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Difference between revisions of "1980 AHSME Problems/Problem 9"

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==Problem==
 
==Problem==
  
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A man walks <math>x</math> miles due west, turns <math>150^\circ</math> to his left and walks 3 miles in the new direction. If he finishes a a point <math>\sqrt{3}</math> from his starting point, then <math>x</math> is
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A man walks <math>x</math> miles due west, turns <math>150^\circ</math> to his left and walks <math>3</math> miles in the new direction. If he finishes a a point <math>\sqrt{3}</math> from his starting point, then <math>x</math> is
  
 
<math>\text{(A)} \ \sqrt 3 \qquad \text{(B)} \ 2\sqrt{5} \qquad \text{(C)} \ \frac 32 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{not uniquely determined}</math>
 
<math>\text{(A)} \ \sqrt 3 \qquad \text{(B)} \ 2\sqrt{5} \qquad \text{(C)} \ \frac 32 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{not uniquely determined}</math>

Revision as of 02:37, 23 July 2025

Problem

A man walks $x$ miles due west, turns $150^\circ$ to his left and walks $3$ miles in the new direction. If he finishes a a point $\sqrt{3}$ from his starting point, then $x$ is

$\text{(A)} \ \sqrt 3 \qquad \text{(B)} \ 2\sqrt{5} \qquad \text{(C)} \ \frac 32 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{not uniquely determined}$

Solution 1

Let us think about this. We only know that he ends up $\sqrt{3}$ away from the origin. However, think about the locus of points $\sqrt{3}$ away from the origin, a circle. However, his path could end on any part of the circle below the $x-$axis, so therefore, the answer is $\fbox{E: not uniquely determined}.$

Solution 2

Let $S$ be his starting point, $T$ be the point where he turns, and $F$ be his finishing point. Since he turned $150^{\circ}$ at $T$, $\angle STF = 30^{\circ}$. By the Law of Cosines, $FS^2 = FT^2 + ST^2 - 2 \cdot FT \cdot ST \cos \angle STF$. That is, $(\sqrt{3})^2 = 3^2 + x^2 - 2 \cdot 3 \cdot x \cdot \cos 30^{\circ}$. Combining all terms on one side yields $x^2 - 3x\sqrt{3} + 6 = 0$, which factors as $(x - \sqrt{3})(x - 2\sqrt{3}) = 0$. Therefore, $x = \sqrt{3}$ and $x = 2\sqrt{3}$ are both possible values of $x$, so the answer is $\fbox{(E)}$.

-j314andrews

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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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