2016 AMC 10A Problem 10

Each colored region is a rectangle with a smaller rectangle cut out of its middle, so find the three rectangle sizes first.

1. Let the inner rectangle be $1$ foot wide and $x$ feet long.
2. A $1$-foot border on all four sides adds $2$ feet to each dimension. So the middle rectangle is $3$ by $x + 2$, and the outer rectangle is $5$ by $x + 4$.
3. Now find the area of each colored region:
- Inner region: $1 \cdot x = x$.
- Middle ring: the $3$ by $(x+2)$ rectangle minus the inner rectangle: $3x + 6 - x = 2x + 6$.
- Outer ring: the $5$ by $(x+4)$ rectangle minus the middle rectangle: $5x + 20 - (3x + 6) = 2x + 14$.
4. In an arithmetic progression, the gap from first to second equals the gap from second to third.
- First gap: $(2x + 6) - x = x + 6$.
- Second gap: $(2x + 14) - (2x + 6) = 8$.
5. So $x + 6 = 8$ and $x = 2$.

Check: the areas are $2$, $10$, $18$, which go up by $8$ each time.

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

The area of a frame is always the outer rectangle minus the inner one. Each area here is a simple expression in $x$, and "arithmetic progression" just means the two gaps are equal, which gives one easy equation.

Test the choices. With $x = 2$ the three rectangles are $1 \times 2$, $3 \times 4$ and $5 \times 6$, with areas $2$, $12$, $30$. The colored regions are $2$, $12 - 2 = 10$ and $30 - 12 = 18$, evenly spaced. With $x = 1$ they would be $1$, $8$, $16$, which are not evenly spaced.

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 $\boxed{\textbf{(B)}\ 2}$.