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\begin{document}

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\section*{five decreasing squares solution}

\hypertarget{solution_to_the_five_decreasing_squares_puzzle}{}\section*{{Solution to the Five Decreasing Squares Puzzle}}\label{solution_to_the_five_decreasing_squares_puzzle}

[[FiveDecreasingSquares.png:pic]]

\begin{quote}%
What fraction of the largest square is shaded?


\end{quote}
\hypertarget{solution_by_dissection}{}\subsection*{{Solution by [[Dissection]]}}\label{solution_by_dissection}

[[FiveDecreasingSquaresDissection.png:pic]]

In the above picture, the area of the orange square is twice the area of the blue square while the area of the green square is twice the area of one orange square and half a blue square. So the orange square has area five times that of the blue square. Therefore in the original puzzle the largest square has area twenty-five times that of the smallest.

\hypertarget{solution_by_pythagoras_theorem}{}\subsection*{{Solution by [[Pythagoras’ Theorem]]}}\label{solution_by_pythagoras_theorem}

[[FiveDecreasingSquaresLabelled.png:pic]]

With the points labelled as above, let $a$ be the length of $A F$, $b$ the length of $A B$, and $c$ of $F C$. Then $A G$ has length $\frac{1}{2}b$ and $F G$ has length $\frac{3}{2}b$ so applying [[Pythagoras’ Theorem]] to triangle $A F G$ shows that

\begin{displaymath}
a^2 = \frac{b^2}{4} + \frac{9b^2}{4} = \frac{10 b^2}{4}
\end{displaymath}
Applying [[Pythagoras’ Theorem]] to triable $D F C$ then shows that $b^2 = 2 c^2$. Putting those together yields $a^2 = 5 c^2$.

This ratio occurs twice in the main diagram, so the area of the largest square is twenty-five times that of the smallest.



\end{document}