Difference between revisions of "2008 AMC 12A Problems/Problem 13"
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<math>\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}</math> | <math>\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}</math> | ||
| − | == Solution | + | == Solution |
<asy>size(200); | <asy>size(200); | ||
defaultpen(fontsize(10)); | defaultpen(fontsize(10)); | ||
Revision as of 10:28, 10 June 2019
- 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 
and 
lie on a circle centered at 
, and 
. A second circle is internally tangent to the first and tangent to both 
and 
. What is the ratio of the area of the smaller circle to that of the larger circle?
== Solution
![[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); 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("\(Q\)",D,SE);[/asy]](http://latex.artofproblemsolving.com/d/3/1/d3142ba25b2b5802e567ff06a53c2a14a3afb759.png)
Let 
be the center of the small circle with radius 
, and let 
be the point where the small circle is tangent to 
. Also, let 
be the point where the small circle is tangent to the big circle with radius 
.
Then 
is a right triangle, and a 
triangle at that. Therefore, 
.
Since 
, we have 
, or 
, or 
.
Then the ratio of areas will be 
squared, or 
.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
