Problem Analysis #4

The problem requires finding the total arc length $L$ of the set of points $P$ on the circle $x^2+y^2=4$ (since $OP=2$ and $P$ lies in the $xy$-plane) that satisfy $\angle AOP \ge \frac{2\pi}{3}$ and $\angle OAP \le \frac{\pi}{3}$, with $A=(0,1,1)$ and $O=(0,0,0)$.

Let $P = (2\cos\varphi, 2\sin\varphi, 0)$.
Compute $\cos\angle AOP = \frac{\overrightarrow{OA}\cdot\overrightarrow{OP}}{|\overrightarrow{OA}||\overrightarrow{OP}|} = \frac{\sin\varphi}{\sqrt{2}}$.
The condition $\angle AOP \ge \frac{2\pi}{3}$ implies $\cos\angle AOP \le -\frac{1}{2}$, so $\frac{\sin\varphi}{\sqrt{2}} \le -\frac{1}{2}$ $\Rightarrow$ $\sin\varphi \le -\frac{\sqrt{2}}{2}$.
On $[0,2\pi)$, this gives $\varphi \in \left[\frac{5\pi}{4}, \frac{7\pi}{4}\right]$.

Next, $\cos\angle OAP = \frac{1 - \sin\varphi}{\sqrt{3 - 2\sin\varphi}}$.
The condition $\angle OAP \le \frac{\pi}{3}$ implies $\cos\angle OAP \ge \frac{1}{2}$. Solving $\frac{1 - t}{\sqrt{3 - 2t}} \ge \frac{1}{2}$ for $t = \sin\varphi$ yields $t \le \frac{3-\sqrt{5}}{4} \approx 0.191$.
Since for $\varphi \in [5\pi/4, 7\pi/4]$ we have $\sin\varphi \in [-1, -\sqrt{2}/2] \subseteq (-\infty, 0.191]$, the second condition is automatically satisfied on the entire first interval. Hence the admissible points are exactly those on the circle with $\varphi \in [5\pi/4, 7\pi/4]$.

The angular measure of this arc is $\frac{7\pi}{4} - \frac{5\pi}{4} = \frac{\pi}{2}$.
The radius is $2$, so the arc length is $L = 2 \cdot \frac{\pi}{2} = \pi$.

Thus $L = \pi = \frac{1 \cdot \pi}{1}$ in lowest terms, giving $p = 1$, $q = 1$, and $p+q = 2$.

$\boxed{2}$