Difference between revisions of "2008 AMC 12A Problems/Problem 13"
m (→Problem) |
Awesome3.14 (talk | contribs) (→Solution) |
||
| Line 5: | Line 5: | ||
<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 1 == |
<asy>size(200); | <asy>size(200); | ||
defaultpen(fontsize(10)); | defaultpen(fontsize(10)); | ||
| Line 34: | Line 34: | ||
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>. | ||
| + | |||
| + | == Solution 2 == | ||
| + | The incenter of an equilateral triangle is the same as the centroid of an equilateral triangle. 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. <math>\frac{1}{3}=\frac{1}{9}</math>. The answer is B. | ||
Then the ratio of areas will be <math>\frac{1}{3}</math> squared, or <math>\frac{1}{9}\Rightarrow \boxed{\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>. | ||
Revision as of 00:49, 24 December 2020
- The following problem is from both the 2008 AMC 12A #13 and 2008 AMC 10A #16, so both problems redirect to this page.
Contents
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 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); 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 
.
Solution 2
The incenter of an equilateral triangle is the same as the centroid of an equilateral triangle. 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. 
. The answer is B.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
