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

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<math> (\mathrm {A}) \ 5 \qquad (\mathrm {B}) \ 6 \qquad (\mathrm {C})\ 8 \qquad (\mathrm {D}) \ 9 \qquad (\mathrm {E})\ 10 </math>
 
<math> (\mathrm {A}) \ 5 \qquad (\mathrm {B}) \ 6 \qquad (\mathrm {C})\ 8 \qquad (\mathrm {D}) \ 9 \qquad (\mathrm {E})\ 10 </math>
 
== Solution ==
 
 
===Solution 1===
 
[[Image:2005_12A_AMC-16b.png]]
 
 
Set <math>s =1</math> so that we only have to find <math>r</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:
 
 
<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 > 1</math>, so <math>r = 9 \boxed{(D)}</math>.
 
 
===Solution 2===
 
Applying [[Wildin's Theorem]] directly 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>
 
 
As before, <math>r > 1</math>, so <math>r = 9 \boxed{(D)}</math>.
 
  
 
== See also ==
 
== See also ==

Latest revision as of 15:22, 19 August 2024

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]

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

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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