Difference between revisions of "2008 AMC 10B Problems/Problem 23"
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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 | + | 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}} | {{MAA Notice}} |
Revision as of 16:43, 2 August 2020
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.