2008 AMC 10B Problems/Problem 23

Problem

A rectangular floor measures $a$ by $b$ feet, where $a$ and $b$ are positive integers and $b > a$ . An artist paints a rectangle on the floor with the sides of the rectangle parallel to the floor. The unpainted part of the floor forms a border of width $1$ foot around the painted rectangle and occupies half the area of the whole 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

Because the painted part of the floor covers half the area, then the unpainted rectangle covers half the area as well. Since the border width is 1 foot, the dimensions of the rectangle are $a-2$ by $b-2$ . With this information we can make the equation:

$\begin{eqnarray*} ab &=& 2\left((a-2)(b-2)\right) \\ ab &=& 2ab - 4a - 4b + 8 \\ ab - 4a - 4b + 8 &=& 0 \end{eqnarray*}$ Applying Simon's Favorite Factoring Trick, we get

$\begin{eqnarray*}ab - 4a - 4b + 16 &=& 8 \\ (a-4)(b-4) &=& 8 \end{eqnarray*}$

Since $b > a$ , then we have the possibilities $(a-4) = 1$ and $(b-4) = 8$ , or $(a-4) = 2$ and $(b-4) = 4$ . This allows for 2 possibilities: $(5,12)$ or $(6,8)$ which gives us $\boxed{\textbf{(B) 2}}$

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2008 AMC 10B Problems/Problem 23

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