2008 AMC 10B Problem 23

Write the area condition as an equation, then rearrange it into a product of two whole numbers.

1. The border is $1$ foot wide on every side, so the painted rectangle loses $1$ foot at each end in both directions. It measures $(a - 2)$ by $(b - 2)$.
2. The border is half the floor, so the painted part is the other half:
$$(a - 2)(b - 2) = \frac{ab}{2}.$$
3. Multiply both sides by $2$ and expand:
$$2ab - 4a - 4b + 8 = ab.$$
4. Subtract $ab$ from both sides:
$$ab - 4a - 4b + 8 = 0.$$
5. The left side almost factors. Notice that $(a - 4)(b - 4) = ab - 4a - 4b + 16$. Our expression has $8$ instead of $16$, so add $8$ to both sides:
$$ab - 4a - 4b + 16 = 8, \qquad\text{that is,}\qquad (a - 4)(b - 4) = 8.$$
6. Now $a - 4$ and $b - 4$ are whole numbers that multiply to $8$, and $b - 4$ is the bigger one because $b > a$. The positive pairs are $1 \times 8$ and $2 \times 4$:
- $a - 4 = 1$, $b - 4 = 8$: $(a, b) = (5, 12)$.
- $a - 4 = 2$, $b - 4 = 4$: $(a, b) = (6, 8)$.
7. Negative pairs like $(-8) \times (-1)$ or $(-4) \times (-2)$ would make $a - 4$ at most $-4$, so $a$ would be $0$ or less. Not allowed.
8. Check: a $5 \times 12$ floor has area $60$ and painted part $3 \times 10 = 30$, which is half. A $6 \times 8$ floor has area $48$ and painted part $4 \times 6 = 24$, which is half.

There are $2$ possible pairs.

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

An equation with $ab$, $a$ and $b$ in it can often be turned into (something) $\times$ (something) $=$ number by adding the right constant to both sides. Once it is a product of whole numbers, the solutions come from listing the factor pairs of that number. Border and frame problems lead to this kind of equation again and again.