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| − | ==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>?
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| − | <asy>
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| − | import graph;
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| − | size(150);
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| − | defaultpen(fontsize(8));
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| − | pair OA=(-1,0),OB=(2/3,8/9),OC=(2/3,-8/9),OD=(0,0);
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| − | draw(Circle(OD,2));
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| − | draw(Circle(OA,1));
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| − | draw(Circle(OB,8/9));
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| − | draw(Circle(OC,8/9));
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| − | label("$A$",OA+expi(pi/2),(0,1));
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| − | label("$B$",OB+8/9*expi(pi/180*130),(-1,0.5));
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| − | label("$C$",OC+8/9*expi(pi/180*230),(-1,-0.5));
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| − | label("$D$",OD+2*expi(2*pi/3),(-1,1));
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| − | </asy>
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| − | | |
| − | <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>
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| − | | |
| − | ==Solution 1==
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| − | | |
| − | <asy>
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| − | import graph;
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| − | size(400);
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| − | defaultpen(fontsize(10));
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| − | pair OA=(-1,0),OB=(2/3,8/9),OC=(2/3,-8/9),OD=(0,0),E=(2/3,0);
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| − | real t = 2.5;
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| − | 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);
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| − | draw(Circle(OD,2));
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| − | draw(Circle(OA,1));
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| − | draw(Circle(OB,8/9));
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| − | draw(Circle(OC,8/9));
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| − | draw(OA--OB--OC--cycle);
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| − | draw(OD--OB--OB+(OB-OD)*4/5);
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| − | draw(OA--E);
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| − | label("$O_{A}$",OA,(-1,0));
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| − | label("$O_{B}$",OB,(-1,1));
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| − | label("$O_{C}$",OC,(-1,-1));
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| − | label("$O_{D}$",OD,(-1,-1));
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| − | label("$E$",E,(0.5,-1));
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| − | label("$r$",OB+(OB-OD)*2/5,(-0.5,1));
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| − | label("$r$",(1*OA+3*OB)/4,(-0.5,1));
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| − | dot(OA^^OB^^OC^^OD^^E);
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| − | draw(OA1--OB1--OC1--cycle);
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| − | draw(OD1--OB1);
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| − | draw(OA1--E1);
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| − | label("$O_{A}$",OA1,(-1,0));
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| − | label("$O_{B}$",OB1,(1,1));
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| − | label("$O_{C}$",OC1,(1,-1));
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| − | label("$O_{D}$",OD1,(0,-1));
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| − | label("$E$",E1,(1,0));
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| − | label("$1+r$",(OA1+OB1)/2,(-0.5,1));
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| − | label("$r$",(E1+OB1)/2,(1,0));
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| − | label("$r$",(E1+OC1)/2,(1,0));
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| − | label("$2-r$",(OB1+OD1)/2,(-1,0));
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| − | label("$1$",(OA1+OD1)/2,(0,-1));
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| − | label("$x$",(E1+OD1)/2,(0,-1));
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| − | dot(OA1^^OB1^^OC1^^OD1^^E1);
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| − | </asy>
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| − | 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>.
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| − | 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,
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| − | <cmath>\begin{eqnarray*}
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| − | r^2 + (1+\sqrt{4-4r})^2 &=& (1+r)^2\\
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| − | 1+4-4r+2\sqrt{4-4r}&=& 2r + 1\\
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| − | 1-r &=& \left(\frac{6r-4}{4}\right)^2\\
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| − | \frac{9}{4}r^2-2r&=& 0\\
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| − | r &=& \frac 89
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| − | \end{eqnarray*}</cmath>
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| − |
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| − | So the answer is <math>\boxed{\mathrm{(D)}\ \frac{8}{9}}</math>.
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| − | ==Solution 2==
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| − | 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>.
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| − | | |
| − | == See also ==
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| − | * <url>viewtopic.php?t=131335 AoPS topic</url>
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| − | {{AMC10 box|year=2004|ab=A|num-b=22|num-a=24}}
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| − | | |
| − | [[Category:Introductory Geometry Problems]]
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| − | {{MAA Notice}}
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