2016 AMC 10A Problem 24

Label the quadrilateral $ABCD$, with $AB=BC=CD=200$. Let each of these equal chords subtend central angle $2\theta$. The chord formula gives $$200=2(200\sqrt2)\sin\theta,$$ so $\sin\theta=\frac{1}{2\sqrt2}$. In particular, $\theta<30^\circ$, so the smaller central angle subtended by $AD$ is $6\theta<180^\circ$. Therefore another use of the chord formula gives $$AD=2(200\sqrt2)\sin(3\theta).$$ Using $\sin(3\theta)=3\sin\theta-4\sin^3\theta$, we obtain $$\begin{aligned} \sin(3\theta) &=\frac{3}{2\sqrt2}-\frac{1}{4\sqrt2} \\ &=\frac{5}{4\sqrt2}. \end{aligned}$$ Hence $$AD=400\sqrt2\cdot\frac{5}{4\sqrt2}=500.$$ Thus, the correct answer is E.