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\begin{document}

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\section*{four squares vii solution}

\hypertarget{solution_to_the_four_squares_vii_puzzle}{}\section*{{Solution to the Four Squares VII Puzzle}}\label{solution_to_the_four_squares_vii_puzzle}

[[FourSquaresVII.png:pic]]

\begin{quote}%
The green area is $7$. What’s the total area of the four squares?


\end{quote}
\hypertarget{solution_by_calculation}{}\subsection*{{Solution by Calculation}}\label{solution_by_calculation}

[[FourSquaresVIILabelled.png:pic]]

In the diagram above, $a$, $b$, $c$, and $d$ are the side lengths of the squares in ascending order. We have $c = a + b$ and $d = c + b$. Our goal is to compute $a^2 + b^2 + c^2 + d^2$ given that the green area has area $7$.

It is possible to rewrite everything in terms of $a$ and $b$, but the expressions turn out to be more straightforward if instead everything is written in terms of $b$ and $c$. This uses $a = c - b$ and $d = c + b$.

This is because of the degenerate version of the [[parallelogram law]] (where the parallelogram lies on a straight line):

\begin{displaymath}
a^2 + d^2 = (c - b)^2 + (c + b)^2 = c^2 - 2 c b + b^2 + c^2 + 2 c b + b^2 = 2 c^2 + 2 b^2
\end{displaymath}
So the full area is:

\begin{displaymath}
a^2 + b^2 + c^2 + d^2 = 3 b^2 + 3 c^2
\end{displaymath}
The white area is easier to compute directly than the green as it is four right-angled triangles. It is:

\begin{displaymath}
\frac{1}{2} d^2 + \frac{1}{2} c(c + d) + \frac{1}{2} (b - a) b + \frac{1}{2} a^2 = \frac{1}{2} \left( a^2 + b^2 + c^2 + d^2 + c d - a b \right)
\end{displaymath}
We already know what $a^2 + b^2 + c^2 + d^2$ is in terms of $b$ and $c$, the other terms simplify as follows:

\begin{displaymath}
c d - a b = c (c + b) - b (c - b) = c^2 + c b - b c + b^2 = c^2 + b^2
\end{displaymath}
So the white area is $2 b^2 + 2 c^2$. The green area is therefore $b^2 + c^2$ and so the total area is:

\begin{displaymath}
3 b^2 + 3 c^2 = 3 \times 7 = 21
\end{displaymath}
\emph{Note: in the interest of completeness, it can be shown that the parts of the green region that lie in the square of side $b$ and of side $c$ have the same area as the lower white triangle in the largest square. However, unless $a$ and $b$ are in the correct proportion they do not fit as they are into that region but need further dissection. So although this can be used for a dissection method for finding the green area, it needs a little algebra to show that it works.}

\hypertarget{solution_by_invariance_principles}{}\subsection*{{Solution by [[Invariance Principles]]}}\label{solution_by_invariance_principles}

The part of this problem that can be varied is the relative sizes of the two small squares. The extremes for this are when the squares are the same size and when the smaller square is non-existant.

[[FourSquaresVIISameSize.png:pic]]

When the two are the same size, the relationships between the sizes are that $b = a$, $c = 2 a$, and $d = 3 a$. The total area is therefore $15 a^2$. The green area can be subdivided as above to show that its area is:

\begin{displaymath}
\frac{1}{2} (a + 3 a) a + \frac{1}{2} \times 3 a \times 2 a = 2 a^2 + 3 a^2 = 5 a^2
\end{displaymath}
This shows that the total area is three times that of the green area, hence is $21$.

[[FourSquaresVIIOneSquare.png:pic]]

When the smallest square vanishes, the remaining squares have sizes $c = b$ and $d = 2 b$. The total area is therefore $6 b^2$. The green area can be subdivided as above to show that its area is:

\begin{displaymath}
\frac{1}{2} \times 2 b \times b + \frac{1}{2} \times 2 b \times b = 2 b^2
\end{displaymath}
So, as before, as the green region has area $7$, the total area is $3 \times 7 = 21$.



\end{document}