The problem asks for the slope of the tangent to the curve $C$ at the point $P$ where it intersects the line $y = 3x - 5e$. The curve is defined by $x(t) = e^{4t}(1 + \sin^2 \pi t)$ and $y(t) = e^{4t}(1 - 3\cos^2 \pi t)$.

First, I will find the parameter $t$ corresponding to the intersection point $P$ by substituting the parametric equations into the line equation and solving for $t$. Then, I will compute the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$ at this value of $t$ to find the slope $m = \frac{dy/dt}{dx/dt}$. Finally, I will express the slope in the form $\frac{a\pi - b}{\pi + c}$ and calculate $a+b+c$.

[PYTHON] import sympy from sympy import symbols, exp, sin, cos, pi, solve, diff

t = symbols('t', real=True) x = exp(4t) * (1 + sin(pit)**2) y = exp(4t) * (1 - 3cos(pi*t)**2)

Line equation: y = 3x - 5e

Substitute x(t) and y(t) into the line equation

equation = y - (3x - 5exp(1))

Solve for t

We suspect t might be a simple fraction based on the problem type.

Let's try to solve it numerically or symbolically.

The manual derivation suggested t = 1/4. Let's verify.

t_sol = solve(equation, t) print(f"Solutions for t: {t_sol}")

Assuming we find a unique real solution, let's call it t_val

If solve returns multiple, we need to check which one is valid or if there's a domain constraint (none stated).

Based on manual derivation, t=0.25 is a candidate.

t_val = sympy.S(1)/4

Verify t_val is a solution

lhs = equation.subs(t, t_val) print(f"Equation value at t=1/4: {lhs.simplify()}")

Calculate derivatives

dx_dt = diff(x, t) dy_dt = diff(y, t)

Evaluate derivatives at t_val

dx_val = dx_dt.subs(t, t_val) dy_val = dy_dt.subs(t, t_val)

slope = dy_val / dx_val print(f"Slope: {slope.simplify()}") [/PYTHON] [THOUGHT] The Python code confirmed that $t = 1/4$ is a solution (the equation evaluates to 0). The calculated slope is $\frac{3\pi - 2}{\pi + 6}$. Comparing this to the form $\frac{a\pi - b}{\pi + c}$: $a = 3$ $b = 2$ $c = 6$

The question asks for $a+b+c$. $a+b+c = 3 + 2 + 6 = 11$.

I should double check if there are other intersection points. The equation was $5e^{1-4t} = 5$, which implies $1-4t = 0 \implies t=1/4$. Since the exponential function is one-to-one, this is the unique solution for real $t$.

The slope calculation: $x'(1/4) = e(6+\pi)$ $y'(1/4) = e(3\pi - 2)$ Ratio: $\frac{e(3\pi - 2)}{e(6+\pi)} = \frac{3\pi - 2}{\pi + 6}$.

The form matches perfectly. $a=3, b=2, c=6$. Sum = 11.

Final check of the steps.

1. Intersection: $y = 3x - 5e$. $e^{4t}(1 - 3\cos^2 \pi t) = 3e^{4t}(1 + \sin^2 \pi t) - 5e$. Divide by $e^{4t}$: $1 - 3\cos^2 \pi t = 3 + 3\sin^2 \pi t - 5e^{1-4t}$. Use $\sin^2 = 1 - \cos^2$: $1 - 3\cos^2 = 3 + 3(1-\cos^2) - 5e^{1-4t}$. $1 - 3\cos^2 = 6 - 3\cos^2 - 5e^{1-4t}$. $1 = 6 - 5e^{1-4t} \implies 5e^{1-4t} = 5 \implies 1-4t=0 \implies t=1/4$. Correct.

2. Derivatives: $x = e^{4t} + e^{4t}\sin^2 \pi t$. $x' = 4e^{4t} + 4e^{4t}\sin^2 \pi t + e^{4t}(2\sin \pi t \cos \pi t)\pi$. At $t=1/4$, $\pi t = \pi/4$. $\sin = \cos = 1/\sqrt{2}$. $\sin^2 = 1/2$. $2\sin\cos = 1$. $x' = 4e(1 + 1/2) + e(1)\pi = e(6 + \pi)$. Correct.

   $y = e^{4t} - 3e^{4t}\cos^2 \pi t$. $y' = 4e^{4t} - [12e^{4t}\cos^2 \pi t + 3e^{4t}(2\cos \pi t (-\sin \pi t)\pi)]$. $y' = 4e^{4t} - 12e^{4t}\cos^2 \pi t + 3\pi e^{4t}\sin(2\pi t)$. At $t=1/4$: $y' = 4e - 12e(1/2) + 3\pi e(1) = 4e - 6e + 3\pi e = e(3\pi - 2)$. Correct.

3. Slope: $\frac{e(3\pi - 2)}{e(\pi + 6)} = \frac{3\pi - 2}{\pi + 6}$. Correct.

4. Coefficients: $a=3, b=2, c=6$. $a+b+c = 11$.

The solution seems robust. [/THOUGHT] [THOUGHT] The slope of the tangent to the curve $C$ at point $P$ is $\frac{3\pi - 2}{\pi + 6}$. Matching this with the form $\frac{a\pi - b}{\pi + c}$, we have $a=3$, $b=2$, and $c=6$. We need to compute $a+b+c$. $a+b+c = 3+2+6 = 11$. [/THOUGHT]

\boxed{11}

The slope of the tangent to the curve $C$ at point $P$ is $\frac{3\pi - 2}{\pi + 6}$. Matching this with the form $\frac{a\pi - b}{\pi + c}$, we identify $a=3$, $b=2$, and $c=6$. We need to compute $a+b+c$. $a+b+c = 3+2+6 = 11$.

\boxed{11}