# Solution to the [[Three Regular Hexagons and a Semi-Circle]] Puzzle +-- {.image} [[ThreeRegularHexagonsandaSemiCircle.png:pic]] > Three regular hexagons. What’s the diameter of the semicircle? =-- ## Solution by Properties of a [[Regular Hexagon]] and [[Angle at the Centre is Twice the Angle at the Circumference]] +-- {.image} [[ThreeRegularHexagonsandaSemiCircleLabelled.png:pic]] =-- In the above diagram, the original semi-circle is extended to a full circle and the hexagons inside the semi-circle are reflected into the lower half. The point labelled $O$ is the centre of the circle. Using angles in a [[regular hexagon]], angle $E \hat{D} A$ is $60^\circ$, as is angle $B \hat{D} F$. Therefore, angle $F \hat{D} C$ is also $60^\circ$ and so $A D C$ is a straight line. Then $B F C$ is also a straight line, and angle $D \hat{C} F$ is $30^\circ$, so angle $A \hat{C} B$ is also $30^\circ$. Then since the [[angle at the centre is twice the angle at the circumference]], angle $A \hat{O} B$ is $60^\circ$. So triangle $A O B$ is an [[isosceles]] triangle with one angle $60^circ$, hence is actually [[equilateral]]. The radius of the semi-circle is therefore the same as the diameter of the hexagon, which is twice the side length, hene is $4.