2016 AMC 12A Problem 21

Attempt Problem 21 of the 2016 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 2016 AMC 12A solutions, or check the answer key.

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21.

A quadrilateral is inscribed in a circle of radius 2002.200\sqrt{2}. Three of the sides of this quadrilateral have length 200.200. What is the length of its fourth side?

200200

2002200\sqrt{2}

2003200\sqrt{3}

3002300\sqrt{2}

500500

Answer: E
Concepts:cyclic quadrilaterallaw of cosinestrigonometric identity
Difficulty rating: 2040
Solution:

Let θ\theta be the central angle subtending a side of length 200,200, with radius R=2002.R=200\sqrt2. By the law of cosines on the isosceles triangle from the center, 2002=2R2(1cosθ)=160000(1cosθ), \begin{gathered} 200^2=2R^2(1-\cos\theta)\\ =160000(1-\cos\theta), \end{gathered} so cosθ=34.\cos\theta=\dfrac34.

The three equal sides use three consecutive arcs of angle θ,\theta, so the fourth arc has angle 2π3θ.2\pi-3\theta. Its chord has the same length as a chord with central angle 3θ,3\theta, and cos3θ=4cos3θ3cosθ=4276494=916. \begin{gathered} \cos 3\theta=4\cos^3\theta-3\cos\theta\\ =4\cdot\dfrac{27}{64}-\dfrac94\\ =-\dfrac{9}{16}. \end{gathered} Its length squared is 2R2(1cos3θ)=160000(1+916)=1600002516=250000, \begin{gathered} 2R^2(1-\cos 3\theta)\\ =160000\left(1+\dfrac{9}{16}\right)\\ =160000\cdot\dfrac{25}{16}\\ =250000, \end{gathered} so the fourth side is 500.500.

Thus, the correct answer is E.

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