance from $(1, 0)$ to our ray. Call this distance $\delta$. Consider the triangle with vertices $(0, 0)$, $(1, 0)$, and $(R, 1)$. Then the area of this triangle is $\frac{1}{2}$. On the other hand, the area is also given by $\frac{1}{2} \delta \sqrt{R^2 + 1}$. So, $\delta = \frac{1}{\sqrt{R^2+1}}$. Therefore, if $r < \frac{1}{\sqrt{R^2+1}}$, we will have a clear line of sight given by this ray.
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}