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]]}} | ||
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== 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>? | |
<center><asy> | <center><asy> | ||
Line 14: | Line 16: | ||
label("\(B\)", B); | label("\(B\)", B); | ||
label("\(C\)", C); | label("\(C\)", C); | ||
− | label("D", (-1.2,1.8)); | + | label("\(D\)", (-1.2,1.8)); |
</asy></center> | </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> | + | <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== |
− | === Solution 1 | + | https://www.youtube.com/watch?v=4-lbEZkFJdc |
+ | |||
+ | == Solution 1 == | ||
<asy> | <asy> | ||
Line 60: | Line 64: | ||
dot(OA1^^OB1^^OC1^^OD1^^E1); | dot(OA1^^OB1^^OC1^^OD1^^E1); | ||
</asy> | </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>. | + | |
+ | 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, | 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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So the answer is <math>\boxed{\mathrm{(D)}\ \frac{8}{9}}</math>. | So the answer is <math>\boxed{\mathrm{(D)}\ \frac{8}{9}}</math>. | ||
− | + | ||
+ | == Solution 2 == | ||
<center><asy> | <center><asy> | ||
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label("\(r\)", (-4/9,-2/3),S); | label("\(r\)", (-4/9,-2/3),S); | ||
label("\(h\)", (0,-1/3), W); | label("\(h\)", (0,-1/3), W); | ||
− | </asy></center> | + | </asy> |
+ | </center> | ||
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>, | 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>, | ||
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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 3 == | |
We can apply [[Descartes' Circle Formula]]. | We can apply [[Descartes' Circle Formula]]. | ||
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2004|ab=A|num-b=18|num-a=20}} | {{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]] | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:29, 31 July 2023
- 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
?
![[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]](http://latex.artofproblemsolving.com/6/9/6/69677da9f9025d93eca8bc2a4d63d4a3b5b11c9c.png)
Video solution by Punxsutawney Phil
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
![[asy] unitsize(15mm); pair A=(0,1),B=(-8/9,-2/3),C=(8/9,-2/3),D=(0,0), E=(0,-2/3); draw(Circle(D,2)); draw(Circle(A,1)); draw(Circle(B,8/9)); draw(Circle(C,8/9)); draw(A--B--C--A); draw(B--D--C); draw(A--E); dot(A);dot(B);dot(C);dot(D);dot(E); label("\(D\)", D,NW); label("\(A\)", A,N); label("\(B\)", B,W); label("\(C\)", C,E); label("\(E\)", E,SE); label("\(1\)",(-.4,.7)); label("\(1\)",(0,0.5),W); label("\(r\)", (-.8,-.1)); label("\(r\)", (-4/9,-2/3),S); label("\(h\)", (0,-1/3), W); [/asy]](http://latex.artofproblemsolving.com/e/2/1/e21e9f6600899f824d87d28a241a4d9bfb5ee11a.png)
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.