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Difference between revisions of "2007 AMC 10A Problems/Problem 24"

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Obviously, <math>\overline{EF}</math> is parallel to <math>\overline{AB}</math>.  Thus, <math>ABFE</math> is a [[rectangle]], and so its area is <math>b\times{h}=2\times{(AO+OB)}=2\times{2(2\sqrt{2})}=8\sqrt{2}</math>.
 
Obviously, <math>\overline{EF}</math> is parallel to <math>\overline{AB}</math>.  Thus, <math>ABFE</math> is a [[rectangle]], and so its area is <math>b\times{h}=2\times{(AO+OB)}=2\times{2(2\sqrt{2})}=8\sqrt{2}</math>.
  
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Since <math>\overline{OC}</math> is tangent to <math>\circ{A}</math>, <math>\triangle{ACO}</math> is a right <math>\triangle</math>.  We know <math>AO=2\sqrt{2}</math> and <math>AC=2</math>, so <math>\triangle{ACO}</math> is isosceles, a <math>45</math>-<math>45</math> right <math>\triangle</math>, and has <math>\overline{CO}</math> with length <math>2</math>.  The area of <math>\triangle{ACO}=\frac{1}{2}bh=2</math>.  For obvious reasons, <math>\triangle{ACO}=\triangle{BDO}</math>, and so the area of <math>\triangle{BDO}</math> is also <math>2</math>.
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Since <math>\overline{OC}</math> is tangent to circle <math>A</math>, <math>\triangle{ACO}</math> is a right <math>\triangle</math>.  We know <math>AO=2\sqrt{2}</math> and <math>AC=2</math>, so <math>\triangle{ACO}</math> is isosceles, a <math>45</math>-<math>45</math> right <math>\triangle</math>, and has <math>\overline{CO}</math> with length <math>2</math>.  The area of <math>\triangle{ACO}=\frac{1}{2}bh=2</math>.  For obvious reasons, <math>\triangle{ACO}=\triangle{BDO}</math>, and so the area of <math>\triangle{BDO}</math> is also <math>2</math>.
  
 
<math>\arc{AEC}</math> (or <math>\arc{BFD}</math>, for that matter) is <math>\frac{1}{8}</math> the area of its circle.  Thus <math>\arc{AEC}</math> and <math>\arc{BFD}</math> both have an area of <math>\frac{\pi}{2}</math>.
 
<math>\arc{AEC}</math> (or <math>\arc{BFD}</math>, for that matter) is <math>\frac{1}{8}</math> the area of its circle.  Thus <math>\arc{AEC}</math> and <math>\arc{BFD}</math> both have an area of <math>\frac{\pi}{2}</math>.

Revision as of 15:33, 27 April 2008

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 $ECODF$?

2007 AMC 10A problem 24.png

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

Solution

The area we are trying to find is simply $ABFE-(\arc{AEC}+\triangle{ACO}+\triangle{BDO}+\arc{BFD})$ (Error compiling LaTeX. Unknown error_msg). Obviously, $\overline{EF}$ is parallel to $\overline{AB}$. 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 $2$.

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

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

See also

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
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