1980 AHSME Problems/Problem 9

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

1980 AHSME (Problems • Answer Key • Resources)
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1980 AHSME Problems/Problem 9

Contents

Problem

Solution 1

Solution 2

See also