Problem Analysis #geometry_26

Motif 3 · Run 1

Python ● execution succeeded
Correct
50.26s
11,711 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

양수 a에 대하여 두 초점이 F, F′인 쌍곡선 x2a2−y2a2=−1\frac{x^2}{a^2}-\frac{y^2}{a^2}=-1 위의 점 (a,2a)(a,\sqrt2a)에서의 접선이 y축과 만나는 점을 P라 하자. PF‾×PF′‾=8\overline{PF}\times\overline{PF\prime}=8일 때, a의 값은?

① 3\sqrt3 ② 433\frac{4\sqrt3}3 ③ 533\frac{5\sqrt3}3 ④ 232\sqrt3 ⑤ 733\frac{7\sqrt3}3

최종 답은 선택지 번호 1~5 중 하나를 \boxed{} 안에 쓰세요.

Ground Truth
2
Recorded model response / solution
Reasoning Process

Recorded Python call:

import sympy as sp
a = sp.symbols('a', positive=True)
c = a*sp.sqrt(2)
P = a/sp.sqrt(2)
PF = c - P
PFp = c + P
prod = sp.simplify(PF*PFp)
sp.solve(sp.Eq(prod, 8), a)
Tool Output
Execution succeeded
Reasoning Process

Recorded Python call:

prod
Tool Output
Execution failed name 'prod' is not defined
Reasoning Process

Recorded Python call:

import sympy as sp
a = sp.symbols('a', positive=True)
c = a*sp.sqrt(2)
P = a/sp.sqrt(2)
PF = c - P
PFp = c + P
prod = sp.simplify(PF*PFp)
prod, sp.solve(sp.Eq(prod, 8), a)
Tool Output
Execution succeeded
Reasoning Process

Recorded Python call:

import sympy as sp
a = sp.symbols('a', positive=True)
c = a*sp.sqrt(2)
P = a/sp.sqrt(2)
PF = c - P
PFp = c + P
prod = sp.simplify(PF*PFp)
print("prod:", prod)
sol = sp.solve(sp.Eq(prod, 8), a)
print("sol:", sol)
Tool Output
Execution succeeded prod: 3*a**2/2 sol: [4*sqrt(3)/3]
Reasoning Process

쌍곡선 x2a2−y2a2=−1\frac{x^2}{a^2}-\frac{y^2}{a^2}=-1는 y2−x2=a2y^2-x^2=a^2이므로, 초점은 F(0,2a), F′(0,−2a)F(0,\sqrt2 a),\ F'(0,-\sqrt2 a)이다.

점 (a,2a)(a,\sqrt2 a)에서의 접선을 구하면,

xaa2−y(2a)a2=−1\frac{x a}{a^2}-\frac{y(\sqrt2 a)}{a^2}=-1

이므로

x−2y=−a.x-\sqrt2 y=-a.

x=0x=0에서 yy-축과의 교점은

−2y=−a⇒y=a2.-\sqrt2 y=-a \Rightarrow y=\frac{a}{\sqrt2}.

따라서 P=(0,a2)P=(0,\frac{a}{\sqrt2})이다.

이제

PF=2a−a2=a2,PF′=2a+a2=3a2.PF=\sqrt2 a-\frac{a}{\sqrt2}=\frac{a}{\sqrt2},\qquad PF'=\sqrt2 a+\frac{a}{\sqrt2}=\frac{3a}{\sqrt2}.

그러므로

PF⋅PF′=a2⋅3a2=3a22.PF\cdot PF'=\frac{a}{\sqrt2}\cdot\frac{3a}{\sqrt2}=\frac{3a^2}{2}.

주어진 조건 3a22=8\frac{3a^2}{2}=8에서

a2=163⇒a=433.a^2=\frac{16}{3}\Rightarrow a=\frac{4\sqrt3}{3}.

이는 선택지 ②에 해당한다.

2\boxed{2}

Final Model Output
2

Comparative Analysis (Pgeometry_26) (Average across runs)

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

Avg Time Taken

Motif 3
56.1
Solar Pro 4
48.1
Grok 4.7
28.7
MiMo V2.6 Pro
25.4
Union Alpha
18.2
Muse Spark 1.3
11.2
DeepSeek V4.1 Flash
8.0
Solar Mini 4 Preview
4.7

Avg Token Usage

Motif 3
13,269
Grok 4.7
8,214
Solar Pro 4
4,897
Muse Spark 1.3
4,791
DeepSeek V4.1 Flash
4,723
MiMo V2.6 Pro
3,849
Union Alpha
3,358
Solar Mini 4 Preview
2,842
    CSAT 2026 · Problem geometry_26 · Motif 3 | EntropyMath