2003 AMC 12B Problems/Problem 21

Revision as of 08:38, 1 September 2022 by Isabelchen (talk | contribs) (Solution 2 (Analytic Geometry))
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 (Analytic Geometry)

2003AMC12BP21.png

$WLOG$, let the object turn clockwise.

Let $B = (0, 0)$, $A = (0, -8)$.

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 $7$ and center $A$. The probability that $AC < 7$ is $\frac{\angle ABO}{180 ^\circ}$.

The function of $\odot B$ is $x^2 + y^2 = 25$, the function of $\odot A$ is $x^2 + (y+8)^2 = 49$.

$O$ is the point that satisfies the system of equations: $\begin{cases} x^2 + y^2 = 25 \\ x^2 + (y+8)^2 = 49 \end{cases}$

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

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

See also

2003 AMC 12B (ProblemsAnswer KeyResources)
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