Difference between revisions of "2008 AMC 10B Problems/Problem 23"
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==Problem== | ==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 | + | 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>? |
| − | A) 1 B) 2 C) 3 D) 4 E) 5 | + | <math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math> |
==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 a-2 by b-2. With this information we can make the equation: | + | 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> | |
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| − | 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 | ||
==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 
by 
feet, where and are positive integers and 
. 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 
foot around the painted rectangle and occupies half the area of the whole floor. How many possibilities are there for the ordered pair 
?
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 
by 
. With this information we can make the equation:
Applying Simon's Favorite Factoring Trick, we get
Since , then we have the possibilities 
and 
, or 
and 
. This allows for 2 possibilities: 
or 
which gives us 
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. 
