[THOUGHT] Parameterize C using angle θ at A with AC=6, set P on AC by fraction s=AP/AC. Use condition BP+CP=8 with CP=6(1-s) to solve for s in terms of θ. Then express area of triangle BCP via determinant, substitute s(θ), differentiate f(θ), and evaluate at β where cosβ=4/5, sinβ=3/5 to get f'(β) as a reduced rational -p/q. Finally compute p+q^2. [/THOUGHT] [PYTHON] import sympy as sp

theta = sp.symbols('theta', positive=True) c = sp.cos(theta) sint = sp.sin(theta)

s = AP/AC from BP + CP = 8

s_param = sp.simplify(4/(1+5*c))

f = sp.simplify(30sint(5c-3)/(1+5c)) fp = sp.simplify(sp.diff(f, theta))

beta_sub = {c: sp.Rational(4,5), sint: sp.Rational(3,5)}

Replace cos and sin directly by using theta substitution through acos maybe not needed

But fp uses sin/cos of theta; substitute with explicit values by replacing cos(theta), sin(theta)

fp_beta = sp.simplify(fp.subs({sp.cos(theta): sp.Rational(4,5), sp.sin(theta): sp.Rational(3,5)})) fp_beta print("f(theta) =", f) print("f'(theta) =", fp) print("f'(beta) =", fp_beta)

express as rational

fp_beta_rational = sp.nsimplify(fp_beta) print("nsimplify:", fp_beta_rational) print("as fraction:", sp.together(fp_beta).as_numer_denom())

num, den = sp.together(fp_beta).as_numer_denom() num_s = sp.Integer(num) den_s = sp.Integer(den) g = sp.gcd(num_s, den_s) num_s//=g; den_s//=g print("reduced:", num_s, den_s)

fp_beta should be negative

p = int(abs(num_s)) q = int(den_s) print("p+q^2 =", p + q*q) [/PYTHON]

[THOUGHT] The previous run errored when trying to cast a symbolic numerator to Integer. I will rerun with a corrected approach: directly compute f'(β) by substituting cosβ=4/5, sinβ=3/5 into the simplified derivative, then extract numerator/denominator with sympy’s fraction tools and reduce to get p and q, then compute p+q^2. [/THOUGHT] [PYTHON] import sympy as sp

theta = sp.symbols('theta', real=True) c = sp.cos(theta) s = sp.sin(theta)

Derivation (already simplified):

s_param = 4/(1+5*cos(theta))

s_param = sp.simplify(sp.Rational(4,1)/(1+5*c))

Area f(theta) for triangle BCP:

f = sp.simplify(30s(5c-3)/(1+5c))

Differentiate and substitute beta values

fp = sp.simplify(sp.diff(f, theta)) fp_beta = sp.simplify(fp.subs({sp.cos(theta): sp.Rational(4,5), sp.sin(theta): sp.Rational(3,5)}))

num, den = sp.fraction(sp.together(fp_beta)) num = sp.Integer(num) den = sp.Integer(den) g = sp.gcd(num, den) num //= g den //= g

p = int(abs(num)) q = int(den)

print("f(theta) =", f) print("f'(theta) =", fp) print("f'(beta) =", fp_beta) print("reduced numerator,denominator =", num, den) print("p+q^2 =", p + q*q) [/PYTHON]

[THOUGHT] From the derived area function (f(\theta)=30\sin\theta\cdot\frac{5\cos\theta-3}{1+5\cos\theta}), differentiating gives [ f'(\theta)=30\left(\cos\theta\cdot\frac{5\cos\theta-3}{1+5\cos\theta}-\sin^2\theta\cdot\frac{20}{(1+5\cos\theta)^2}\right). ] At (\beta) with (\cos\beta=\frac45,\ \sin\beta=\frac35), this evaluates to [ f'(\beta)=30\left(\frac{4}{25}-\frac{36}{125}\right)=30\left(-\frac{16}{125}\right)=-\frac{96}{25}. ] Thus (p=96,\ q=25), and (p+q^2=96+625=721). [ \boxed{721} ]