[THOUGHT] I'll set up equations for a and b using:

1. A on its curve: (b = \log_{16}(8a+2)) → (16^b = 8a+2).
2. Midpoint: (B_x = \frac{77}{4} - a,\quad B_y = \frac{133}{4} - b).
3. Collinearity of O, A', B, where A' is reflection of A across (y=x):
   A' = (b, a). Collinearity with B and O gives determinant zero: (b B_y - a B_x = 0).
4. B on its curve: (B_y = 4^{B_x-1} - \frac{1}{2}).

So the system in a,b is:

• (b - \log_{16}(8a+2) = 0)
• (b\left(\frac{133}{4} - b\right) - a\left(\frac{77}{4} - a\right) = 0)
• (\frac{133}{4} - b - \left(4^{\left(\frac{77}{4}-a\right)-1} - \frac12\right) = 0)

I'll solve numerically with sympy, check for consistency, and then simplify (a \cdot b) to a rational (\frac{p}{q}), then output (p+q). [/THOUGHT] [PYTHON] from sympy import symbols, Eq, nsolve, log, Rational, nsimplify

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

Equations

eq1 = b - log(8a+2,16) Bx = Rational(77,4) - a By = Rational(133,4) - b eq2 = bBy - a*Bx eq3 = By - (4**(Bx-1) - Rational(1,2))

Try multiple initial guesses for robustness

solutions = set() guesses = [(1,1),(5,5),(10,10),(5,10),(10,5),(8,9),(9,8),(12,10),(10,12)] for g in guesses: try: sol = nsolve((eq1,eq2,eq3),(a,b),g, tol=1e-30, maxsteps=100) sol = tuple([sol[0].evalf(20), sol[1].evalf(20)]) solutions.add(tuple(round(float(v),12) for v in sol)) except Exception as e: pass

print("Numeric solutions (rounded):", solutions)

Take one solution and refine symbolically

Use the first solution found

if solutions: a_num,b_num = list(solutions)[0] print("Chosen approx a,b:", a_num, b_num)

# Refine via nsolve using that as guess
sol_refined = nsolve((eq1,eq2,eq3),(a,b),(a_num,b_num), tol=1e-40, maxsteps=200, prec=80)
print("Refined solution:", sol_refined)

a_sol, b_sol = sol_refined
prod = a_sol*b_sol
print("a*b numeric:", prod.evalf(50))
# Try to recognize as rational
prod_rat = nsimplify(prod, [2])
print("a*b nsimplify:", prod_rat)
if prod_rat.is_rational:
    p = prod_rat.as_numer_denom()[0]
    q = prod_rat.as_numer_denom()[1]
    print("p,q:", p, q, "p+q:", p+q)

else: print("No solutions found")

[THOUGHT] Using the system constraints and verified Python solution, we have (a \cdot b = \frac{441}{16}), so (p=441, q=16), and (p+q=457). [ \boxed{457} ]