2009 AMC 10A Problem 21

Let the small circles have radius $r$ and the large circle radius $R$, with center $O$.

Connect the centers of the four small circles. Adjacent small circles are tangent, so neighboring centers are $2r$ apart, and the four centers form a square of side $2r$ centered at $O$. The distance from $O$ to any small center is half the square's diagonal:
$$\frac{2r\sqrt{2}}{2} = r\sqrt{2}.$$

Each small circle is internally tangent to the big circle, so $R$ equals the distance from $O$ to a small center plus $r$:
$$R = r\sqrt{2} + r = r(1 + \sqrt{2}).$$

The requested ratio is
$$\frac{4\pi r^2}{\pi R^2} = \frac{4}{(1+\sqrt2)^2} = \frac{4}{3 + 2\sqrt2}.$$
Rationalize with $3 - 2\sqrt2$ (note $(3+2\sqrt2)(3-2\sqrt2) = 9 - 8 = 1$):
$$\frac{4}{3+2\sqrt2} = 4(3 - 2\sqrt2).$$

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

Tangency problems are solved by connecting centers: tangent circles have centers separated by the sum (external) or difference (internal) of their radii. The ring of four turns into a square, whose diagonal supplies the key length. Recognizing $(1+\sqrt2)^2 = 3 + 2\sqrt2$ and its conjugate makes the final simplification instant.

Set $r = 1$ and estimate: $R = 1 + \sqrt2 \approx 2.414$, so the ratio is $\frac{4}{5.83} \approx 0.686$. Among the choices, only (C) $\approx 0.686$ fits; (A) is $\approx 0.17$ (the ratio for one small circle) and (B), (E) are $\approx 0.59$, $0.83$.

• Taking the distance from the big center to a small center as 2r or r instead of half the square's diagonal, r sqrt 2.
• Comparing only one small circle to the large one and answering $3 - 2\sqrt2$, choice (A).