2008 AMC 12B Problems/Problem 16

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 $8=1*8$ and $8=2*4$ are the only positive factorings of $8$ .

$(a,b)=(5,12)$ or $(a,b)=(6,8)$ yielding $2\Rightarrow \textbf{(B)}$ solutions. Notice that because $b>a$ , the reversed pairs are invalid.