AMC12 2012 B

We can assume $Y$ coincides with $D$ and $CD\parallel AF$ as before. In which case, we will have $BC=EF=41(\sqrt{3}-1)$. So we have square $AXDZ$ inscribed in equiangular hexagon $ABCDEF$ with $X$ on $\overline{BC}$ and $Z$ on $\overline{EF}$. Let $\angle BXA = \theta$; then $\angle BAX=60^\circ -\theta$. Let $BX=u$. In $\triangle ABX$ we have \begin{align} \frac{2s}{\sqrt{3}}=\frac{u}{\sin(60^\circ-\theta)}=\frac {40}{\sin\theta} \end{align} We also have $\angle CXD=90^\circ - \theta$ and $\angle CDX = \theta-30^\circ$. Let $CX=v$. In $\triangle CDX$ we have \begin{align}\tag{2} \frac{2s}{\sqrt{3}}=\frac{v}{\sin(\theta-30^\circ)}=\frac {CD}{\cos\theta} \end{align} Now $BC=u+v=41(\sqrt{3}-1)$. From $(1)$ and $(2)$ we get\begin{align*} 41(\sqrt{3}-1) &= \frac{2s}{\sqrt{3}}\left(\sin(60^\circ-\theta)+\sin(\theta-30^\circ)\right) \\ &= \frac{2s}{\sqrt{3}} \cdot \frac{\sqrt{3}-1}2\cdot (\sin\theta + \cos\theta) \end{align*} From $(1)$ we get $s\sin\theta = 20\sqrt{3}$ and therefore $s\cos\theta = \sqrt{s^2-3\cdot 20^2}$. Thus \[41(\sqrt{3}-1) = \frac{\sqrt{3}-1}{\sqrt{3}}(20\sqrt{3}+\sqrt{s^2-3\cdot 20^2})\]which simplifies to\[3\cdot 21^2 = s^2-3\cdot 20^2.\]Since $(20, 21, 29)$ is a Pythagorean triple, we get $s=29\sqrt{3}$, i.e. $\framebox{A}$.