Problem Analysis #4

To find the total arc length $L$ of the set of all possible points $P$ on the $xy$-plane, we analyze the given conditions.

Let the coordinates of the points be:

• $O = (0, 0, 0)$
• $A = (0, 1, 1)$
• $P = (x, y, 0)$

Since $P$ lies on the $xy$-plane and $OP = 2$, we have:

$$x^2 + y^2 = 4$$

This means $P$ lies on a circle of radius $R = 2$ centered at the origin in the $xy$-plane.

Condition 1: $\angle AOP \ge \frac{2\pi}{3}$

Using the dot product of vectors $\vec{OA} = (0, 1, 1)$ and $\vec{OP} = (x, y, 0)$:

$$\cos(\angle AOP) = \frac{\vec{OA} \cdot \vec{OP}}{OA \cdot OP} = \frac{y}{\sqrt{2} \cdot 2} = \frac{y}{2\sqrt{2}}$$

Since $\angle AOP \ge \frac{2\pi}{3}$, we must have:

$$\cos(\angle AOP) \le \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$$

Thus,

$$\frac{y}{2\sqrt{2}} \le -\frac{1}{2} \implies y \le -\sqrt{2}$$

Condition 2: $\angle OAP \le \frac{\pi}{3}$

Let $d = AP$ be the distance between $A$ and $P$. Using the coordinates of $A$ and $P$:

$$d^2 = x^2 + (y-1)^2 + (0-1)^2 = x^2 + y^2 - 2y + 2$$

Since $x^2 + y^2 = 4$, we have:

$$d^2 = 6 - 2y$$

By the Law of Cosines in $\triangle OAP$:

$$\cos(\angle OAP) = \frac{d^2 + OA^2 - OP^2}{2 \cdot OA \cdot d} = \frac{d^2 + 2 - 4}{2\sqrt{2}d} = \frac{d^2 - 2}{2\sqrt{2}d}$$

Since $\angle OAP \le \frac{\pi}{3}$, we must have:

$$\cos(\angle OAP) \ge \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$

This gives:

$$\frac{d^2 - 2}{2\sqrt{2}d} \ge \frac{1}{2} \implies d^2 - \sqrt{2}d - 2 \ge 0$$

Solving the quadratic inequality for $d > 0$:

$$d \ge \frac{\sqrt{2} + \sqrt{10}}{2} \implies d^2 \ge 3 + \sqrt{5}$$

Substituting $d^2 = 6 - 2y$:

$$6 - 2y \ge 3 + \sqrt{5} \implies y \le \frac{3 - \sqrt{5}}{2}$$

Combining the Conditions

We have two upper bounds for $y$:

1. $y \le -\sqrt{2} \approx -1.414$
2. $y \le \frac{3 - \sqrt{5}}{2} \approx 0.382$

The intersection of these two conditions is:

$$y \le -\sqrt{2}$$

Calculating the Arc Length

The set of points $P$ lies on the circle $x^2 + y^2 = 4$ with $y \le -\sqrt{2}$. The intersection of the line $y = -\sqrt{2}$ with the circle $x^2 + y^2 = 4$ gives:

$$x^2 + 2 = 4 \implies x = \pm \sqrt{2}$$

The points of intersection are $(\sqrt{2}, -\sqrt{2})$ and $(-\sqrt{2}, -\sqrt{2})$, which correspond to angles of $-\frac{\pi}{4}$ and $-\frac{3\pi}{4}$ respectively.

The central angle $\theta$ of this arc is:

$$\theta = -\frac{\pi}{4} - \left(-\frac{3\pi}{4}\right) = \frac{\pi}{2}$$

Since the radius of the circle is $R = 2$, the total arc length $L$ is:

$$L = R\theta = 2 \cdot \frac{\pi}{2} = \pi$$

Writing $L$ in the form $p\pi/q$ in lowest terms gives $p = 1$ and $q = 1$. Thus,

$$p + q = 1 + 1 = 2$$

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