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Difference between revisions of "2005 AMC 10B Problems/Problem 7"

 
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== Problem ==
 
== Problem ==
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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?
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<math>\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} </math>
 
== Solution ==
 
== Solution ==
 +
Let the side of the largest square be <math>x</math>. It follows that the diameter of the inscribed circle is also <math>x</math>. Therefore, the diagonal of the square inscribed inscribed in the circle is <math>x</math>. The side length of the smaller square is <math>\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}</math>. Similarly, the diameter of the smaller inscribed circle is <math>\dfrac{x\sqrt{2}}{2}</math>. Hence, its radius is <math>\dfrac{x\sqrt{2}}{4}</math>. The area of this circle is <math>\left(\dfrac{x\sqrt{2}}{4}\right)^2\pi=\dfrac{2\pi x^2}{16}=\dfrac{x^2\pi}{8}</math>, and the area of the largest square is <math>x^2</math>. The ratio of the areas is <math>\dfrac{\dfrac{x^2\pi}{8}}{x^2}=\boxed{\mathrm{(B)}\ \dfrac{\pi}{8}}</math>.
 
== See Also ==
 
== See Also ==
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*[[2005 AMC 10B Problems]]
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{{AMC10 box|year=2005|ab=B|num-b=6|num-a=8}}

Revision as of 17:35, 9 July 2011

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

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 $\dfrac{\dfrac{x^2\pi}{8}}{x^2}=\boxed{\mathrm{(B)}\ \dfrac{\pi}{8}}$.

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
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