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2011 AMC 12A Problems/Problem 11: Difference between revisions

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Added solution.
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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 $A, B,$ and $C$ each have radius 1. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}.$ What is the area inside circle $C$ but outside circle $A$ and circle $B?$

$\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad \textbf{(B)}\ \frac{\pi}{2} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{3\pi}{4} \qquad \textbf{(E)}\ 1+\frac{\pi}{2}}$ (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]

The requested area is the area of $C$ minus the area shared between circles $A$, $B$ and $C$.

Let $M$ be the midpoint of $\overline{AB}$ and $D$ be the other intersection of circles $C$ and $B$.

Then area shared between $C$, $A$ and $B$ is $4$ of the regions between arc $\widehat {MD}$ and line $\overline{MD}$, which is (considering the arc on circle $B$) a quarter of the circle $B$ minus $\triangle MDB$:

$\frac{\pi r^2}{4}-\frac{bh}{2}$

$b = h = r = 1$

(We can assume this because $\angle DBM$ is 90 degrees, since $CDBM$ is a square, due the application of the tangent chord theorem at point $M$)

So the area of the small region is

$\frac{\pi}{4}-\frac{1}{2}$

The requested area is area of circle $C$ minus 4 of this area:

$\pi 1^2 - 4(\frac{\pi}{4}-\frac{1}{2}) = \pi - \pi + 2 = 2$

$\boxed{\textbf{C}}$.

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
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