2007 AMC 10A Problem 24

Attempt Problem 24 of the 2007 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 2007 AMC 10A solutions, or check the answer key.

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24.

Circles centered at AA and BB each have radius 2,2, as shown. Point OO is the midpoint of AB,\overline{AB}, and OA=22.OA = 2\sqrt{2}. Segments OCOC and ODOD are tangent to the circles centered at AA and B,B, respectively, and EF\overline{EF} is a common tangent. What is the area of the shaded region ECODF?ECODF?

823\dfrac{8\sqrt{2}}{3}

824π8\sqrt{2} - 4 - \pi

424\sqrt{2}

42+π84\sqrt{2} + \dfrac{\pi}{8}

822π28\sqrt{2} - 2 - \dfrac{\pi}{2}

Answer: B
Concepts:tangent linesectorarea decomposition
Difficulty rating: 1960
Solution:

Rectangle ABFEABFE has area AEAB=242=82.AE \cdot AB = 2 \cdot 4\sqrt2 = 8\sqrt2.

Right triangles ACOACO and BDOBDO each have hypotenuse 222\sqrt2 and a leg of 2,2, so each is isosceles right with area 2.2.

Angles CAECAE and DBFDBF are each 45,45^\circ, so sectors CAECAE and DBFDBF each have area 18π22=π2.\tfrac18 \pi \cdot 2^2 = \tfrac{\pi}{2}.

The shaded area is 82222π2=824π. \begin{gathered} 8\sqrt2 - 2 \cdot 2 - 2 \cdot \tfrac{\pi}{2} \\ = 8\sqrt2 - 4 - \pi. \end{gathered}

Thus, the correct answer is B.

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