2009 AMC 10A Problem 21

Call the radius of each small circle $r$ and the radius of the big circle $R$. Let $O$ be the center of the big circle.

1. Neighboring small circles touch, so their centers are $r + r = 2r$ apart. The four small centers form a square with side $2r$, and $O$ is at the middle of this square.
2. The diagonal of a square with side $2r$ is $2r\sqrt{2}$ (Pythagorean theorem: $\sqrt{(2r)^2 + (2r)^2} = 2r\sqrt2$). The distance from $O$ to a small center is half of it, $r\sqrt{2}$.
3. Each small circle touches the big circle from inside. Go from $O$ to a small center ($r\sqrt2$), then keep going outward one more small radius ($r$) to reach the big circle:
$$R = r\sqrt{2} + r = r(1 + \sqrt{2}).$$
4. The ratio of areas:
$$\frac{4 \times \pi r^2}{\pi R^2} = \frac{4r^2}{r^2(1 + \sqrt2)^2} = \frac{4}{(1 + \sqrt2)^2}.$$
5. Expand: $(1 + \sqrt2)^2 = 1 + 2\sqrt2 + 2 = 3 + 2\sqrt2$.
6. To clear the square root from the bottom, use $(3 + 2\sqrt2)(3 - 2\sqrt2) = 9 - 8 = 1$. Multiply top and bottom by $3 - 2\sqrt2$:
$$\frac{4}{3 + 2\sqrt2} = \frac{4(3 - 2\sqrt2)}{1} = 4(3 - 2\sqrt2).$$

The answer is $\boxed{\textbf{(C)}\ 4(3-2\sqrt2)}$.

For circles that touch, connect their centers. Touching circles have centers exactly the sum of their radii apart (or the difference, if one is inside the other). Here the four centers make a square, and the square's diagonal gives the length you need.

Estimate. Take $r = 1$, so $R = 1 + \sqrt2 \approx 2.414$ and $R^2 \approx 5.83$. The ratio is about $\frac{4}{5.83} \approx 0.69$. The choices are about (A) $0.17$, (B) $0.59$, (C) $0.69$, (D) $0.79$ and (E) $0.83$, so only (C) fits.