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

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$",(-1.5,4)); draw((-1,4)--midpoint(O3--P3));[/asy]$

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

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.

Page

Toolbox

Search

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

Revision as of 15:22, 19 August 2024

Problem

See also

@@ Line 2: / Line 2: @@
 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);
+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>
-[[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 ==
-[[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:
-<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>.
 == See also ==
-{{AMC12 box|year=2005|ab=A|num-b=15|num-a=17}}
+{{AMC12 box|year=2005|num-b=15|num-a=17|ab=A}}
 [[Category:Introductory Geometry Problems]]
+{{MAA Notice}}