Problem Analysis #29

Solar-Pro-2 (31B)(high) · Run 1

67.20s
10,423 toks
Problem Statement

Consider the point A(2,0)A(-2,0) in the coordinate plane. Suppose two points PP and QQ satisfy the following conditions:

  • [(i)] The vectors AP\overrightarrow{AP} and AQ\overrightarrow{AQ} are parallel.
  • [(ii)] The dot product satisfies APAQ=9\overrightarrow{AP} \cdot \overrightarrow{AQ} = 9. Let the two circles
C1:x2+y2=1,\qquadC2:(x+1)2+y2=1C_{1} : x^{2} + y^{2} = 1, \qquadC_{2} : (x+1)^{2} + y^{2} = 1

intersect at two distinct points R1R_{1} and R2R_{2}.Consider the region R\mathcal{R} consisting of the arc R1OR2R_{1}O'R_{2} of circle C1C_{1},which contains the point O(1,0)O'(-1,0), together with the arc R1OR2R_{1}OR_{2} of circle C2C_{2},which contains the point O(0,0)O(0,0).When the point PP moves along the region R\mathcal{R}, let kk denote the total lengthof the locus traced by the corresponding point QQ.For any real number ss, let s\lfloor s \rfloor denote the greatest integer less than or equal to ss.Determine the value of k\lfloor k \rfloor.

Ground Truth
17
Execution Trace
Reasoning Process

[THOUGHT] To solve the problem, we need to determine the total length of the locus traced by point QQ as PP moves along the region R\mathcal{R}. The key steps involve understanding the transformation from PP to QQ and the geometry of the region R\mathcal{R}.

  1. Transformation from PP to QQ:

    • Given that vectors AP\overrightarrow{AP} and AQ\overrightarrow{AQ} are parallel and their dot product is 9, we find that QQ is the inversion of PP with respect to a circle centered at A(2,0)A(-2, 0) with radius 3, followed by a translation by (2,0)(-2, 0).
  2. Region R\mathcal{R}:

    • The region R\mathcal{R} consists of two arcs: one from circle C1C_1 (unit circle centered at (0,0)(0, 0)) and another from circle C2C_2 (centered at (1,0)(-1, 0) with radius 1). These arcs intersect at points R1R_1 and R2R_2.
  3. Image of R\mathcal{R} under Inversion:

    • The image of the arc on C1C_1 under inversion is a circular arc of a circle centered at (4,0)(4, 0) with radius 3, subtending an angle of 240240^\circ. The length of this arc is 240360×2π×3=4π\frac{240}{360} \times 2\pi \times 3 = 4\pi.
    • The image of the arc on C2C_2 under inversion is a vertical line segment from (52,332)\left( \frac{5}{2}, \frac{3\sqrt{3}}{2} \right) to (52,332)\left( \frac{5}{2}, -\frac{3\sqrt{3}}{2} \right). The length of this segment is 333\sqrt{3}.
  4. Total Length kk:

    • Summing the lengths of the two images gives k=4π+33k = 4\pi + 3\sqrt{3}.
  5. Numerical Calculation:

    • Using Python, we calculate 4π12.5664\pi \approx 12.566 and 335.1963\sqrt{3} \approx 5.196. Their sum is approximately 17.76217.762, so the floor value is 17.

[PYTHON] import math

Calculate the total length k

k = 4 * math.pi + 3 * math.sqrt(3) print(math.floor(k)) [/PYTHON]

Tool Output
[PYTHON OUTPUT] 17 [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] The Python code calculates k=4π+3317.762k = 4\pi + 3\sqrt{3} \approx 17.762, and the floor value is 17. This matches the observation, so the final answer is 17\boxed{17}. [/THOUGHT] 17\boxed{17}

Final Model Output
17

Comparative Analysis (P29) (Average across runs)

Avg Time Taken

K-EXAONE-236B-A23B
946.4
EXAONE-4.0.1-32B (high)
298.6
GPT-5.1 (high)
227.4
Gemini-3-Pro-Preview
139.0
Solar-Open-100B
118.5
Deepseek-V3.2
109.1
Kanana-2-30B-Thinking-2601
95.8
Llama-VARCO-8B-Instruct
74.3
Kanana-2-30B-Thinking
73.3
K-EXAONE-236B-A23B
72.2
Solar-Pro-2 (31B)(high)
67.2
GPT-5.2 (high)
64.2
Grok-4.1-fast
38.9
Claude-Opus-4.5
38.3
HCX-007(high)
28.4
A.X-4.0 (72B)
21.1

Avg Token Usage

K-EXAONE-236B-A23B
79,802
Solar-Open-100B
19,908
Kanana-2-30B-Thinking-2601
19,089
GPT-5.1 (high)
14,833
Kanana-2-30B-Thinking
12,633
Claude-Opus-4.5
11,516
K-EXAONE-236B-A23B
10,828
Solar-Pro-2 (31B)(high)
10,423
Grok-4.1-fast
9,673
Deepseek-V3.2
8,977
Gemini-3-Pro-Preview
8,560
GPT-5.2 (high)
8,089
EXAONE-4.0.1-32B (high)
6,140
Llama-VARCO-8B-Instruct
5,349
HCX-007(high)
4,304
A.X-4.0 (72B)
2,252