2011 AMC 12A Problem 17

Attempt Problem 17 of the 2011 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 2011 AMC 12A solutions, or check the answer key.

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17.

Circles with radii 1,1, 2,2, and 33 are mutually externally tangent. What is the area of the triangle determined by the points of tangency?

35\dfrac{3}{5}

45\dfrac{4}{5}

11

65\dfrac{6}{5}

43\dfrac{4}{3}

Answer: D
Concepts:tangent circlescoordinate geometryshoelace formula
Difficulty rating: 1920
Solution:

The centers are separated by the sums of radii: 3,3, 4,4, and 5,5, a right triangle with the right angle at the radius-11 center. Place that center at (0,0),(0,0), the radius-22 center at (3,0),(3,0), and the radius-33 center at (0,4).(0,4).

The tangency points lie on the segments at distances equal to the radii: (1,0),(1, 0), (0,1),(0, 1), and on the hypotenuse at (3,0)+2(3,4)5=(95,85).(3,0) + 2 \cdot \tfrac{(-3,4)}{5} = \left(\tfrac95, \tfrac85\right).

By the shoelace formula the area is 121(185)+0+95(01)=12125=65. \begin{gathered} \tfrac12\left| 1\left(1 - \tfrac85\right) + 0 + \tfrac95(0 - 1) \right| \\ = \tfrac12 \cdot \tfrac{12}{5} = \tfrac65. \end{gathered}

Thus, the correct answer is D.

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