[Geometry]
Problem: In equilateral hexagon $A B C D E F, \mathrm{~m} \angle A=2 \mathrm{~m} \angle C=2 \mathrm{~m} \angle E=5 \mathrm{~m} \angle D=10 \mathrm{~m} \angle B=10 \mathrm{~m} \angle F$, and diagonal $B E=3$. Compute $[A B C D E F]$, that is, the area of $A B C D E F$.

Solution: Let $\mathrm{m} \angle B=\alpha$. Then the sum of the measures of the angles in the hexagon is:

$$
\begin{aligned}
720^{\circ} & =\mathrm{m} \angle A+\mathrm{m} \angle C+\mathrm{m} \angle E+\mathrm{m} \angle D+\mathrm{m} \angle B+\mathrm{m} \angle F \\
& =10 \alpha+5 \alpha+5 \alpha+2 \alpha+\alpha+\alpha=24 \alpha .
\end{aligned}
$$



Thus $30^{\circ}=\alpha$ and $\mathrm{m} \angle A=300^{\circ}$, so the exterior angle at $A$ has measure $60^{\circ}=\mathrm{m} \angle D$. Further, because $A B=C D$ and $D E=A F$, it follows that $\triangle C D E \cong \triangle B A F$. Thus

$$
[A B C D E F]=[A B C E F]+[C D E]=[A B C E F]+[A B F]=[B C E F] .
$$

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To compute $[B C E F]$, notice that because $\mathrm{m} \angle D=60^{\circ}, \triangle C D E$ is equilateral. In addition,

$$
\begin{aligned}
150^{\circ} & =\mathrm{m} \angle B C D \\
& =\mathrm{m} \angle B C E+\mathrm{m} \angle D C E=\mathrm{m} \angle B C E+60^{\circ} .
\end{aligned}
$$

Therefore $\mathrm{m} \angle B C E=90^{\circ}$. Similarly, because the hexagon is symmetric, $\mathrm{m} \angle C E F=90^{\circ}$, so quadrilateral $B C E F$ is actually a square with side length 3 . Thus $C E=\frac{B E}{\sqrt{2}}=\frac{3}{\sqrt{2}}$, and $[A B C D E F]=[B C E F]=\frac{9}{2}$.

Alternate Solution: Calculate the angles of the hexagon as in the first solution. Then proceed as follows.

First, $A B C D E F$ can be partitioned into four congruent triangles. Because the hexagon is equilateral and $\mathrm{m} \angle A B C=\mathrm{m} \angle A F E=30^{\circ}$, it follows that $\triangle A B C$ and $\triangle A F E$ are congruent isosceles triangles whose base angles measure $75^{\circ}$. Next, $\mathrm{m} \angle A B C+\mathrm{m} \angle B C D=30^{\circ}+150^{\circ}=$ $180^{\circ}$, so $\overline{A B} \| \overline{C D}$. Because these two segments are also congruent, quadrilateral $A B C D$ is a parallelogram. In particular, $\triangle C D A \cong \triangle A B C$. Similarly, $\triangle E D A \cong \triangle A F E$.

Now let $a=A C=A E$ be the length of the base of these isosceles triangles, and let $b=A B$ be the length of the other sides (or of the equilateral hexagon). Because the four triangles are congruent, $[A B C D E F]=[A B C]+[A C D]+[A D E]+[A E F]=4[A B C]=4 \cdot \frac{1}{2} b^{2} \sin 30^{\circ}=b^{2}$. Applying the Law of Cosines to $\triangle A B C$ gives $a^{2}=b^{2}+b^{2}-2 b^{2} \cos 30^{\circ}=(2-\sqrt{3}) b^{2}$. Because $4-2 \sqrt{3}=(\sqrt{3}-1)^{2}$, this gives $a=\left(\frac{\sqrt{3}-1}{\sqrt{2}}\right) b$. Using the given length $B E=3$ and applying the Law of Cosines to $\triangle A B E$ gives

$$
\begin{aligned}
9 & =a^{2}+b^{2}-2 a b \cos 135^{\circ} \\
& =a^{2}+b^{2}+\sqrt{2} a b \\
& =(2-\sqrt{3}) b^{2}+b^{2}+(\sqrt{3}-1) b^{2} \\
& =2 b^{2} .
\end{aligned}
$$

Thus $[A B C D E F]=b^{2}=\frac{9}{2}$.

Final answer: $\frac{9}{2}$