# 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>? | ||

− | < | + | <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> | ||

− | + | <math> (\mathrm {A}) \ 5 \qquad (\mathrm {B}) \ 6 \qquad (\mathrm {C})\ 8 \qquad (\mathrm {D}) \ 9 \qquad (\mathrm {E})\ 10 </math> | |

== Solution == | == Solution == | ||

+ | |||

+ | ===Solution 1=== | ||

[[Image:2005_12A_AMC-16b.png]] | [[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> | <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 > | + | Quite obviously <math>r > 1</math>, so <math>r = 9 \boxed{(D)}</math>. |

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## Latest revision as of 01:56, 10 January 2021

## Problem

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

## Solution

### Solution 1

Set so that we only have to find . Draw the segment between the center of the third circle and the large circle; this has length . 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 and hypotenuse . The Pythagorean Theorem yields:

Quite obviously , so .