Tenth Philippine Mathematical Olympiad

Solution:
$27 \sqrt{3}$ square units

Note that $\triangle A C E$ is equilateral. Each interior angle of $A B C D E F$ measures $\frac{1}{6}(6-2)\left(180^{\circ}\right)=120^{\circ}$. Using a property of a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, we have
$$\frac{1}{2} A C=\frac{\sqrt{3}}{2} \cdot 6 \quad \text{or} \quad A C=6 \sqrt{3}$$
The height of $\triangle A C E$ is $\frac{\sqrt{3}}{2} \cdot 6 \sqrt{3}=9$, so that
$$\text{area of } \triangle A C E=\frac{1}{2} \cdot 6 \sqrt{3} \cdot 9=27 \sqrt{3}.$$