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

(Created page with "==Problem== In the <math>xy</math>-plane, consider the L-shaped region bounded by horizontal and vertical segments with vertices at <math>(0,0), (0,3), (3,3), (3,1), (5,1)</math>...")
 
(→‎Solution)
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<math> \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} </math>
 
<math> \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} </math>
 
==Solution==
 
==Solution==
 +
Let the vertices be <math>A=(0,0),B=(0,3),C=(3,3),D=(3,1),E=(5,1),F=(5,0)</math>. It is easy to see that the line must pass through <math>CD</math>. Let the line intersect <math>CD</math> at the point <math>G=(3,3-x)</math> (i.e. the point <math>x</math> units below <math>C</math>). Since the quadrilateral <math>ABCG</math> and pentagon <math>GDEFA</math> must have the same area, we have the equation <math>\frac{3}\times\frac{1}{2}\times(x+3)=\frac{1}{2}\times3\times(3-x)+2</math>. This simplifies into <math>3x=2</math>, or <math>x=\frac{2}{3}</math>, so <math>G=(3,\frac{7}{3})</math>. Therefore the slope of the line is <math>\boxed{\textbf{(E)}\ \frac{7}{9}}</math>

Revision as of 12:50, 15 February 2016

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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