Difference between revisions of "1994 AHSME Problems/Problem 23"

Problem

In the $xy$-plane, consider the L-shaped region bounded by horizontal and vertical segments with vertices at $(0,0), (0,3), (3,3), (3,1), (5,1)$ and $(5,0)$. The slope of the line through the origin that divides the area of this region exactly in half is $[asy] Label l; l.p=fontsize(6); xaxis("x",0,6,Ticks(l,1.0,0.5),EndArrow); yaxis("y",0,4,Ticks(l,1.0,0.5),EndArrow); draw((0,3)--(3,3)--(3,1)--(5,1)--(5,0)--(0,0)--cycle,black+linewidth(2));[/asy]$ $\textbf{(A)}\ \frac{2}{7} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{2}{3} \qquad\textbf{(D)}\ \frac{3}{4} \qquad\textbf{(E)}\ \frac{7}{9}$

Solution

Let the vertices be $A=(0,0),B=(0,3),C=(3,3),D=(3,1),E=(5,1),F=(5,0)$. It is easy to see that the line must pass through $CD$. Let the line intersect $CD$ at the point $G=(3,3-x)$ (i.e. the point $x$ units below $C$). Since the quadrilateral $ABCG$ and pentagon $GDEFA$ must have the same area, we have the equation $3\times\frac{1}{2}\times(x+3)=\frac{1}{2}\times3\times(3-x)+2$. This simplifies into $3x=2$, or $x=\frac{2}{3}$, so $G=(3,\frac{7}{3})$. Therefore the slope of the line is $\boxed{\textbf{(E)}\ \frac{7}{9}}$

Solution 2

Consider the small rectangle between $x=3$ and $x=5$ with area $2$. 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 $x=0$ and $x=3$ and $y=3$ and $y=c$ for some unknown $c$. For the area to be the same we need $(3-c) \cdot 3 = 2$, or $c=7/3$. After excluding our two offsetting areas, we're left with a rectangle from $x=0$ to $x=3$ and $y=0$ to $y=c$. The area of this region is clearly bisected by its diagonal line. The slope of this line is $c/3 = 7/9$ and the answer is $\fbox{E}$