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\begin{document}

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\section*{four equilateral triangles solution}

\hypertarget{solution_to_the_four_equilateral_triangles_puzzle}{}\section*{{Solution to the Four Equilateral Triangles Puzzle}}\label{solution_to_the_four_equilateral_triangles_puzzle}

\emph{Posted on Aug 13, 2018}

[[FourEquilateralTriangles.png:pic]]

\begin{quote}%
Here are $4$ equilateral triangles. Find the missing area.


\end{quote}
\hypertarget{solution_by_properties_of_equilateral_triangles}{}\subsection*{{Solution by Properties of [[Equilateral Triangles]]}}\label{solution_by_properties_of_equilateral_triangles}

[[FourEquilateralTrianglesLabelled.png:pic]]

With the points labelled as above, the [[equilateral triangle]] $C E G$ is formed by taking four copies of $C D H$ and assembling them into a single triangle. Its area is therefore $4 \times 5 = 20$, so it is [[congruent]] to triangle $A I J$. Since $C H$ and $H G$ have the same length, this means that the height of $H$ above $A C$ is half the height of $I$.

The point $K$ is so that $H K$ is parallel to $A C$. Since the height of $H$ above $A C$ is half that of $I$, $K$ is also the midpoint of $I B$. As triangle $B H I$ is [[equilateral]], angle $B K H$ is a [[right-angle]] and so $I B$ is [[perpendicular]] to $A C$. The length of $I B$ is thus the height of the triangle $A I J$. From [[lengths in an equilateral triangle]], the length of $I B$ is $\frac{\sqrt{3}}{2}$ times the length of $A I$, so the area of $B H I$ is $\frac{3}{4}$ that of triangle $A I J$, namely $15$.

\hypertarget{solution_by_dissection_and_symmetry}{}\subsection*{{Solution by [[Dissection]] and [[Symmetry]]}}\label{solution_by_dissection_and_symmetry}

[[FourEquilateralTrianglesDissected.png:pic]]

In the above diagram, the original diagram has been extended so that the outer shape is an equilateral triangle. The outer triangle consists of four copies of the middle triangle of the original diagram. Looking at the upper right triangle, the area of one of these four is $20 + 5\times 5 = 45$.

In the central triangle, the three [[right-angled triangles]] are all half of the pink triangle, so have area $10$. This leaves $45 - 3 \times 10 = 15$ for the red triangle.



\end{document}