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2004 AMC 12A Problems/Problem 19: Difference between revisions

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== Problem 19 ==
== Problem 19 ==
Circles <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>?
[[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>,


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>r^2 + h^2 = (2-r)^2</math>
Now using the [[Pythagorean Theorem]] on <math>\triangle BAE</math>,


<math>r^2 + h^2 = 4 - 4r + r^2</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^2 = 4 - 4r</math>
Substituting <math>h</math>,


<math>h = 2\sqrt{1-r}</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>


Now Using the pythagorean theorem on <math>\triangle BAE</math>
==See Also==
 
{{AMC12 box|year=2004|ab=A|num-b=18|num-a=20}}
<math>r^2 + (h+1)^2 = (r+1)^2</math>
 
<math>r^2 + h^2 + 2h + 1 = r^2 + 2r + 1</math>
 
<math>h^2 + 2h = 2r</math>


Substituting <math>h</math> in
[[Category:Introductory Geometry Problems]]
 
<math>(4-4r) + 4\sqrt{1-r} = 2r</math>
 
<math>4\sqrt{1-r} = 6r - 4</math>
 
<math>16-16r = 36r^2 - 48r + 16</math>
 
<math>0 = 36r^2 - 32r</math>
 
<math>r = \frac{32}{36} = \frac{8}{9} \Rightarrow \qquad \textbf{(D)}</math>
 
==See Also==
{{AMC12 box|year=2004|ab=A|num-b=18|num-a=19}}

Revision as of 15:40, 15 August 2008

Problem 19

Circles $A, B$ and $C$ are externally tangent to each other, and internally tangent to circle $D$. Circles $B$ and $C$ are congruent. Circle $A$ has radius $1$ and passes through the center of $D$. What is the radius of circle $B$?

$\text {(A)} \frac23 \qquad \text {(B)} \frac {\sqrt3}{2} \qquad \text {(C)}\frac78 \qquad \text {(D)}\frac89 \qquad \text {(E)}\frac {1 + \sqrt3}{3}$

Solution

[asy] unitsize(20mm); 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,N); label("\(A\)", A,N); label("\(B\)", B,W); label("\(C\)", C,E); label("\(E\)", E,S); label("\(1\)",(-.4,.7)); label("\(1\)",(0,.5),E); label("\(r\)", (-.8,-.1)); label("\(r\)", (-4/9,-2/3),S); label("\(h\)", (0,-1/3),E); [/asy]

Note that $BD= 2-r$ since $D$ is the center of the larger circle of radius $2$. Using the Pythagorean Theorem on $\triangle BDE$,

\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*}

Now using the Pythagorean Theorem on $\triangle BAE$,

\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*}

Substituting $h$,

\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*}

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
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