2009 AMC 10A Problem 14

Call each rectangle's longer side $a$ and its shorter side $b$.

1. Walk along one side of the outer square. You pass the long side of one rectangle and then the short side of the next one. So the outer square has side $a + b$.
2. Walk along one side of the inner square. It runs along the long side of one rectangle, except for the part covered by the short side of its neighbor. So the inner square has side $a - b$.
3. The outer area is $4$ times the inner area. Since $4 = 2 \times 2$, the outer side is $2$ times the inner side:
$$a + b = 2(a - b).$$
4. Expand and solve: $a + b = 2a - 2b$, so $3b = a$. The ratio of long side to short side is $\dfrac{a}{b} = 3$.
5. Check with $a = 3$, $b = 1$: the outer square has side $4$ and area $16$; the inner square has side $2$ and area $4$. And $16 = 4 \times 4$.

The answer is $\boxed{\textbf{(A)}\ 3}$.

In this pinwheel of four rectangles, the outer square's side is the sum of the rectangle's sides and the inner square's side is their difference. When areas are compared, take square roots to compare sides: an area $4$ times as big means sides $2$ times as long.

Use areas instead. The outer square is the inner square plus the four rectangles. With the numbers from the check above ($a = 3$, $b = 1$): $16 = 4 + 4 \times 3$. In general, $(a+b)^2 = 4(a-b)^2$ can be expanded to $a^2 + 2ab + b^2 = 4a^2 - 8ab + 4b^2$, which simplifies to $3a^2 - 10ab + 3b^2 = 0$, or $(3a - b)(a - 3b) = 0$. Since $a$ is the longer side, $a = 3b$.