Difference between revisions of "2004 AMC 12A Problems/Problem 19"
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== Problem 19 == | == Problem 19 == | ||
− | + | [[Circle]]s <math>A, B</math> and <math>C</math> are externally tangent to each other, and internally tangent to circle <math>D</math>. Circles <math>B</math> and <math>C</math> are congruent. Circle <math>A</math> has radius <math>1</math> and passes through the center of <math>D</math>. What is the [[radius]] of circle <math>B</math>? | |
<math>\text {(A)} \frac23 \qquad \text {(B)} \frac {\sqrt3}{2} \qquad \text {(C)}\frac78 \qquad \text {(D)}\frac89 \qquad \text {(E)}\frac {1 + \sqrt3}{3}</math> | <math>\text {(A)} \frac23 \qquad \text {(B)} \frac {\sqrt3}{2} \qquad \text {(C)}\frac78 \qquad \text {(D)}\frac89 \qquad \text {(E)}\frac {1 + \sqrt3}{3}</math> | ||
==Solution== | ==Solution== | ||
− | <asy> | + | <center><asy> |
unitsize(20mm); | unitsize(20mm); | ||
pair A=(0,1),B=(-8/9,-2/3),C=(8/9,-2/3),D=(0,0), E=(0,-2/3); | pair A=(0,1),B=(-8/9,-2/3),C=(8/9,-2/3),D=(0,0), E=(0,-2/3); | ||
Line 29: | Line 29: | ||
label("\(r\)", (-4/9,-2/3),S); | label("\(r\)", (-4/9,-2/3),S); | ||
label("\(h\)", (0,-1/3),E); | label("\(h\)", (0,-1/3),E); | ||
− | </asy> | + | </asy></center> |
− | Note that <math>BD= 2-r</math> since D is the center of the larger circle of radius 2 | + | Note that <math>BD= 2-r</math> since <math>D</math> is the center of the larger circle of radius <math>2</math>. Using the Pythagorean Theorem on <math>\triangle BDE</math>, |
− | + | <cmath> | |
+ | \begin{align*} | ||
+ | r^2 + h^2 &= (2-r)^2 \\ | ||
+ | r^2 + h^2 &= 4 - 4r + r^2 \\ | ||
+ | h^2 &= 4 - 4r \\ | ||
+ | h &= 2\sqrt{1-r} \end{align*}</cmath> | ||
− | <math> | + | Now using the [[Pythagorean Theorem]] on <math>\triangle BAE</math>, |
− | < | + | <cmath> |
+ | \begin{align*} | ||
+ | r^2 + (h+1)^2 &= (r+1)^2 \\ | ||
+ | r^2 + h^2 + 2h + 1 &= r^2 + 2r + 1 \\ | ||
+ | h^2 + 2h &= 2r \end{align*} </cmath> | ||
− | <math>h | + | Substituting <math>h</math>, |
− | < | + | <cmath> |
+ | \begin{align*} | ||
+ | (4-4r) + 4\sqrt{1-r} &= 2r \\ | ||
+ | 4\sqrt{1-r} &= 6r - 4 \\ | ||
+ | 16-16r &= 36r^2 - 48r + 16 \\ | ||
+ | 0 &= 36r^2 - 32r \\ | ||
+ | r &= \frac{32}{36} = \frac{8}{9} \Longrightarrow \qquad \textbf{(D)} \end{align*}</cmath> | ||
− | + | ==See Also== | |
− | + | {{AMC12 box|year=2004|ab=A|num-b=18|num-a=20}} | |
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− | + | [[Category:Introductory Geometry Problems]] | |
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Revision as of 15:40, 15 August 2008
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 |