Difference between revisions of "2008 AMC 12B Problems/Problem 16"

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==Problem==
 
==Problem==
A rectangular floor measures <math>a</math> by <math>b</math> feet, where <math>a</math> and <math>b</math> are positive integers with <math>b > a</math>. 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 <math>2</math> feet around the painted rectangle and occupies half of the area of the entire floor. How many possibilities are there for the ordered pair <math>(a,b)</math>?
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A rectangular floor measures <math>a</math> by <math>b</math> feet, where <math>a</math> and <math>b</math> are positive integers with <math>b > a</math>. 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 <math></math> foot around the painted rectangle and occupies half of the area of the entire floor. How many possibilities are there for the ordered pair <math>(a,b)</math>?
  
 
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math>
 
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math>

Revision as of 20:18, 23 January 2012

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 $$ (Error compiling LaTeX. Unknown error_msg) 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 $8=1*8$ 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.

See Also

2008 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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 12 Problems and Solutions