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

(Redirected page to 2004 AMC 12A Problems/Problem 19)
(Tag: New redirect)
 
(11 intermediate revisions by 8 users not shown)
Line 1: Line 1:
==Problem==
+
#redirect [[2004 AMC 12A Problems/Problem 19]]
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>?
 
 
 
<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>
 
 
 
==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 <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*}
 
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>\mathrm{(D)}</math>.
 
 
 
== See also ==
 
* <url>viewtopic.php?t=131335 AoPS topic</url>
 
{{AMC10 box|year=2004|ab=A|num-b=22|num-a=24}}
 
 
 
[[Category:Introductory Geometry Problems]]
 
{{MAA Notice}}
 

Latest revision as of 13:52, 17 August 2020