AMC 12A 2025 (Problem 8)Pentagon ABCDEABCDEABCDE is inscribed in a circle, and ∠BEC=∠CED=30∘\angle BEC=\angle CED=30^\circ∠BEC=∠CED=30∘. Let line ACACAC and line BDBDBD intersect at point FFF, and suppose that AB=9AB=9AB=9 and AD=24AD=24AD=24. What is BFBFBF?(A) 5711\text{(A)}\;\frac{57}{11}(A)1157(B) 5911\text{(B)}\;\frac{59}{11}(B)1159(C) 6011\text{(C)}\;\frac{60}{11}(C)1160(D) 6111\text{(D)}\;\frac{61}{11}(D)1161(E) 6311\text{(E)}\;\frac{63}{11}(E)1163Related TopicsCoreToolkit 42 — Angle Bisector TheoremToolkit 58 — Law of CosinesMajorToolkit 61 — 4 Rules of Cyclic QuadrilateralsMinorToolkit 57 — Famous Trigonometric ValuesToolkit 96 — Ratio ManipulationHints (7)Hint 1Find ∠DAC\angle DAC∠DAC and ∠CAB\angle CAB∠CAB.Hint 2By Toolkit 61 — 4 Rules of Cyclic Quadrilaterals:∠DAC=∠DEC=30∘\angle DAC=\angle DEC=30^\circ∠DAC=∠DEC=30∘∠CAB=∠CEB=30∘\angle CAB=\angle CEB=30^\circ∠CAB=∠CEB=30∘Hint 3Find BFFD\frac{BF}{FD}FDBF.Hint 4By Toolkit 42 — Angle Bisector Theorem, Angle Bisector Theorem in △ABD\triangle ABD△ABD:BFDF=ABAD=924=38\frac{BF}{DF}=\frac{AB}{AD}=\frac{9}{24}=\frac{3}{8}DFBF=ADAB=249=83Hint 5Find BDBDBD.Hint 6By Toolkit 58 — Law of Cosines, Law of Cosines in △ABD\triangle ABD△ABD:BD2=AB2+AD2−2(AB)(AD)cos60∘BD^2=AB^2+AD^2-2(AB)(AD)\cos60^\circBD2=AB2+AD2−2(AB)(AD)cos60∘=92+242−2(9)(24)(12)=9^2+24^2-2(9)(24)\left(\frac12\right)=92+242−2(9)(24)(21)=81+576−216=441=81+576-216=441=81+576−216=441⟹BD=21\Longrightarrow BD=21⟹BD=21Hint 7From Hints 4 and 6:BFDF=38\frac{BF}{DF}=\frac38DFBF=83⟹BFBF+DF=33+8\Longrightarrow \frac{BF}{BF+DF}=\frac{3}{3+8}⟹BF+DFBF=3+83⟹BFBD=311\Longrightarrow \frac{BF}{BD}=\frac{3}{11}⟹BDBF=113⟹BF=21(311)=6311\Longrightarrow BF=21\left(\frac{3}{11}\right)=\frac{63}{11}⟹BF=21(113)=1163Final Answer(E) 6311\frac{63}{11}1163Related Problems (3)AMC 12A 2024 (Problem 19)AMC 10A/12A 2025 (Problem 23/16)AMC 12A 2022 (Problem 12)
