Difference between revisions of "2008 AMC 12A Problems/Problem 13"
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| + | {{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #13]] and [[2008 AMC 10A Problems/Problem 16|2008 AMC 10A #16]]}} | ||
== Problem == | == Problem == | ||
Points <math>A</math> and <math>B</math> lie on a [[circle]] centered at <math>O</math>, and <math>\angle AOB = 60^\circ</math>. A second circle is internally [[tangent]] to the first and tangent to both <math>\overline{OA}</math> and <math>\overline{OB}</math>. What is the ratio of the area of the smaller circle to that of the larger circle? | Points <math>A</math> and <math>B</math> lie on a [[circle]] centered at <math>O</math>, and <math>\angle AOB = 60^\circ</math>. A second circle is internally [[tangent]] to the first and tangent to both <math>\overline{OA}</math> and <math>\overline{OB}</math>. What is the ratio of the area of the smaller circle to that of the larger circle? | ||
| − | <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 == | ||
| Line 33: | Line 34: | ||
== See also == | == See also == | ||
| − | {{AMC12 box|year=2008|num-b=12|num-a=14|ab=A}} | + | {{AMC12 box|year=2008|ab=A|num-b=12|num-a=14}} |
| + | {{AMC10 box|year=2008|ab=A|num-b=15|num-a=17}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
Revision as of 00:22, 26 April 2008
- 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(300); defaultpen(0.8); 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("\(30^{\circ}\)",(.53,.1),O); label("\(r\)",(C+D)/2,E); label("\(2r\)",(O+C)/2,SE); 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/e/f/0/ef0f01a2ab884db72631c2145d635dbeda4e4cc0.png)
Let 
be the center of the small circle with radius 
, and let 
be the point where the small circle is tangent to 
, and finally, let 
be the point where the small circle is tangent to the big circle with radius 
. Then 
is a right triangle, and a 30-60-90 triangle at that. So, 
. 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 | ||
