AMC8 2003

(C) For Figure A, the area of the square is $2^2 = 4\ \text{cm}^2$. The diameter of the circle is $2\ \text{cm}$, so the radius is $1\ \text{cm}$ and the area of the circle is $\pi\ \text{cm}^2$. So the area of the shaded region is $4 - \pi\ \text{cm}^2$. For Figure B, the area of the square is also $4\ \text{cm}^2$. The radius of each of the four circles is $\frac{1}{2}\ \text{cm}$, and the area of each circle is $\left(\frac{1}{2}\right)^2\pi=\frac{1}{4}\pi\ \text{cm}^2$. The combined area of all four circles is $\pi\ \text{cm}^2$. So the shaded regions in A and B have the same area. For Figure C, the radius of the circle is $1\ \text{cm}$, so the area of the circle is $\pi\ \text{cm}^2$. Because the diagonal of the inscribed square is the hypotenuse of a right triangle with legs of equal lengths, use the Pythagorean Theorem to determine the length $s$ of one side of the inscribed square. That is, $s^2+s^2=2^2=4$. So $s^2=2\ \text{cm}^2$, the area of the square. Therefore, the area of the shaded region is $\pi-2\ \text{cm}^2$. Because $\pi-2>1$ and $4-\pi<1$, the shaded region in Figure C has the largest area. Note that the second figure consists of four small copies of the first figure. Because each of the four small squares has sides half the length of the sides of the big square, the area of each of the four small figures is $\frac{1}{4}$ the area of Figure A. Because there are four such small figures in Figure B, the shaded regions in A and B have the same area.