# 2005 AMC 10B Problems/Problem 7

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

## 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 $$\frac{\pi r^2}{\left(2\sqrt{2}r\right)^2} =\frac{\pi}{8}\implies \boxed{\mathrm B}.$$ $[asy] draw(Circle((0,0),10),linewidth(0.7)); draw(Circle((0,0),14.1),linewidth(0.7)); draw((0,14.1)--(14.1,0)--(0,-14.1)--(-14.1,0)--cycle,linewidth(0.7)); draw((-14.1,14.1)--(14.1,14.1)--(14.1,-14.1)--(-14.1,-14.1)--cycle,linewidth(0.7)); draw((0,0)--(-14.1,0),linewidth(0.7)); draw((-7.1,7.1)--(0,0),linewidth(0.7)); label("\sqrt{2}r",(-6,0),S); label("r",(-3.5,3.5),NE); label("2r",(-7.1,7.1),W); label("2\sqrt{2}r",(0,14.1),N); [/asy]$ When facing a geometry problem, it is very helpful to draw a diagram.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 