Difference between revisions of "2008 AMC 10B Problems/Problem 23"
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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 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 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>\text{A) 1 B) 2 C) 3 D) 4 E) 5}</math> | + | <math>\text{(A) 1 (B) 2 (C) 3 (D) 4 (E) 5}</math> |
==Solution== | ==Solution== |
Revision as of 01:19, 1 February 2015
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 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.