# Solution to the [[Two Octagons and Two Squares]] Puzzle +-- {.image} [[TwoOctagonsandTwoSquares.png:pic]] > The two regular octagons are concentric. Find [the] relationship between the two squares. =-- ## Solution by [[Pythagoras' Theorem]] and Lengths in an [[Isosceles]] [[Right-Angled Triangle]] +-- {.image} [[TwoOctagonsandTwoSquaresLabelled.png:pic]] =-- 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: $$ \sqrt{ (1 + \sqrt{2})^2 + 1^2 } = \sqrt{1 + 2 \sqrt{2} + 2 + 1} = \sqrt{4 + 2 \sqrt{2}} = \sqrt{ 2 \sqrt{2}(\sqrt{2} + 1) } $$ 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: $$ \frac{1}{\sqrt{2}} \times \sqrt{ 2 \sqrt{2}(\sqrt{2} + 1) } = \sqrt{ \sqrt{2}(\sqrt{2} + 1)} $$ The ratio of lengths of the sides of the squares is therefore: $$ \sqrt{2}(\sqrt{2} + 1) : \sqrt{ \sqrt{2}(\sqrt{2} + 1)} = \sqrt{ \sqrt{2}(\sqrt{2} + 1)} : 1 $$ In terms of area, the larger square has area $2 + \sqrt{2}$ times that of the smaller.