1994 AHSME Problems/Problem 23
Contents
Problem
In the 
-plane, consider the L-shaped region bounded by horizontal and vertical segments with vertices at 
and 
. The slope of the line through the origin that divides the area of this region exactly in half is
![[asy] Label l; l.p=fontsize(6); xaxis("$x$",0,6,Ticks(l,1.0,0.5),EndArrow); yaxis("$y$",0,4,Ticks(l,1.0,0.5),EndArrow); draw((0,3)--(3,3)--(3,1)--(5,1)--(5,0)--(0,0)--cycle,black+linewidth(2));[/asy]](http://latex.artofproblemsolving.com/6/c/7/6c7ce65a9896a7e0f4331e18f31d49d530a402aa.png)
Solution
Let the vertices be 
. It is easy to see that the line must pass through 
. Let the line intersect at the point 
(i.e. the point 
units below 
). Afgtfddeddftgyhujukik,pplkjhdsqqazcxseedvbgrtuiookmmlokhgfffffcddfcrdxsxsxsdcdcfvfvfvtgtbgnhnumjmj,jmhbgvdxsxsx the equation 
. This simplifies into 
, or 
, so 
. Therefore the slope of the line is 
Solution 2
Consider the small rectangle between 
and 
with area 
. If we exclude that area then the shape becomes a much simpler, its just a rectangle. In order to offset the exclusion of that area from the shape below the line, we must also exclude a shape with the same area above the line. We want the remainder after excluding both areas to be simple, so exclude a rectangle between 
and and 
and 
for some unknown 
. For the area to be the same we need 
, or 
. After excluding our two offsetting areas, we're left with a rectangle from to and 
to . The area of this region is clearly bisected by its diagonal line. The line passes through 
and 
so its slope is 
and the answer is 
See Also
| 1994 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
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| All AHSME Problems and Solutions | ||
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