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\begin{document}

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\section*{regular hexagon and rectangle solution}

\hypertarget{solution_to_the_regular_hexagon_and_rectangle_puzzle}{}\section*{{Solution to the Regular Hexagon and Rectangle Puzzle}}\label{solution_to_the_regular_hexagon_and_rectangle_puzzle}

[[RegularHexagonandRectangle.png:pic]]

\begin{quote}%
Here’s a regular hexagon and a rectangle. What’s the angle?


\end{quote}
\hypertarget{solution_by_angle_at_the_circumference_is_half_the_angle_at_the_centre}{}\subsection*{{Solution by [[Angle at the Circumference is Half the Angle at the Centre]]}}\label{solution_by_angle_at_the_circumference_is_half_the_angle_at_the_centre}

[[RegularHexagonandRectangleLabelled.png:pic]]

The circle in the above diagram is the [[circumcircle]] through the vertices of the [[hexagon]] and the point labelled $O$ is its centre. This is also the centre of the rectangle and so the vertices of the rectangle also lie on the circle. The angle $A \hat{D} B$ is then half of angle $A \hat{O} B$ since the [[angle at the circumference is half the angle at the centre]]. Triangle $A O B$ is an [[equilateral triangle]] as it is formed from the [[regular hexagon]] so angle $A \hat{O} B$ is $60^\circ$ meaning that angle $A \hat{D} B$ is $30^\circ$.

\hypertarget{solution_by_invariance_principle}{}\subsection*{{Solution by [[Invariance Principle]]}}\label{solution_by_invariance_principle}

[[RegularHexagonandRectangleExtreme.png:pic]]

Under the assumption that the angle doesn't depend on the size of the rectangle (providing it shares a diagonal with the hexagon) there are two configurations where the angle can be easily deduced. These are both shown in the above diagram. In one, the rectangle is viewed as actually being the line $E A$ so the angle in question is $A \hat{E} B$. In the other, the rectangle is $E G A C$ so the angle in question is $A \hat{C} B$.

Triangle $A B C$ is [[isosceles]] and angle $C \hat{B} A$ is the [[interior angle]] of a [[regular hexagon]] so is $120^\circ$, meaning that angle $A \hat{C} B$ is $\frac{180^\circ - 120^\circ}{2} = 30^\circ$. Then angle $B \hat{E} C$ is also $30^\circ$, while angle $A \hat{E} C$ is $\frac{120^\circ}{2}$, meaning that angle $A \hat{E} B$ is $\frac{120^\circ}{2} - 30^\circ = 30^\circ$.



\end{document}