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\begin{document}

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\section*{rectangle in a hexagon solution}

\hypertarget{solutions_to_the_rectangle_in_a_hexagon_problem}{}\section*{{Solutions to the Rectangle in a Hexagon Problem}}\label{solutions_to_the_rectangle_in_a_hexagon_problem}

\hypertarget{dissection_or_shearing_solution}{}\subsection*{{Dissection or Shearing Solution}}\label{dissection_or_shearing_solution}

One strategy for solving this problem is to [[dissection|dissect]] the rectangle and reconstruct it so that it fits more simply in the hexagon. This can also be achieved by [[shear|shearing]] the rectangle twice.

The stages of the dissection are as follows.

\begin{enumerate}%
\item Remove the triangle at the top and paste it at the bottom.


\item Remove the side triangle on the right and paste it at the left.



\end{enumerate}
At this stage, the rectangle now occupies the central portion of the hexagon. To see how to calculate this area, we divide the hexagon into equilateral triangles and then further split each triangle into half. There are twelve of these halves in total, meaning that each has area $2.5$. Eight of them constitute the central strip, which is the rectangle, so the rectangle has area $20$.

\hypertarget{agg_invariance_principle}{}\subsection*{{Agg Invariance Principle}}\label{agg_invariance_principle}

Using the [[Agg invariance principle]], we can assume that the rectangle is vertical and so occupies the central strip. From there, the deduction follows the above argument following from the dissection.



\end{document}