2016 AMC 12A Problem 21

Let $\theta$ be the central angle subtending a side of length $200,$ with radius $R=200\sqrt2.$ By the law of cosines on the isosceles triangle from the center, $$\begin{gathered} 200^2=2R^2(1-\cos\theta)\\ =160000(1-\cos\theta), \end{gathered}$$ so $\cos\theta=\dfrac34.$ The three equal sides use three consecutive arcs of angle $\theta,$ so the fourth arc has angle $2\pi-3\theta.$ Its chord has the same length as a chord with central angle $3\theta,$ and $$\begin{gathered} \cos 3\theta=4\cos^3\theta-3\cos\theta\\ =4\cdot\dfrac{27}{64}-\dfrac94\\ =-\dfrac{9}{16}. \end{gathered}$$ Its length squared is $$\begin{gathered} 2R^2(1-\cos 3\theta)\\ =160000\left(1+\dfrac{9}{16}\right)\\ =160000\cdot\dfrac{25}{16}\\ =250000, \end{gathered}$$ so the fourth side is $500.$ Thus, the correct answer is E.