2007 AMC 10A Problem 15

By symmetry the big circle is at the center of the square, and each small circle sits in a corner, with its center on a diagonal. So measure half of one diagonal, piece by piece.

1. Center of square to center of a small circle. The circles just touch, so the distance between centers is the sum of the radii: $2 + 1 = 3$.
2. Center of the small circle to the corner. The small circle touches both sides at that corner, so its center is $1$ away from each side. The center and the corner are opposite corners of a $1 \times 1$ square, so this distance is $\sqrt{1^2 + 1^2} = \sqrt2$.
3. Half the diagonal is $3 + \sqrt2$, so the whole diagonal is $6 + 2\sqrt2$.
4. Side of the square. A square's diagonal is its side times $\sqrt2$, so
$$\text{side} = \frac{6 + 2\sqrt2}{\sqrt2} = \frac{6}{\sqrt2} + 2 = 3\sqrt2 + 2.$$
5. Area.
$$(3\sqrt2 + 2)^2 = (3\sqrt2)^2 + 2 \cdot 3\sqrt2 \cdot 2 + 2^2 = 18 + 12\sqrt2 + 4 = 22 + 12\sqrt2.$$

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

Every "tangent" in the problem turns into a distance: two touching circles have centers apart by the sum of their radii, and a circle touching a side has its center one radius from that side. Lining these pieces up along a line of symmetry, here the diagonal, turns the picture into a single sum.

Join the four small centers; they form a smaller square with the same center. Its diagonal runs from one small center through the big circle to the opposite small center: $1 + 2 + 2 + 1 = 6$. So its side is $\frac{6}{\sqrt2} = 3\sqrt2$. The big square sticks out by one radius ($1$) on each side, so its side is $3\sqrt2 + 2$ and its area is $22 + 12\sqrt2$.

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