2023 AMC 12A Problem 18

Attempt Problem 18 of the 2023 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 12A solutions, or check the answer key.

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18.

Circle C1C_1 and C2C_2 each have radius 1,1, and the distance between their centers is 12.\tfrac12. Circle C3C_3 is the largest circle internally tangent to both C1C_1 and C2.C_2. Circle C4C_4 is internally tangent to both C1C_1 and C2C_2 and externally tangent to C3.C_3. What is the radius of C4?C_4?

114\dfrac{1}{14}

112\dfrac{1}{12}

110\dfrac{1}{10}

328\dfrac{3}{28}

19\dfrac{1}{9}

Answer: D
Concepts:tangent circlescoordinate geometry
Difficulty rating: 1990
Solution:

Put the centers at O1=(14,0)O_1=\left(-\tfrac14,0\right) and O2=(14,0).O_2=\left(\tfrac14,0\right). By symmetry C3C_3 is centered at the origin, and internal tangency to C1C_1 gives radius 114=34.1-\tfrac14=\tfrac34.

Let C4C_4 have radius r,r, centered at (0,k)(0,k) on the axis of symmetry. External tangency to C3C_3 gives k=34+r,k=\tfrac34+r, and internal tangency to C1C_1 gives 116+k2=1r.\sqrt{\tfrac{1}{16}+k^2}=1-r.

Substituting, 116+(34+r)2=(1r)2,\tfrac{1}{16}+\left(\tfrac34+r\right)^2=(1-r)^2, which simplifies to 72r=38,\tfrac72 r=\tfrac38, so r=328.r=\dfrac{3}{28}.

Thus, the correct answer is D.

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