Difference between revisions of "2004 AMC 12A Problems/Problem 19"
(→Problem 19) |
(→Solution) |
||
Line 34: | Line 34: | ||
dot(A);dot(B);dot(C);dot(D);dot(E); | dot(A);dot(B);dot(C);dot(D);dot(E); | ||
− | label("\(D\)", D, | + | label("\(D\)", D,NW); |
label("\(A\)", A,N); | label("\(A\)", A,N); | ||
label("\(B\)", B,W); | label("\(B\)", B,W); | ||
Line 40: | Line 40: | ||
label("\(E\)", E,S); | label("\(E\)", E,S); | ||
label("\(1\)",(-.4,.7)); | label("\(1\)",(-.4,.7)); | ||
− | label("\(1\)",(0,.5), | + | label("\(1\)",(0,0.5),W); |
label("\(r\)", (-.8,-.1)); | label("\(r\)", (-.8,-.1)); | ||
label("\(r\)", (-4/9,-2/3),S); | label("\(r\)", (-4/9,-2/3),S); | ||
− | label("\(h\)", (0,-1/3), | + | label("\(h\)", (0,-1/3), W); |
</asy></center> | </asy></center> | ||
Revision as of 20:52, 26 December 2011
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 ?
Solution
Note that since is the center of the larger circle of radius . Using the Pythagorean Theorem on ,
Now using the Pythagorean Theorem on ,
Substituting ,
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 |