Difference between revisions of "2004 AMC 12A Problems/Problem 19"
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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 | + | <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> | ||
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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 | + | === Solution 3 === |
We can apply [[Descartes' Circle Formula]]. | We can apply [[Descartes' Circle Formula]]. |
Revision as of 18:32, 5 November 2017
Problem 19
Circles and are externally tangent to each other, and internally tangent to circle . Circles and are congruent. Circle has radius and passes through the center of . What is the radius of circle ?
Solution
Solution 1
Let be the center of circle for all and let be the tangent point of . Since the radius of is the diameter of , the radius of is . Let the radius of be and let . If we connect , we get an isosceles triangle with lengths . Then right triangle has legs and hypotenuse . Solving for , we get .
Also, right triangle has legs , and hypotenuse . Solving,
So the answer is .
Solution 2
Note that since is the center of the larger circle of radius . Using the Pythagorean Theorem on ,
Now using the Pythagorean Theorem on ,
Substituting ,
Solution 3
We can apply Descartes' Circle Formula.
The four circles have curvatures , and .
We have
Simplifying, we get
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.