Difference between revisions of "2005 AMC 12A Problems/Problem 16"

 
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== Problem ==
 
== Problem ==
Three [[circle]]s of [[radius]] <math>s</math> are drawn in the first [[quadrant]] of the <math>xy</math>-[[plane]]. The first circle is tangent to both axes, the second is [[tangent (geometry)|tangent]] to the first circle and the <math>x</math>-axis, and the third is tangent to the first circle and the <math>y</math>-axis. A circle of radius <math>r > s</math> is tangent to both axes and to the second and third circles. What is <math>\frac{r}{s}</math>?
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Three [[circle]]s of [[radius]] <math>1</math> are drawn in the first [[quadrant]] of the <math>xy</math>-[[plane]]. The first circle is tangent to both axes, the second is [[tangent (geometry)|tangent]] to the first circle and the <math>x</math>-axis, and the third is tangent to the first circle and the <math>y</math>-axis. A circle of radius <math>r > 1</math> is tangent to both axes and to the second and third circles. What is <math>r</math>?
  
 
<asy>
 
<asy>
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[[Image:2005_12A_AMC-16b.png]]
 
[[Image:2005_12A_AMC-16b.png]]
  
[[Without loss of generality]], let <math>s = 1</math>. Draw the segment between the center of the third circle and the large circle; this has length <math>r+1</math>. We then draw the [[radius]] of the large circle that is perpendicular to the [[x-axis]], and draw the perpendicular from this radius to the center of the third circle. This gives us a [[right triangle]] with legs <math>r-3,r-1</math> and [[hypotenuse]] <math>r+1</math>. The [[Pythagorean Theorem]] yields:
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Draw the segment between the center of the third circle and the large circle; this has length <math>r+1</math>. We then draw the [[radius]] of the large circle that is perpendicular to the [[x-axis]], and draw the perpendicular from this radius to the center of the third circle. This gives us a [[right triangle]] with legs <math>r-3,r-1</math> and [[hypotenuse]] <math>r+1</math>. The [[Pythagorean Theorem]] yields:
  
 
<div style="text-align:center;"><math>(r-3)^2 + (r-1)^2 = (r+1)^2</math><br /><math>r^2 - 10r + 9 = 0</math><br /><math>r = 1, 9</math></div>
 
<div style="text-align:center;"><math>(r-3)^2 + (r-1)^2 = (r+1)^2</math><br /><math>r^2 - 10r + 9 = 0</math><br /><math>r = 1, 9</math></div>
  
Quite obviously <math>r > s = 1</math>, so <math>r = 9</math> and <math>\frac rs = \frac 91 = 9 \Longrightarrow \mathrm{(D)}</math>.
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Quite obviously <math>r > 1</math>, so <math>r = 9 \boxed{(D)}</math>.
  
 
===Solution 2===
 
===Solution 2===
  
Don't do this unless really really desperate. But I actually solved this with a ruler (try and see!!). Let <math>s=1</math> and find <math>r</math> in terms of <math>s</math>. The rest is easy.  
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Don't do this unless really really desperate. But I actually solved this with a ruler (try and see!!). Find <math>r</math>. The rest is easy.  
  
 
Solution by franzliszt
 
Solution by franzliszt

Latest revision as of 13:28, 19 September 2020

Problem

Three circles of radius $1$ are drawn in the first quadrant of the $xy$-plane. The first circle is tangent to both axes, the second is tangent to the first circle and the $x$-axis, and the third is tangent to the first circle and the $y$-axis. A circle of radius $r > 1$ is tangent to both axes and to the second and third circles. What is $r$?

[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; pair O0=(9,9), O1=(1,1), O2=(3,1), O3=(1,3); pair P0=O0+9*dir(-45), P3=O3+dir(70); pair[] ps={O0,O1,O2,O3}; dot(ps); draw(Circle(O0,9)); draw(Circle(O1,1)); draw(Circle(O2,1)); draw(Circle(O3,1)); draw(O0--P0,linetype("3 3")); draw(O3--P3,linetype("2 2")); draw((0,0)--(18,0)); draw((0,0)--(0,18)); label("$r$",midpoint(O0--P0),NE); label("$s$",(-1.5,4)); draw((-1,4)--midpoint(O3--P3));[/asy]

$(\mathrm {A}) \ 5 \qquad (\mathrm {B}) \ 6 \qquad (\mathrm {C})\ 8 \qquad (\mathrm {D}) \ 9 \qquad (\mathrm {E})\ 10$

Solution

Solution 1

2005 12A AMC-16b.png

Draw the segment between the center of the third circle and the large circle; this has length $r+1$. We then draw the radius of the large circle that is perpendicular to the x-axis, and draw the perpendicular from this radius to the center of the third circle. This gives us a right triangle with legs $r-3,r-1$ and hypotenuse $r+1$. The Pythagorean Theorem yields:

$(r-3)^2 + (r-1)^2 = (r+1)^2$
$r^2 - 10r + 9 = 0$
$r = 1, 9$

Quite obviously $r > 1$, so $r = 9 \boxed{(D)}$.

Solution 2

Don't do this unless really really desperate. But I actually solved this with a ruler (try and see!!). Find $r$. The rest is easy.

Solution by franzliszt

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