Square $AXYZ$ is inscribed in equiangular hexagon $ABCDEF$ with $X$ on $\overline{BC}$, $Y$ on $\overline{DE}$, and $Z$ on $\overline{EF}$. Suppose that $AB=40$, and $EF=41(\sqrt{3}-1)$. What is the side-length of the square?
$\textbf{(A)}\ 29\sqrt{3} \qquad\textbf{(B)}\ \frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3} \qquad\textbf{(C)}\ \ 20\sqrt{3}+16 \qquad\textbf{(D)}\ 20\sqrt{2}+13 \sqrt{3} \qquad\textbf{(E)}\ 21\sqrt{6}$

(diagram by djmathman)
We can, $\textsc{wlog}$, 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. $\framebox{A}$.