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

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(Added diagram for this solution.)
 
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==Problem==
 
==Problem==
  
−
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
+
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>
  
 
== Solution 1 ==
 
== Solution 1 ==
−
Let us think about this. We only know that he ends up <math>\sqrt{3}</math> away from the origin. However, think about the locus of points <math>\sqrt{3}</math> away from the origin, a circle. However, his path could end on any part of the circle below the <math>x-</math>axis, so therefore, the answer is
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<asy>
−
<math>\fbox{E: not uniquely determined}.</math>
+
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);
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label("$30^\circ$", T, 1.5S+5.4E);
 +
</asy>
 +
 
 +
Let <math>\ell</math> be the line he was initially walking on, <math>T</math> be his turning point, <math>F</math> be his finishing point, and <math>H</math> be the foot of the perpendicular from <math>F</math> to <math>\ell</math>. Since he turned <math>150^\circ</math>,  <math>\angle HTF = 30^{\circ}</math>.  Therefore, <math>HF = \frac{3}{2}</math> miles.  Since <math>F</math> is <math>\sqrt{3}</math> miles from his starting point, his starting point lies on a circle of radius <math>\sqrt{3}</math> miles centered at <math>F</math>.  Since <math>\frac{3}{2} < \sqrt{3} < 3</math>, this circle must intersect <math>\ell</math> once between <math>T</math> and <math>H</math> and once east of <math>H</math>. Let <math>S_1</math> and <math>S_2</math> be these two intersections, respectively.  Both <math>S_1</math> and <math>S_2</math> are possible starting points, and clearly <math>TS_1 \neq TS_2</math>, so <math>x</math> is <math>\fbox{(\textbf{E}) not uniquely determined}</math>.
  
 
== Solution 2 ==  
 
== Solution 2 ==  
  
−
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>(\sqrt{3})^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>(x - \sqrt{3})(x - 2\sqrt{3}) = 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{(E)}</math>.
+
<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);
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label("$F$", F, SE);
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label("$x$", T--S1, N);
 +
label("$\sqrt{3}$", S1--F, 0.6N+0.3E);
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label("$3$", T--F, SW);
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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);
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label("$S$", S2, NE);
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label("$T$", T2, NW);
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label("$F$", F2, S);
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label("$\sqrt{3}$", S2--F2, 0.1E);
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label("$x$", T2--S2, N);
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label("$3$", T2--F2, SW);
 +
label("$30^\circ$", T2, 1.5S+5.4E);
 +
</asy>
 +
 
 +
 
 +
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>.
  
−
-j4andrews
+
-j314andrews
  
 
== See also ==
 
== See also ==

Latest revision as of 20:13, 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

[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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