[THOUGHT] Let me analyze this problem step by step.

1. Point $P_1(1+\cos t, \sin t)$ moves on a circle centered at $(1,0)$ with radius 1.
2. Point $P_2(2\cos\theta(t), 2\sin\theta(t))$ moves on a circle centered at $(0,0)$ with radius 2.
3. Both start at $(2,0)$ and move counterclockwise.
4. The speed ratio is $|P_1'(t)| : |P_2'(t)| = 1 : 4$.

First, let me find the relationship between $t$ and $\theta(t)$.

For $P_1$: At $t=0$, we need $P_1 = (2,0)$, so $1+\cos(0) = 2$ and $\sin(0) = 0$. This works.

The velocity of $P_1$ is $P_1'(t) = (-\sin t, \cos t)$, so $|P_1'(t)| = 1$.

For $P_2$: At $t=0$, we need $\theta(0) = 0$ so that $P_2 = (2,0)$.

The velocity of $P_2$ is $P_2'(t) = (-2\sin\theta \cdot \theta', 2\cos\theta \cdot \theta')$, so $|P_2'(t)| = 2|\theta'(t)|$.

Given the speed ratio $1:4$, we have $1 : 2|\theta'(t)| = 1 : 4$, so $|\theta'(t)| = 2$.

Since both move counterclockwise, $\theta'(t) = 2$, so $\theta(t) = 2t$ (with $\theta(0) = 0$).

Now I need to find the distance between $P_1$ and $P_2$ and maximize it.

$P_1 = (1+\cos t, \sin t)$ $P_2 = (2\cos(2t), 2\sin(2t))$

Distance squared: $D^2 = (2\cos(2t) - 1 - \cos t)^2 + (2\sin(2t) - \sin t)^2$

Let me compute this and find when it's maximized. [/THOUGHT] [PYTHON] import numpy as np from sympy import *

t = symbols('t', real=True)

P1 coordinates

x1 = 1 + cos(t) y1 = sin(t)

P2 coordinates (theta = 2t)

x2 = 2cos(2t) y2 = 2sin(2t)

Distance squared

D_sq = (x2 - x1)**2 + (y2 - y1)**2 D_sq_expanded = expand(D_sq) print("D^2 =", D_sq_expanded)

Simplify using trig identities

D_sq_simplified = simplify(D_sq) print("D^2 simplified =", D_sq_simplified)

Take derivative and set to zero

dD_sq = diff(D_sq, t) dD_sq_simplified = simplify(dD_sq) print("d(D^2)/dt =", dD_sq_simplified)

Solve for critical points

critical_points = solve(dD_sq, t) print("Critical points:", critical_points) [/PYTHON]

[THOUGHT] From the derivative $\frac{d(D^2)}{dt} = 2(8\cos t + 1)\sin t = 0$, the critical points in $[0, 2\pi]$ are:

• $\sin t = 0$: gives $t = 0, \pi$
• $\cos t = -1/8$: gives $t = \arccos(-1/8)$ and $t = 2\pi - \arccos(-1/8)$

At $t = 0$: $D^2 = 8(0) - 2(1) + 2 = 0$ (minimum, both points at same location) At $t = \pi$: $D^2 = 8(0) - 2(-1) + 2 = 4$ At $\cos t = -1/8$: $D^2 = 8\sin^2 t - 2\cos t + 2 = 8(1 - \cos^2 t) - 2\cos t + 2 = 8(1 - 1/64) + 1/4 + 2 = 8(63/64) + 9/4 = 63/8 + 9/4 = 63/8 + 18/8 = 81/8$

So maximum is at $\cos t = -1/8$, which means $\sin t = \sqrt{1 - 1/64} = \sqrt{63/64} = \frac{3\sqrt{7}}{8}$ (positive since counterclockwise in first half).

Now I need to compute: $\sin t \cdot 2\sin\theta(t) - (1+\cos t) \cdot 2\cos\theta(t)$ where $\theta = 2t$.

With $\cos t = -1/8$ and $\sin t = \frac{3\sqrt{7}}{8}$:

• $\cos(2t) = 2\cos^2 t - 1 = 2(1/64) - 1 = 1/32 - 1 = -31/32$
• $\sin(2t) = 2\sin t \cos t = 2 \cdot \frac{3\sqrt{7}}{8} \cdot (-\frac{1}{8}) = -\frac{6\sqrt{7}}{64} = -\frac{3\sqrt{7}}{32}$

The expression: $\sin t \cdot 2\sin(2t) - (1+\cos t) \cdot 2\cos(2t)$ $= \frac{3\sqrt{7}}{8} \cdot 2 \cdot (-\frac{3\sqrt{7}}{32}) - (1 - \frac{1}{8}) \cdot 2 \cdot (-\frac{31}{32})$ $= \frac{3\sqrt{7}}{8} \cdot (-\frac{6\sqrt{7}}{32}) - \frac{7}{8} \cdot (-\frac{62}{32})$ $= -\frac{18 \cdot 7}{256} + \frac{7 \cdot 62}{256}$ $= \frac{-126 + 434}{256}$ $= \frac{308}{256} = \frac{77}{64}$

So $p = 77$, $q = 64$, and $\gcd(77, 64) = 1$.

Therefore $p + q = 77 + 64 = 141$. [/THOUGHT]

\boxed{141}