2016 AMC 10A Problem 10

Let the inner rectangle be $1$ by $x$. Adding a $1$-foot border on every side gives a middle rectangle of size $3$ by $x + 2$, and another border gives the outer rectangle of size $5$ by $x + 4$.

The three colored regions have areas
$$A_1 = x, \qquad A_2 = 3(x+2) - x = 2x + 6, \qquad A_3 = 5(x+4) - 3(x+2) = 2x + 14.$$

For an arithmetic progression the consecutive differences must agree:
$$A_2 - A_1 = x + 6, \qquad A_3 - A_2 = 8.$$
So $x + 6 = 8$ and $x = 2$. (Check: areas $2, 10, 18$.)

The answer is $\boxed{\textbf{(B)}\ 2}$.

Nested rectangles turn ring areas into differences of rectangle areas, and each ring is linear in the unknown length. The arithmetic-progression condition is simply "middle minus first equals last minus middle," one linear equation. Frame-shaped regions always come from subtracting nested areas, never from multiplying the frame's width by a single side.

Test the choices: with $x = 2$ the rectangles are $1 \times 2$, $3 \times 4$, $5 \times 6$ with areas $2, 12, 30$, so the rings are $2, 10, 18$, evenly spaced. With $x = 1$ the rings are $1, 8, 16$, not evenly spaced.

• Using the full rectangle areas 3(x+2) and 5(x+4) as the ring areas instead of subtracting the inner rectangles.
• Growing each dimension by $1$ instead of $2$ when adding a border on all four sides.