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

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==Problem==
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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>1</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>?
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<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math>
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==Solution==
 
==Solution==
 
<math>A{outer}=ab</math>
 
<math>A{outer}=ab</math>
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<math>(a,b)=(5,12)</math> or <math>(a,b)=(6,8)</math> yielding <math>2\Rightarrow \textbf{(B)}</math> solutions. Notice that because <math>b>a</math>, the reversed pairs are invalid.
 
<math>(a,b)=(5,12)</math> or <math>(a,b)=(6,8)</math> yielding <math>2\Rightarrow \textbf{(B)}</math> solutions. Notice that because <math>b>a</math>, the reversed pairs are invalid.
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==See Also==
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{{AMC12 box|year=2008|ab=B|num-b=15|num-a=17}}

Revision as of 17:15, 3 March 2008

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 $8=1*8$ 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 (Problems • Answer Key • Resources)
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
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