2008 AMC 12B Problems/Problem 16

Problem

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 $(a,b)$ ?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

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 $2\Rightarrow \textbf{(B)}$ solutions. Notice that because $b>a$ , the reversed pairs are invalid.

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AMC 12 Problems and Solutions

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2008 AMC 12B Problems/Problem 16

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