Difference between revisions of "2008 AMC 10B Problems/Problem 23"
m (Made "SFFT" a link) |
|||
| Line 1: | Line 1: | ||
| − | ==Problem== | + | [[[http://www.example.com link title]]]==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 occupied half the area of the whole floor. How many possibilities are there for the ordered pair (a,b)? | 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 occupied half the area of the whole floor. How many possibilities are there for the ordered pair (a,b)? | ||
| Line 14: | Line 14: | ||
\end{eqnarray*} | \end{eqnarray*} | ||
</cmath> | </cmath> | ||
| − | Applying | + | Applying [[Simon's Favorite Factoring Trick]], we get |
<cmath>\begin{eqnarray*}ab - 4a - 4b + 16 &=& 8 \\ (a-4)(b-4) &=& 8 \end{eqnarray*}</cmath> | <cmath>\begin{eqnarray*}ab - 4a - 4b + 16 &=& 8 \\ (a-4)(b-4) &=& 8 \end{eqnarray*}</cmath> | ||
Revision as of 17:21, 31 January 2014
[[link title]]==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 occupied half the area of the whole floor. How many possibilities are there for the ordered pair (a,b)?
A) 1 B) 2 C) 3 D) 4 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 
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 gives 2 possibilities: (5,12) or (6,8), So the answer is 
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. 
