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Difference between revisions of "2003 AMC 12B Problems/Problem 21"

(fake solution is not a solution)
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\qquad\mathrm{(E)}\ \frac{1}{2}</math>
 
\qquad\mathrm{(E)}\ \frac{1}{2}</math>
  
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== Solution ==
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== Solution 1 ==
 
By the [[Law of Cosines]],
 
By the [[Law of Cosines]],
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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It follows that <math>0 < \alpha < \frac {\pi}3</math>, and the probability is <math>\frac{\pi/3}{\pi} = \frac 13 \Rightarrow \mathrm{(D)}</math>.
 
It follows that <math>0 < \alpha < \frac {\pi}3</math>, and the probability is <math>\frac{\pi/3}{\pi} = \frac 13 \Rightarrow \mathrm{(D)}</math>.
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==Solution 2==
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[[File:2003AMC12BP21.png|500px]]
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
== See also ==
 
== See also ==

Revision as of 09:55, 31 August 2022

Problem

An object moves $8$ cm in a straight line from $A$ to $B$, turns at an angle $\alpha$, measured in radians and chosen at random from the interval $(0,\pi)$, and moves $5$ cm in a straight line to $C$. What is the probability that $AC < 7$?

$\mathrm{(A)}\ \frac{1}{6} \qquad\mathrm{(B)}\ \frac{1}{5} \qquad\mathrm{(C)}\ \frac{1}{4} \qquad\mathrm{(D)}\ \frac{1}{3} \qquad\mathrm{(E)}\ \frac{1}{2}$

Solution 1

By the Law of Cosines, \begin{align*} AB^2 + BC^2 - 2 AB \cdot BC \cos \alpha = 89 - 80 \cos \alpha = AC^2 &< 49\\ \cos \alpha &> \frac 12\\ \end{align*}

It follows that $0 < \alpha < \frac {\pi}3$, and the probability is $\frac{\pi/3}{\pi} = \frac 13 \Rightarrow \mathrm{(D)}$.

Solution 2

2003AMC12BP21.png


~isabelchen

See also

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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