Problem Analysis #common_14

Recorded Python call:

import sympy as sp
import numpy as np
import math

r = sp.Rational(2)
C = sp.Point(3, 4)
# D divides AB with AD:DB = 2:1 -> AD = 2, D=(2,0); circle center A radius 2
E = sp.Point(sp.Rational(6,5), sp.Rational(8,5))   # E on AC, AE = 2
F = sp.Point(-2, 0)
print("AD =", math.dist((0,0),(2,0)), " DB =", math.dist((2,0),(3,0)))
print("AE =", float(sp.sqrt(E.x**2+E.y**2)), " (E on segment AC)")

# G on circle x^2+y^2=4 with CG=2√6: subtracting gives 6x+8y=5
yv = sp.symbols('yv', real=True)
xexpr = (sp.Rational(5) - 8*yv)/6
sols = sp.solve(sp.Eq(xexpr**2 + yv**2, 4), yv)
angE = math.degrees(math.atan2(1.6, 1.2))
print("angle of E (deg):", angE, " F at 180 deg")
for s in sols:
    xv = sp.simplify(xexpr.subs(yv, s)); yy = sp.simplify(s)
    ang = math.degrees(math.atan2(float(yy), float(xv)))
    print("  candidate: x =", xv, ", y =", yy, ", angle =", round(ang,4),
          ", |CG| =", round(math.dist((float(xv),float(yy)),(3,4)),6))

yG = (4 + 3*sp.sqrt(15))/10
xG = sp.simplify((sp.Rational(5) - 8*yG)/6)
G = sp.Point(xG, yG)
print("chosen G =", G, " angle(deg) =", round(math.degrees(math.atan2(float(yG), float(xG))),4))
print("CG^2 =", sp.simplify((G.x-3)**2 + (G.y-4)**2), " G on circle A:", sp.simplify(G.x**2+G.y**2))

circ = sp.Circle(C, E, G)
O = circ.center; R = sp.simplify(circ.radius)
print("circle CEG: O =", sp.simplify(O.x), sp.simplify(O.y), " R =", R, " R^2 =", sp.simplify(R**2))

# independent radius check: R = |CE||EG||CG| / (4*area)
EG = sp.sqrt(sp.simplify((G.x-E.x)**2 + (G.y-E.y)**2))
cr = sp.simplify((E.x-C.x)*(G.y-C.y) - (E.y-C.y)*(G.x-C.x))
area = abs(cr)/2
Rind = sp.simplify(3*EG*2*sp.sqrt(6)/(4*area))
print("CE =", sp.simplify(sp.sqrt((E.x-3)**2+(E.y-4)**2)), " EG =", EG, " CG =", 2*sp.sqrt(6),
      " area =", sp.simplify(area), " R(independent) =", sp.simplify(Rind))

sinBAC = sp.Rational(4,5)
GH = sp.simplify(2*R*sinBAC)
print("GH = 2R·sin∠BAC =", GH, "≈", float(GH))
choices = [6*sp.sqrt(15)/5, 38*sp.sqrt(10)/25, 14*sp.sqrt(3)/5, 32*sp.sqrt(15)/25, 8*sp.sqrt(10)/5]
for i, ch in enumerate(choices, start=1):
    print(f"  choice {i}: {ch} ≈ {float(ch):.6f}   equals GH: {sp.simplify(ch-GH)==0}")

# numeric location of H on the circle with ∠HCG = ∠BAC, and GH measurement
Of = (float(O.x), float(O.y)); Rf = float(R)
Cf = np.array([3.0, 4.0]); Gf = np.array([float(G.x), float(G.y)])
theta = math.acos(0.6)
phis = np.linspace(0, 2*np.pi, 4000001)
Hs = np.stack([Of[0] + Rf*np.cos(phis), Of[1] + Rf*np.sin(phis)], axis=1)
v1 = Hs - Cf; v2 = Gf - Cf
cosang = (v1 @ v2)/(np.linalg.norm(v1, axis=1)*np.linalg.norm(v2))
angs = np.arccos(np.clip(cosang, -1, 1))
idx = np.argmin(np.abs(angs - theta))
phi = phis[idx]; H = (Of[0] + Rf*math.cos(phi), Of[1] + Rf*math.sin(phi))
print("numeric H ≈", (round(H[0],6), round(H[1],6)), " phi(deg) =", round(math.degrees(phi),4))
print("∠HCG =", round(math.degrees(angs[idx]),6), "target =", round(math.degrees(theta),6))
print("GH numeric =", round(math.dist(H,(float(G.x),float(G.y))),6),
      "  2R sinθ =", round(2*Rf*math.sin(theta),6))
def ang(P): return math.atan2(P[1]-Of[1], P[0]-Of[0]) % (2*math.pi)
aC, aEo, aGc, aH = ang((3.0,4.0)), ang((float(E.x),float(E.y))), ang((float(G.x),float(G.y))), ang(H)
print("center-angles(deg): C=%.3f E=%.3f G=%.3f H=%.3f" %
      (math.degrees(aC), math.degrees(aEo), math.degrees(aGc), math.degrees(aH)))
in_minor = min(aGc,aC) < aEo < max(aGc,aC)
print("E on minor arc CG:", in_minor, " -> H on arc CG not containing E:",
      not (min(aGc,aC) < aH < max(aGc,aC)) if in_minor else (min(aGc,aC) < aH < max(aGc,aC)))

설정(좌표): A=(0,0), B=(3,0), C=(3,4). AD:DB=2:1이므로 AD=2, D=(2,0) → 원 A: x²+y²=4.

점 E: AC(단위벡터 (3/5, 4/5)) 위에서 AE=2 → E=(6/5, 8/5). (검산: AE=2, CE=3, E는 선분 AC 내부 ✓)

점 G: 원 A 위에서 CG=2√6 → (x−3)²+(y−4)²=24, x²+y²=4를 빼면 6x+8y=5. 두 후보 중 작은 호 EF(각도 53°~180°)에 있는 점은 G=(3/10−2√15/5, 2/5+3√15/10) (편각 128.65°). 검산: CG²=24, G는 원 위 ✓

원 CEG의 반지름: 세 변 CE=3, EG=√6, CG=2√6, 면적 = ½·CE·CG·cos∠ECG = ½·3·2√6·(4/5) = 3√15/4이므로

(심파이 원 계산 O=(21/10−14√15/25, 14/5+21√15/50), R=4√15/5와 일치, 독립 재계산도 일치 ✓)

GH: H가 원 CEG 위에 있으므로 ∠CHG = ∠CEG = ∠BAC (동호). 사인법칙으로

수치 검증: 원 CEG 위에서 ∠HCG=∠BAC=53.13°를 만족하는 H≈(−2.4885, 6.3619)을 수치 탐색으로 찾았고, 중심각 확인 결과 E는 작은 호 CG 위, H는 E를 포함하지 않는 호 CG 위(그림 조건 일치)이며 GH≈4.9574 = 2R sinθ와 일치. 이 값은 ④ $\frac{32\sqrt{15}}{25}$≈4.9574와 정확히 같고 나머지 선택지와는 다름 ✓

Recorded Python call:

import sympy as sp
# final check: GH = 32√15/25 equals choice 4, others differ
GH = sp.Rational(32,25)*sp.sqrt(15)
choices = {1: 6*sp.sqrt(15)/5, 2: 38*sp.sqrt(10)/25, 3: 14*sp.sqrt(3)/5,
           4: 32*sp.sqrt(15)/25, 5: 8*sp.sqrt(10)/5}
matches = [k for k,v in choices.items() if sp.simplify(v-GH)==0]
print("GH =", GH, "-> matching choice index:", matches)

답: 선택지 ④

$\boxed{4}$