# Solution to the [[Tangential Hexagon and Circle]] Puzzle +-- {.image} [[TangentialHexagonandCircle.png:pic]] > A unit circle is tangent to a regular hexagon. What’s the side length? =-- ## Solution by Properties of [[Isosceles|Isosceles Triangle]] and [[Equilateral Triangles]], [[Hexagon|Angles in a Hexagon]], [[Angle Between a Circle and a Tangent]], and [[Angles at a Point on a Straight Line]] +-- {.image} [[TangentialHexagonandCircleLabelled.png:pic]] =-- 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 triangle|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.