2011 AMC 12A Problems/Problem 11

Problem

Circles $A, B,$ and $C$ each has 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 $B?$

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

Solution 1

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

The 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 to 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:

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

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

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$",F,N); [/asy]$

We can move the area above the part of the circle above the segment $EF$ down, and similarly for the other side. Then, we have a square, whose diagonal is $2$ , so the area is then just $\left(\frac{2}{\sqrt{2}}\right)^2 = \boxed{\textbf{2 = C}}$ .

~ Minor Edits, Challengees24

Video Solution

https://www.youtube.com/watch?v=u23iWcqbJlE ~Shreyas S

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=olRZuK11mAI

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

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search