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

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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>r/s</math>?
 
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>r/s</math>?
  
<math>
+
<asy>
(\mathrm {A}) \ 5 \qquad (\mathrm {B}) \ 6 \qquad (\mathrm {C})\ 8 \qquad (\mathrm {D}) \ 9 \qquad (\mathrm {E})\ 10
+
unitsize(3mm);
</math>
+
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);
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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));
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label("$r$",midpoint(O0--P0),NE);
 +
label("$s$",(-1.5,4));
 +
draw((-1,4)--midpoint(O3--P3));</asy>
  
[[Image:2005_12A_AMC-16.png]]
+
<math> (\mathrm {A}) \ 5 \qquad (\mathrm {B}) \ 6 \qquad (\mathrm {C})\ 8 \qquad (\mathrm {D}) \ 9 \qquad (\mathrm {E})\ 10 </math>
  
 
== Solution ==
 
== Solution ==

Revision as of 16:29, 24 November 2011

Problem

Three circles of radius $s$ 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 $r > s$ is tangent to both axes and to the second and third circles. What is $r/s$?

[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

2005 12A AMC-16b.png

Without loss of generality, let $s = 1$. 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 $r > s = 1$, so $r = 9$ and $\frac rs = \frac 91 = 9 \Longrightarrow \mathrm{(D)}$.

See also

2005 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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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