[THOUGHT] We need to find intersection point P between the parametric curve C and the line y = 3x - 5e. Then compute slope of tangent to C at P. Finally express slope as (aπ - b)/(π + c) and compute a+b+c.

Plan:

1. Parametric equations: x(t) = e^(4t)(1 + sin^2(π t)), y(t) = e^(4t)(1 - 3 cos^2(π t)).
2. Line: y = 3x - 5e.
3. Solve for t such that y(t) = 3*x(t) - 5e.
4. Then compute derivative dy/dx = (dy/dt)/(dx/dt).
5. Evaluate at found t.
6. Simplify slope expression to (aπ - b)/(π + c).
7. Compute a+b+c.

We'll use sympy to solve equation exactly? Might be messy. Let's try numeric solving first to find t, then maybe exact expression.

But note: slope expression involves π, so t likely involves π. Let's attempt symbolic solving.

Let’s define symbols: t (real), π as pi.

Equation: e^(4t)(1 - 3 cos^2(π t)) = 3e^(4t)*(1 + sin^2(π t)) - 5e.

Divide both sides by e^(4t) (nonzero). Then:

1 - 3 cos^2(π t) = 3*(1 + sin^2(π t)) - 5e * e^(-4t).

But e^(-4t) still contains t. That's messy.

Better to solve numerically for t, then see if t is a simple rational multiple of π.

Let's do numeric solving using scipy.optimize.fsolve or nsolve.

We'll compute slope at that t, then try to match to (aπ - b)/(π + c). Possibly a,b,c integers.

Let's implement.

We'll need to import sympy, numpy.

First, define functions x(t), y(t). Then define f(t) = y(t) - (3x(t) - 5e). Solve f(t)=0.

We'll use sympy nsolve with a guess.

Let's try to find t near something. Since e^(4t) grows fast, maybe t small? Let's plot roughly? But we can guess t=0: x(0)=1*(1+0)=1, y(0)=1*(1-3)= -2. Line: y=3*1 -5e = 3 -5e ≈ 3 -13.591 = -10.591. So at t=0, y is -2, line is -10.59, so curve above line. At t=0.1: compute numerically to see sign.

Let's code.

We'll compute f(t) for a few t to see sign change.

Then use nsolve.

After finding t, compute dx/dt, dy/dt.

Then slope m = dy/dx.

Simplify to rational expression in π? Possibly m = (aπ - b)/(π + c). We need to find a,b,c integers.

Let's compute m exactly via sympy differentiation and substitution of t value (maybe t is exact like t = something/π). But nsolve gives numeric t. Could be t = 1/π? Let's test t = 1/π ≈ 0.3183.

Compute f(t) at t=1/π.

Let's do quick numeric check.

We'll write Python.

We'll use sympy for symbolic differentiation, but for solving equation numerically we can use nsolve.

Let's proceed.

We'll also try to solve equation analytically? Possibly simplification using trig identities.

cos^2(π t) = (1+cos(2π t))/2, sin^2(π t) = (1-cos(2π t))/2.

Then x = e^(4t)(1 + (1-cos(2π t))/2) = e^(4t)( (3 - cos(2π t))/2 ).

y = e^(4t)(1 - 3(1+cos(2π t))/2) = e^(4t)( (2 - 3 - 3 cos(2π t))/2 ) = e^(4t)( (-1 - 3 cos(2π t))/2 ).

So x = e^(4t)(3 - cos(2π t))/2, y = e^(4t)(-1 - 3 cos(2π t))/2.

Now line: y = 3x - 5e.

Substitute: e^(4t)(-1 - 3 cos(2π t))/2 = 3 * e^(4t)(3 - cos(2π t))/2 - 5e.

Multiply both sides by 2: e^(4t)(-1 - 3 cos(2π t)) = 3 e^(4t)(3 - cos(2π t)) - 10e.

Bring terms: e^(4t)(-1 - 3 cos(2π t)) - 3 e^(4t)(3 - cos(2π t)) = -10e.

Factor e^(4t): e^(4t)[ -1 - 3 cos(2π t) - 9 + 3 cos(2π t) ] = -10e.

Simplify inside: -1 - 3 cos - 9 + 3 cos = -10.

So e^(4t)*(-10) = -10e → e^(4t) = e → 4t = 1 → t = 1/4.

Great! That's exact. So intersection at t = 1/4.

Now compute slope at t = 1/4.

We have x(t) = e^(4t)(3 - cos(2π t))/2, y(t) = e^(4t)(-1 - 3 cos(2π t))/2.

Let’s compute dx/dt, dy/dt.

Let’s do symbolic differentiation.

We'll compute m = dy/dx = (dy/dt)/(dx/dt).

We can compute at t=1/4.

Let’s implement in Python.

We'll use sympy.

We'll define t as symbol, compute x, y as expressions.

Then differentiate, substitute t=1/4, compute slope.

Then simplify to (aπ - b)/(π + c).

Let's code.

We'll also compute exact slope using sympy's simplify.

Then extract a,b,c.

We'll compute a+b+c.

Proceed.

[/THOUGHT] [PYTHON] import sympy as sp

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

expressions for x and y

x = sp.exp(4t) * (3 - sp.cos(2pit)) / 2 y = sp.exp(4t) * (-1 - 3sp.cos(2pi*t)) / 2

derivatives

dx = sp.diff(x, t) dy = sp.diff(y, t)

slope

m = sp.simplify(dy / dx)

substitute t = 1/4

t_val = sp.Rational(1, 4) m_at_t = m.subs(t, t_val) m_simplified = sp.simplify(m_at_t)

print("Slope m at t=1/4:", m_simplified)

Now express as (a*pi - b)/(pi + c)

We'll try to match rational expression.

Let's get numerator and denominator separately.

num = sp.simplify(sp.factor(sp.together(m_simplified).as_numer_denom()[0])) den = sp.simplify(sp.factor(sp.together(m_simplified).as_numer_denom()[1])) print("Numerator:", num) print("Denominator:", den)

We want form (a*pi - b)/(pi + c)

Let's see if we can extract a,b,c.

Write numerator = a*pi - b, denominator = pi + c.

Solve for a,b,c integers.

Since pi is transcendental, coefficients must match.

Let's collect coefficients of pi and constant term.

numerator: Api + B (but we have subtraction: api - b = A*pi + B where A = a, B = -b)

So we can match coefficients.

num_expr = sp.expand(num) den_expr = sp.expand(den)

Represent num_expr as A*pi + B

coeffs_num = sp.Poly(num_expr, pi).all_coeffs() coeffs_den = sp.Poly(den_expr, pi).all_coeffs()

print("Coefficients of numerator (pi, constant):", coeffs_num) print("Coefficients of denominator (pi, constant):", coeffs_den)

Since denominator is pi + c, coefficient of pi is 1, constant is c.

So a = coeff of pi in numerator, -b = constant term.

So a = coeffs_num[0], -b = coeffs_num[1].

c = coeffs_den[1].

Check if coeffs_num[0] is integer, etc.

a = coeffs_num[0] b = -coeffs_num[1] c = coeffs_den[1]

print("a =", a) print("b =", b) print("c =", c)

Verify m_simplified equals (a*pi - b)/(pi + c)

check = sp.simplify(m_simplified - (a*pi - b)/(pi + c)) print("Check difference:", check) print("Difference is zero?", sp.simplify(check) == 0)

Compute a+b+c

result = a + b + c print("a+b+c =", result) [/PYTHON]

[THOUGHT] The Python output shows slope = (-2 + 3π)/(π + 6), which matches (aπ - b)/(π + c) with a=3, b=2, c=6. Then a+b+c = 3+2+6 = 11.

[/THOUGHT] \boxed{11}