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 10B Problems/Problem 23"

(New page: ==Problem== {{problem}} ==Solution== {{solution}} ==See also== {{AMC10 box|year=2008|ab=B|num-b=22|num-a=24}})
 
m (→‎Solution)
 
(23 intermediate revisions by 14 users not shown)
Line 1: Line 1:
 
==Problem==
 
==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 and <math>b > a</math>. 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 <math>1</math> foot around the painted rectangle and occupies half the area of the whole 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>
  
 
==Solution==
 
==Solution==
−
{{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 <math>a-2</math> by <math>b-2</math>. With this information we can make the equation:
 +
 
 +
<cmath>
 +
\begin{eqnarray*}
 +
ab &=& 2\left((a-2)(b-2)\right) \\
 +
ab &=& 2ab - 4a - 4b + 8 \\
 +
ab - 4a - 4b + 8 &=& 0
 +
\end{eqnarray*}
 +
</cmath>
 +
Applying [[Simon's Favorite Factoring Trick]], we get
 +
 
 +
<cmath>\begin{eqnarray*}ab - 4a - 4b + 16 &=& 8 \\ (a-4)(b-4) &=& 8 \end{eqnarray*}</cmath>
 +
 
 +
Since <math>b > a</math>, then we have the possibilities <math>(a-4) = 1</math> and <math>(b-4) = 8</math>, or <math>(a-4) = 2</math> and <math>(b-4) = 4</math>. This allows for 2 possibilities: <math>(5,12)</math> or <math>(6,8)</math> which gives us <math>\boxed{\textbf{(B)} \: 2}</math>
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=22|num-a=24}}
 
{{AMC10 box|year=2008|ab=B|num-b=22|num-a=24}}
 +
{{MAA Notice}}

Latest revision as of 22:47, 17 October 2023

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 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 (Problems • Answer Key • Resources)
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

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_10B_Problems/Problem_23&oldid=199791"