[THOUGHT] Let me analyze this problem step by step.

1. Point $A(a,b)$ is on the curve $y = \log_{16}(8x + 2)$, so $b = \log_{16}(8a + 2)$.

2. Point $B$ is on the curve $y = 4^{x-1} - \frac{1}{2}$.

3. The reflection of $A$ across the line $y = x$ is the point $(b, a)$. This point lies on the line $OB$ (the line through the origin and $B$).

4. The midpoint of segment $AB$ is $\left(\frac{77}{8}, \frac{133}{8}\right)$.

Let me denote $B = (c, d)$ where $d = 4^{c-1} - \frac{1}{2}$.

From the midpoint condition:

• $\frac{a + c}{2} = \frac{77}{8}$, so $a + c = \frac{77}{4}$
• $\frac{b + d}{2} = \frac{133}{8}$, so $b + d = \frac{133}{4}$

The reflection $(b, a)$ lies on line $OB$, which has slope $\frac{d}{c}$. So: $\frac{a}{b} = \frac{d}{c}$, which means $ac = bd$.

Let me also note that $y = \log_{16}(8x + 2)$ and $y = 4^{x-1} - \frac{1}{2}$ might be inverse functions.

Let me check: If $y = \log_{16}(8x + 2)$, then $16^y = 8x + 2$, so $x = \frac{16^y - 2}{8}$.

For the inverse: $x = \frac{16^y - 2}{8} = \frac{4^{2y} - 2}{8} = \frac{4^{2y}}{8} - \frac{1}{4}$.

Hmm, let me check if $y = 4^{x-1} - \frac{1}{2}$ is the inverse.

If $y = 4^{x-1} - \frac{1}{2}$, then $y + \frac{1}{2} = 4^{x-1} = \frac{4^x}{4}$, so $4^x = 4(y + \frac{1}{2}) = 4y + 2$.

Then $x = \log_4(4y + 2)$.

So the inverse of $y = 4^{x-1} - \frac{1}{2}$ is $y = \log_4(4x + 2)$.

But we have $y = \log_{16}(8x + 2)$. Let me convert: $\log_{16}(8x + 2) = \frac{\log_4(8x + 2)}{2}$.

These aren't quite inverses. Let me solve this system numerically. [/THOUGHT] [PYTHON] from sympy import symbols, log, solve, Rational, Eq, simplify, nsimplify

Define variables

a, b, c, d = symbols('a b c d', real=True, positive=True)

Equations:

1. b = log_16(8a + 2)

2. d = 4^(c-1) - 1/2

3. a + c = 77/4

4. b + d = 133/4

5. ac = bd (from reflection condition)

Let's express everything in terms of a and b

c = 77/4 - a

d = 133/4 - b

From equation 1: 16^b = 8a + 2, so a = (16^b - 2)/8

From equation 2: d + 1/2 = 4^(c-1), so 4^(c-1) = 133/4 - b + 1/2 = 135/4 - b

Let me set up the equations properly

from sympy import log, exp, Rational, symbols, solve, Eq, simplify

a, b = symbols('a b', real=True, positive=True)

c and d in terms of a and b

c = Rational(77, 4) - a d = Rational(133, 4) - b

Equation 1: b = log_16(8a + 2)

This means 16^b = 8a + 2

eq1 = Eq(16**b, 8*a + 2)

Equation 2: d = 4^(c-1) - 1/2

d + 1/2 = 4^(c-1)

eq2 = Eq(d + Rational(1, 2), 4**(c - 1))

Equation 3: ac = bd

eq3 = Eq(a * c, b * d)

Substitute c and d

eq2_sub = eq2.subs([(c, Rational(77, 4) - a), (d, Rational(133, 4) - b)]) eq3_sub = eq3.subs([(c, Rational(77, 4) - a), (d, Rational(133, 4) - b)])

print("Equation 1:", eq1) print("Equation 2 (substituted):", eq2_sub) print("Equation 3 (substituted):", eq3_sub) [/PYTHON]

[THOUGHT] Let me continue solving this system of equations numerically to find the values of a and b. [/THOUGHT] [PYTHON] from sympy import symbols, solve, Eq, Rational, nsolve, log, N import numpy as np from scipy.optimize import fsolve

Define the system of equations numerically

def equations(vars): a, b = vars c = 77/4 - a d = 133/4 - b

# eq1: 16^b = 8a + 2
eq1 = 16**b - (8*a + 2)

# eq2: d + 1/2 = 4^(c-1)
eq2 = (d + 0.5) - 4**(c - 1)

# eq3: a*c = b*d
eq3 = a*c - b*d

return [eq1, eq3]  # We have 2 unknowns, need 2 equations

But we have 3 equations and 2 unknowns - let me check which two to use

Actually eq1 and eq2 should determine a and b, and eq3 should be automatically satisfied

def equations_v2(vars): a, b = vars c = 77/4 - a d = 133/4 - b

# eq1: 16^b = 8a + 2
eq1 = 16**b - (8*a + 2)

# eq2: d + 1/2 = 4^(c-1)
eq2 = (d + 0.5) - 4**(c - 1)

return [eq1, eq2]

Initial guess

initial_guess = [5, 2] solution = fsolve(equations_v2, initial_guess, full_output=True) a_sol, b_sol = solution[0] print(f"a = {a_sol}, b = {b_sol}")

Verify the solution

c_sol = 77/4 - a_sol d_sol = 133/4 - b_sol print(f"c = {c_sol}, d = {d_sol}")

Check all equations

print(f"Check eq1: 16^b = {16b_sol}, 8a+2 = {8*a_sol + 2}") print(f"Check eq2: d + 0.5 = {d_sol + 0.5}, 4^(c-1) = {4(c_sol - 1)}") print(f"Check eq3: ac = {a_sol * c_sol}, bd = {b_sol * d_sol}")

Calculate a * b

ab = a_sol * b_sol print(f"a * b = {ab}") [/PYTHON]

[THOUGHT] The numerical solution gives us approximately:

• a = 15.75 = 63/4
• b = 1.75 = 7/4

Let me verify: a × b = (63/4) × (7/4) = 441/16

So p = 441 and q = 16. Let me check if they are coprime: 441 = 21² = 9 × 49 = 3² × 7², and 16 = 2⁴. They share no common factors, so they are coprime.

Therefore p + q = 441 + 16 = 457. [/THOUGHT]

\boxed{457}