AMC12 2008 12B, Problem 16

A rectangular floor measures $a$ by $b$ feet, where $a$ and $b$ are positive integers with $b > a$. An artist paints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor. The unpainted part of the floor forms a border of width $1$ foot around the painted rectangle and occupies half of the area of the entire floor. How many possibilities are there for the ordered pair $(a,b)$?

Solution: $A_{outer}=ab$ $A_{inner}=(a-2)(b-2)$ $A_{outer}=2A_{inner}$ $ab=2(a-2)(b-2)=2ab-4a-4b+8$ $0=ab-4a-4b+8$ By Simon's Favorite Factoring Trick: $8=ab-4a-4b+16=(a-4)(b-4)$ Since $8=1\times8$ and $8=2\times4$ are the only positive factorings of $8$. $(a,b)=(5,12)$ or $(a,b)=(6,8)$ yielding $\Rightarrow$ $2$ solutions. Notice that because $b>a$, the reversed pairs are invalid. $\fbox{2}$.

Answer: 2