2023 AMC 10A Problem 22

Attempt Problem 22 of the 2023 AMC 10A 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 10A solutions, or check the answer key.

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22.

Circle C1C_1 and C2C_2 each have radius 1,1, and the distance between their centers is 12.\frac{1}{2}. 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 geometrysymmetry
Difficulty rating: 2270
Solution:

Put the centers of C1,C2C_1, C_2 at (±14,0).\left(\pm\frac14, 0\right). By symmetry the largest circle inside both sits at the origin with radius r3,r_3, where 1r3=14,1 - r_3 = \frac14, so r3=34.r_3 = \frac34. Let C4C_4 be centered at (0,y)(0, y) with radius r.r. Internal tangency to C1C_1 gives 116+y2=1r,\sqrt{\frac1{16} + y^2} = 1 - r, and external tangency to C3C_3 gives y=34+r.y = \frac34 + r. Substitute the second into the first: 116+(34+r)2=(1r)2.\frac1{16} + \left(\frac34 + r\right)^2 = (1 - r)^2. This collapses to 72r=38,\frac72 r = \frac38, so r=328.r = \frac{3}{28}. Therefore, the answer is D.

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