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),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("<math>\sqrt{2}r</math>",(-6,0),S); label("<math>r</math>",(-3.5,3.5),NE); label("<math>2r</math>",(-7.1,7.1),W); label("<math>2\sqrt{2}r</math>",(0,14.1),N); [/asy]
+
[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("<math>\sqrt{2}r</math>",(-6,0),S);
 +
label("<math>r</math>",(-3.5,3.5),NE);
 +
label("<math>2r</math>",(-7.1,7.1),W);
 +
label("<math>2\sqrt{2}r</math>",(0,14.1),N);
 +
[/asy]
  
 
== See Also ==
 
== See Also ==

Revision as of 22:15, 26 April 2016

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?

$\mathrm{(A)} \frac{\pi}{16} \qquad \mathrm{(B)} \frac{\pi}{8} \qquad \mathrm{(C)} \frac{3\pi}{16} \qquad \mathrm{(D)} \frac{\pi}{4} \qquad \mathrm{(E)} \frac{\pi}{2}$

Solution 1

Let the side of the largest square be $x$. It follows that the diameter of the inscribed circle is also $x$. Therefore, the diagonal of the square inscribed inscribed in the circle is $x$. The side length of the smaller square is $\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}$. Similarly, the diameter of the smaller inscribed circle is $\dfrac{x\sqrt{2}}{2}$. Hence, its radius is $\dfrac{x\sqrt{2}}{4}$. The area of this circle is $\left(\dfrac{x\sqrt{2}}{4}\right)^2\pi=\dfrac{2\pi x^2}{16}=\dfrac{x^2\pi}{8}$, and the area of the largest square is $x^2$. The ratio of the areas is $\dfrac{\dfrac{x^2\pi}{8}}{x^2}=\boxed{\mathrm{(B)}\ \dfrac{\pi}{8}}$.

Solution 2

Let the radius of the smaller circle be $r$. Then the side length of the smaller square is $2r$. The radius of the larger circle is half the length of the diagonal of the smaller square, so it is $\sqrt{2}r$. Hence the larger square has sides of length $2\sqrt{2}r$. The ratio of the area of the smaller circle to the area of the larger square is therefore \[\frac{\pi r^2}{\left(2\sqrt{2}r\right)^2} =\boxed{\frac{\pi}{8}}.\]

[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]

See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
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. AMC logo.png