On a circle are placed nonnegative real numbers
with indices taken modulo (so , , etc.). They satisfy the local constraint
For such a configuration define
Let be the maximal possible value of over all admissible configurations satisfying the above condition.
It can be shown that for each positive integer , the maximum is a rational number. For the specific case , write
in lowest terms, where and are coprime positive integers.
Determine the integer .
[THOUGHT] I will explore the optimization numerically for small n using nonlinear programming (SLSQP) to guess the exact maximal pattern and value, then infer the formula for even n=2026. I’ll compare the numeric maxima to candidate closed forms (like n/8) and inspect the maximizing configuration structure. [/THOUGHT] [PYTHON] import numpy as np from math import isclose from scipy.optimize import minimize
def max_S_numeric(n, trials=50, seed=0): rng = np.random.default_rng(seed) best = -1 bestx=None
# constraints: x_i>=0, triple sums <=1
cons=[]
for i in range(n):
cons.append({
'type':'ineq',
'fun': lambda x, i=i: 1 - (x[i] + x[(i+1)%n] + x[(i+2)%n])
})
bounds=[(0,1) for _ in range(n)]
def S(x):
return sum(x[i]*x[(i+2)%n] for i in range(n))
def obj(x):
return -S(x)
for t in range(trials):
# random x, then scale down to satisfy triple constraints
x = rng.random(n)
# iterative scaling to satisfy constraints
for _ in range(10):
viol = np.array([x[i]+x[(i+1)%n]+x[(i+2)%n] for i in range(n)])
m = viol.max()
if m <= 1+1e-9:
break
x = x/(m+1e-9)
res = minimize(obj, x, method='SLSQP', bounds=bounds, constraints=cons,
options={'maxiter':2000, 'ftol':1e-12, 'disp':False})
if res.success:
val = -res.fun
if val > best + 1e-7:
best=val
bestx=res.x
return best, bestx
def analyze(n): best, x = max_S_numeric(n, trials=120, seed=42+n) print("n", n, "best", best) if x is not None: # print rounded pattern xr = np.round(x, 6) print("x rounded:", xr) # check max triple sum and min triples = np.array([x[i]+x[(i+1)%n]+x[(i+2)%n] for i in range(n)]) print("triple max", triples.max(), "min", triples.min()) # compute S S = sum(x[i]*x[(i+2)%n] for i in range(n)) print("S recomputed", S) print("compare n/8", n/8, "diff", S - n/8) print("compare n/9", n/9, "diff", S - n/9) print("---")
for n in range(4,17): analyze(n) [/PYTHON]
[THOUGHT] From the computed maxima for many even (n), the optimum consistently equals (n/8) (achieved by the alternating configuration (0,\tfrac12,0,\tfrac12,\dots)), hence for (n=2026), [ M_{2026}=\frac{2026}{8}=\frac{1013}{4}\Rightarrow p+q=1013+4=1017. ] \boxed{1017}