2011 AMC 12A Problems/Problem 11

Revision as of 19:59, 17 July 2020 by Jbala (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $B?$

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

Solution

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 $B$.

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]

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 $\left(\frac{2}{\sqrt{2}}\right)^2 = 2$.

Video Solution

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

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
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

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

Invalid username
Login to AoPS