The problem asks us to find the value of $p+q$ where $p/q$ is the value of a specific expression at the time $t_0$ when the distance between two moving points $P_1$ and $P_2$ is maximized.

First, let's determine the positions of $P_1$ and $P_2$. $P_1$ moves on a circle with center $(1,0)$ and radius 1. $P_1(t) = (1+\cos t, \sin t)$. $P_1'(t) = (-\sin t, \cos t)$, so speed $|P_1'(t)| = 1$.

$P_2$ moves on a circle with center $(0,0)$ and radius 2. $P_2(t) = (2\cos\theta(t), 2\sin\theta(t))$. Speed $|P_2'(t)| = 2|\theta'(t)|$. Given the ratio of speeds is $1:4$, we have $1 : 2|\theta'(t)| = 1:4$, which implies $|\theta'(t)| = 2$. Since the motion is counterclockwise, $\theta'(t) = 2$. With the starting point $(2,0)$ at $t=0$, we have $\theta(0)=0$, so $\theta(t) = 2t$. Thus, $P_2(t) = (2\cos(2t), 2\sin(2t))$.

Next, we find the time $t_0$ that maximizes the distance between $P_1$ and $P_2$. Let $D^2(t)$ be the squared distance. $D^2(t) = (1+\cos t - 2\cos 2t)^2 + (\sin t - 2\sin 2t)^2$. Expanding this, we get: $D^2(t) = 6 - 2\cos t - 4\cos 2t$. To maximize distance, we minimize $f(t) = 2\cos t + 4\cos 2t$. Let $x = \cos t$. Then $\cos 2t = 2x^2 - 1$. We minimize $h(x) = 2x + 4(2x^2 - 1) = 8x^2 + 2x - 4$ for $x \in [-1, 1]$. The minimum occurs at $x = -2/(2*8) = -1/8$. So at $t=t_0$, $\cos t_0 = -1/8$.

Finally, we evaluate the given expression at $t_0$: Expression $= \sin t \cdot 2\sin(2t) - (1+\cos t) \cdot 2\cos(2t)$. Using trigonometric identities, this simplifies to $-2\cos 3t - 2\cos 2t$. In terms of $x = \cos t$, this is $-8x^3 - 4x^2 + 6x + 2$. Substituting $x = -1/8$, we get the value $77/64$.

Thus $p=77, q=64$. They are relatively prime. $p+q = 77+64 = 141$.

[PYTHON] import sympy from sympy import symbols, cos, sin, diff, solve, simplify, Rational

t = symbols('t')

Define positions

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

Squared distance

dist_sq = (x1 - x2)**2 + (y1 - y2)**2 dist_sq = simplify(dist_sq) print(f"Squared distance simplified: {dist_sq}")

To maximize distance, we minimize the negative of it, or just find critical points

Let's differentiate with respect to t

deriv = diff(dist_sq, t)

We want to solve deriv = 0

The expression involves sin(t) and sin(2t) etc.

Let's substitute cos(t) = x to find the algebraic minimum

x = symbols('x')

dist_sq in terms of x

From manual derivation: 6 - 2cos(t) - 4cos(2t)

cos(2t) = 2x^2 - 1

dist_x = 6 - 2x - 4(2*x**2 - 1) print(f"Distance squared in terms of x=cos(t): {dist_x}")

Find critical points for x in [-1, 1]

deriv_x = diff(dist_x, x) crit_points = solve(deriv_x, x) print(f"Critical points for x: {crit_points}")

Check which maximizes the distance

max_dist = -1 best_x = None for cp in crit_points: val = dist_x.subs(x, cp) if val > max_dist: max_dist = val best_x = cp

Check endpoints

for ep in [-1, 1]: val = dist_x.subs(x, ep) if val > max_dist: max_dist = val best_x = ep

print(f"x that maximizes distance: {best_x}")

Calculate the target expression at best_x

Expression: sin(t)*2sin(2t) - (1+cos(t))*2cos(2t)

Simplify this expression to terms of cos(t)

expr = sin(t) * 2 * sin(2t) - (1 + cos(t)) * 2 * cos(2t) expr_simplified = simplify(expr) print(f"Target expression simplified: {expr_simplified}")

Convert to polynomial in x

expr_simplified should be -2cos(2t) - 2cos(3t)

cos(2t) = 2x^2 - 1

cos(3t) = 4x^3 - 3x

poly_x = -2*(2x**2 - 1) - 2(4x**3 - 3x) print(f"Polynomial in x: {poly_x}")

final_val = poly_x.subs(x, best_x) print(f"Final value: {final_val}") [/PYTHON] [THOUGHT] The Python code confirmed the manual derivation. The squared distance is $10 - 8\cos(t)^2 - 2\cos(t)$. Wait, the Python output says 10 - 8*cos(t)**2 - 2*cos(t). My manual derivation was $6 - 2\cos t - 4\cos 2t = 6 - 2\cos t - 4(2\cos^2 t - 1) = 6 - 2\cos t - 8\cos^2 t + 4 = 10 - 2\cos t - 8\cos^2 t$. This matches the Python output. The critical point for $x = \cos t$ is found by maximizing $10 - 2x - 8x^2$. Derivative is $-2 - 16x = 0 \implies x = -1/8$. This matches. The target expression simplified to $-2\cos(2t) - 2\cos(3t)$. In terms of $x$, this is $-8x^3 - 4x^2 + 6x + 2$. Substituting $x = -1/8$: Result is $77/64$. $p=77, q=64$. $p+q = 141$. The integers are relatively prime.

Final check of the question details:

• $P_1$ center $(1,0)$, radius 1. Correct.
• $P_2$ center $(0,0)$, radius 2. Correct.
• Start at $(2,0)$. Correct.
• Counterclockwise. Correct.
• Speed ratio $1:4$. Correct.
• Expression: $\sin t \cdot 2\sin\theta(t) - (1+\cos t)\cdot 2\cos\theta(t)$. Correct.
• $p, q$ relatively prime positive integers. $77/64$ is positive. $gcd(77, 64) = 1$. Correct.

The value is $77/64$. $p+q = 77+64 = 141$. [/THOUGHT]

\boxed{141}

The value of the expression is $\frac{77}{64}$. Thus $p=77$ and $q=64$. The sum $p+q = 77+64 = 141$.

\boxed{141}