Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
Line 2: Line 2:
 
Circles <math>A</math>, <math>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>?
 
Circles <math>A</math>, <math>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>?
  
−
<center>[[Image:AMC10_2004A_23.png]]</center>
+
<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>
  
 
<math> \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} </math>
 
<math> \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} </math>
  
 
==Solution==
 
==Solution==
−
Let <math>O</math> be the center of <math>D</math>, and <math>E</math> be the intersection 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>. If we connect the centers of the circles <math>A, B, C</math> (we will denote these as <math>A_1, B_1, C_1</math>, we get an [[isosceles triangle]] with lengths <math>1 + r, 2r</math>. Also, <math>B_1E</math> is the difference between the radius of <math>D</math>, <math>2</math>, and <math>r</math>, so right <math>\triangle OB_1E</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 \Longrightarow x = \sqrt{4-4r}</math>.
 
  
−
Also, right triangle <math>A_1B_1E</math> has legs <math>r, 1+x</math>, and hypotenuse <math>1+r</math>. Solving,  
+
<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}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}E</math> has legs <math>r, 1+x</math>, and hypotenuse <math>1+r</math>. Solving,  
  
 
<cmath>\begin{eqnarray*}
 
<cmath>\begin{eqnarray*}

Revision as of 12:43, 13 January 2012

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

[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 $O_{D}O_{B}E$ 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 $O_{A}O_{B}E$ 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 $\mathrm{(D)}$.

See also

  • <url>viewtopic.php?t=131335 AoPS topic</url>
2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 10 Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10A_Problems/Problem_23&oldid=44229"