# Solution to the Four Squares III Puzzle +-- {.image} [[FourSquaresIII.png:pic]] > Four squares. The blue area is $2$. What’s the green area? =-- ## Solution by [[Transformations]] +-- {.image} [[FourSquaresIIILabelled.png:pic]] =-- With the points labelled as in the above diagram, point $E$ moves along the line $F G$. With respect to the point $C$, $B$ can be obtained from $E$ by a rotation by $45^\circ$ and a scaling by $\frac{1}{\sqrt{2}}$. As this is a fixed [[transformation]], $B$ also moves along a straight line. When $E$ is at $F$, $B$ is at $G$, and when $E$ is at $G$, $B$ is at $H$. Hence $B$ moves along the line $G H$. As $H$ is the midpoint of the side $A C$, the lengths of $A B$ and $B C$ are equal. Since $A B$ is the diagonal of the blue square and $B C$ the side of the green square, the green square is twice the area of the blue square (this can be seen by [[dissecting]] each square into [[congruent]] [[isosceles]] [[right-angled triangles]]; the green into $4$ and the blue into $2$). Hence the green square has area $4$. ## Solution by [[Invariance Principle]] When the puzzle is in one of the two configurations where $E$ is at $F$ or where $E$ is at $G$, the relationship between the green and blue squares is straightforward to see. +-- {.image} [[FourSquaresIIISpecialA.png:pic]] =-- +-- {.image} [[FourSquaresIIISpecialB.png:pic]] =--