, 2013

Solution:
Because $B_{6} A_{3} B_{3} A_{6}$ and $B_{1} A_{4} B_{4} A_{1}$ are parallelograms, $B_{6} A_{3}=A_{6} B_{3}$ and $A_{1} B_{1}=A_{4} B_{4}$. By the congruence of the large triangles $A_{1} A_{3} A_{5}$ and $A_{2} A_{4} A_{6}$, $A_{1} A_{3}=A_{4} A_{6}$. Thus, $B_{6} A_{3}+A_{1} B_{1}-A_{1} A_{3}=A_{6} B_{3}+A_{4} B_{4}-A_{4} A_{6}$, so $B_{6} B_{1}=B_{3} B_{4}$. Similarly, opposite sides of hexagon $B_{1} B_{2} B_{3} B_{4} B_{5} B_{6}$ are equal, and implying that the triangles opposite each other on the outside of this hexagon are congruent.

Furthermore, by definition $B_{5} B_{6}\parallel A_{3} A_{5}$, $B_{3} B_{4}\parallel A_{1} A_{3}$, $B_{6} B_{1} \parallel A_{4} A_{6}$ and $B_{1} B_{2} \parallel A_{1} A_{5}$. Let the area of triangle $A_{1} A_{3} A_{5}$ and triangle $A_{2} A_{4} A_{6}$ be $k^{2}$. Then, by similar triangles,

$$\begin{aligned} & \sqrt{\frac{1}{k^{2}}}=\frac{A_{1} B_{6}}{A_{1} A_{3}} \\ & \sqrt{\frac{4}{k^{2}}}=\frac{B_{6} B_{1}}{A_{4} A_{6}}=\frac{B_{1} B_{6}}{A_{1} A_{3}} \\ & \sqrt{\frac{9}{k^{2}}}=\frac{A_{3} B_{1}}{A_{1} A_{3}} \end{aligned}$$

Summing yields $6 / k=1$, so $k^{2}=36$. To finish, the area of $B_{1} B_{2} B_{3} B_{4} B_{5} B_{6}$ is equivalent to the area of the triangle $A_{1} A_{3} A_{5}$ minus the areas of the smaller triangles provided in the hypothesis. Thus, our answer is $36-1-4-9=22$.