Problem Analysis #2

K-EXAONE-236B-A23B · Run 1

Correct
8.43s
1,463 toks

API · no tools. This is the prompt and response recorded for this attempt; a full solution may not have been returned.

Problem Statement

Return a single integer only. The problem has been revised so that no final answer is an irrational number, a parameterized expression, a range, a proof, a drawing, a graph, a locus equation, or a region description.

In the setting of the original midpoint-angle problem, the condition q=1/3q=1/3 leads to a positive value pp satisfying

p=322.p=3-2\sqrt2.

Instead of giving this irrational value, compute the integer

N=p+1p.N=p+\frac1p.
Ground Truth
6
Recorded model response / solution
Reasoning Process

To compute the integer value of

N=p+1pN = p + \frac{1}{p}

where

p=322,p = 3 - 2\sqrt{2},

we proceed as follows.


Step 1: Compute the reciprocal $$\frac{1}{p}$$

Given:

p=322p = 3 - 2\sqrt{2}

We rationalize $$\frac{1}{p}$$ by multiplying numerator and denominator by the conjugate of the denominator:

1322=13223+223+22=3+22(3)2(22)2\frac{1}{3 - 2\sqrt{2}} = \frac{1}{3 - 2\sqrt{2}} \cdot \frac{3 + 2\sqrt{2}}{3 + 2\sqrt{2}} = \frac{3 + 2\sqrt{2}}{(3)^2 - (2\sqrt{2})^2}

Now calculate the denominator:

(3)2(22)2=942=98=1(3)^2 - (2\sqrt{2})^2 = 9 - 4 \cdot 2 = 9 - 8 = 1

So,

1p=3+22\frac{1}{p} = 3 + 2\sqrt{2}

Step 2: Add pp and $$\frac{1}{p}$$

N=p+1p=(322)+(3+22)N = p + \frac{1}{p} = (3 - 2\sqrt{2}) + (3 + 2\sqrt{2})

Grouping terms:

N=(3+3)+(22+22)=6+0=6N = (3 + 3) + (-2\sqrt{2} + 2\sqrt{2}) = 6 + 0 = 6

Final Answer

6\boxed{6}
Final Model Output
6

Comparative Analysis (P2) (Average across runs)

Averages use reported measurements only. Missing usage is not treated as zero.

Avg Time Taken

DeepSeek V4 Pro
10.2
K-EXAONE-236B-A23B
8.7
Gemini 3.5 Flash
4.3
GPT-5.5
4.0
Solar Pro 3
2.5
Claude Opus 4.8
2.4
KT Mi:dm 2.0 Base Instruct
1.0

Avg Token Usage

K-EXAONE-236B-A23B
1484.7
KT Mi:dm 2.0 Base Instruct
969.0
Gemini 3.5 Flash
892.3
Solar Pro 3
511.7
DeepSeek V4 Pro
400.0
Claude Opus 4.8
348.0
GPT-5.5
294.7
    Tokyo · Problem 2 · K-EXAONE-236B-A23B | EntropyMath