Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution 2)
Line 11: Line 11:
 
== Solution 2 ==  
 
== Solution 2 ==  
  
−
Let <math>S</math> be his starting point, <math>T</math> be 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>.
+
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>.
 +
 
 +
-j4andrews
  
 
== See also ==
 
== See also ==

Revision as of 03:39, 25 June 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 $\fbox{(E)}$.

-j4andrews

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1980_AHSME_Problems/Problem_9&oldid=252203"