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

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Let <math>S</math> be his starting point, <math>T</math> be the point where he turns, and <math>F</math> be his finishing point.  Since he turned <math>150^{\circ}</math> at <math>T</math>, <math>\angle STF = 30^{\circ}</math>.  By the Law of Cosines, <math>FS^2 = FT^2 + ST^2 - 2 \cdot FT \cdot ST \cos \angle STF</math>. That is, <math>\left(\sqrt{3}\right)^2 = 3^2 + x^2 - 2 \cdot 3 \cdot x \cdot \cos 30^{\circ}</math>.  Combining all terms on one side yields <math>x^2 - 3x\sqrt{3} + 6 = 0</math>, which factors as <math>\left(x - \sqrt{3}\right)\left(x - 2\sqrt{3}\right) = 0</math>.  Therefore, <math>x = \sqrt{3}</math> and <math>x = 2\sqrt{3}</math> are both possible values of <math>x</math>, so the answer is <math>\fbox{(\textbf{E}) not uniquely determined}</math>.
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Let <math>S</math> be his starting point, <math>T</math> be the point where he turns, and <math>F</math> be his finishing point.  Since he turned <math>150^{\circ}</math> at <math>T</math>, <math>\angle STF = 30^{\circ}</math>.  By the Law of Cosines, <math>FS^2 = FT^2 + ST^2 - 2 \cdot FT \cdot ST \cos \angle STF</math>. That is, <math>\left(\sqrt{3}\right)^2 = 3^2 + x^2 - 2 \cdot 3 \cdot x \cdot \cos 30^{\circ}</math>.  Combining all terms on one side yields <math>x^2 - 3x\sqrt{3} + 6 = 0</math>, which factors as <math>\left(x - \sqrt{3}\right)\left(x - 2\sqrt{3}\right) = 0</math>.  Therefore, <math>x = \sqrt{3}</math> and <math>x = 2\sqrt{3}</math> are both possible values of <math>x</math>, so <math>x</math> is <math>\fbox{(\textbf{E}) not uniquely determined}</math>.
  
 
-j314andrews
 
-j314andrews

Revision as of 18:20, 25 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

Since he turned $150^\circ$, the two segments of his walk must form a $30^{\circ}$ angle. Therefore, his end point must be $\frac{3}{2}$ miles away from the line in which the first segment of his walk lies. Since his end point was $\sqrt{3}$ miles from his start point, his start point must lie on a circle of radius $\sqrt{3}$ centered at his end point. Since $\sqrt{3} > \frac{3}{2}$, a line $\frac{3}{2}$ miles away from his end point and a circle of radius $\sqrt{3}$ centered at his end point must intersect twice. These are both possible start points, so the answer 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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