2005 AMC 10B Problem 7

The problem gives no sizes, so choose a convenient one. Let the largest square have side $2$, so its area is $2 \cdot 2 = 4$.

1. Big circle. A circle inside a square touches all four sides, so its diameter equals the side of the square: $2$. Its radius is $1$.
2. Inner square. A square inside a circle touches the circle at its four corners. So the square's diagonal is a diameter of the circle: $2$.
3. Side of the inner square. The diagonal cuts the square into two right triangles with legs $s$ and $s$ and hypotenuse $2$. By the Pythagorean theorem, $s^2 + s^2 = 4$, so $s^2 = 2$ and $s = \sqrt{2}$.
4. Small circle. It sits inside the inner square, so its diameter is $\sqrt{2}$ and its radius is $\frac{\sqrt{2}}{2}$. Its area is
$$\pi \cdot \left(\frac{\sqrt{2}}{2}\right)^2 = \pi \cdot \frac{2}{4} = \frac{\pi}{2}.$$
5. Ratio. Small circle over big square: $\dfrac{\pi/2}{4} = \dfrac{\pi}{8}$.

The answer is $\boxed{\textbf{(B)}\ \frac{\pi}{8}}$.

There are only two rules for shapes inside each other. A circle inside a square: diameter equals the square's side. A square inside a circle: the square's diagonal equals the diameter. When a problem asks only for a ratio, choose a size that keeps the numbers simple; the ratio will not depend on your choice.

Use two quick facts. First, a circle inside a square covers $\frac{\pi}{4}$ of the square (with side $2$: area $\pi$ out of $4$). Second, the inner square has area $2$, which is half of the outer square. So the small circle is $\frac{\pi}{4}$ of a square that is itself half the big square: $\frac{\pi}{4} \cdot \frac{1}{2} = \frac{\pi}{8}$.