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Difference between revisions of "2004 AMC 10A Problems/Problem 23"

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So the answer is <math>\boxed{\mathrm{(D)}\ \frac{8}{9}}</math>.
 
So the answer is <math>\boxed{\mathrm{(D)}\ \frac{8}{9}}</math>.
 
==Solution 2==
 
==Solution 2==
−
Using [[Descartes' Circle Formula]], <math>\left(1-\frac{1}{2}(\frac{1}{r}+\frac{1}{r}\right)^2=2\left(\frac{1}{4}+1+\frac{1}{r^2}+\frac{1}{r^2}\right)</math>. Solving this gives us linear equation with <math>r=\boxed{\mathrm{(D)}\ \frac{8}{9}}</math>.
+
Using [[Descartes' Circle Formula]], <math>\left(1-\frac{1}{2}+\frac{1}{r}+\frac{1}{r}\right)^2=2\left(\frac{1}{4}+1+\frac{1}{r^2}+\frac{1}{r^2}\right)</math>. Solving this gives us linear equation with <math>r=\boxed{\mathrm{(D)}\ \frac{8}{9}}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 12:59, 8 February 2016

Problem

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] import graph; size(150); defaultpen(fontsize(8)); pair OA=(-1,0),OB=(2/3,8/9),OC=(2/3,-8/9),OD=(0,0); draw(Circle(OD,2)); draw(Circle(OA,1)); draw(Circle(OB,8/9)); draw(Circle(OC,8/9)); label("$A$",OA+expi(pi/2),(0,1)); label("$B$",OB+8/9*expi(pi/180*130),(-1,0.5)); label("$C$",OC+8/9*expi(pi/180*230),(-1,-0.5)); label("$D$",OD+2*expi(2*pi/3),(-1,1)); [/asy]

$\mathrm{(A) \ } \frac{2}{3} \qquad \mathrm{(B) \ } \frac{\sqrt{3}}{2} \qquad \mathrm{(C) \ } \frac{7}{8} \qquad \mathrm{(D) \ } \frac{8}{9} \qquad \mathrm{(E) \ } \frac{1+\sqrt{3}}{3}$

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

Using Descartes' Circle Formula, $\left(1-\frac{1}{2}+\frac{1}{r}+\frac{1}{r}\right)^2=2\left(\frac{1}{4}+1+\frac{1}{r^2}+\frac{1}{r^2}\right)$. Solving this gives us linear equation with $r=\boxed{\mathrm{(D)}\ \frac{8}{9}}$.

See also

  • <url>viewtopic.php?t=131335 AoPS topic</url>
2004 AMC 10A (Problems • Answer Key • Resources)
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