AMC 12A 2025 (Problem 24)A circle of radius rrr is surrounded by 121212 circles of radius 111, externally tangent to the central circle and sequentially tangent to each other, as shown. Then rrr can be written as a+b+c\sqrt{a}+\sqrt{b}+ca+b+c, where a,b,ca,b,ca,b,c are integers. What is a+b+ca+b+ca+b+c?(A) 3\text{(A)}\;3(A)3(B) 5\text{(B)}\;5(B)5(C) 7\text{(C)}\;7(C)7(D) 9\text{(D)}\;9(D)9(E) 11\text{(E)}\;11(E)11Related TopicsCoreToolkit 68 — Circles, Tangents, and Radical AxisMajorToolkit 58 — Law of CosinesHints (4)Hint 1360∘12=30∘\frac{360^\circ}{12}=30^\circ12360∘=30∘Hint 2By Toolkit 58 — Law of Cosines, Law of Cosines in △OO1O2\triangle OO_1O_2△OO1O2:22=(r+1)2+(r+1)2−2(r+1)(r+1)cos30∘2^2=(r+1)^2+(r+1)^2-2(r+1)(r+1)\cos30^\circ22=(r+1)2+(r+1)2−2(r+1)(r+1)cos30∘4=(r+1)2(2−3)4=(r+1)^2(2-\sqrt3)4=(r+1)2(2−3)(r+1)2=42−3⋅2+32+3=4(2+3)(1)(r+1)^2=\frac{4}{2-\sqrt3}\cdot\frac{2+\sqrt3}{2+\sqrt3}=4(2+\sqrt3)\qquad(1)(r+1)2=2−34⋅2+32+3=4(2+3)(1)Hint 3(1+3)2=1+23+3=4+23(2)(1+\sqrt3)^2=1+2\sqrt3+3=4+2\sqrt3\qquad(2)(1+3)2=1+23+3=4+23(2)Hint 4From (1)(1)(1) and (2)(2)(2):(r+1)2=4(1+32)2=2(1+3)2(r+1)^2=4\left(\frac{1+\sqrt3}{\sqrt2}\right)^2=2(1+\sqrt3)^2(r+1)2=4(21+3)2=2(1+3)2⇒r+1=2(1+3)=2+6\Rightarrow r+1=\sqrt2(1+\sqrt3)=\sqrt2+\sqrt6⇒r+1=2(1+3)=2+6r=2+6−1=a+b+cr=\sqrt2+\sqrt6-1=\sqrt a+\sqrt b+cr=2+6−1=a+b+ca+b+c=2+6−1=7a+b+c=2+6-1=7a+b+c=2+6−1=7Final Answer(C) 777Related Problems (7)AMC 10A/12A 2023 (Problem 18/22)AMC 8 2026 (Problem 23)AMC 10A 2025 (Problem 10)AMC 12A 2024 (Problem 19)AIME I 2025 (Problem 6)AMC 12A 2022 (Problem 12)AMC 12A 2025 (Problem 8)
