2014 AMC 12B Problem 21

Let $x = BE = GH = CF$ and $\theta = \angle DHG = \angle AGJ$ $= \angle FKH,$ with $AD = GJ = HK = 1.$ In right triangle $GDH,$ $x \sin\theta = DG = 1 - \cos\theta,$ so $x = \dfrac{1 - \cos\theta}{\sin\theta}.$ Along side $CD,$ $$\begin{gathered} 1 = CF + FH + HD \\ = x + \sin\theta + x\cos\theta. \end{gathered}$$ Substituting for $x$ gives $$\begin{gathered} 1 = \dfrac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta} \\ {}+ \sin\theta \\ = \dfrac{\sin^2\theta}{\sin\theta} \\ {}+ \sin\theta \\ = 2\sin\theta. \end{gathered}$$ Hence $\sin\theta = \tfrac12,$ so $\theta = 30^\circ$ and $$x = \dfrac{1 - \frac{\sqrt3}{2}}{\frac12} = 2 - \sqrt3.$$ Thus, the correct answer is C.