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\begin{document}

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\section*{three semi-circles around a triangle solution}

\hypertarget{solution_to_the_three_semicircles_around_a_triangle_puzzle}{}\section*{{Solution to the Three Semi-Circles Around a Triangle Puzzle}}\label{solution_to_the_three_semicircles_around_a_triangle_puzzle}

[[ThreeSemiCirclesAroundaTriangle.png:pic]]

\begin{quote}%
What's the area of the red triangle?


\end{quote}
\hypertarget{solution_by_area_of_a_triangle}{}\subsection*{{Solution by [[Area of a Triangle]]}}\label{solution_by_area_of_a_triangle}

[[ThreeSemiCirclesAroundaTriangleLabelled.png:pic]]

The decomposition shown in the above diagram splits the red triangle into seven regions. Other than the central one, the regions can be paired up with triangles $F C B$ and $F B E$ comprising one pair and continuing this around the central triangle.

That triangles $F C B$ and $F B E$ have the same area comes from the fact that $C B$ and $B E$ are radii of the same semi-circle so have the same length, and then the triangles share $F$ as apex above this line. Then $F C$ is the same length as $C A$ as both are radii of another semi-circle, and they are the same perpendicular distance from $C B$ (albeit on different sides). Therefore triangles $F C B$ and $A B C$ have the same area.

This argument continues round the pairs, showing that each of the extra triangles has the same area as the central one. The area of the red triangle is therefore $7 \times 6 = 42$.



\end{document}