Difference between revisions of "2008 AMC 12A Problems/Problem 13"
I like pie (talk | contribs) (Duplicate, standardized answer choices, added AMC 10 box, style) |
I like pie (talk | contribs) (Solution - Minor edits) |
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== Solution == | == Solution == | ||
− | + | <asy>size(200); | |
− | size( | + | defaultpen(fontsize(10)); |
− | defaultpen( | ||
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); | ||
picture p = new picture; | picture p = new picture; | ||
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draw(D-(0.2,0)--D-(0.2,-0.2)--D-(0,-0.2)); | draw(D-(0.2,0)--D-(0.2,-0.2)--D-(0,-0.2)); | ||
draw(Circle(C,1)); | draw(Circle(C,1)); | ||
− | label(" | + | label(" |
label(" | label(" | ||
− | label(" | + | label(" |
label(" | label(" | ||
label(" | label(" | ||
label(" | label(" | ||
label(" | label(" | ||
− | label(" | + | label(" |
− | </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> | + | 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>. |
+ | |||
+ | Then <math>PQO</math> is a right triangle, and a <math>30-60-90</math> triangle at that. Therefore, <math>OP=2PQ</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 B</math>. | ||
== See also == | == See also == |
Revision as of 00:28, 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
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 |