Notes
four congruent squares inside a larger square solution

Solution to the Four Congruent Squares Inside a Larger Square Puzzle

Solution by Similar Triangles and Pythagoras' Theorem

With the points labelled as above, let $a$, $b$, $c$ be the side lengths of the right-angled triangle $G F H$ in increasing order, so that $a$ is the length of $F H$, $b$ of $G F$, and $c$ of $G H$. Pythagoras' theorem gives the relationship $c^2 = a^2 + b^2$.

Triangle $H I C$ is congruent to triangle $G F H$ since angle $C \hat{H} I$ is equal to angle $F \hat{G} H$ and the lengths of $I H$ and $G F$ are equal. So $C H$ has length $c$ and thus the side length of the outer square is $a + b + c$.

This means that $A E$ has length $a + c$. Triangle $A E G$ is similar to triangle $G F H$, so the ratios $b : a$ and $a + c : b$ are equal. This gives the relationship $b^2 = a(a + c)$. Putting this in to the equation above from Pythagoras’ theorem yields $c^2 = 2 a^2 + a c$, or $c^2 - a c - 2 a^2 = 0$. This has solutions $c = -a$ or $c = 2 a$. Since $c$ and $a$ are lengths (and so positive), it must be that $c = 2 a$.

The total area of purple regions is then $a b + (a + c) b = 4 a b$.

To find the red area, note that triangle $A B H$ is similar to triangle $G F H$ and so the length of $A H$ is double that of $B H$, which is $a + b$. So the length of $A H$ is $2 a + 2 b$. The sum of the two red segments is therefore $2 a$. The red regions can therefore be put together to form a rectangle with side lengths $2a$ and $b$, so the area of the red regions is $2 a b$.

Therefore the ratio of red to purple is $1 : 2$.

Solution by Similar Triangles and Hidden Point

This solution is similar to the above, except that it uses properties of similar triangles instead of Pythagoras' theorem and so avoids factorising a quadratic. Having noted, as above, that the length of $A E$ is $a + c$, let $J$ be the point on $A E$ so that $A J$ has length $c$ and $J E$ length $a$. Then $J E G$, $A K J$, and $G K J$ are all congruent triangles, and so the length of $A G$ is twice that of $E G$. The rest of the proof continues as above.