2009 AMC 12A Problem 13

Attempt Problem 13 of the 2009 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12A solutions, or check the answer key.

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13.

A ship sails 1010 miles in a straight line from AA to B,B, turns through an angle between 4545^\circ and 60,60^\circ, and then sails another 2020 miles to C.C. Let ACAC be measured in miles. Which of the following intervals contains AC2?AC^2?

[400,500][400, 500]

[500,600][500, 600]

[600,700][600, 700]

[700,800][700, 800]

[800,900][800, 900]

Answer: D
Concepts:law of cosinesbounding to limit cases
Difficulty rating: 1770
Solution:

By the Law of Cosines, AC2=102+20221020cos(ABC)=500400cos(ABC). \begin{aligned} AC^2 &= 10^2 + 20^2 \\ &\quad {}- 2\cdot 10\cdot 20\cos(\angle ABC) \\ &= 500 - 400\cos(\angle ABC). \end{aligned}

The ship turns through an angle between 4545^\circ and 60,60^\circ, so the interior angle ABC\angle ABC lies between 120120^\circ and 135.135^\circ.

Since cos120=12\cos 120^\circ = -\dfrac{1}{2} and cos135=22,\cos 135^\circ = -\dfrac{\sqrt{2}}{2}, 700=500+200AC2500+2002<800. \begin{aligned} 700 &= 500 + 200 \\ &\le AC^2 \le 500 + 200\sqrt{2} \\ &\lt 800. \end{aligned}

So AC2AC^2 lies in [700,800].[700, 800].

Thus, the correct answer is D.

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