[THOUGHT] We have two points: P1 = (1+cos t, sin t) on circle center (1,0) radius 1. P2 = (2 cos θ(t), 2 sin θ(t)) on circle center (0,0) radius 2. Both start at intersection (2,0) at t=0. So at t=0: P1(1+cos0, sin0) = (2,0) correct. P2(2 cos θ(0), 2 sin θ(0)) = (2,0) => cos θ(0)=1, sin θ(0)=0 => θ(0)=0. They move counterclockwise. Speeds satisfy |P1'(t)| : |P2'(t)| = 1 : 4. We need to find t = t0 when distance between P1 and P2 is maximized. At that moment, the expression sin t * 2 sin θ(t) - (1+cos t) * 2 cos θ(t) = p/q (rational). Find p+q.

First, we need to relate θ(t) and t via speed condition. Compute derivatives: P1(t) = (1+cos t, sin t) P1'(t) = (-sin t, cos t) Speed |P1'(t)| = sqrt(sin^2 t + cos^2 t) = 1. So speed of P1 is constant 1.

P2(t) = (2 cos θ(t), 2 sin θ(t)) P2'(t) = (-2 θ'(t) sin θ, 2 θ'(t) cos θ) Speed |P2'(t)| = sqrt(4 (θ')^2 sin^2 θ + 4 (θ')^2 cos^2 θ) = 2 |θ'(t)|. Given ratio |P1'| : |P2'| = 1 : 4 => 1 / (2 |θ'|) = 1/4 => 2 |θ'| = 4 => |θ'| = 2. Since both moving counterclockwise, θ'(t) positive? Let's check: At start, moving counterclockwise from (2,0). For P1, increasing t moves counterclockwise? At t=0, P1=(2,0). For small t>0, cos t decreases, sin t increases, so point moves upward and leftward, which is counterclockwise on circle centered (1,0) radius 1. Yes. For P2, counterclockwise from (2,0) means increasing θ from 0. So θ'(t) > 0. So θ'(t) = 2. Thus θ(t) = 2t + C. At t=0, θ(0)=0 => C=0. So θ(t) = 2t.

Thus P2(t) = (2 cos(2t), 2 sin(2t)).

Now distance squared between P1 and P2: D^2(t) = (1+cos t - 2 cos(2t))^2 + (sin t - 2 sin(2t))^2. We need to maximize D(t) over t in [0,2π]. Since both periodic, we can consider t in [0,2π].

We need to find t0 where D is maximum. Then compute expression: E(t) = sin t * 2 sin θ(t) - (1+cos t) * 2 cos θ(t) = 2 sin t sin(2t) - 2 (1+cos t) cos(2t). Simplify: E(t) = 2[ sin t sin(2t) - (1+cos t) cos(2t) ]. We can expand sin(2t)=2 sin t cos t, cos(2t)=cos^2 t - sin^2 t = 2 cos^2 t -1 = 1 - 2 sin^2 t. But maybe we can simplify E further: E(t) = 2[ sin t * 2 sin t cos t - (1+cos t)(cos^2 t - sin^2 t) ] = 2[ 2 sin^2 t cos t - (1+cos t)(cos^2 t - sin^2 t) ]. Alternatively, we can express E(t) in terms of dot product? Actually note that E(t) = 2[ (sin t) sin(2t) - (1+cos t) cos(2t) ] = -2[ (1+cos t) cos(2t) - sin t sin(2t) ] = -2 * something like dot product? Actually (1+cos t, sin t) dot (cos(2t), -sin(2t))? Not exactly.

Maybe we can relate E to distance? Or we can compute directly after finding t0.

We need to find t0 that maximizes D(t). Let's compute D^2(t) explicitly.

We'll use Python to compute D^2(t), find its maximum, then compute E(t0) and see if it's rational.

We'll use sympy or numeric optimization. Since we need exact rational? The expression E(t0) is given as p/q, so likely rational. So t0 might be a special angle. Let's compute D^2(t) symbolically and find critical points.

Let's do in Python. [/THOUGHT] [PYTHON] import sympy as sp

define variable

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

define coordinates

x1 = 1 + sp.cos(t) y1 = sp.sin(t) x2 = 2 * sp.cos(2t) y2 = 2 * sp.sin(2t)

squared distance

D2 = (x1 - x2)**2 + (y1 - y2)**2

simplify D2

D2_simplified = sp.simplify(D2) print("D^2(t) =", D2_simplified) [/PYTHON]

[THOUGHT] The squared distance simplifies to $D^2(t) = 8\sin^2 t - 2\cos t + 2$. To maximize distance, we can maximize $D^2(t)$. Since $\sin^2 t = 1 - \cos^2 t$, we can write $D^2(t) = 8(1 - \cos^2 t) - 2\cos t + 2 = 8 - 8\cos^2 t - 2\cos t + 2 = 10 - 8\cos^2 t - 2\cos t$. Let $u = \cos t$, then $D^2 = 10 - 8u^2 - 2u = -8u^2 - 2u + 10$. This is a quadratic in $u$ with maximum at vertex since coefficient of $u^2$ is negative. Vertex at $u = -\frac{b}{2a} = -\frac{-2}{2(-8)} = \frac{2}{-16} = -\frac{1}{8}$. So maximum occurs when $\cos t = -\frac{1}{8}$. Then $\sin^2 t = 1 - \frac{1}{64} = \frac{63}{64}$, so $\sin t = \pm \frac{\sqrt{63}}{8} = \pm \frac{3\sqrt{7}}{8}$. Since both points move counterclockwise and start at (2,0), for maximum distance, we need to determine the correct sign of $\sin t$. At $t=0$, distance is 0 (since both at same point). As $t$ increases, distance increases. The maximum likely occurs for $t$ in $(0, \pi)$ where $\sin t > 0$. So take $\sin t = \frac{3\sqrt{7}}{8}$.

Now compute $E(t) = 2[ \sin t \sin(2t) - (1+\cos t) \cos(2t) ]$. Use identities: $\sin(2t) = 2\sin t \cos t$, $\cos(2t) = 2\cos^2 t - 1$. Substitute $\cos t = -\frac{1}{8}$, $\sin t = \frac{3\sqrt{7}}{8}$.

Compute stepwise: $\sin(2t) = 2 \cdot \frac{3\sqrt{7}}{8} \cdot \left(-\frac{1}{8}\right) = -\frac{6\sqrt{7}}{64} = -\frac{3\sqrt{7}}{32}$. $\cos(2t) = 2 \cdot \frac{1}{64} - 1 = \frac{2}{64} - 1 = \frac{1}{32} - 1 = -\frac{31}{32}$. Now compute $\sin t \sin(2t) = \frac{3\sqrt{7}}{8} \cdot \left(-\frac{3\sqrt{7}}{32}\right) = -\frac{9 \cdot 7}{256} = -\frac{63}{256}$. Compute $(1+\cos t) = 1 - \frac{1}{8} = \frac{7}{8}$. Then $(1+\cos t) \cos(2t) = \frac{7}{8} \cdot \left(-\frac{31}{32}\right) = -\frac{217}{256}$. Thus $\sin t \sin(2t) - (1+\cos t) \cos(2t) = -\frac{63}{256} - \left(-\frac{217}{256}\right) = -\frac{63}{256} + \frac{217}{256} = \frac{154}{256} = \frac{77}{128}$. Multiply by 2: $E(t) = 2 \cdot \frac{77}{128} = \frac{77}{64}$. So $p/q = 77/64$, with $p=77, q=64$ relatively prime. Then $p+q = 77+64 = 141$.

Thus final answer is 141. [/THOUGHT] $\boxed{141}$