[THOUGHT] I will compute the area enclosed by the polar curve r = 2 + cosθ using the polar area formula A = ½∫₀^{2π} r² dθ. Then I will find the slope dy/dx at θ = π/4 using the polar derivative formulas: dx/dθ = dr/dθ·cosθ − r·sinθ, dy/dθ = dr/dθ·sinθ + r·cosθ, with dr/dθ = −sinθ. I will simplify the resulting expression to the form b + √c. Finally, I will compute abc/π where a = area, b and c are the integers from the slope expression. [/THOUGHT] [PYTHON] import sympy as sp

Define theta

theta = sp.symbols('theta', real=True)

Part (1): Area

r = 2 + sp.cos(theta) area = sp.Rational(1,2) * sp.integrate(r**2, (theta, 0, 2*sp.pi)) area_simplified = sp.simplify(area)

Part (2): Slope dy/dx

dr_dtheta = sp.diff(r, theta) # -sin(theta) dx_dtheta = dr_dtheta * sp.cos(theta) - r * sp.sin(theta) dy_dtheta = dr_dtheta * sp.sin(theta) + r * sp.cos(theta) slope = sp.simplify(dy_dtheta / dx_dtheta)

Evaluate at theta = pi/4

theta_val = sp.pi/4 r_val = r.subs(theta, theta_val) slope_at_point = sp.simplify(slope.subs(theta, theta_val))

Express slope as b + sqrt(c)

Extract b and c

b = sp.nsimplify(slope_at_point - sp.sqrt(slope_at_point2 - sp.simplify(slope_at_point2 - sp.simplify(slope_at_point**2)))) # placeholder, we will manually identify

Instead, we can match pattern b + sqrt(c)

Use sp.collect to see if sqrt term present

collected = sp.collect(slope_at_point, sp.sqrt(2), evaluate=False)

Since sqrt(2) appears, we can write slope = b + sqrt(c)

Let's solve for b and c

slope = b + sqrt(c)

We know slope = sqrt(2) - 2, so b = -2, c = 2

b_val = -2 c_val = 2

Compute abc/pi

abc_over_pi = (area_simplified * b_val * c_val) / sp.pi

Print results

print("Area a:", area_simplified) print("Slope at theta=pi/4:", slope_at_point) print("b:", b_val) print("c:", c_val) print("abc/pi:", abc_over_pi) [/PYTHON]

[THOUGHT] The Python output provides the exact values: area ( a = \frac{9\pi}{2} ), slope expression ( b = -2 ), ( c = 2 ), and ( \frac{abc}{\pi} = -18 ). The calculation confirms this result. [/THOUGHT] \boxed{-18}