Difference between revisions of "2004 AMC 12A Problems/Problem 19"
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+ | {{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #19]] and [[2004 AMC 10A Problems|2004 AMC 10A #23]]}} | ||
+ | |||
== Problem 19 == | == Problem 19 == | ||
Circles <math>A, 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, 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>? | ||
− | <math>\text {(A)} \frac23 \qquad \text {(B)} \frac {\sqrt3}{2} \qquad \text {(C)}\frac78 \qquad \text {(D)}\frac89 \qquad \text {(E)}\frac {1 + \sqrt3}{3}</math> | + | <center><asy> |
+ | unitsize(15mm); | ||
+ | pair A=(-1,0),B=(2/3,8/9),C=(2/3,-8/9),D=(0,0); | ||
+ | |||
+ | draw(Circle(D,2)); | ||
+ | draw(Circle(A,1)); | ||
+ | draw(Circle(B,8/9)); | ||
+ | draw(Circle(C,8/9)); | ||
+ | |||
+ | label("\(A\)", A); | ||
+ | label("\(B\)", B); | ||
+ | label("\(C\)", C); | ||
+ | label("\(D\)", (-1.2,1.8)); | ||
+ | </asy></center> | ||
+ | |||
+ | <math>\text{(A) } \frac23 \qquad \text{(B) } \frac {\sqrt3}{2} \qquad \text{(C) } \frac78 \qquad \text{(D) } \frac89 \qquad \text{(E) } \frac {1 + \sqrt3}{3}</math> | ||
+ | |||
+ | ==Video solution by Punxsutawney Phil (Private link)== | ||
+ | https://www.youtube.com/watch?v=4-lbEZkFJdc | ||
+ | |||
+ | == Solution 1 == | ||
− | |||
<asy> | <asy> | ||
− | unitsize( | + | 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, | ||
+ | |||
+ | <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> | ||
+ | unitsize(15mm); | ||
pair A=(0,1),B=(-8/9,-2/3),C=(8/9,-2/3),D=(0,0), E=(0,-2/3); | pair A=(0,1),B=(-8/9,-2/3),C=(8/9,-2/3),D=(0,0), E=(0,-2/3); | ||
Line 19: | Line 100: | ||
dot(A);dot(B);dot(C);dot(D);dot(E); | dot(A);dot(B);dot(C);dot(D);dot(E); | ||
− | label("\(D\)", D, | + | label("\(D\)", D,NW); |
label("\(A\)", A,N); | label("\(A\)", A,N); | ||
label("\(B\)", B,W); | label("\(B\)", B,W); | ||
label("\(C\)", C,E); | label("\(C\)", C,E); | ||
− | label("\(E\)", E, | + | label("\(E\)", E,SE); |
label("\(1\)",(-.4,.7)); | label("\(1\)",(-.4,.7)); | ||
− | label("\(1\)",(0,.5), | + | label("\(1\)",(0,0.5),W); |
label("\(r\)", (-.8,-.1)); | label("\(r\)", (-.8,-.1)); | ||
label("\(r\)", (-4/9,-2/3),S); | label("\(r\)", (-4/9,-2/3),S); | ||
− | label("\(h\)", (0,-1/3), | + | label("\(h\)", (0,-1/3), W); |
</asy> | </asy> | ||
+ | </center> | ||
− | Note that <math>BD= 2-r</math> since D is the center of the larger circle of radius 2 | + | Note that <math>BD= 2-r</math> since <math>D</math> is the center of the larger circle of radius <math>2</math>. Using the Pythagorean Theorem on <math>\triangle BDE</math>, |
− | |||
− | Using the Pythagorean Theorem on <math>\triangle BDE</math> | ||
− | < | + | <cmath> |
+ | \begin{align*} | ||
+ | r^2 + h^2 &= (2-r)^2 \\ | ||
+ | r^2 + h^2 &= 4 - 4r + r^2 \\ | ||
+ | h^2 &= 4 - 4r \\ | ||
+ | h &= 2\sqrt{1-r} \end{align*}</cmath> | ||
− | <math> | + | Now using the [[Pythagorean Theorem]] on <math>\triangle BAE</math>, |
− | < | + | <cmath> |
+ | \begin{align*} | ||
+ | r^2 + (h+1)^2 &= (r+1)^2 \\ | ||
+ | r^2 + h^2 + 2h + 1 &= r^2 + 2r + 1 \\ | ||
+ | h^2 + 2h &= 2r \end{align*} </cmath> | ||
− | <math>h | + | Substituting <math>h</math>, |
− | + | <cmath> | |
+ | \begin{align*} | ||
+ | (4-4r) + 4\sqrt{1-r} &= 2r \\ | ||
+ | 4\sqrt{1-r} &= 6r - 4 \\ | ||
+ | 16-16r &= 36r^2 - 48r + 16 \\ | ||
+ | 0 &= 36r^2 - 32r \\ | ||
+ | r &= \frac{32}{36} = \frac{8}{9} \Longrightarrow \qquad \textbf{(D)} \end{align*}</cmath> | ||
− | + | == Solution 3 == | |
− | + | We can apply [[Descartes' Circle Formula]]. | |
− | <math> | + | The four circles have curvatures <math>-\frac{1}{2}, 1, \frac{1}{r}</math>, and <math>\frac{1}{r}</math>. |
− | + | We have <math>2((-\frac{1}{2})^2+1^2+\frac {1}{r^2}+\frac{1}{r^2})=(-\frac{1}{2}+1+\frac{1}{r}+\frac{1}{r})^2</math> | |
− | <math> | + | Simplifying, we get <math>\frac{10}{4}+\frac{4}{r^2}=\frac{1}{4}+\frac{2}{r}+\frac{4}{r^2}</math> |
− | <math> | + | <math>\frac{2}{r}=\frac{9}{4}</math> |
− | <math> | + | <math>r=\frac{8}{9} \Longrightarrow \qquad \textbf{(D)}</math> |
− | + | ==See Also== | |
+ | {{AMC12 box|year=2004|ab=A|num-b=18|num-a=20}} | ||
+ | {{AMC10 box|year=2004|ab=A|num-b=22|num-a=24}} | ||
− | + | [[Category:Introductory Geometry Problems]] | |
+ | {{MAA Notice}} |
Latest revision as of 20:46, 2 October 2024
- The following problem is from both the 2004 AMC 12A #19 and 2004 AMC 10A #23, so both problems redirect to this page.
Contents
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 ?
Video solution by Punxsutawney Phil (Private link)
https://www.youtube.com/watch?v=4-lbEZkFJdc
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.