*Posted by Saketh in : potd , trackback
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A (very small) house is to be constructed at the center of a circular plot of land with radius $20$ meters. The developer plans to place trees around the house in a rather peculiar manner. Taking the house to be the origin, a sapling will be planted at every lattice point within the circle. The unit distance is to be $1$ meter.

Due to recent developments in genetic engineering, the trees can be modified to stop growing at any desired girth. To reduce construction costs, however, all of the trees will be identical.

The inhabitants of the house wish to be able to see outside of the plot. That is, there must exist some ray we can draw from the origin to the perimeter of the circle which does not intersect any of the trees. Determine the greatest $R$ such that for any tree radius $r < R$, the view is not blocked.

Assume for the purposes of this problem that the house is a single point at the origin.

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Solution:

We begin by considering the segment drawn from the origin to the point $(20,1)$. Clearly, we only need to look at the the lattice points in the first quadrant with y-value $0$ or $1$ to determine the the value of $R$ for this particular ray. In fact, it is sufficient to consider the trees at $(1,0)$ and $(49,1)$.

We can determine the distance from the points $(1,0)$ and $(49,1)$ to our segment by employing the dot product. The unit vector normal to our segment is $\hat{r}=<\frac{1}{\sqrt{401}},\frac{-50}{\sqrt{401}}>$. Thus, the distances to our two points are given by $\hat{r} \cdot <1,0> = \frac{1}{\sqrt{401}}$ and $\hat{r} \cdot <49,1>$. Clearly, the first value has a smaller magnitude. Thus, for any tree radius $r$ less than $\frac{1}{\sqrt{401}}$ we can observe the point $(50,1)$ from the origin.

We shall now show that for any radius $r$ greater than or equal to $\frac{1}{\sqrt{401}}$ it is not possible to see outside the circle. Suppose we draw some arbitrary diameter of the circle. Consider the rectangle produced by extending the line horizontally by $\frac{1}{20}$ units on each side. This rectangle has area $\frac{2}{20} \cdot {40} = 4 = 2^2$ and therefore by Minkowski’s theorem contains two lattice points symmetric about the origin. This is equivalent to saying that for any tree radius greater than or equal to $\frac{1}{20}$ our diameter intersects the trees centered at these points. Additionally, each of the two radii making up this diameter must intersect exactly one of these trees.

We can strengthen this by shrinking the width of the rectangle to $\frac{2}{\sqrt{401}}$ and increasing its length to $2\sqrt{401}$. Since there are no lattice points that have a distance between $20$ and $\sqrt{401}$ from the origin, we can re-employ our previous observation with Minkowski’s theorem. Thus, for any radius $r$ greater than or equal to $\frac{1}{\sqrt{401}}$, any ray drawn from the house to the perimeter of the plot will intersect some tree.

The desired value of $R$ is therefore $\boxed{\frac{1}{\sqrt{401}}}$.

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