Notes
tangential hexagon and circle solution

Solution to the Tangential Hexagon and Circle Puzzle

Solution by Properties of Isosceles Triangle and Equilateral Triangles, Angles in a Hexagon, Angle Between a Circle and a Tangent, and Angles at a Point on a Straight Line

With the diagram labelled as above, triangle $A B G$ is isosceles, and angle $A \hat{B} G = 120^\circ$ as it is the interior angle in a regular hexagon. So angle $B \hat{G} A$ is $30^\circ$. Therefore, angle $B \hat{G} E$ is $150^\circ$ by angles at a point on a straight line.

Triangles $D H G$ and $D F G$ are both right-angled and share a side, so are congruent. This means that $G D$ bisects angle $H \hat{G} F$, so angle $H \hat{G} D$ is $75^\circ$. Similarly, angle $C \hat{B} G$ is $60^\circ$ as it is the exterior angle of a regular hexagon so angle $D \hat{B} G$ is $30^\circ$.

The angles in triangle $B D G$ are therefore $30^\circ$, $75^\circ$, and $75^\circ$, meaning that it is isosceles. Therefore $D B$ and $B G$ have the same length.

Triangle $B D H$ is half an equilateral triangle, so the length of $B D$ is twice that of $D H$, hence is $2$. Therefore $B G$ has length $2$, and this is the side length of the hexagon.