Difference between revisions of "2004 AMC 12A Problems/Problem 19"
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label("\(B\)", B); | label("\(B\)", B); | ||
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Revision as of 19:29, 17 July 2021
- The following problem is from both the 2004 AMC 12A #19 and 2004 AMC 10A #23, so both problems redirect to this page.
Problem 19
Circles and
are externally tangent to each other, and internally tangent to circle
. Circles
and
are congruent. Circle
has radius
and passes through the center of
. What is the radius of circle
?
![[asy] unitsize(15mm); pair A=(-1,0),B=(2/3,8/9),C=(2/3,-8/9),D=(0,0); draw(Circle(D,2)); draw(Circle(A,1)); draw(Circle(B,8/9)); draw(Circle(C,8/9)); label("\(A\)", A); label("\(B\)", B); label("\(C\)", C); label("\(D\)", (-1.2,1.8)); [/asy]](http://latex.artofproblemsolving.com/6/9/6/69677da9f9025d93eca8bc2a4d63d4a3b5b11c9c.png)
Solution 1
Let be the center of circle
for all
and let
be the tangent point of
. Since the radius of
is the diameter of
, the radius of
is
.
Let the radius of be
and let
. If we connect
, we get an isosceles triangle with lengths
.
Then right triangle has legs
and hypotenuse
. Solving for
, we get
.
Also, right triangle has legs
, and hypotenuse
. Solving,
So the answer is .
Solution 2
![[asy] unitsize(15mm); pair A=(0,1),B=(-8/9,-2/3),C=(8/9,-2/3),D=(0,0), E=(0,-2/3); draw(Circle(D,2)); draw(Circle(A,1)); draw(Circle(B,8/9)); draw(Circle(C,8/9)); draw(A--B--C--A); draw(B--D--C); draw(A--E); dot(A);dot(B);dot(C);dot(D);dot(E); label("\(D\)", D,NW); label("\(A\)", A,N); label("\(B\)", B,W); label("\(C\)", C,E); label("\(E\)", E,SE); label("\(1\)",(-.4,.7)); label("\(1\)",(0,0.5),W); label("\(r\)", (-.8,-.1)); label("\(r\)", (-4/9,-2/3),S); label("\(h\)", (0,-1/3), W); [/asy]](http://latex.artofproblemsolving.com/e/2/1/e21e9f6600899f824d87d28a241a4d9bfb5ee11a.png)
Note that since
is the center of the larger circle of radius
. Using the Pythagorean Theorem on
,
Now using the Pythagorean Theorem on ,
Substituting ,
Solution 3
We can apply Descartes' Circle Formula.
The four circles have curvatures , and
.
We have
Simplifying, we get
See Also
2004 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 |
2004 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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.