2005 AMC 10B Problem 7

Choose the outer square to have side $2$, so its area is $4$.

The first circle is inscribed in it, so its diameter is $2$ and its radius is $1$.

The second square is inscribed in that circle, so its diagonal is $2$. A square with diagonal $d$ has side $d/\sqrt{2}$, so its side is $\sqrt{2}$.

The smallest circle is inscribed in this square, so its diameter is $\sqrt{2}$ and its radius is $\frac{\sqrt{2}}{2}$. Its area is
$$\pi \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{\pi}{2}.$$

The ratio is $\frac{\pi/2}{4} = \frac{\pi}{8}$.

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

Inscribing alternates between "diameter equals side" (circle in square) and "diameter equals diagonal" (square in circle). A square inscribed in a circle inscribed in a square has exactly half the outer square's area, so every two steps of nesting halve the area. Picking a concrete outer side length keeps the radicals manageable.

Ratio chaining: the big circle occupies $\frac{\pi}{4}$ of the big square; the inner square is half the big square, so the inner circle is $\frac{\pi}{4} \cdot \frac{1}{2} = \frac{\pi}{8}$ of the big square.

• Confusing a square's diagonal with its side when it is inscribed in a circle, losing a factor of 2.
• Comparing the smallest circle to the smaller square (ratio $\frac{\pi}{4}$) instead of the largest square.