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2003 AMC 12B Problem 21

Below is the professionally curated solution for Problem 21 of the 2003 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 12B solutions, or check the answer key.

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Concepts:law of cosinesgeometric probabilitytrigonometry

Difficulty rating: 1910

21.

An object moves 88 cm in a straight line from AA to B,B, turns at an angle α,\alpha, measured in radians and chosen at random from the interval (0,π),(0, \pi), and moves 55 cm in a straight line to C.C. What is the probability that AC<7?AC \lt 7?

16\dfrac{1}{6}

15\dfrac{1}{5}

14\dfrac{1}{4}

13\dfrac{1}{3}

12\dfrac{1}{2}

Solution:

Let β=π−α\beta = \pi - \alpha be the interior angle of △ABC\triangle ABC at B.B. By the Law of Cosines, AC2=82+52−2(8)(5)cos⁡β=89−80cos⁡β. AC^2 = 8^2 + 5^2 - 2(8)(5)\cos\beta = 89 - 80\cos\beta.

Then AC<7AC \lt 7 means 89−80cos⁡β<49,89 - 80\cos\beta \lt 49, i.e. cos⁡β>12,\cos\beta \gt \dfrac{1}{2}, i.e. β<π3.\beta \lt \dfrac{\pi}{3}.

As α\alpha is uniform on (0,π),(0, \pi), so is β.\beta. The probability is π/3π=13. \frac{\pi/3}{\pi} = \frac{1}{3}.

Thus, the correct answer is D.

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