2008 AMC 10B Problems/Problem 23

Revision as of 21:23, 24 January 2010 by Albert1993 (talk | contribs) (Solution)

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
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All AMC 10 Problems and Solutions