45 lines
3.5 KiB
TeX
45 lines
3.5 KiB
TeX
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\section{Polya's Orchard Problem}
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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:
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\begin{center}
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\includegraphics[width=0.5\linewidth]{orchard.png}
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\end{center}
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\problem{}
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Show that if $r < \frac{1}{\sqrt{R^2 + 1}}$, you have at least one directon with a clear line of sight. \\
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\hint{Take a look at the ray through the point $(R, 1)$ and calculate the distance from the closest integer points to the ray.}
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\begin{solution}
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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 $12$. 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.
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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Show that there is no line of sight through the orchard if $r > \frac{1}{R}$. You may want to use the following steps:
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\begin{itemize}
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\item Show that there is no line of sight if $r \geq 1$.
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\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
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to rule out any lattice points that sit outside the orchard but inside the rectangle.
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\end{itemize}
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% 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
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\begin{solution}
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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.
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\end{solution}
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\vfill
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\problem{Challenge}
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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. \\
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\hint{Use the orchard.}
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\vfill
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\pagebreak
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