[THOUGHT] We have points A(5,0), B(-5,0). Let C be such that AC = 6. Let P lie on segment AC. Condition: BP + CP = 8. Let ∠A = θ in triangle ABC. Area of triangle BCP is f(θ). Given cos β = 4/5, 0<β<π/2, and f'(β) = -p/q. Need p+q^2.

We need to express f(θ) in terms of θ, then differentiate and evaluate at β where cos β = 4/5.

Set up coordinates: Place A at (5,0), B at (-5,0). Let C be at distance 6 from A. Let ∠A = θ, which is angle at A in triangle ABC. That angle is between AB and AC. Since AB is horizontal from A to B? Actually A(5,0), B(-5,0) so vector AB = (-10,0). But angle at A is between AB and AC. We can define C such that AC length 6, and direction making angle θ with AB. But careful: ∠A = θ is angle BAC. So angle between AB and AC. AB is from A to B: vector B - A = (-10,0). So AB points left. Let's define AC vector making angle θ with AB. So we can param.

Let’s set coordinates: A = (5,0). Let AC length = 6. Let direction of AC be angle φ measured from positive x-axis. Then C = (5 + 6 cos φ, 0 + 6 sin φ). But angle ∠A = θ is angle between AB and AC. Vector AB = (-10,0). Vector AC = (6 cos φ, 6 sin φ). The angle between them satisfies cos θ = (AB·AC)/(|AB||AC|) = ((-10)(6 cos φ) + 0(6 sin φ))/(10*6) = (-60 cos φ)/60 = -cos φ. So cos θ = -cos φ. Also sin θ = sqrt(1 - cos^2 θ). Since 0 < θ < π? Possibly θ can be acute? Actually ∠A in triangle ABC could be acute or obtuse? But given cos β = 4/5 positive, so β is acute. So θ = β is acute, so cos θ positive. Then cos θ = -cos φ positive implies cos φ negative. So φ is in (π/2, π). So we can set φ = π - θ? Because cos(π - θ) = -cos θ. Indeed if φ = π - θ, then cos φ = -cos θ, sin φ = sin θ. That matches cos θ = -cos φ. So we can param: C = (5 + 6 cos(π - θ), 6 sin(π - θ)) = (5 - 6 cos θ, 6 sin θ). Because cos(π - θ) = -cos θ, sin(π - θ) = sin θ.

Thus C = (5 - 6 cos θ, 6 sin θ). Check AC distance: sqrt(( -6 cos θ)^2 + (6 sin θ)^2) = 6 sqrt(cos^2 θ + sin^2 θ) = 6. Good.

Now P lies on segment AC. Let AP = t * AC, where t ∈ [0,1). Since P distinct from C, t < 1. So P = A + t*(C - A) = (5,0) + t*(-6 cos θ, 6 sin θ) = (5 - 6t cos θ, 6t sin θ).

Now condition (iii): BP + CP = 8. Compute distances.

B = (-5,0). So BP = distance between P and B: sqrt((5 - 6t cos θ +5)^2 + (6t sin θ)^2) = sqrt((10 - 6t cos θ)^2 + (6t sin θ)^2).

CP = distance between P and C: C - P = (5 - 6 cos θ - (5 - 6t cos θ), 6 sin θ - 6t sin θ) = (-6 cos θ + 6t cos θ, 6 sin θ - 6t sin θ) = (6(t-1) cos θ, 6(1-t) sin θ). Wait sign: 6(t-1) cos θ = -6(1-t) cos θ. So CP length = sqrt((6(t-1) cos θ)^2 + (6(1-t) sin θ)^2) = sqrt(36(1-t)^2 cos^2 θ + 36(1-t)^2 sin^2 θ) = 6|1-t|. Since t<1, 1-t>0, so CP = 6(1-t).

Thus CP = 6(1-t). That's simple.

Now BP = sqrt((10 - 6t cos θ)^2 + (6t sin θ)^2). Expand: = sqrt(100 - 120 t cos θ + 36 t^2 cos^2 θ + 36 t^2 sin^2 θ) = sqrt(100 - 120 t cos θ + 36 t^2 (cos^2 θ + sin^2 θ)) = sqrt(100 - 120 t cos θ + 36 t^2).

So condition: sqrt(100 - 120 t cos θ + 36 t^2) + 6(1-t) = 8.

We need to solve for t in terms of θ. Let’s denote S = sqrt(100 - 120 t cos θ + 36 t^2). Then S = 8 - 6(1-t) = 8 - 6 + 6t = 2 + 6t.

Thus sqrt(100 - 120 t cos θ + 36 t^2) = 2 + 6t.

Square both sides: 100 - 120 t cos θ + 36 t^2 = (2 + 6t)^2 = 4 + 24t + 36 t^2.

Cancel 36 t^2 both sides: 100 - 120 t cos θ = 4 + 24t.

So 100 - 120 t cos θ - 4 - 24t = 0 => 96 - 120 t cos θ - 24t = 0 => 96 = t(120 cos θ + 24) => t = 96 / (120 cos θ + 24) = 96 / (24(5 cos θ + 1)) = 4 / (5 cos θ + 1).

Thus t = 4/(5 cos θ + 1). Note t must be in [0,1). Since cos θ positive (θ acute), denominator > 1, so t < 4/(5*1+1)=4/6=2/3 <1. Also t>0. Good.

Now we have P coordinates: P = (5 - 6t cos θ, 6t sin θ).

Now area of triangle BCP: f(θ) = area of triangle with vertices B, C, P. We can compute using coordinates.

Let B = (-5,0), C = (5 - 6 cos θ, 6 sin θ), P = (5 - 6t cos θ, 6t sin θ).

Area = 1/2 * | (B_x (C_y - P_y) + C_x (P_y - B_y) + P_x (B_y - C_y) ) |.

Compute using determinant formula: area = 0.5 * | x_B(y_C - y_P) + x_C(y_P - y_B) + x_P(y_B - y_C) |.

Plug in:

B_x = -5, B_y = 0. C_x = 5 - 6 cos θ, C_y = 6 sin θ. P_x = 5 - 6t cos θ, P_y = 6t sin θ.

Compute differences: y_C - y_P = 6 sin θ - 6t sin θ = 6(1-t) sin θ. y_P - y_B = 6t sin θ - 0 = 6t sin θ. y_B - y_C = 0 - 6 sin θ = -6 sin θ.

Now sum: -5 * (6(1-t) sin θ) + (5 - 6 cos θ)(6t sin θ) + (5 - 6t cos θ)(-6 sin θ).

Factor 6 sin θ: = 6 sin θ [ -5(1-t) + t(5 - 6 cos θ) - (5 - 6t cos θ) ].

Simplify inside brackets: -5 + 5t + 5t - 6t cos θ - 5 + 6t cos θ = (-5 -5) + (5t+5t) + (-6t cos θ + 6t cos θ) = -10 + 10t + 0 = 10(t - 1).

Thus sum = 6 sin θ * 10(t - 1) = 60 sin θ (t - 1).

Area = 0.5 * | 60 sin θ (t - 1) | = 30 | sin θ (t - 1) |.

Since t < 1, t-1 negative, sin θ positive (θ acute), so sin θ (t-1) negative, absolute value = - sin θ (t-1) = sin θ (1 - t). So f(θ) = 30 sin θ (1 - t).

Now substitute t = 4/(5 cos θ + 1). Then 1 - t = 1 - 4/(5 cos θ + 1) = (5 cos θ + 1 - 4)/(5 cos θ + 1) = (5 cos θ - 3)/(5 cos θ + 1).

Thus f(θ) = 30 sin θ * (5 cos θ - 3)/(5 cos θ + 1).

Simplify: f(θ) = 30 sin θ (5 cos θ - 3) / (5 cos θ + 1).

Now we need f'(θ) and evaluate at β where cos β = 4/5, 0<β<π/2. Also sin β = sqrt(1 - (4/5)^2) = sqrt(1 - 16/25) = sqrt(9/25) = 3/5.

Compute derivative f'(θ). Let’s differentiate using quotient rule or product rule.

Let u = 30 sin θ (5 cos θ - 3), v = (5 cos θ + 1). Then f = u/v.

Compute u' and v'.

u = 30 sin θ (5 cos θ - 3). Let’s differentiate: u' = 30 [ cos θ (5 cos θ - 3) + sin θ * (-5 sin θ) ] = 30 [ 5 cos^2 θ - 3 cos θ - 5 sin^2 θ ].

But sin^2 θ = 1 - cos^2 θ, so -5 sin^2 θ = -5(1 - cos^2 θ) = -5 + 5 cos^2 θ. So inside: 5 cos^2 θ - 3 cos θ -5 + 5 cos^2 θ = 10 cos^2 θ - 3 cos θ -5.

Thus u' = 30 (10 cos^2 θ - 3 cos θ -5).

v = 5 cos θ + 1, v' = -5 sin θ.

Then f' = (u' v - u v') / v^2.

Compute u v': u v' = 30 sin θ (5 cos θ - 3) * (-5 sin θ) = -150 sin^2 θ (5 cos θ - 3).

Compute u' v: u' v = 30 (10 cos^2 θ - 3 cos θ -5) * (5 cos θ + 1).

Thus f' = [30 (10 cos^2 θ - 3 cos θ -5)(5 cos θ + 1) - (-150 sin^2 θ (5 cos θ - 3))] / (5 cos θ + 1)^2 = [30 (10 cos^2 θ - 3 cos θ -5)(5 cos θ + 1) + 150 sin^2 θ (5 cos θ - 3)] / (5 cos θ + 1)^2.

We can simplify by factoring 30: = 30 [ (10 cos^2 θ - 3 cos θ -5)(5 cos θ + 1) + 5 sin^2 θ (5 cos θ - 3) ] / (5 cos θ + 1)^2.

Now plug θ = β, cos β = 4/5, sin β = 3/5.

Compute numeric values.

First compute t at β: t = 4/(5*(4/5) + 1) = 4/(4 + 1) = 4/5 = 0.8.

Now compute f(β) = 30 sin β (5 cos β - 3)/(5 cos β + 1) = 30*(3/5)(5(4/5)-3)/(5*(4/5)+1) = 30*(3/5)(4-3)/(4+1) = 30(3/5)(1)/(5) = 30(3/5)(1/5) = 30(3/25) = 90/25 = 18/5 = 3.6.

Now compute f'(β). Let's compute using Python to avoid algebra mistakes.

We'll compute the expression numerically.

[/THOUGHT] [PYTHON] import sympy as sp

define variable theta

theta = sp.symbols('theta')

define f(theta)

f = 30sp.sin(theta)(5sp.cos(theta)-3)/(5sp.cos(theta)+1)

compute derivative

fprime = sp.diff(f, theta)

simplify

fprime_simplified = sp.simplify(fprime) print("f'(θ) =", fprime_simplified)

evaluate at beta where cos(beta)=4/5, sin(beta)=3/5

define beta symbolically? We can substitute cos(theta)=4/5, sin(theta)=3/5.

But careful: derivative expression may involve sin and cos. We'll substitute.

subs = {sp.cos(theta): sp.Rational(4,5), sp.sin(theta): sp.Rational(3,5)} fprime_at_beta = fprime_simplified.subs(subs) print("f'(β) =", fprime_at_beta) print("Numerical value:", fprime_at_beta.evalf()) [/PYTHON]

[THOUGHT] The Python output gave f'(θ) in a form that still contains cos(2θ) and cos(3θ) after substitution, because the substitution didn't replace those. I need to compute cos(2β) and cos(3β) explicitly given cos β = 4/5, sin β = 3/5.

cos(2β) = 2 cos²β - 1 = 2*(16/25) - 1 = 32/25 - 1 = 7/25. cos(3β) = 4 cos³β - 3 cos β = 4*(64/125) - 3*(4/5) = 256/125 - 12/5 = 256/125 - 300/125 = -44/125.

Now substitute into f'(β) expression from output: f'(β) = 6cos(2β) + (15/2)cos(3β) - 72/25. Compute: 6(7/25) = 42/25. (15/2)(-44/125) = (15 * -44) / (2 * 125) = (-660) / 250 = -66/25. So sum: 42/25 - 66/25 - 72/25 = (42 - 66 - 72)/25 = (-96)/25 = -96/25.

Thus f'(β) = -96/25, so p = 96, q = 25 (relatively prime positive integers). Then p + q² = 96 + 625 = 721.

Check: 96 and 25 are coprime. So answer is 721. [/THOUGHT] \boxed{721}