2009 AMC 10A Problem 14

Let each rectangle have longer side $a$ and shorter side $b$.

Along one side of the outer square we see a long side of one rectangle followed by a short side of the next, so the outer square has side $a + b$. Along one side of the inner square we see a long side minus the short side that overlaps it, so the inner square has side $a - b$.

The area condition gives
$$(a + b)^2 = 4(a - b)^2 \quad\Longrightarrow\quad a + b = 2(a - b),$$
taking positive square roots. Then $a + b = 2a - 2b$, so $a = 3b$ and $\dfrac{a}{b} = 3$.

Check with $a = 3$, $b = 1$: outer square $4^2 = 16$, inner square $2^2 = 4$, ratio $4$.

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

This pinwheel of rectangles is a classic configuration: the outer side is $a + b$ and the inner side is $a - b$, so the two squares encode the sum and difference of the rectangle's sides. A ratio of areas becomes a ratio of sides by a square root, and then one linear equation determines $a : b$.

Area bookkeeping: the outer square equals the inner square plus four rectangles, so $(a+b)^2 = (a-b)^2 + 4ab$, an identity. Combined with $(a+b)^2 = 4(a-b)^2$ this gives $3(a-b)^2 = 4ab$, i.e. $3a^2 - 10ab + 3b^2 = 0$, so $(3a - b)(a - 3b) = 0$ and $a = 3b$.

• Taking the inner square's side to be a or b rather than a - b, or turning the area ratio 4 into a side ratio 4 instead of 2.
• Solving $3a^2 - 10ab + 3b^2 = 0$ and reporting the root $a/b = \frac{1}{3}$ instead of the ratio of longer to shorter.