Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
 
Line 21: Line 21:
 
Since <math>8=1\times8</math> and <math>8=2\times4</math> are the only positive factorings of <math>8</math>.
 
Since <math>8=1\times8</math> and <math>8=2\times4</math> are the only positive factorings of <math>8</math>.
  
−
<math>(a,b)=(5,12)</math> or <math>(a,b)=(6,8)</math> yielding \Rightarrow \textbf{(B)}<math> </math>2<math> solutions. Notice that because </math>b>a$, the reversed pairs are invalid.
+
<math>(a,b)=(5,12)</math> or <math>(a,b)=(6,8)</math> yielding <math>\Rightarrow\textbf{(B)}</math> <math>2</math> solutions. Notice that because <math>b>a</math>, the reversed pairs are invalid.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2008|ab=B|num-b=15|num-a=17}}
 
{{AMC12 box|year=2008|ab=B|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:03, 12 October 2021

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 $\Rightarrow\textbf{(B)}$ $2$ 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_16&oldid=163444"