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Difference between revisions of "2011 AMC 12A Problems/Problem 11"

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== Problem ==
 
== Problem ==
−
Circles <math>A, B,</math> and <math>C</math> each have radius 1. Circles <math>A</math> and <math>B</math> share one point of tangency. Circle <math>C</math> has a point of tangency with the midpoint of <math>\overline{AB}.</math> What is the area inside circle <math>C</math> but outside circle <math>A</math> and circle <math>B?</math>
+
Circles <math>A, B,</math> and <math>C</math> each has radius 1. Circles <math>A</math> and <math>B</math> share one point of tangency. Circle <math>C</math> has a point of tangency with the midpoint of <math>\overline{AB}.</math> What is the area inside circle <math>C</math> but outside circle <math>A</math> and circle <math>B?</math>
  
 
<math>
 
<math>
Line 7: Line 7:
 
\textbf{(C)}\  2 \qquad
 
\textbf{(C)}\  2 \qquad
 
\textbf{(D)}\ \frac{3\pi}{4} \qquad
 
\textbf{(D)}\ \frac{3\pi}{4} \qquad
−
\textbf{(E)}\ 1+\frac{\pi}{2}} </math>
+
\textbf{(E)}\ 1+\frac{\pi}{2} </math>
 +
 
 +
==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 <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>.
 +
 
 +
The 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 to 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\left(\frac{\pi}{4}-\frac{1}{2}\right)
 +
= \pi - \pi + 2
 +
= 2</math>
  
−
== Solution ==
 
 
<math>\boxed{\textbf{C}}</math>.
 
<math>\boxed{\textbf{C}}</math>.
 +
 +
== 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 <math>EF</math> down, and similarly for the other side. Then, we have a square, whose diagonal is <math>2</math>, so the area is then just <math>\left(\frac{2}{\sqrt{2}}\right)^2 = \boxed{\textbf{2 = C}}</math>.
 +
 +
~ 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
  
 
== 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}}
 +
{{AMC10 box|year=2011|num-b=17|num-a=19|ab=A}}
 +
{{MAA Notice}}

Latest revision as of 21:24, 23 October 2024

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

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
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

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