2007 AMC 10A Problems/Problem 24

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Problem

Circles centered at $A$ and $B$ each have radius $2$ , as shown. Point $O$ is the midpoint of $\overline{AB}$ , and $OA = 2\sqrt {2}$ . Segments $OC$ and $OD$ are tangent to the circles centered at $A$ and $B$ , respectively, and $EF$ is a common tangent. What is the area of the shaded region $ECODF$ ?

$\text{(A)}\ \frac {8\sqrt {2}}{3} \qquad \text{(B)}\ 8\sqrt {2} - 4 - \pi \qquad \text{(C)}\ 4\sqrt {2} \qquad \text{(D)}\ 4\sqrt {2} + \frac {\pi}{8} \qquad \text{(E)}\ 8\sqrt {2} - 2 - \frac {\pi}{2}$

Solution

The area we are trying to find is simply $ABFE-(\arc{AEC}+\triangle{ACO}+\triangle{BDO}+\arc{BFD})$ . Obviously, $\overline{EF}\parallel\overline{AB}$ . Thus, $ABFE$ is a rectangle, and so its area is $b\times{h}=2\times{(AO+OB)}=2\times{2(2\sqrt{2})}=8\sqrt{2}$ .

Since $\overline{OC}$ is tangent to circle $A$ , $\triangle{ACO}$ is a right $\triangle$ . We know $AO=2\sqrt{2}$ and $AC=2$ , so $\triangle{ACO}$ is isosceles, a $45$ - $45$ right $\triangle$ , and has $\overline{CO}$ with length $2$ . The area of $\triangle{ACO}=\frac{1}{2}bh=2$ . For obvious reasons, $\triangle{ACO}=\triangle{BDO}$ , and so the area of $\triangle{BDO}$ is also $2$ .

$\arc{AEC}$ (or $\arc{BFD}$ , for that matter) is $\frac{1}{8}$ the area of its circle. Thus $\arc{AEC}$ and $\arc{BFD}$ both have an area of $\frac{\pi}{2}$ .

Plugging all of these areas back into the original equation yields $8\sqrt{2}-(\frac{\pi}{2}+2+2+\frac{\pi}{2})=8\sqrt{2}-(4+\pi)=\boxed{8\sqrt{2}-4-\pi}\ \mathrm{(B)}$ .

2007 AMC 10A (Problems)
Preceded by Problem 23	Followed by Problem 25
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2007 AMC 10A Problems/Problem 24

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Problem

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