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 TopicsCoreCircle TangencyMajorLaw of CosinesCheck AnswerYour answer:ABCDECheckHints (4)Hint 1360∘12=30∘\frac{360^\circ}{12}=30^\circ12360∘=30∘Hint 2By 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 (14)AMC 8 2026 (Problem 23)AMC 10A 2025 (Problem 10)AMC 10B 2025 (Problem 12)AMC 10B 2025 (Problem 20)AMC 12A 2024 (Problem 19)AIME I 2025 (Problem 6)AMC 12A 2022 (Problem 12)AMC 10A 2025 (Problem 22)AMC 12A 2025 (Problem 8)AIME I 2026 (Problem 3)AIME I 2026 (Problem 5)AMC 12B 2025 (Problem 21)AMC 12B 2025 (Problem 25)AIME I 2026 (Problem 14)View all related problems →
