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? $\textbf{(A)}\ 29\sqrt{3} \qquad\textbf{(B)}\ \frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3}\qquad\textbf{(C)}\ 20\sqrt{3}+16$ $\textbf{(D)}\ 20\sqrt{2}+13\sqrt{3} \qquad\textbf{(E)}\ 21\sqrt{6}$
Solution
Draw $YH\bot BC, AG\bot BC$. Assume $BX=x$, Let $a=40=AB, b=41(\sqrt3-1)=EF$.
Notice $\triangle ABG$ is a $30-60-90$ triangle, so $BG=\frac{a}2, AG=\frac{\sqrt3 a}2$.
$\triangle AGX\cong \triangle XHY$. So $YH=XG=x+\frac{a}2, HX=AG=\frac{\sqrt3 a}2$.
Draw $YW//CD$. Then we have $\triangle ABX\cong \triangle YEZ, \triangle YWX\cong \triangle AFZ$, so $BW=EF=b$. $XW=b-x$. $WH=XH-WX=\frac{\sqrt3 a}2-b+x$.
$\triangle YHW$ is a $30-60-90$ triangle, so $YH:HW=\sqrt 3$, so $(x+\frac{a}2): (\frac{\sqrt 3 a}2-b+x)=\sqrt3$.
Square $AXYZ$ is inscribed in equiangular hexagon $ABCDEF$ with $X$ on $\overline{BC}$, $Y$ on $\overline{DE}$, and $Z$ on $\overline{EF}$. Suppose that $AB=30$, and $EF=6+8\sqrt3$. What is the side-length of the square?
$\textbf{(A)}\ 29\sqrt{3} \qquad\textbf{(B)}\ \frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3}\qquad\textbf{(C)}\ 20\sqrt{3}+16$ $\textbf{(D)}\ 20\sqrt{2}+13\sqrt{3} \qquad\textbf{(E)}\ 21\sqrt{6}$ 
