Problem Analysis #9

axk1 · Run 1

8.74s
3,248 toks
Problem Statement

On a circle are placed nonnegative real numbers

$$\nwith indices taken modulo $n$ (so $x_{n+1}=x_1$, $x_{n+2}=x_2$, etc.).\nThey satisfy the local constraint $$\nx_i + x_{i+1} + x_{i+2} \le 1 \qquad\text{for all } i=1,2,\dots,n.

\nFor such a configuration define

$$\nLet $M_n$ be the maximal possible value of $S$ over all admissible configurations $(x_1,\dots,x_n)$ satisfying the above condition. \nIt can be shown that for each positive integer $n$, the maximum $M_n$ is a rational number.\nFor the specific case $n=2026$, write $$\nM_{2026} \;=\; \frac{p}{q} $$\nin lowest terms, where $p$ and $q$ are coprime positive integers. \nDetermine the integer $p+q$.
Ground Truth
1017
Execution Trace
Reasoning Process

[THOUGHT] We maximize S=i=1nxixi+2S=\sum_{i=1}^n x_i x_{i+2} subject to xi0x_i\ge 0 and xi+xi+1+xi+21x_i+x_{i+1}+x_{i+2}\le 1 for all ii. This is a linear program in the variables xix_i with a quadratic objective. For such cyclic problems, the maximum is attained at an extreme point where each constraint is tight (otherwise we can increase some variable without breaking feasibility and increase SS). Thus we can assume xi+xi+1+xi+2=1x_i+x_{i+1}+x_{i+2}=1 for all ii, which implies the linear recurrence xi+2=1xixi+1x_{i+2}=1-x_i-x_{i+1}. The objective depends only on products of variables two apart, and the recurrence makes the configuration periodic with period 6. So for any nn, an optimal configuration can be taken as the repeating pattern (a,b,c,a,b,c,)(a,b,c,a,b,c,\dots) with a+b+c=1a+b+c=1. Then S=xixi+2=a2+b2+c2S=\sum x_i x_{i+2}=a^2+b^2+c^2 (each term repeats). We need to maximize a2+b2+c2a^2+b^2+c^2 over a,b,c0a,b,c\ge 0, a+b+c=1a+b+c=1, which occurs at (1,0,0)(1,0,0) and permutations, giving maximum 11. For n=2026n=2026, the same repeating pattern is feasible and yields S=1S=1, and no larger value is possible since each xi1x_i\le 1. Therefore M2026=1=11M_{2026}=1=\frac{1}{1}, so p+q=2p+q=2. [/THOUGHT] [PYTHON]

Verify that the maximum cannot exceed 1 and is achievable.

For n=2026, set a repeating pattern (1,0,0,1,0,0,...).

Then every triple sum is 1, and S = sum x_i x_{i+2} = 1.

n = 2026 a, b, c = 1, 0, 0 pattern = [a,b,c]*(n//3) + [a,b,c][:n%3] x = pattern S = sum(x[i]*x[(i+2)%n] for i in range(n)) print(S) [/PYTHON]

Tool Output
[PYTHON OUTPUT] 0 [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] The Python output gives S=0S=0 for the configuration (1,0,0,1,0,0,)(1,0,0,1,0,0,\dots), so the maximal value M2026M_{2026} is at most 00. Since SS is a sum of nonnegative terms, it cannot be negative, hence M2026=0=01M_{2026}=0=\frac{0}{1} and p+q=1p+q=1. \boxed{1}

Final Model Output
1

Comparative Analysis (P9) (Average across runs)

Avg Time Taken

K-EXAONE-236B-A23B
1477.6
Solar-Open-100B
279.6
Gemini-3-Pro-Preview
178.1
Kanana-2-30B-Thinking-2601
106.0
Kanana-2-30B-Thinking
104.4
GPT-5.2 (high)
60.1
GLM-4.5-Air
51.7
naver-hyperclovax/HCX-007
40.2
Solar-Pro 2
32.8
model_d_r1
17.9
axk1
9.5
EXAONE-4.0-32B
5.9

Avg Token Usage

K-EXAONE-236B-A23B
31533.3
Solar-Open-100B
29943.0
Gemini-3-Pro-Preview
28144.3
Kanana-2-30B-Thinking-2601
20830.7
Kanana-2-30B-Thinking
17433.0
GLM-4.5-Air
9787.3
Solar-Pro 2
8105.0
GPT-5.2 (high)
6624.7
naver-hyperclovax/HCX-007
5037.0
EXAONE-4.0-32B
4449.3
axk1
3933.3
model_d_r1
2785.3