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 $(a,b)$?

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

Solution

Because the unpainted part of the floor covers half the area, then the painted 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 $\boxed{\textbf{(B) 2}}$

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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