Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2008 AMC 12A Problems/Problem 13"

(→‎Solution 1)
(→‎Solution 1)
 
(13 intermediate revisions by 2 users not shown)
Line 8: Line 8:
 
<asy>size(200);
 
<asy>size(200);
 
defaultpen(fontsize(10));
 
defaultpen(fontsize(10));
−
pair O=(0,0), A=(3,0), B=(3/2,3/2*3^.5), C=(3^.5,1), D=(3^.5,0), F=(1.5*3^.5,1.5);
+
pair O=(0,0), A=(3,0), B=(3/2,3/2*3^.5), C=(3^.5,1), D=(3^.5,0), F=(1.5*3^.5,1.5), G=(2*3^.5,2);
 
picture p = new picture;  
 
picture p = new picture;  
 
draw(p,Circle(O,0.2));
 
draw(p,Circle(O,0.2));
Line 27: Line 27:
 
label("\(R\)",(O+A)/2-(0,0.3),S);
 
label("\(R\)",(O+A)/2-(0,0.3),S);
 
label("\(P\)",C,NW);
 
label("\(P\)",C,NW);
 +
label("\(A\)",(3,0), A);
 +
label("\(B\)",(3/2,3/2*3^.5), B);
 +
label("\(C\)",(1.5*3^.5,1.5), G);
 
label("\(Q\)",D,SE);</asy>
 
label("\(Q\)",D,SE);</asy>
−
 
 
Let <math>P</math> be the center of the small circle with radius <math>r</math>, and let <math>Q</math> be the point where the small circle is [[tangent]] to <math>OA</math>. Also, let <math>C</math> be the point where the small circle is tangent to the big circle with radius <math>R</math>.
 
Let <math>P</math> be the center of the small circle with radius <math>r</math>, and let <math>Q</math> be the point where the small circle is [[tangent]] to <math>OA</math>. Also, let <math>C</math> be the point where the small circle is tangent to the big circle with radius <math>R</math>.
  
Line 35: Line 37:
 
Since <math>OP=OC-PC=OC-r=R-r</math>, we have <math>R-r=2PQ</math>, or <math>R-r=2r</math>, or <math>\frac{1}{3}=\frac{r}{R}</math>.
 
Since <math>OP=OC-PC=OC-r=R-r</math>, we have <math>R-r=2PQ</math>, or <math>R-r=2r</math>, or <math>\frac{1}{3}=\frac{r}{R}</math>.
  
−
Ratio of areas of circle is ratio of radii squared, so the answer is <math>{\frac{1}{3}}^2 = \frac{1}{9}</math>
+
Ratio of areas of circles is ratio of radii squared, so the answer is <math>\left(\frac{1}{3}\right)^2 = \frac{1}{9} \Rightarrow \boxed{B}</math>
  
 
== Solution 2 ==
 
== Solution 2 ==

Latest revision as of 19:32, 4 June 2021

The following problem is from both the 2008 AMC 12A #13 and 2008 AMC 10A #16, so both problems redirect to this page.

Problem

Points $A$ and $B$ lie on a circle centered at $O$, and $\angle AOB = 60^\circ$. A second circle is internally tangent to the first and tangent to both $\overline{OA}$ and $\overline{OB}$. What is the ratio of the area of the smaller circle to that of the larger circle?

$\mathrm{(A)}\ \frac{1}{16}\qquad\mathrm{(B)}\ \frac{1}{9}\qquad\mathrm{(C)}\ \frac{1}{8}\qquad\mathrm{(D)}\ \frac{1}{6}\qquad\mathrm{(E)}\ \frac{1}{4}$

Solution 1

[asy]size(200); defaultpen(fontsize(10)); pair O=(0,0), A=(3,0), B=(3/2,3/2*3^.5), C=(3^.5,1), D=(3^.5,0), F=(1.5*3^.5,1.5), G=(2*3^.5,2); picture p = new picture; draw(p,Circle(O,0.2)); clip(p,O--C--A--cycle); add(p); draw(Circle(O,3)); dot(A); dot(B); dot(C); dot(O); draw(A--O--B); draw(O--C--D); draw(C--F); draw(D-(0.2,0)--D-(0.2,-0.2)--D-(0,-0.2)); draw(Circle(C,1)); label("\(30^{\circ}\)",(0.65,0.15),O); label("\(r\)",(C+D)/2,E); label("\(2r\)",(O+C)/2,NNE); label("\(O\)",O,SW); label("\(r\)",(C+F)/2,SE); label("\(R\)",(O+A)/2-(0,0.3),S); label("\(P\)",C,NW); label("\(A\)",(3,0), A); label("\(B\)",(3/2,3/2*3^.5), B); label("\(C\)",(1.5*3^.5,1.5), G); label("\(Q\)",D,SE);[/asy] Let $P$ be the center of the small circle with radius $r$, and let $Q$ be the point where the small circle is tangent to $OA$. Also, let $C$ be the point where the small circle is tangent to the big circle with radius $R$.

Then $PQO$ is a right triangle. Angle $POQ$ is $30$ degrees because line $OP$ bisects angle $AOB$ (this can be proved by dropping a perpendicular line from $P$ to line $OB$, letting their intersection be point $S$, and proving triangles $PQO$ and $PSO$ congruent), meaning that $PQO$ is a $30-60-90$ triangle. Therefore, $OP=2PQ$.

Since $OP=OC-PC=OC-r=R-r$, we have $R-r=2PQ$, or $R-r=2r$, or $\frac{1}{3}=\frac{r}{R}$.

Ratio of areas of circles is ratio of radii squared, so the answer is $\left(\frac{1}{3}\right)^2 = \frac{1}{9} \Rightarrow \boxed{B}$

Solution 2

Like in Solution 1, let $P$ be the center of the small circle with radius $r$, and let $Q$ be the point where the small circle is tangent to $OA$.

Let $N$ be the tangency point of the two circles. As shown in Solution 1, $POQ = 30$ degrees, so angle $NOA$ is also $30$ degrees. Let the line tangent to the two circles at $N$ intersect $OA$ and $OB$ at points $C$ and $D$, respectively. Since line $CD$ is tangent to circles $O$ and $P$, it must be perpendicular to $ON$, meaning that angle $ONC$ must be $90$ degrees. Because angle $NOA$ is $30$ degrees, angle $DCO$ is $180-30-90 = 60$ degrees. Angle $DOC$ is also $60$ degrees, so triangle $DOC$ is equilateral.

Note that an equilateral triangle's incenter is also its centroid. This means the center of the inscribed circle is also the centroid. From properties of median lengths, the radius of the large circle is 3 times the radius of the small circle.

Then the ratio of areas will be $\frac{1}{3}$ squared, or $\frac{1}{9}\Rightarrow \boxed{\text{B}}$.

See also

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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
2008 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_13&oldid=155277"