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# 2011 AMC 8 Problems/Problem 25

## Problem

A circle with radius $1$ is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?

$[asy] filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,white,black); filldraw(Circle((0,0),1), mediumgray,black); filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);[/asy]$

$\textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2$

## Solution 1

The area of the smaller square is the one half of the product of its diagonals. Note that the distance from a corner of the smaller square to the center is equivalent to the circle's radius so the diagonal is equal to the diameter: $2*2*1/2=2.$

The circle's shaded area is the area of the smaller square subtracted from the area of the circle: $\pi - 2.$

If you draw the diagonals of the smaller square, you will see that the larger square is split $4$ congruent half-shaded squares. The area between the squares is equal to the area of the smaller square: $2.$

Approximating $\pi$ to $3.14,$ the ratio of the circle's shaded area to the area between the two squares is about

$$\frac{\pi-2}{2} \approx \frac{3.14-2}{2} = \frac{1.14}{2} \approx \boxed{\textbf{(A)}\ \frac12}$$

## Solution 2

For the ratio of the circle's shaded area to the area between the squares to be $1,$ they would have to be approximately the same size. For any ratio larger than that, the circle's shaded area must be greater. However, we can clearly see that the circle's shaded area is part of the area between the squares, and is approximately $\boxed{\textbf{(A)}\ \frac12}$.

Note that this solution is not rigorous, because we still should show that the ratio is less than $\frac{3}{4}$.

## See Also

 2011 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast Problem 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 AJHSME/AMC 8 Problems and Solutions

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

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