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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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
Small hint:
Put the centers of C1,C2C_1, C_2 at (±14,0);\left(\pm\frac14, 0\right); C3C_3 is centered at the origin with radius from 1−r3=141 - r_3 = \frac14
Big hint:
Let C4C_4 be centered at (0,y)(0, y) with radius r;r; internal tangency to C1C_1 and external tangency to C3C_3 give two equations
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 1−r3=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=1−r,\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=(1−r)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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