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 $\frac{3}\times\frac{1}{2}\times(x+3)=\frac{1}{2}\times3\times(3-x)+2$ . This simplifies into $3x=2$ , or $x=\frac{2}{3}$ , so $G=(3,\frac{7}{3})$ . Therefore the slope of the line is $\boxed{\textbf{(E)}\ \frac{7}{9}}$

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1994 AHSME Problems/Problem 23

Problem

Solution