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Difference between revisions of "2004 AMC 12A Problems/Problem 19"

(→‎See Also)
(→‎Solution)
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==Solution==
 
==Solution==
 +
=== Solution 1 ===
 +
 +
<asy>
 +
import graph;
 +
size(400);
 +
defaultpen(fontsize(10));
 +
pair OA=(-1,0),OB=(2/3,8/9),OC=(2/3,-8/9),OD=(0,0),E=(2/3,0);
 +
real t = 2.5;
 +
pair OA1=(-2+2*t,0),OB1=(4/3+2*t,16/9),OC1=(4/3+2*t,-16/9),OD1=(0+2*t,0),E1=(4/3+2*t,0);
 +
draw(Circle(OD,2));
 +
draw(Circle(OA,1));
 +
draw(Circle(OB,8/9));
 +
draw(Circle(OC,8/9));
 +
draw(OA--OB--OC--cycle);
 +
draw(OD--OB--OB+(OB-OD)*4/5);
 +
draw(OA--E);
 +
label("$O_{A}$",OA,(-1,0));
 +
label("$O_{B}$",OB,(-1,1));
 +
label("$O_{C}$",OC,(-1,-1));
 +
label("$O_{D}$",OD,(-1,-1));
 +
label("$E$",E,(0.5,-1));
 +
label("$r$",OB+(OB-OD)*2/5,(-0.5,1));
 +
label("$r$",(1*OA+3*OB)/4,(-0.5,1));
 +
dot(OA^^OB^^OC^^OD^^E);
 +
draw(OA1--OB1--OC1--cycle);
 +
draw(OD1--OB1);
 +
draw(OA1--E1);
 +
label("$O_{A}$",OA1,(-1,0));
 +
label("$O_{B}$",OB1,(1,1));
 +
label("$O_{C}$",OC1,(1,-1));
 +
label("$O_{D}$",OD1,(0,-1));
 +
label("$E$",E1,(1,0));
 +
label("$1+r$",(OA1+OB1)/2,(-0.5,1));
 +
label("$r$",(E1+OB1)/2,(1,0));
 +
label("$r$",(E1+OC1)/2,(1,0));
 +
label("$2-r$",(OB1+OD1)/2,(-1,0));
 +
label("$1$",(OA1+OD1)/2,(0,-1));
 +
label("$x$",(E1+OD1)/2,(0,-1));
 +
dot(OA1^^OB1^^OC1^^OD1^^E1);
 +
</asy>
 +
Let <math>O_{i}</math> be the center of circle <math>i</math> for all <math>i \in \{A,B,C,D\}</math> and let <math>E</math> be the tangent point of <math>B,C</math>. Since the radius of <math>D</math> is the diameter of <math>A</math>, the radius of <math>D</math> is <math>2</math>. Let the radius of <math>B,C</math> be <math>r</math> and let <math>O_{D}E = x</math>. If we connect <math>O_{A},O_{B},O_{C}</math>, we get an [[isosceles triangle]] with lengths <math>1 + r, 2r</math>. Then right triangle <math>O_{D}O_{B}O_{E}</math> has legs <math>r, x</math> and [[hypotenuse]] <math>2-r</math>. Solving for <math>x</math>, we get <math>x^2 = (2-r)^2 - r^2 \Longrightarrow x = \sqrt{4-4r}</math>.
 +
 +
Also, right triangle <math>O_{A}O_{B}O_{E}</math> has legs <math>r, 1+x</math>, and hypotenuse <math>1+r</math>. Solving,
  
−
=== Solution 1 ===
+
<cmath>\begin{eqnarray*}
 +
r^2 + (1+\sqrt{4-4r})^2 &=& (1+r)^2\
 +
1+4-4r+2\sqrt{4-4r}&=& 2r + 1\
 +
1-r &=& \left(\frac{6r-4}{4}\right)^2\
 +
\frac{9}{4}r^2-2r&=& 0\
 +
r &=& \frac 89
 +
\end{eqnarray*}</cmath>
 +
 +
So the answer is <math>\boxed{\mathrm{(D)}\ \frac{8}{9}}</math>.
 +
=== Solution 2 ===
  
 
<center><asy>
 
<center><asy>
Line 76: Line 128:
 
r &= \frac{32}{36} = \frac{8}{9} \Longrightarrow \qquad \textbf{(D)} \end{align*}</cmath>
 
r &= \frac{32}{36} = \frac{8}{9} \Longrightarrow \qquad \textbf{(D)} \end{align*}</cmath>
  
−
=== Solution 2 ===
+
=== Solution 3 ===
  
 
We can apply [[Descartes' Circle Formula]].
 
We can apply [[Descartes' Circle Formula]].

Revision as of 19:32, 5 November 2017

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 $B$?

[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]

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

Solution

Solution 1

[asy] import graph; size(400); defaultpen(fontsize(10)); pair OA=(-1,0),OB=(2/3,8/9),OC=(2/3,-8/9),OD=(0,0),E=(2/3,0); real t = 2.5; pair OA1=(-2+2*t,0),OB1=(4/3+2*t,16/9),OC1=(4/3+2*t,-16/9),OD1=(0+2*t,0),E1=(4/3+2*t,0); draw(Circle(OD,2)); draw(Circle(OA,1)); draw(Circle(OB,8/9)); draw(Circle(OC,8/9)); draw(OA--OB--OC--cycle); draw(OD--OB--OB+(OB-OD)*4/5); draw(OA--E); label("$O_{A}$",OA,(-1,0)); label("$O_{B}$",OB,(-1,1)); label("$O_{C}$",OC,(-1,-1)); label("$O_{D}$",OD,(-1,-1)); label("$E$",E,(0.5,-1)); label("$r$",OB+(OB-OD)*2/5,(-0.5,1)); label("$r$",(1*OA+3*OB)/4,(-0.5,1)); dot(OA^^OB^^OC^^OD^^E); draw(OA1--OB1--OC1--cycle); draw(OD1--OB1); draw(OA1--E1); label("$O_{A}$",OA1,(-1,0)); label("$O_{B}$",OB1,(1,1)); label("$O_{C}$",OC1,(1,-1)); label("$O_{D}$",OD1,(0,-1)); label("$E$",E1,(1,0)); label("$1+r$",(OA1+OB1)/2,(-0.5,1)); label("$r$",(E1+OB1)/2,(1,0)); label("$r$",(E1+OC1)/2,(1,0)); label("$2-r$",(OB1+OD1)/2,(-1,0)); label("$1$",(OA1+OD1)/2,(0,-1)); label("$x$",(E1+OD1)/2,(0,-1)); dot(OA1^^OB1^^OC1^^OD1^^E1); [/asy] Let $O_{i}$ be the center of circle $i$ for all $i \in \{A,B,C,D\}$ and let $E$ be the tangent point of $B,C$. Since the radius of $D$ is the diameter of $A$, the radius of $D$ is $2$. Let the radius of $B,C$ be $r$ and let $O_{D}E = x$. If we connect $O_{A},O_{B},O_{C}$, we get an isosceles triangle with lengths $1 + r, 2r$. Then right triangle $O_{D}O_{B}O_{E}$ has legs $r, x$ and hypotenuse $2-r$. Solving for $x$, we get $x^2 = (2-r)^2 - r^2 \Longrightarrow x = \sqrt{4-4r}$.

Also, right triangle $O_{A}O_{B}O_{E}$ has legs $r, 1+x$, and hypotenuse $1+r$. Solving,

\begin{eqnarray*} r^2 + (1+\sqrt{4-4r})^2 &=& (1+r)^2\\ 1+4-4r+2\sqrt{4-4r}&=& 2r + 1\\ 1-r &=& \left(\frac{6r-4}{4}\right)^2\\ \frac{9}{4}r^2-2r&=& 0\\ r &=& \frac 89 \end{eqnarray*}

So the answer is $\boxed{\mathrm{(D)}\ \frac{8}{9}}$.

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]

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*}

Solution 3

We can apply Descartes' Circle Formula.

The four circles have curvatures $-\frac{1}{2}, 1, \frac{1}{r}$, and $\frac{1}{r}$.

We have $2((-\frac{1}{2})^2+1^2+\frac {1}{r^2}+\frac{1}{r^2})=(-\frac{1}{2}+1+\frac{1}{r}+\frac{1}{r})^2$

Simplifying, we get $\frac{10}{4}+\frac{4}{r^2}=\frac{1}{4}+\frac{2}{r}+\frac{4}{r^2}$

$\frac{2}{r}=\frac{9}{4}$

$r=\frac{8}{9} \Longrightarrow \qquad \textbf{(D)}$

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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