Notes
three hexagons on a circle solution

Solution to the Three Hexagons on a Circle Puzzle

Solution by Properties of a Regular Hexagon and Angles in the Same Segment

In the above image, $Q$ is the centre of the yellow hexagon and $B$ is where the continuation of line segment $E G$ meets the circle. Angle $A \hat{F} G$ is the interior angle of a regular hexagon so is $120^\circ$ and angle $G \hat{F} Q$ is half that, so is $60^\circ$. Therefore, angle $A \hat{F} Q$ is $180^\circ$ so $Q F A$ is a straight line. Similarly, $D H C$ is a straight line.

Since angles in the same segment are equal, angles $D \hat{A} E$ and $D \hat{C} E$ are the same. Using angles in a regular hexagon, angles $E \hat{Q} A$ and $E \hat{D} C$ are both seen to be equal to $120^\circ$. So triangles $E D C$ and $E Q A$ have the same interior angles and are similar. Then from lengths in a regular hexagon, line segments $E Q$ and $E D$ have the same length, so triangles $E D C$ and $E Q A$ are actually congruent.

This means that line segments $D C$ and $Q A$ have the same length, then since $Q F$ and $D H$ have the same length as each other, so also $H C$ and $F A$ have the same length as each other. Since the length of $H C$ is twice the side length of the smaller hexagon, this means that the scale factor from the orange to pink hexagons is $2$, and so the area scale factor is $4$.

Hence the pink hexagon has area $4 \times 6 = 24$.

Solution by Symmetry

With the diagram labelled as above, the circle and yellow hexagon share a line of symmetry which is the perpendicular bisector of $D E$. This carries $D A$ to $E B$ and so establishes that $F A$ and $G B$ have the same length.

Then reflection in the perpendicular bisector of $D H$ brings $D E$ to $H G$, and reflection in the perpendicular bisector of $D C$ brings $D E$ to $C B$. Since these perpendicular bisectors are parallel, this establishes $H G$ and $C B$ are parallel. Hence $G B C H$ is a parallelogram and so $H C$ and $G B$ have the same length.

Thus $F A$ and $H C$ have the same length so, as above, the hexagons are related by a scale factor of $2$ so the pink hexagon has area $4 \times 6 = 24$.

Solution by Invariance Principle

The size of the yellow hexagon is not specified, so it can be varied. Although the size of the orange hexagon is fixed in the puzzle, the key is the relationship between the orange and pink hexagons, so the sizes of all the hexagons can be regarded as variable.

Thus consider the outer circle to be fixed, with $E$ a fixed point on its circumference. Let $D$ be another point on the circumference. The choice of $D$ then determines the rest of the diagram since $E D$ is a side of the yellow regular hexagon, $C$ is found by continuing $D H$ until it meets the circle, and similarly $A$ by continuing $D F$.

As $D$ moves, the angle $E \hat{D} C$ remains $120^circ$. By the converse to angles in the same segment are equal, this means that $C$ remains fixed in place. A similar argument applied to angle $E \hat{D} A$ shows that $A$ remains fixed in place. So triangle $E C A$ does not move.

Placing $D$ half way round the arc from $E$ to $C$ puts $H$ at $C$ and $F$ at $A$ showing that $E C A$ is an equilateral triangle. Then placing $D$ at $E$ so that the yellow hexagon has no size also brings $H$ and $F$ to $E$. In this configuration, $E C$ is the diagonal of the orange hexagon and $E A$ the side of the pink.

This shows that the scale factor from the orange to pink hexagons is $2$, so the area scale factor is $4$ as before.