Difference between revisions of "2011 AMC 12A Problems/Problem 11"
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== Solution == | == Solution == | ||
| + | |||
| + | <asy> | ||
| + | unitsize(1.1cm); | ||
| + | defaultpen(linewidth(.8pt)); | ||
| + | dotfactor=4; | ||
| + | |||
| + | pair A=(0,0), B=(2,0), C=(1,-1); | ||
| + | pair M=(1,0); | ||
| + | pair D=(2,-1); | ||
| + | dot (A); | ||
| + | dot (B); | ||
| + | dot (C); | ||
| + | dot (D); | ||
| + | dot (M); | ||
| + | |||
| + | draw(Circle(A,1)); | ||
| + | draw(Circle(B,1)); | ||
| + | draw(Circle(C,1)); | ||
| + | |||
| + | draw(A--B); | ||
| + | draw(M--D); | ||
| + | draw(D--B); | ||
| + | |||
| + | label("$A$",A,W); | ||
| + | label("$B$",B,E); | ||
| + | label("$C$",C,W); | ||
| + | label("$M$",M,NE); | ||
| + | label("$D$",D,SE); | ||
| + | </asy> | ||
| + | |||
| + | The requested area is the area of <math>C</math> minus the area shared between circles <math>A</math>, <math>B</math> and <math>C</math>. | ||
| + | |||
| + | Let <math>M</math> be the midpoint of <math>\overline{AB}</math> and <math>D</math> be the other intersection of circles <math>C</math> and <math>B</math>. | ||
| + | |||
| + | Then area shared between <math>C</math>, <math>A</math> and <math>B</math> is <math>4</math> of the regions between arc <math>\widehat {MD}</math> and line <math>\overline{MD}</math>, which is (considering the arc on circle <math>B</math>) a quarter of the circle <math>B</math> minus <math>\triangle MDB</math>: | ||
| + | |||
| + | <math>\frac{\pi r^2}{4}-\frac{bh}{2}</math> | ||
| + | |||
| + | <math>b = h = r = 1</math> | ||
| + | |||
| + | (We can assume this because <math>\angle DBM</math> is 90 degrees, since <math>CDBM</math> is a square, due the application of the tangent chord theorem at point <math>M</math>) | ||
| + | |||
| + | So the area of the small region is | ||
| + | |||
| + | <math>\frac{\pi}{4}-\frac{1}{2}</math> | ||
| + | |||
| + | The requested area is area of circle <math>C</math> minus 4 of this area: | ||
| + | |||
| + | <math>\pi 1^2 - 4(\frac{\pi}{4}-\frac{1}{2}) | ||
| + | = \pi - \pi + 2 | ||
| + | = 2</math> | ||
| + | |||
<math>\boxed{\textbf{C}}</math>. | <math>\boxed{\textbf{C}}</math>. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=10|num-a=12|ab=A}} | {{AMC12 box|year=2011|num-b=10|num-a=12|ab=A}} | ||
Revision as of 23:39, 21 February 2011
Problem
Circles 
and 
each have radius 1. Circles 
and 
share one point of tangency. Circle has a point of tangency with the midpoint of 
What is the area inside circle but outside circle and circle 
$\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad \textbf{(B)}\ \frac{\pi}{2} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{3\pi}{4} \qquad \textbf{(E)}\ 1+\frac{\pi}{2}}$ (Error compiling LaTeX. Unknown error_msg)
Solution
![[asy] unitsize(1.1cm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), B=(2,0), C=(1,-1); pair M=(1,0); pair D=(2,-1); dot (A); dot (B); dot (C); dot (D); dot (M); draw(Circle(A,1)); draw(Circle(B,1)); draw(Circle(C,1)); draw(A--B); draw(M--D); draw(D--B); label("$A$",A,W); label("$B$",B,E); label("$C$",C,W); label("$M$",M,NE); label("$D$",D,SE); [/asy]](http://latex.artofproblemsolving.com/f/8/4/f84b9e326bd427c8608100174e752c08642c3129.png)
The requested area is the area of minus the area shared between circles , and .
Let 
be the midpoint of 
and 
be the other intersection of circles and .
Then area shared between , and is 
of the regions between arc 
and line 
, which is (considering the arc on circle ) a quarter of the circle minus 
:
(We can assume this because 
is 90 degrees, since 
is a square, due the application of the tangent chord theorem at point )
So the area of the small region is
The requested area is area of circle minus 4 of this area:
.
See also
| 2011 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 10 |
Followed by Problem 12 |
| 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 12 Problems and Solutions | |
