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

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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==
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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>
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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>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>
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== Solution 2==
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Consider the small rectangle between <math>x=3</math> and <math>x=5</math> with area <math>2</math>.  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 <math>x=0</math> and <math>x=3</math> and <math>y=3</math> and <math>y=c</math> for some unknown <math>c</math>.  For the area to be the same we need <math>(3-c) \cdot 3 = 2</math>, or <math>c=7/3</math>.  After excluding our two offsetting areas, we're left with a rectangle from <math>x=0</math> to <math>x=3</math> and <math>y=0</math> to <math>y=c</math>.  The area of this region is clearly bisected by its diagonal line.  The line passes through <math>(0,0)</math> and <math>(3,c)</math> so its slope is <math>c/3 = 7/9</math> and the answer is <math>\fbox{E}</math>
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==See Also==
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{{AHSME box|year=1994|num-b=22|num-a=24}}
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{{MAA Notice}}

Latest revision as of 03:22, 28 May 2021

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 line passes through $(0,0)$ and $(3,c)$ so its slope is $c/3 = 7/9$ and the answer is $\fbox{E}$

See Also

1994 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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