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

Revision as of 11:45, 25 February 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

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 $\frac{\pi/3}{\pi} = \frac 13 \Rightarrow \mathrm{(D)}$ .

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

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

@@ Line 16: / Line 16: @@
 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>.
-==Fake-solve Solution==
-Notice that if the angle chosen is <math>0</math> radians, then the distance <math>AC</math> is <math>8+5=13</math>. If the angle chosen is <math>2\pi</math> radians, then the distance <math>AC</math> is <math>8-5=3</math>. Now, note that <math>|7-13|=6</math> and <math>|7-3|=4</math>. Finally, since <math>\frac 46 = \frac 23</math>, our answer is <math>1-\frac 23 = \frac 13\Rightarrow \mathrm{(D)}</math>.
-Solution by franzliszt
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Difference between revisions of "2003 AMC 12B Problems/Problem 21"

Revision as of 11:45, 25 February 2022

Problem

Solution

See also