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Difference between revisions of "1965 AHSME Problems/Problem 4"

(→‎Problem)
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\textbf{(D) }\ 4 \qquad  
 
\textbf{(D) }\ 4 \qquad  
 
\textbf{(E) }\ 8  </math>
 
\textbf{(E) }\ 8  </math>
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== Solution ==
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<asy>
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draw((-36,0)--(36,0), arrow=Arrows);
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label("$\ell_2$", (40,0));
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draw((14,-32)--(30,32), arrow=Arrows);
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label("$\ell_1$", (32,36));
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draw((-4,-32)--(12,32), arrow=Arrows);
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label("$\ell_3$", (14,36));
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draw((5,-32)--(21,32), dotted, arrow=Arrows);
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label("$\ell_4$", (23,36));
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</asy>
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The lines are coplanar, <math>\ell_1 \parallel \ell_3</math>, and <math>\ell_1</math> intersects <math>\ell_2</math>. Therefore, <math>\ell_3</math> also intersects <math>\ell_2</math>. The  [[locus]] of all points equidistant from parallel lines <math>\ell_1</math> and <math>\ell_3</math> is a third parallel line in between them. Let this line be <math>\ell_4</math>, and let the distance from <math>\ell_4</math> to either <math>\ell_1</math> or <math>\ell_3</math> be <math>d</math>. The points equidistant from lines <math>\ell_1</math>, <math>\ell_2</math>, and <math>\ell_3</math> must all lie on <math>\ell_4</math> and be a distance <math>d</math> from line <math>\ell_2</math>. There are only 2 points, on either side of <math>\ell_2</math>, which satisfy these conditions. Thus, our answer is <math>\fbox{(C) 2}</math>.
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==See Also==
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{{AHSME 40p box|year=1965|num-b=3|num-a=5}}
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[[Category:Introductory Geometry Problems]]

Revision as of 08:49, 18 July 2024

Problem

Line $\ell_2$ intersects line $\ell_1$ and line $\ell_3$ is parallel to $\ell_1$. The three lines are distinct and lie in a plane. The number of points equidistant from all three lines is:

$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 4 \qquad \textbf{(E) }\ 8$


Solution

[asy] draw((-36,0)--(36,0), arrow=Arrows); label("$\ell_2$", (40,0)); draw((14,-32)--(30,32), arrow=Arrows); label("$\ell_1$", (32,36)); draw((-4,-32)--(12,32), arrow=Arrows); label("$\ell_3$", (14,36)); draw((5,-32)--(21,32), dotted, arrow=Arrows); label("$\ell_4$", (23,36)); [/asy]

The lines are coplanar, $\ell_1 \parallel \ell_3$, and $\ell_1$ intersects $\ell_2$. Therefore, $\ell_3$ also intersects $\ell_2$. The locus of all points equidistant from parallel lines $\ell_1$ and $\ell_3$ is a third parallel line in between them. Let this line be $\ell_4$, and let the distance from $\ell_4$ to either $\ell_1$ or $\ell_3$ be $d$. The points equidistant from lines $\ell_1$, $\ell_2$, and $\ell_3$ must all lie on $\ell_4$ and be a distance $d$ from line $\ell_2$. There are only 2 points, on either side of $\ell_2$, which satisfy these conditions. Thus, our answer is $\fbox{(C) 2}$.

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions
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