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1980 AHSME Problems/Problem 9

Revision as of 20:13, 25 July 2025 by J314andrews (talk | contribs) (Added diagram for this solution.)
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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

[asy] size(200); pair T=origin, H=(sqrt(3)*1.5, 0), F=(sqrt(3)*1.5, -1.5), S1=(sqrt(3), 0), S2=(2*sqrt(3), 0); draw(H--T--F--cycle); draw((-1,0)--(5,0)); draw(arc(F, sqrt(3), 0, 180)); label("$H$", H, SE); label("$\ell$", (5,0), E); label("$T$", T, N); label("$F$", F, SE); label("$S_1$", S1, NW); label("$S_2$", S2, NE); label("$30^\circ$", T, 1.5S+5.4E); [/asy]

Let $\ell$ be the line he was initially walking on, $T$ be his turning point, $F$ be his finishing point, and $H$ be the foot of the perpendicular from $F$ to $\ell$. Since he turned $150^\circ$, $\angle HTF = 30^{\circ}$. Therefore, $HF = \frac{3}{2}$ miles. Since $F$ is $\sqrt{3}$ miles from his starting point, his starting point lies on a circle of radius $\sqrt{3}$ miles centered at $F$. Since $\frac{3}{2} < \sqrt{3} < 3$, this circle must intersect $\ell$ once between $T$ and $H$ and once east of $H$. Let $S_1$ and $S_2$ be these two intersections, respectively. Both $S_1$ and $S_2$ are possible starting points, and clearly $TS_1 \neq TS_2$, so $x$ is $\fbox{(\textbf{E}) not uniquely determined}$.

Solution 2

[asy] size(320); pair T=origin, S1=(sqrt(3), 0), F=(sqrt(3)*1.5, -1.5); draw(S1--T--F--cycle); label("$S$", S1, NE); label("$T$", T, NW); label("$F$", F, SE); label("$x$", T--S1, N); label("$\sqrt{3}$", S1--F, 0.6N+0.3E); label("$3$", T--F, SW); label("$30^\circ$", T, 1.5S+5.4E); pair T2=(3, 0), S2=(3+2*sqrt(3), 0), F2=(3+sqrt(3)*1.5, -1.5); draw(S2--T2--F2--cycle); label("$S$", S2, NE); label("$T$", T2, NW); label("$F$", F2, S); label("$\sqrt{3}$", S2--F2, 0.1E); label("$x$", T2--S2, N); label("$3$", T2--F2, SW); label("$30^\circ$", T2, 1.5S+5.4E); [/asy]


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, $\left(\sqrt{3}\right)^2 = 3^2 + x^2 - 2 \cdot 3 \cdot x \cdot \cos 30^{\circ}$. Combining all terms on one side yields $x^2 - 3x\sqrt{3} + 6 = 0$, which factors as $\left(x - \sqrt{3}\right)\left(x - 2\sqrt{3}\right) = 0$. Therefore, $x = \sqrt{3}$ and $x = 2\sqrt{3}$ are both possible values of $x$, so $x$ is $\fbox{(\textbf{E}) not uniquely determined}$.

-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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