Difference between revisions of "2007 AMC 10A Problems/Problem 24"
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The area we are trying to find is simply <math>ABFE-(\arc{AEC}+\triangle{ACO}+\triangle{BDO}+\arc{BFD}</math>. | The area we are trying to find is simply <math>ABFE-(\arc{AEC}+\triangle{ACO}+\triangle{BDO}+\arc{BFD}</math>. | ||
| − | Obviously, <math>EF</math> is parallel to <math>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>EF</math> is parallel to <math>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>. |
Since <math>OC</math> is tangent to <math>\circle{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>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>. | Since <math>OC</math> is tangent to <math>\circle{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>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 15:08, 27 April 2008
Problem
Circles centered at 
and 
each have radius 
, as shown. Point 
is the midpoint of 
, and 
. Segments 
and 
are tangent to the circles centered at and , respectively, and 
is a common tangent. What is the area of the shaded region 
?
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, is parallel to 
. Thus, 
is a rectangle, and so its area is 
.
Since is tangent to $\circle{A}$ (Error compiling LaTeX. Unknown error_msg), 
is a right 
. We know 
and 
, so is isosceles, a 
- right , and has 
with length . The area of 
. For obvious reasons, 
, and so the area of 
is also .
$\arc{AEC}$ (Error compiling LaTeX. Unknown error_msg) (or $\arc{BFD}$ (Error compiling LaTeX. Unknown error_msg), for that matter) is 
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 
.
Plugging all of these areas back into the original equation yields 
.
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 | ||
