Difference between revisions of "2008 AMC 12A Problems/Problem 13"
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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>. | ||
− | Then the ratio of areas will be <math>\frac{1}{3}</math> squared, or <math>\frac{1}{9}\Rightarrow \text{B}</math>. | + | Then the ratio of areas will be <math>\frac{1}{3}</math> squared, or <math>\frac{1}{9}\Rightarrow \boxed{\text{B}}</math>. |
== See also == | == See also == |
Revision as of 10:55, 3 June 2011
- 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
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 |