Difference between revisions of "2001 AMC 12 Problems/Problem 18"
(New page: == Problem == A circle centered at <math>A</math> with a radius of 1 and a circle centered at <math>B</math> with a radius of 4 are externally tangent. A third circle is tangent to the fi...) |
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== Problem == | == Problem == | ||
− | A circle centered at <math>A</math> with a radius of 1 and a circle centered at <math>B</math> with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. | + | A circle centered at <math>A</math> with a radius of 1 and a circle centered at <math>B</math> with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. What is the radius of the third circle? |
<asy> | <asy> | ||
Line 32: | Line 32: | ||
\text{(E) }\frac {1}{2} | \text{(E) }\frac {1}{2} | ||
</math> | </math> | ||
+ | |||
+ | [[Category: Introductory Geometry Problems]] | ||
== Solution == | == Solution == | ||
+ | |||
+ | === Solution 1 === | ||
<asy> | <asy> | ||
unitsize(1cm); | unitsize(1cm); | ||
− | pair A=(0,1), B=(4,4), C=(4,1); | + | pair A=(0,1), B=(4,4), C=(4,1), S=(12/9,4/9); |
dot(A); dot(B); | dot(A); dot(B); | ||
draw( circle(A,1) ); | draw( circle(A,1) ); | ||
Line 47: | Line 51: | ||
label("$B$",B,N); | label("$B$",B,N); | ||
label("$C$",C,E); | label("$C$",C,E); | ||
+ | label("$S$",S,N); | ||
− | filldraw( circle( | + | filldraw( circle(S,4/9), lightgray, black ); |
− | dot( | + | dot(S); |
draw( rightanglemark(A,C,B) ); | draw( rightanglemark(A,C,B) ); | ||
+ | draw( S -- A ); | ||
+ | draw( S -- B ); | ||
</asy> | </asy> | ||
Line 83: | Line 90: | ||
Now there are two possibilities: either <math>\frac{4-s}s=-2</math>, or <math>\frac{4-s}s=2</math>. In the first case clearly <math>s<0</math>, hence this is not the correct case. (Note: This case corresponds to the other circle that is tangent to both given circles and the <math>x</math> axis - a large circle whose center is somewhere to the left of <math>A</math>.) The second case solves to <math>s=\frac 43</math>. We then have <math>4r = s^2 = \frac {16}9</math>, hence <math>r = \boxed{\frac 49}</math>. | Now there are two possibilities: either <math>\frac{4-s}s=-2</math>, or <math>\frac{4-s}s=2</math>. In the first case clearly <math>s<0</math>, hence this is not the correct case. (Note: This case corresponds to the other circle that is tangent to both given circles and the <math>x</math> axis - a large circle whose center is somewhere to the left of <math>A</math>.) The second case solves to <math>s=\frac 43</math>. We then have <math>4r = s^2 = \frac {16}9</math>, hence <math>r = \boxed{\frac 49}</math>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | |||
+ | The horizontal line is the equivalent of a circle of curvature <math>0</math>, thus we can apply [[Descartes' Circle Formula]]. | ||
+ | |||
+ | The four circles have curvatures <math>0, 1, \frac 14</math>, and <math>\frac 1r</math>. | ||
+ | |||
+ | We have <math>2\left(0^2+1^2+\frac {1}{4^2}+\frac{1}{r^2}\right)=\left(0+1+\frac 14+\frac 1r\right)^2</math> | ||
+ | |||
+ | Simplifying, we get <math>\frac{34}{16}+\frac{2}{r^2}=\frac{25}{16}+\frac{5}{2r}+\frac{1}{r^2}</math> | ||
+ | |||
+ | <cmath>\frac{1}{r^2}-\frac{5}{2r}+\frac{9}{16}=0</cmath> | ||
+ | <cmath>\frac{16}{r^2}-\frac{40}{r}+9=0</cmath> | ||
+ | <cmath>\left(\frac{4}{r}-9\right)\left(\frac{4}{r}-1\right)=0</cmath> | ||
+ | |||
+ | Obviously <math>r</math> cannot equal <math>4</math>, therefore <math>r = \boxed{\frac 49}</math>. | ||
+ | |||
+ | === Video Solution === | ||
+ | https://youtu.be/zOwYoFOUg2U | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2001|num-b=17|num-a=19}} | {{AMC12 box|year=2001|num-b=17|num-a=19}} | ||
+ | {{MAA Notice}} |
Latest revision as of 11:55, 30 November 2021
Problem
A circle centered at with a radius of 1 and a circle centered at with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. What is the radius of the third circle?
Solution
Solution 1
In the triangle we have and , thus by the Pythagorean theorem we have .
We can now pick a coordinate system where the common tangent is the axis and lies on the axis. In this coordinate system we have and .
Let be the radius of the small circle, and let be the -coordinate of its center . We then know that , as the circle is tangent to the axis. Moreover, the small circle is tangent to both other circles, hence we have and .
We have and . Hence we get the following two equations:
Simplifying both, we get
As in our case both and are positive, we can divide the second one by the first one to get .
Now there are two possibilities: either , or . In the first case clearly , hence this is not the correct case. (Note: This case corresponds to the other circle that is tangent to both given circles and the axis - a large circle whose center is somewhere to the left of .) The second case solves to . We then have , hence .
Solution 2
The horizontal line is the equivalent of a circle of curvature , thus we can apply Descartes' Circle Formula.
The four circles have curvatures , and .
We have
Simplifying, we get
Obviously cannot equal , therefore .
Video Solution
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 |
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