2004 AMC 12A Problem 19

Attempt Problem 19 of the 2004 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 2004 AMC 12A solutions, or check the answer key.

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19.

Circles A,A, B,B, and CC are externally tangent to each other and internally tangent to circle D.D. Circles BB and CC are congruent. Circle AA has radius 11 and passes through the center of D.D. What is the radius of circle B?B?

23\dfrac{2}{3}

32\dfrac{\sqrt{3}}{2}

78\dfrac{7}{8}

89\dfrac{8}{9}

1+33\dfrac{1 + \sqrt{3}}{3}

Answer: D
Concepts:tangent circlesPythagorean Theoremsystem of equations
Difficulty rating: 1920
Solution:

Circle AA has radius 11 and passes through the center of DD while being internally tangent to D,D, so DD has radius 2.2.

Place the center of DD at the origin, with AA centered at (1,0).(-1, 0). By symmetry, BB has center (x,y)(x, y) and radius r,r, with CC its mirror image across the horizontal axis, so the two congruent circles touch on that axis and y=r.y = r.

Internal tangency to DD gives x2+y2=(2r)2,x^2 + y^2 = (2 - r)^2, and external tangency to AA gives (x+1)2+y2=(1+r)2.(x + 1)^2 + y^2 = (1 + r)^2.

Subtracting and using y=ry = r yields x=23x = \tfrac23 and r=89.r = \tfrac89. The radius of circle BB is 89.\tfrac89.

Thus, the correct answer is D.

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