Notes
trisected hexagon solution

Solution to the Trisected Hexagon Puzzle

Solution by Trigonometry, Regions in a Regular Hexagon, and Area of a Triangle

With the points labelled as above, the regions $A H G$ and $A B C$ are each one sixth of the hexagon. Therefore, the triangle $G F A$ is one sixth of the hexagon. Triangle $F E A$ is half of the blue region so is also one sixth of the hexagon. Therefore, by the area of a triangle, triangles $F E A$ and $G F A$ must have the same “base”, so $G F$ and $F E$ have the same length.

Angle $F \hat{A} G$ can then be computed using trigonometry, since $G A$ has length $\sqrt{3}$ times that of $G E$, so angle $F \hat{A} G$ is given by $\tan^{-1}\left(\frac{1}{2\sqrt{3}}\right)$. Then angle $C \hat{A} D$ is the same, and so since angle $C \hat{A} G = 60^\circ$, we have that angle $D \hat{A} F$ is (to $3$ decimal places):