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\begin{document}

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\section*{four equilateral triangles inside a hexagon solution}

\hypertarget{solution_to_the_four_equilateral_triangles_inside_a_hexagon_puzzle}{}\section*{{Solution to the [[four equilateral triangles inside a hexagon]] Puzzle}}\label{solution_to_the_four_equilateral_triangles_inside_a_hexagon_puzzle}

[[FourEquilateralTrianglesInsideaHexagon.png:pic]]

\begin{quote}%
Four equilateral triangles inside a regular hexagon. What fraction is shaded?


\end{quote}
\hypertarget{solution_by_symmetry}{}\subsection*{{Solution by [[Symmetry]]}}\label{solution_by_symmetry}

[[FourEquilateralTrianglesInsideaHexagonLabelled.png:pic]]

The solution to this is perhaps easier to see if some of the lines are removed, as in the above diagram. The angles at the centre of the white regions are all $60^\circ$. As the hexagon is invariant under rotating by $60^\circ$, this means that when rotating anticlockwise by $60^\circ$ then $O B$ maps to $O D$, $O E$ to $O G$, and $O I$ to $O K$.

Therefore, rotating triangle $O L K$ by $60^\circ$ clockwise brings $O K$ along $O I$, creating a new region $O G H J$. Rotating this region by $60^\circ$ clockwise brings $O G to O E$, creating a new region $O D F H O$. Lastly, rotating this region by $60^\circ$ brings $O D$ to $O B$ with the final region $O A C F O$.

This is $\frac{1}{3}$ of the area of the hexagon.



\end{document}