Problem Analysis #common_18

Motif 3 · Run 1

Python ● execution succeeded
Correct
7.47s
2,861 toks

Infron · Motif 3 · 46문항 × 3회 · Python TAR. 131,072 output tokens · 기본 1,800초 · 6+1 turns. ACC는 전체 138회, Pass@1은 각 문항의 첫 유효 시도, Pass@3는 3회 내 정답으로 계산합니다.

Problem Statement

AB‾=5\overline{AB}=5, AC‾=6\overline{AC}=6이고 cos⁡(∠BAC)=−35\cos(\angle BAC)=-\frac35인 삼각형 ABC의 넓이를 구하시오.

최종 정수 답을 \boxed{} 안에 쓰세요.

Ground Truth
12
Recorded model response / solution
Reasoning Process

Recorded Python call:

import math
AB=5
AC=6
cosA=-3/5
sinA=math.sqrt(1-cosA**2)
area=0.5*AB*AC*sinA
area
Tool Output
Execution succeeded
Reasoning Process

Recorded Python call:

import math
AB=5; AC=6; cosA=-3/5
sinA=math.sqrt(1-cosA**2)
area=0.5*AB*AC*sinA
print(sinA, area)
Tool Output
Execution succeeded 0.8 12.0
Reasoning Process
sin⁡(∠BAC)=1−cos⁡2(∠BAC)=1−(−35)2=1625=45\sin(\angle BAC)=\sqrt{1-\cos^2(\angle BAC)} =\sqrt{1-\left(-\frac35\right)^2} =\sqrt{\frac{16}{25}}=\frac45

삼각형 ABCABC의 넓이는

12⋅AB⋅AC⋅sin⁡(∠BAC)=12⋅5⋅6⋅45=12\frac12 \cdot AB \cdot AC \cdot \sin(\angle BAC) = \frac12 \cdot 5 \cdot 6 \cdot \frac45 = 12

따라서 삼각형 ABCABC의 넓이는

12\boxed{12}
Final Model Output
12

Comparative Analysis (Pcommon_18) (Average across runs)

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

Avg Time Taken

Union Alpha
96.0
Solar Pro 4
28.7
Grok 4.7
10.9
MiMo V2.6 Pro
8.5
Motif 3
7.8
Muse Spark 1.3
5.6
DeepSeek V4.1 Flash
4.1
Solar Mini 4 Preview
1.5

Avg Token Usage

Grok 4.7
6,949
Solar Pro 4
3,872
Muse Spark 1.3
3,192
Union Alpha
2,841
Motif 3
2,286
DeepSeek V4.1 Flash
2,236
Solar Mini 4 Preview
1,800
MiMo V2.6 Pro
976
    CSAT 2026 · Problem common_18 · Motif 3 | EntropyMath