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</math>. |
== See also == | == See also == | ||
− | {{AMC12 box|year=2005 | + | {{AMC12 box|year=2005|num-b=15|num-a=17|ab=A}} |
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 15:30, 30 March 2024
Contents
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:
![$(r-3)^2 + (r-1)^2 = (r+1)^2$](http://latex.artofproblemsolving.com/a/2/3/a234f93c17da5e55cc6d61dcb2dd96c94b967a8b.png)
![$r^2 - 10r + 9 = 0$](http://latex.artofproblemsolving.com/c/f/5/cf5947e2aa4bf9c1f4187105b5fcc854f8bddee5.png)
![$r = 1, 9$](http://latex.artofproblemsolving.com/4/8/2/482f67dafc8ed1baa5cf823440067133e5dc45fa.png)
Quite obviously , so
.
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.