AMC 10A 2025 (Problem 22)A circle of radius rrr is surrounded by three circles, whose radii are 111, 222, and 333, all externally tangent to the inner circle and externally tangent to each other, as shown below. What is rrr?(A) 14\text{(A)}\;\frac{1}{4}(A)41(B) 623\text{(B)}\;\frac{6}{23}(B)236(C) 311\text{(C)}\;\frac{3}{11}(C)113(D) 517\text{(D)}\;\frac{5}{17}(D)175(E) 310\text{(E)}\;\frac{3}{10}(E)103Related TopicsCoreToolkit 95 — Coordinate GeometryMinorToolkit 81 — SubstitutionToolkit 21 — Quadratic FormulaHints (7)Hint 1Hint 2Use coordinate geometry.Hint 3Let O1(0,0)O_1(0,0)O1(0,0), O2(3,0)O_2(3,0)O2(3,0), O3(4,0)O_3(4,0)O3(4,0), and O(x,y)O(x,y)O(x,y).OO12=x2+y2=(1+r)2=1+r2+2r(1)OO_1^2=x^2+y^2=(1+r)^2=1+r^2+2r \qquad (1)OO12=x2+y2=(1+r)2=1+r2+2r(1)OO22=x2+(y−3)2=(2+r)2=4+r2+4r(2)OO_2^2=x^2+(y-3)^2=(2+r)^2=4+r^2+4r \qquad (2)OO22=x2+(y−3)2=(2+r)2=4+r2+4r(2)OO32=(x−4)2+y2=(3+r)2=9+r2+6r(3)OO_3^2=(x-4)^2+y^2=(3+r)^2=9+r^2+6r \qquad (3)OO32=(x−4)2+y2=(3+r)2=9+r2+6r(3)Hint 4From (1) and (2):(y−3)2−y2=2r+3(y-3)^2-y^2=2r+3(y−3)2−y2=2r+39−6y=2r+39-6y=2r+39−6y=2r+3y=3−r3(4)y=\frac{3-r}{3} \qquad (4)y=33−r(4)Hint 5From (1) and (3):(x−4)2−x2=4r+8(x-4)^2-x^2=4r+8(x−4)2−x2=4r+816−8x=4r+816-8x=4r+816−8x=4r+8x=2−r2(5)x=\frac{2-r}{2} \qquad (5)x=22−r(5)Hint 6From (1), (4), and (5):(3−r3)2+(2−r2)2=1+r2+2r\left(\frac{3-r}{3}\right)^2+\left(\frac{2-r}{2}\right)^2=1+r^2+2r(33−r)2+(22−r)2=1+r2+2r⇒4(3−r)2+9(2−r)2=36(1+r2+2r)\Rightarrow 4(3-r)^2+9(2-r)^2=36(1+r^2+2r)⇒4(3−r)2+9(2−r)2=36(1+r2+2r)⇒4(9−6r+r2)+9(4−4r+r2)=36(1+r2+2r)\Rightarrow 4(9-6r+r^2)+9(4-4r+r^2)=36(1+r^2+2r)⇒4(9−6r+r2)+9(4−4r+r2)=36(1+r2+2r)⇒23r2+132r−36=0\Rightarrow 23r^2+132r-36=0⇒23r2+132r−36=0Hint 7Use Quadratic Formula:r=−132±1322−4(23)(−36)46r=\frac{-132\pm\sqrt{132^2-4(23)(-36)}}{46}r=46−132±1322−4(23)(−36)=−66±14423=\frac{-66\pm\sqrt{144}}{23}=23−66±144=−66+12123=\frac{-66+12\sqrt{1}}{23}=23−66+121=623=\frac{6}{23}=236Final Answer(B) 623\frac{6}{23}236Related Problems (1)AMC 10A 2025 (Problem 20)
