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

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Plugging all of these areas back into the original equation yields <math>8\sqrt{2}-(\frac{\pi}{2}+2+2+\frac{\pi}{2})=8\sqrt{2}-(4+\pi)=\boxed{8\sqrt{2}-4-\pi}\ \mathrm{(B)}</math>.
 
Plugging all of these areas back into the original equation yields <math>8\sqrt{2}-(\frac{\pi}{2}+2+2+\frac{\pi}{2})=8\sqrt{2}-(4+\pi)=\boxed{8\sqrt{2}-4-\pi}\ \mathrm{(B)}</math>.
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==Video Solution==
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https://www.youtube.com/watch?v=2RG7G4ODG9A      ~David
  
 
==See also==
 
==See also==
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 
{{MAA Notice}}
 
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== Video Solution ==
 
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https://www.youtube.com/watch?v=2RG7G4ODG9A    ~David
 

Latest revision as of 20:26, 1 January 2025

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 $ECODF$?

[asy] size(5cm); pair A=(-2*sqrt(2),0), B = (2*sqrt(2),0), C = A+2*dir(45), D = B+2*dir(135), E = A+2*dir(90), F = B+2*dir(90); fill((0,0)--C--E--F--D--cycle,gray(0.6)); unfill(circle(A,2)); unfill(circle(B,2)); draw(circle(A,2)); draw(circle(B,2)); draw(E--F); draw(C--(0,0)--D); draw(A+2*dir(300)--A--B--B+2*dir(240)); dot((0,0)); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); label("$A$",A,dir(180)); label("$B$",B,dir(0)); label("$C$",C,dir(240)); label("$D$",D,dir(300)); label("$E$",E,dir(90)); label("$F$",F,dir(90)); label("$O$",(0,0),dir(270)); label("$2$",A+dir(300),dir(210)); label("$2$",B+dir(240),dir(330)); [/asy]

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

Solution

The area we are trying to find is simply $ABFE-(\overarc{AEC}+\triangle{ACO}+\triangle{BDO}+\overarc{BFD}).\overline{EF}\parallel\overline{AB}$. 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 an isosceles right triangle, and has $\overline{CO}$ with length $2$. The area of $\triangle{ACO}=\frac{1}{2}bh=2$. By symmetry, $\triangle{ACO}\cong\triangle{BDO}$, and so the area of $\triangle{BDO}$ is also $2$.

$\overarc{AEC}$ (or $\overarc{BFD}$, for that matter) is $\frac{1}{8}$ the area of its circle since $\angle{OAC}$ is 45 degrees and $\angle{OAE}$ forms a right triangle. Thus $\overarc{AEC}$ and $\overarc{BFD}$ both have an area of $\frac{\pi}{2}$.

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

Video Solution

https://www.youtube.com/watch?v=2RG7G4ODG9A ~David

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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