2007 AMC 10A Problem 15

By symmetry the big circle sits at the center of the square and each small circle's center lies on a diagonal. Trace one diagonal from the center to a corner.

From the center to a small circle's center: the circles are externally tangent, so this distance is $2 + 1 = 3$.

From the small circle's center to the corner: the small circle is tangent to both sides meeting at that corner, so its center is $1$ unit from each side, i.e. at the far vertex of a $1 \times 1$ square in the corner. That distance is $\sqrt{2}$.

Half the diagonal is $3 + \sqrt{2}$, so the full diagonal is $6 + 2\sqrt{2}$ and the side is
$$s = \frac{6 + 2\sqrt2}{\sqrt2} = 3\sqrt2 + 2.$$
The area is
$$s^2 = (3\sqrt2 + 2)^2 = 18 + 12\sqrt2 + 4 = 22 + 12\sqrt2.$$

The answer is $\boxed{\textbf{(B)}\ 22 + 12\sqrt {2}}$.

Tangency statements translate into center distances: externally tangent circles have centers separated by the sum of the radii, and a circle tangent to two perpendicular sides has its center $r$ from each. Lining these pieces up along a line of symmetry (here the diagonal) converts the whole picture into a single one-dimensional sum.

Connect the four small centers: they form a square whose diagonal is $1 + 4 + 1 = 6$, so its side is $3\sqrt2$. The big square is wider by one radius on each side, so its side is $3\sqrt2 + 2$, and its area is $22 + 12\sqrt2$.

• Taking the distance from a small circle's center to the corner as $1$ (the radius) instead of $\sqrt2$, the diagonal of a $1$-by-$1$ square.
• Squaring $3\sqrt2 + 2$ incorrectly (dropping the cross term $12\sqrt2$) and picking a choice without a radical.