smc

We can, $\text{textsc}wlog$, 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 $△ABX$ we have $$\begin{array}{r}\frac{2s}{\sqrt{3}}=\frac{u}{\sin (60^{\circ }-\theta )}=\frac{40}{\sin \theta }\end{array}$$ We also have $\angle CXD=90^{\circ }-\theta$ and $\angle CDX=\theta -30^{\circ }$. Let $CX=v$. In $△CDX$ we have $$\begin{array}{r}\frac{2s}{\sqrt{3}}=\frac{v}{\sin (\theta -30^{\circ })}=\frac{CD}{\cos \theta }\end{array}\text{\hspace{1em}(2)}$$ Now $BC=u+v=41(\sqrt{3}-1)$. From $(1)$ and $(2)$ we get$$\begin{array}{rl}41(\sqrt{3}-1)& =\frac{2s}{\sqrt{3}}\left(\sin (60^{\circ }-\theta )+\sin (\theta -30^{\circ })\right)\\ & =\frac{2s}{\sqrt{3}}\cdot \frac{\sqrt{3}-1}{2}\cdot (\sin \theta +\cos \theta )\end{array}$$ 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. $\boxed{29\sqrt{3}}$.