AMC 12B 2025 (Problem 25)Three concentric circles have radii 111, 222, 333. An equilateral triangle with side length sss has one vertex on each circle. What is s2s^2s2?(A) 6\text{(A)}\;6(A)6(B) 254\text{(B)}\;\frac{25}{4}(B)425(C) 132\text{(C)}\;\frac{13}{2}(C)213(D) 274\text{(D)}\;\frac{27}{4}(D)427(E) 7\text{(E)}\;7(E)7Related TopicsCoreToolkit 59 — Ptolemy's TheoremMajorToolkit 58 — Law of CosinesMinorToolkit 61 — 4 Rules of Cyclic QuadrilateralsHints (4)Hint 1Hint 23s=OB⋅AC=OA⋅BC+OC⋅AB=2s+s=3s3s=OB\cdot AC=OA\cdot BC+OC\cdot AB=2s+s=3s3s=OB⋅AC=OA⋅BC+OC⋅AB=2s+s=3sBy Toolkit 59 — Ptolemy's Theorem,OABC is cyclic.OABC\text{ is cyclic.}OABC is cyclic.Hint 3By Toolkit 61 — 4 Rules of Cyclic Quadrilaterals,∠BOC=∠BAC=60∘\angle BOC=\angle BAC=60^\circ∠BOC=∠BAC=60∘Hint 4By Toolkit 58 — Law of Cosines in △BOC\triangle BOC△BOC:s2=32+12−2(3)(1)cos60∘s^2=3^2+1^2-2(3)(1)\cos60^\circs2=32+12−2(3)(1)cos60∘=10−6(12)=7=10-6\left(\frac12\right)=7=10−6(21)=7Final Answer(E) 777Related Problems (5)AMC 12A 2024 (Problem 19)AMC 12A 2022 (Problem 12)AMC 12A 2025 (Problem 8)AMC 12A 2025 (Problem 24)AMC 12B 2025 (Problem 21)
