3. Let $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ be a convex hexagon such that $A_{i} A_{i+2} \| A_{i+3} A_{i+5}$ for $i=1,2,3$ (we take $A_{i+6}=A_{i}$ for each $i$ ). Segment $A_{i} A_{i+2}$ intersects segment $A_{i+1} A_{i+3}$ at $B_{i}$, for $1 \leq i \leq 6$, as shown. Furthermore, suppose that $\triangle A_{1} A_{3} A_{5} \cong \triangle A_{4} A_{6} A_{2}$. Given that $\left[A_{1} B_{5} B_{6}\right]=1,\left[A_{2} B_{6} B_{1}\right]=$ 4 , and $\left[A_{3} B_{1} B_{2}\right]=9$ (by $[X Y Z]$ we mean the area of $\triangle X Y Z$ ), determine the area of hexagon $B_{1} B_{2} B_{3} B_{4} B_{5} B_{6}$.
Answer: 22 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 $B_{5} B_{6}\left\|A_{3} A_{5}, B_{3} B_{4}\right\| A_{1} A_{3}, B_{6} B_{1} \| A_{4} A_{6}$ and $B_{1} B_{2} \| A_{1} A_{5}$. 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 $36-1-4-9=22$.