2008 AMC 10A Problem 10

1. $S_1$ has area $16$, so its side is $4$.
2. Mark the midpoint of each side and join them. This makes a tilted square $S_2$ and cuts off four triangles, one at each corner of $S_1$.
3. Each corner triangle has a right angle and two legs of length $2$ (half of a side). Its area is
$$\frac{1}{2} \times 2 \times 2 = 2.$$
4. The four corners together have area $4 \times 2 = 8$. What is left is $S_2$, with area $16 - 8 = 8$.
5. Notice what happened: $S_2$ has exactly half the area of $S_1$. This is true for any square. If you fold the four corner triangles inward, they exactly cover the middle square, so the middle square is half.
6. Doing the same thing to $S_2$ halves the area again:
$$\text{area of } S_3 = \frac{1}{2} \times 8 = 4.$$

The answer is $\boxed{\textbf{(E)}\ 4}$.

"The midpoint square has half the area" is a fact worth remembering. The four corner triangles fold in over the inner square and cover it exactly, so inner square and corners are equal halves. Once you know it, repeated steps are easy: $16$, $8$, $4$, $2$, and so on.

Use side lengths. A side of $S_2$ is the long side of a corner triangle with legs $2$ and $2$, so by the Pythagorean theorem it is $\sqrt{2^2 + 2^2} = \sqrt{8}$. Its area is $(\sqrt8)^2 = 8$. Then $S_2$'s half-sides are $\frac{\sqrt8}{2}$, and a side of $S_3$ is $\sqrt{\left(\frac{\sqrt8}{2}\right)^2 + \left(\frac{\sqrt8}{2}\right)^2} = \sqrt{2 + 2} = 2$. So $S_3$ has area $4$.