2012 AMC 12B Problem 21

Extend $EF$ and $CB$ to a line through $A$ perpendicular to both, meeting them at $H$ and $J.$ Since $\angle ABJ=60^\circ,$ we have $BJ=20$ and $AJ=20\sqrt3.$ With $u=BX,$ the Pythagorean theorem gives $s^2=(20+u)^2+(20\sqrt3)^2.$ The equiangular angles make the four corner triangles congruent, and chasing the equal segments along $EF$ yields $$\begin{aligned} u+20\sqrt3 &= 41(\sqrt3-1) \\ &\quad {}+\frac{20+u}{\sqrt3}, \end{aligned}$$ so $u=21\sqrt3-20.$ Since $20+u=21\sqrt3,$ we get $$\begin{aligned} s^2 &= (21\sqrt3)^2+(20\sqrt3)^2 \\ &= 3(441+400)=3\cdot29^2, \end{aligned}$$ giving $s=29\sqrt3.$ Thus, the correct answer is A.