Difference between revisions of "2005 AMC 10B Problems/Problem 7"
(→Solution 2) |
(→Solution 2) |
||
| Line 9: | Line 9: | ||
Let the radius of the smaller circle be <math>r</math>. Then the side length of the smaller square is <math>2r</math>. The radius of the larger circle is half the length of the diagonal of the smaller square, so it is <math>\sqrt{2}r</math>. Hence the larger square has sides of length <math>2\sqrt{2}r</math>. The ratio of the area of the smaller circle to the area of the larger square is therefore <cmath>\frac{\pi r^2}{\left(2\sqrt{2}r\right)^2} =\boxed{\frac{\pi}{8}}.</cmath> | Let the radius of the smaller circle be <math>r</math>. Then the side length of the smaller square is <math>2r</math>. The radius of the larger circle is half the length of the diagonal of the smaller square, so it is <math>\sqrt{2}r</math>. Hence the larger square has sides of length <math>2\sqrt{2}r</math>. The ratio of the area of the smaller circle to the area of the larger square is therefore <cmath>\frac{\pi r^2}{\left(2\sqrt{2}r\right)^2} =\boxed{\frac{\pi}{8}}.</cmath> | ||
| − | + | <asy> | |
draw(Circle((0,0),10),linewidth(0.7)); | draw(Circle((0,0),10),linewidth(0.7)); | ||
draw(Circle((0,0),14.1),linewidth(0.7)); | draw(Circle((0,0),14.1),linewidth(0.7)); | ||
| Line 16: | Line 16: | ||
draw((0,0)--(-14.1,0),linewidth(0.7)); | draw((0,0)--(-14.1,0),linewidth(0.7)); | ||
draw((-7.1,7.1)--(0,0),linewidth(0.7)); | draw((-7.1,7.1)--(0,0),linewidth(0.7)); | ||
| − | label(" | + | label("$\sqrt{2}r$",(-6,0),S); |
| − | label(" | + | label("$r$",(-3.5,3.5),NE); |
| − | label(" | + | label("$2r$",(-7.1,7.1),W); |
| − | label(" | + | label("$2\sqrt{2}r$",(0,14.1),N); |
| − | + | </asy> | |
== See Also == | == See Also == | ||
Revision as of 18:39, 5 October 2016
Contents
Problem
A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smaller circle to the area of the larger square?
Solution 1
Let the side of the largest square be 
. It follows that the diameter of the inscribed circle is also . Therefore, the diagonal of the square inscribed inscribed in the circle is . The side length of the smaller square is 
. Similarly, the diameter of the smaller inscribed circle is 
. Hence, its radius is 
. The area of this circle is 
, and the area of the largest square is 
. The ratio of the areas is 
.
Solution 2
Let the radius of the smaller circle be 
. Then the side length of the smaller square is 
. The radius of the larger circle is half the length of the diagonal of the smaller square, so it is 
. Hence the larger square has sides of length 
. The ratio of the area of the smaller circle to the area of the larger square is therefore ![\[\frac{\pi r^2}{\left(2\sqrt{2}r\right)^2} =\boxed{\frac{\pi}{8}}.\]](http://latex.artofproblemsolving.com/f/5/b/f5bdcf6e704b50cde17dc3e41b8e9137a6e49ac6.png)
![[asy] draw(Circle((0,0),10),linewidth(0.7)); draw(Circle((0,0),14.1),linewidth(0.7)); draw((0,14.1)--(14.1,0)--(0,-14.1)--(-14.1,0)--cycle,linewidth(0.7)); draw((-14.1,14.1)--(14.1,14.1)--(14.1,-14.1)--(-14.1,-14.1)--cycle,linewidth(0.7)); draw((0,0)--(-14.1,0),linewidth(0.7)); draw((-7.1,7.1)--(0,0),linewidth(0.7)); label("$\sqrt{2}r$",(-6,0),S); label("$r$",(-3.5,3.5),NE); label("$2r$",(-7.1,7.1),W); label("$2\sqrt{2}r$",(0,14.1),N); [/asy]](http://latex.artofproblemsolving.com/7/a/7/7a7cd0405ef0ae3c41933c15d7c8e2157d14ef4d.png)
See Also
| 2005 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
