Difference between revisions of "2008 AMC 10B Problems/Problem 23"

(Solution)
Line 8: Line 8:
 
==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 a-2 by b-2. With this information we can make the equation:
 +
 
ab = 2[(a-2)(b-2)]
 
ab = 2[(a-2)(b-2)]
 +
 
ab = 2ab - 4a - 4b + 8
 
ab = 2ab - 4a - 4b + 8
 +
 
ab - 4a - 4b + 16 = 8
 
ab - 4a - 4b + 16 = 8
 +
 
(a-4)(b-4) = 8
 
(a-4)(b-4) = 8
 +
 
Since b > a, then we have the possibilities (a-4) = 1 and (b-4) = 8, or (a-4) = 2 and (b-4) = 4. This gives 2 (B) possibilities: (5,12) or (6,8).
 
Since b > a, then we have the possibilities (a-4) = 1 and (b-4) = 8, or (a-4) = 2 and (b-4) = 4. This gives 2 (B) possibilities: (5,12) or (6,8).
  
 
==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}}

Revision as of 12:48, 13 February 2009

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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 a-2 by b-2. With this information we can make the equation:

ab = 2[(a-2)(b-2)]

ab = 2ab - 4a - 4b + 8

ab - 4a - 4b + 16 = 8

(a-4)(b-4) = 8

Since b > a, then we have the possibilities (a-4) = 1 and (b-4) = 8, or (a-4) = 2 and (b-4) = 4. This gives 2 (B) possibilities: (5,12) or (6,8).

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
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