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Difference between revisions of "2011 AMC 12A Problems/Problem 11"
Line 70: Line 70:
 
dotfactor=4;
 
dotfactor=4;
  
−
pair A=(0,0), B=(2,0), C=(1,1)
+
pair A=(0,0), B=(2,0), C=(1,1);
 
pair D=(2,1);
 
pair D=(2,1);
 
pair E=(0,1);
 
pair E=(0,1);
Line 87: Line 87:
 
draw(Circle(C,1));
 
draw(Circle(C,1));
  
−
draw(A--B);
+
draw (D--F--E--M--D);
−
draw(M--E);
 
−
draw(E--B);
 
−
draw (D--F--E--M);
 
  
 
label("$A$",A,W);
 
label("$A$",A,W);
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label("$C$",C,W);
 
label("$C$",C,W);
 
label("$M$",M,NE);
 
label("$M$",M,NE);
−
label("$D$",D,SE);
+
label("$D$",D,E);
−
label("$E$",E,SE);
+
label("$E$",E,W);
−
label("$F$",F,SE);
+
label("$F$",F,N);
 
</asy>
 
</asy>
 +
 +
Instead, we can move the area above the region
  
 
== 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 21:45, 24 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:

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

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

Solution 2

[asy] unitsize(1.1cm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), B=(2,0), C=(1,1); pair D=(2,1); pair E=(0,1); pair F = (1, 2); pair M = (1, 0); dot (A); dot (B); dot (C); dot (D); dot (E); dot (F); dot (M); draw(Circle(A,1)); draw(Circle(B,1)); draw(Circle(C,1)); draw (D--F--E--M--D); label("$A$",A,W); label("$B$",B,E); label("$C$",C,W); label("$M$",M,NE); label("$D$",D,E); label("$E$",E,W); label("$F$",F,N); [/asy]

Instead, we can move the area above the region

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_11&oldid=37108"