Notes
two octagons and two squares solution

Solution to the Two Octagons and Two Squares Puzzle

Solution by Pythagoras' Theorem and Lengths in an Isosceles Right-Angled Triangle

Label the points as in the above diagram. Take the length of line segment $B C$ as one unit.

Triangle $B C D$ is an isosceles right-angled triangle, so line segment $B D$ has length $\sqrt{2}$. Therefore, line segment $A C$ has length $1 + \sqrt{2}$ and the outer square has side length $2 + \sqrt{2} = \sqrt{2}(\sqrt{2} + 1)$.

Applying Pythagoras' theorem to triangle $A C B$ shows that line segment $A B$ has length:

Similarly, the length of $A B$ is $2 + \sqrt{2}$ times the length of $A F$ and the length of $A E$ is $1 + \sqrt{2}$ times the length of $A F$. Therefore, the length of $A E$ is $\frac{1 + \sqrt{2}}{2 + \sqrt{2}} = \frac{1 + \sqrt{2}}{\sqrt{2}(\sqrt{2} + 1)} = \frac{1}{\sqrt{2}}$ times the length of $A B$, so is:

The ratio of lengths of the sides of the squares is therefore:

In terms of area, the larger square has area $2 + \sqrt{2}$ times that of the smaller.