[THOUGHT] We need to solve for point A = (a,b) on curve y = log_{16}(8x+2) in first quadrant, and point B = (c,d) on curve y = 4^{x-1} - 1/2 in first quadrant, such that:

1. Reflection of A across line y = x lies on line OB. Reflection of A across y=x is (b,a). So (b,a) lies on line through O(0,0) and B(c,d). That means (b,a) is collinear with O and B, i.e., (b,a) = t*(c,d) for some t>0. Equivalently, a/c = b/d? Actually collinearity: points O, B, and (b,a) are collinear, so slope from O to B equals slope from O to (b,a): d/c = a/b => ad = bc? Wait careful: slope OB = d/c, slope O to (b,a) = a/b. So a/b = d/c => ac = bd? No: a/b = d/c => cross multiply: ac = bd. So ac = bd. That's one condition.

2. Midpoint of AB is M = ((a+c)/2, (b+d)/2) = (77/8, 133/8). So: a + c = 77/4 b + d = 133/4

3. A lies on curve y = log_{16}(8x+2). So b = log_{16}(8a+2). Since log base 16, we can write b = ln(8a+2)/ln(16). But maybe easier: 16^b = 8a+2 => 8a = 16^b - 2 => a = (16^b - 2)/8.

4. B lies on curve y = 4^{x-1} - 1/2. So d = 4^{c-1} - 1/2.

We have unknowns a,b,c,d. Equations: (1) a + c = 77/4 (2) b + d = 133/4 (3) ac = bd (4) b = log_{16}(8a+2) (5) d = 4^{c-1} - 1/2

We can try to solve numerically using Python.

Let's denote: From (1): c = 77/4 - a From (2): d = 133/4 - b From (4): b = log_{16}(8a+2) From (5): d = 4^{c-1} - 1/2

Plug into (3): ac = bd.

We can define function f(a) = ac - bd, with c, b, d expressed in terms of a.

We need to find a>0 such that f(a)=0.

Also b>0, d>0, c>0 (first quadrant). So a>0, b>0, c>0, d>0.

Let's solve numerically.

We'll use sympy or numerical root finding.

We'll also need to compute ab = ab, then express as p/q reduced, compute p+q.

Let's implement. </think>

[PYTHON] import sympy as sp

Define variable a

a = sp.symbols('a', real=True, positive=True)

c = 77/4 - a

c = 77/4 - a

b = log_16(8a + 2)

log_16(x) = ln(x)/ln(16)

b = sp.log(8*a + 2) / sp.log(16)

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

d = 4**(c - 1) - sp.Rational(1, 2)

Equation ac = bd

eq = a * c - b * d

Solve for a

solutions = sp.solve(eq, a) print("Solutions for a:") for sol in solutions: print(sol)

Filter real positive solutions and check if b,c,d positive

for sol in solutions: if sol.is_real and sol > 0: a_val = float(sol) c_val = 77/4 - a_val b_val = sp.log(8a_val + 2) / sp.log(16) d_val = 4**(c_val - 1) - 0.5 print(f"\na = {a_val}, c = {c_val}, b = {b_val}, d = {d_val}") # Check positivity if b_val > 0 and d_val > 0 and c_val > 0: print("All positive, valid.") else: print("Not all positive.") # Compute ab ab = a_val * b_val print(f"a*b = {ab}") # Convert to rational approximation from fractions import Fraction frac = Fraction(ab).limit_denominator(1000000) print(f"Fraction approx: {frac}") p = frac.numerator q = frac.denominator print(f"p+q = {p+q}") break else: print("No positive solution found.") [/PYTHON]

[THOUGHT] The Python output indicates that the equation is too complex for sympy to solve directly. However, the problem likely has a neat solution. Let's instead use the given midpoint and the reflection condition more directly.

Given midpoint M = ((a+c)/2, (b+d)/2) = (77/8, 133/8). So: a + c = 77/4 ... (1) b + d = 133/4 ... (2)

Also, A lies on y = log_16(8x+2) so b = log_16(8a+2). B lies on y = 4^{x-1} - 1/2 so d = 4^{c-1} - 1/2.

Reflection of A across y=x is A' = (b, a). This lies on line OB (O is origin). So A' is collinear with O and B, meaning (b, a) is proportional to (c, d). So a/d = b/c? Wait: slope OB = d/c, slope OA' = a/b. But A' lies on OB, so slope OB = slope OA' => d/c = a/b => ad = bc.

Thus we have two equations: (1) a + c = 77/4 (2) b + d = 133/4 (3) ad = bc (4) b = log_16(8a+2) (5) d = 4^{c-1} - 1/2

We can substitute (4) and (5) into (3) and (2). Let's solve numerically.

We'll use Python's fsolve or solve numerically.

Plan:

1. Let a be unknown.
2. c = 77/4 - a.
3. b = log(8a+2)/log(16).
4. d = 4**(c-1) - 0.5.
5. Equation ad - bc = 0.

Solve for a > 0, with c > 0, b > 0, d > 0.

Then compute a*b, express as fraction p/q, output p+q.

Let's implement. </think>

[THOUGHT] We'll solve numerically using sympy's nsolve.

We need initial guess. Since a and c are positive and sum to 77/4 ≈ 19.25, guess a ≈ 10.

Also b and d positive.

Let's code. </think>

[PYTHON] import sympy as sp

Define variable a

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

c from (1)

c = 77/4 - a

b from (4)

b = sp.log(8*a + 2) / sp.log(16)

d from (5)

d = 4**(c - 1) - sp.Rational(1, 2)

Equation ad = bc -> ad - bc = 0

eq = a * d - b * c

Numerical solve with initial guess a = 10

sol = sp.nsolve(eq, a, 10) print("a =", sol)

Compute c, b, d

a_val = float(sol) c_val = 77/4 - a_val b_val = sp.log(8*a_val + 2) / sp.log(16) d_val = 4**(c_val - 1) - 0.5

print("c =", c_val) print("b =", b_val) print("d =", d_val)

Check positivity

if a_val > 0 and c_val > 0 and b_val > 0 and d_val > 0: print("All positive.") else: print("Not all positive.")

Compute a*b

ab = a_val * b_val print("a*b =", ab)

Convert to fraction

from fractions import Fraction frac = Fraction(ab).limit_denominator(1000000) print("Fraction p/q =", frac) p = frac.numerator q = frac.denominator print("p+q =", p+q) [/PYTHON]

[THOUGHT] The previous code had an error in using Fraction directly on a float. Let's fix that by using sympy's rational approximation or direct fraction from the exact symbolic solution if possible. But given the error, I'll rewrite the numerical approach carefully.

We need to solve the system:

1. ( c = \frac{77}{4} - a )
2. ( b = \log_{16}(8a + 2) )
3. ( d = 4^{c-1} - \frac12 )
4. ( ad = bc ) (since A reflected over y=x is (b,a), and O=(0,0), so O, B, (b,a) collinear → slopes OB = slope O(b,a) → d/c = a/b → ad = bc)
5. ( a, b, c, d > 0 ).

Let's solve numerically with high precision and find a*b as rational.

[PYTHON] import sympy as sp

Define variable a

a = sp.symbols('a', real=True, positive=True)

c from (1)

c = 77/4 - a

b from (4)

b = sp.log(8*a + 2) / sp.log(16)

d from (5)

d = 4**(c - 1) - sp.Rational(1, 2)

Equation ad = bc

eq = ad - bc

Try solving numerically with different initial guesses

for guess in [5, 10, 15]: sol = sp.nsolve(eq, a, guess) a_val = float(sol) c_val = 77/4 - a_val b_val = sp.log(8a_val + 2) / sp.log(16) d_val = 4**(c_val - 1) - 0.5 # Check positivity if a_val > 0 and c_val > 0 and b_val > 0 and d_val > 0: print(f"Found a={a_val}, b={b_val}, c={c_val}, d={d_val}") # Compute ab ab = a_val * b_val # Use sympy's rational approximation ab_sym = sp.nsimplify(ab, [sp.log(2)]) print(f"a*b (simplified) = {ab_sym}") # Convert to fraction frac = sp.nsimplify(ab_sym, [sp.Rational]) print(f"Fraction p/q = {frac}") # If frac is rational, extract p,q if frac.is_Rational: p = frac.p q = frac.q print(f"p = {p}, q = {q}") print(f"p+q = {p+q}") break [/PYTHON]

[THOUGHT] The numerical solver is failing. Let's try a more systematic approach: from ( ad = bc ), substitute expressions for b, c, d in terms of a.

We have: ( c = 77/4 - a ) ( b = \frac{\ln(8a+2)}{\ln 16} ) ( d = 4^{c-1} - \frac12 = 4^{77/4 - a - 1} - \frac12 = 4^{73/4 - a} - \frac12 )

Equation: ( a \left( 4^{73/4 - a} - \frac12 \right) = \left( \frac{\ln(8a+2)}{\ln 16} \right) \left( \frac{77}{4} - a \right) ).

Let’s solve numerically with a good initial guess. Since a, b, c, d positive, and c = 77/4 - a > 0 → a < 19.25. Also b>0 → 8a+2 > 1 → a > -0.125, trivial. d>0 → ( 4^{73/4 - a} > 0.5 ) → ( 73/4 - a > \log_4(0.5) = -0.5 ) → ( 18.25 - a > -0.5 ) → ( a < 18.75 ). So a in (0, 18.75).

Let's try a around 10–18.

[PYTHON] import mpmath as mp

Define functions

def eq_func(a): c = 77/4 - a b = mp.log(8a + 2, 16) # log base 16 d = 4**(c - 1) - 0.5 return ad - b*c

Scan for a sign change

for a_test in [i/10 for i in range(1, 188)]: f = eq_func(a_test) if abs(f) < 1e-10: print(f"Root near a={a_test}, f={f}") if f == 0: print(f"Exact root a={a_test}") break # Check sign changes if a_test > 0 and a_test < 18.75: if f * eq_func(a_test - 0.1) <= 0: print(f"Sign change in [{a_test-0.1}, {a_test}]: f={eq_func(a_test-0.1)}, {f}") # Refine root = mp.findroot(eq_func, (a_test-0.1, a_test)) print(f"Refined root a={root}") a_val = float(root) c_val = 77/4 - a_val b_val = mp.log(8a_val + 2, 16) d_val = 4**(c_val - 1) - 0.5 print(f"c={c_val}, b={b_val}, d={d_val}") ab = a_val * b_val print(f"ab = {ab}") # Try to rationalize from fractions import Fraction # Use high precision to approximate frac = Fraction(ab).limit_denominator(1000000) print(f"Fraction p/q = {frac}") p = frac.numerator q = frac.denominator print(f"p+q = {p+q}") break [/PYTHON]

[THOUGHT] The error is due to mixing sympy and mpmath. Let's stick to pure mpmath for numerical solving and rational approximation.

We solve ( a(4^{73/4 - a} - 1/2) = (\log_{16}(8a+2))(77/4 - a) ).

We'll find a root in a reasonable range, compute a*b, and approximate as fraction.

[PYTHON] import mpmath as mp

mp.mp.dps = 30

Define equation

def f(a): c = 77/4 - a b = mp.log(8a + 2, 16) d = mp.power(4, c - 1) - 0.5 return ad - b*c

Find root by scanning

for a_val in mp.linspace(5, 18, 1000): if f(a_val) == 0: root = a_val break if f(a_val) * f(a_val + 0.01) < 0: # Refine root = mp.findroot(f, (a_val, a_val + 0.01)) break

a_val = float(root) c_val = 77/4 - a_val b_val = mp.log(8*a_val + 2, 16) d_val = mp.power(4, c_val - 1) - 0.5

print(f"a = {a_val}") print(f"b = {b_val}") print(f"c = {c_val}") print(f"d = {d_val}")

Check positivity

if a_val > 0 and c_val > 0 and b_val > 0 and d_val > 0: print("All positive.") else: print("Not all positive.")

ab = a_val * b_val print(f"a*b = {ab}")

Rational approximation

from fractions import Fraction frac = Fraction(ab).limit_denominator(1000000) print(f"Fraction p/q = {frac}") p = frac.numerator q = frac.denominator print(f"p+q = {p+q}") [/PYTHON]

[THOUGHT] The error is from using Fraction on a float that might be a sympy object. Let's compute everything as floats and approximate the fraction.

We have: ( a(4^{73/4 - a} - 1/2) = \frac{\ln(8a+2)}{\ln 16} \cdot (77/4 - a) ).

Let's solve numerically with mpmath, then compute a*b and use mp.nsimplify with known constants to get a rational expression.

[PYTHON] import mpmath as mp

mp.mp.dps = 40

Define equation

def f(a): c = 77/4 - a b = mp.log(8a + 2, 16) d = mp.power(4, c - 1) - 0.5 return ad - b*c

Find root in (0, 18.75)

Try bisection

low, high = 1, 18.7 for _ in range(100): mid = (low + high) / 2 if f(low) * f(mid) <= 0: high = mid else: low = mid root = (low + high) / 2 print(f"Root a = {root}")

a_val = root c_val = 77/4 - a_val b_val = mp.log(8*a_val + 2, 16) d_val = mp.power(4, c_val - 1) - 0.5

print(f"a = {a_val}") print(f"b = {b_val}") print(f"c = {c_val}") print(f"d = {d_val}")

Check positivity

if a_val > 0 and c_val > 0 and b_val > 0 and d_val > 0: print("All positive.") else: print("Not all positive.")

ab = a_val * b_val print(f"a*b = {ab}")

Try to rationalize using nsimplify

ab_sym = mp.nsimplify(ab, [mp.log(2)]) print(f"a*b simplified: {ab_sym}")

If ab_sym is rational, extract numerator/denominator

if 'Rational' in str(type(ab_sym)): # ab_sym might be Rational-like num = float(ab_sym.numerator()) den = float(ab_sym.denominator()) print(f"Rational p/q = {num}/{den}") p = int(num) q = int(den) print(f"p+q = {p+q}") else: # Approximate as fraction with denominator up to 1e6 from fractions import Fraction frac = Fraction(float(ab_sym)).limit_denominator(1000000) p = frac.numerator q = frac.denominator print(f"Approximate p/q = {p}/{q}") print(f"p+q = {p+q}") [/PYTHON]

[THOUGHT] The error is that nsimplify is from sympy, not mpmath. Let's use sympy for symbolic simplification after finding a numerical root. We'll compute the root numerically, then use sympy's nsimplify to get a rational expression for a*b.

[PYTHON] import sympy as sp

Define a as a symbol

a_sym = sp.symbols('a', real=True, positive=True)

Expressions

c_sym = sp.Rational(77, 4) - a_sym b_sym = sp.log(8*a_sym + 2) / sp.log(16) d_sym = 4**(c_sym - 1) - sp.Rational(1, 2)

Equation ad = bc

eq_sym = a_sym * d_sym - b_sym * c_sym

Solve numerically using nsolve

a_sol = sp.nsolve(eq_sym, a_sym, 10, prec=50) print(f"a = {a_sol}")

Convert to high-precision float

a_val = float(a_sol) c_val = 77/4 - a_val b_val = sp.log(8*a_val + 2, 16).evalf() d_val = 4**(c_val - 1) - 0.5

print(f"b = {b_val}") print(f"c = {c_val}") print(f"d = {d_val}")

Check positivity

if a_val > 0 and c_val > 0 and b_val > 0 and d_val > 0: print("All positive.") else: print("Not all positive.")

ab_val = a_val * b_val print(f"a*b = {ab_val}")

Use sympy's nsimplify

ab_sym = sp.nsimplify(ab_val, [sp.log(2)]) print(f"a*b simplified: {ab_sym}")

If rational, extract p, q

if ab_sym.is_Rational: p = ab_sym.p q = ab_sym.q print(f"p = {p}, q = {q}") print(f"p+q = {p+q}") else: # Approximate as fraction from fractions import Fraction frac = Fraction(float(ab_sym)).limit_denominator(1000000) p = frac.numerator q = frac.denominator print(f"Approximate p/q = {p}/{q}") print(f"p+q = {p+q}") [/PYTHON]

[THOUGHT] The Python output gives a*b = 42261266267569 / 1250000000000, so p = 42261266267569, q = 1250000000000. They are coprime? Let's check gcd.

[PYTHON] import math p = 42261266267569 q = 1250000000000 g = math.gcd(p, q) print("gcd =", g) if g == 1: print("p and q are coprime.") print("p+q =", p+q) else: # Reduce fraction p //= g q //= g print("Reduced: p =", p, "q =", q) print("p+q =", p+q) [/PYTHON]

[THOUGHT] The computed p+q is 43511266267569. This is the final answer.

\boxed{43511266267569}