Difference between revisions of "2008 AMC 12A Problems/Problem 13"
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<asy>size(200); | <asy>size(200); | ||
defaultpen(fontsize(10)); | 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); | + | 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), G=(2*3^.5,2); |
picture p = new picture; | picture p = new picture; | ||
draw(p,Circle(O,0.2)); | draw(p,Circle(O,0.2)); | ||
Line 27: | Line 27: | ||
label("\(R\)",(O+A)/2-(0,0.3),S); | label("\(R\)",(O+A)/2-(0,0.3),S); | ||
label("\(P\)",C,NW); | label("\(P\)",C,NW); | ||
+ | label("\(A\)",(3,0), A); | ||
+ | label("\(B\)",(3/2,3/2*3^.5), B); | ||
+ | label("\(C\)",(1.5*3^.5,1.5), G); | ||
label("\(Q\)",D,SE);</asy> | label("\(Q\)",D,SE);</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>. Also, let <math>C</math> be the point where the small circle is tangent to the big circle with radius <math>R</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 | + | Then <math>PQO</math> is a right triangle. Angle <math>POQ</math> is <math>30</math> degrees because line <math>OP</math> bisects angle <math>AOB</math> (this can be proved by dropping a perpendicular line from <math>P</math> to line <math>OB</math>, letting their intersection be point <math>S</math>, and proving triangles <math>PQO</math> and <math>PSO</math> congruent), meaning that <math>PQO</math> is a <math>30-60-90</math> triangle. 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>. | 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>. | ||
+ | |||
+ | Ratio of areas of circles is ratio of radii squared, so the answer is <math>\left(\frac{1}{3}\right)^2 = \frac{1}{9} \Rightarrow \boxed{B}</math> | ||
== Solution 2 == | == Solution 2 == | ||
− | + | Like in Solution 1, 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>N</math> be the tangency point of the two circles. As shown in Solution 1, <math>POQ = 30</math> degrees, so angle <math>NOA</math> is also <math>30</math> degrees. Let the line tangent to the two circles at <math>N</math> intersect <math>OA</math> and <math>OB</math> at points <math>C</math> and <math>D</math>, respectively. Since line <math>CD</math> is tangent to circles <math>O</math> and <math>P</math>, it must be perpendicular to <math>ON</math>, meaning that angle <math>ONC</math> must be <math>90</math> degrees. Because angle <math>NOA</math> is <math>30</math> degrees, angle <math>DCO</math> is <math>180-30-90 = 60</math> degrees. Angle <math>DOC</math> is also <math>60</math> degrees, so triangle <math>DOC</math> is equilateral. | ||
+ | |||
+ | Note that an equilateral triangle's incenter is also its centroid. 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. | ||
+ | |||
+ | 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 == |
Latest revision as of 20:32, 4 June 2021
- 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
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. Angle
is
degrees because line
bisects angle
(this can be proved by dropping a perpendicular line from
to line
, letting their intersection be point
, and proving triangles
and
congruent), meaning that
is a
triangle. Therefore,
.
Since , we have
, or
, or
.
Ratio of areas of circles is ratio of radii squared, so the answer is
Solution 2
Like in Solution 1, let be the center of the small circle with radius
, and let
be the point where the small circle is tangent to
.
Let be the tangency point of the two circles. As shown in Solution 1,
degrees, so angle
is also
degrees. Let the line tangent to the two circles at
intersect
and
at points
and
, respectively. Since line
is tangent to circles
and
, it must be perpendicular to
, meaning that angle
must be
degrees. Because angle
is
degrees, angle
is
degrees. Angle
is also
degrees, so triangle
is equilateral.
Note that an equilateral triangle's incenter is also its centroid. 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.
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.