2003 AMC 12B Problems/Problem 21

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 (Trigonometry)

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

$WLOG$ , let the object turn clockwise.

Note that the possible points of $C$ create a semi-circle of radius $5$ and center $B$ . The area where $AC < 7$ is enclosed by a circle of radius $7$ center $A$ .

Let $B = (0, 0)$ . Function of $\odot B = x^2 + y^2 = 25$ , function of $\odot A = x^2 + (y+8)^2 = 49$ .

$O$ is the point that satisfies both functions.

$x^2 + (y+8)^2 - x^2 - y^2 = 49 - 25$ , $64 + 16y =24$ , $y = - \frac52$ , $x = \frac{5 \sqrt{3}}{2}$ , $O = (\frac{5 \sqrt{3}}{2}, - \frac52)$

Note that $\triangle BDO$ is a $30-60-90$ triangle, as $BO = 5$ , $BD = \frac{5 \sqrt{3}}{2}$ , $DO = \frac52$ . As a result $\angle CBO = 30 ^\circ$ , $\angle ABO = 60 ^\circ$ .

Therefore the probability that $AC < 7$ is $\frac{\angle ABO}{180 ^\circ} = \frac{60 ^\circ}{180 ^\circ} = \boxed{\textbf{(D) } \frac13 }$

~isabelchen

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2003 AMC 12B Problems/Problem 21

Contents

Problem

Solution 1 (Trigonometry)

Solution 2

See also