Edits

Advanced/Lattices/parts/2 orchard.tex

@@ -12,7 +12,7 @@ Show that if $r < \frac{1}{\sqrt{R^2 + 1}}$, you have at least one directon with
 \hint{Take a look at the ray through the point $(R, 1)$ and 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 $12$. 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.
+	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.
 \end{solution}
 
 
@@ -39,6 +39,12 @@ Show that there is no line of sight through the orchard if $r > \frac{1}{R}$. Yo
 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