Notes
three triangles in a hexagon solution

Solution to the Three Triangles in a Hexagon Puzzle

Solution by Transformations

As the two equilateral triangles are the same size, the point $F$ in the above diagram is equidistant from $B$ and $O$ and so moves on the line that passes through $A$ and $D$. Since $G$ can be obtained from $F$ by rotating it by $60^\circ$ clockwise about $B$, it also moves on a straight line. To find the line, we consider two special cases of the diagram.

In this first special case, $F$ is at the centre of the hexagon and $G$ is coincident with $C$.

In the second special case, $F$ is coincident with $A$ and $G$ is at the centre of the hexagon.

The line that $G$ moves along is therefore the diagonal from $C$ to $O$. This line is parallel to the side $D E$ and so the “height” of the triangle $G D E$ above its “base” of $D E$ is constant, relative to the length of $D E$. This means that the area of the triangle is a fixed fraction of the area of the hexagon. By considering the second special case it is clear that the triangle is $\frac{1}{6}$th of the hexagon and so the hexagon has area $36$.

Solution by Invariance Principle

Either of the special cases described above shows that the area of the purple triangle is one sixth of the area of the hexagon.