2003 AMC 12B Problem 21

Question

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 \lt 7?

\dfrac{1}{6}

\dfrac{1}{5}

\dfrac{1}{4}

\dfrac{1}{3}

\dfrac{1}{2}

Let

\beta = \pi - \alpha

be the interior angle at

B;

apply the Law of Cosines to

\triangle ABC

AC^2 = 89 - 80\cos\beta,

so

AC \lt 7

is equivalent to

\cos\beta \gt \dfrac{1}{2}

Let

\beta = \pi - \alpha

be the interior angle of

\triangle ABC

at

B.

By the Law of Cosines,

\begin{aligned} &AC^2 = 8^2 + 5^2 \\ &\quad {}- 2(8)(5)\cos\beta \\ &= 89 - 80\cos\beta. \end{aligned}

Then

AC \lt 7

means

89 - 80\cos\beta \lt 49,

i.e.

\cos\beta \gt \dfrac{1}{2},

i.e.

\beta \lt \dfrac{\pi}{3}.

As

\alpha

is uniform on

(0, \pi),

so is

\beta.

The probability is

\frac{\frac{\pi}{3}}{\pi} = \frac{1}{3}.

Thus, the correct answer is D.

Problem 21 in other years