# Solution to the Four Squares IV Puzzle +-- {.image} [[FourSquaresIV.png:pic]] > Four squares. What’s the angle? =-- ## Solution by [[Angle at the Centre is Twice the Angle at the Circumference]] and [[Angle in a Semi-Circle]] +-- {.image} [[FourSquaresIVLabelled.png:pic]] =-- With the points labelled as in the above diagram, the circle is centred at $G$ and passes through the vertices $A$, $E$, and $F$ of the light blue squares. As point $E$ lies on the line segment $C B$, angle $A \hat{C} E$ is $45^\circ$, while angle $A \hat{G} E$ is $90^\circ$. By the converse to the [[Angle at the Centre is Twice the Angle at the Circumference]], this means that $C$ lies on the circle. Since $F A$ is a [[diameter]] of the circle, angle $F \hat{C} A$ is the [[angle in a semi-circle]] and so is $90^\circ$. Then as angle $D \hat{C} A$ is $45^\circ$, angle $F \hat{C} D$ is $45^\circ$. ## Solution by the [[Invariance Principle]] +-- {.image} [[FourSquaresIVRotated.png:pic]] =-- Consider the rotated diagram above and view the dark blue square, $A B C D$, as fixed. In this scenario, the point labelled $E$ is free to move on the line segment $B C$. As it does so, the point labelled $F$ also moves on a straight line (relative to the dark square). The extremes are when $E$ is at $B$ and at $C$ and are shown below. From this, $F$ moves on a line from $C$ at an angle of $45^\circ$ to the line segment $D C$. +-- {.image} [[FourSquaresIVExtremeB.png:pic]] =-- +-- {.image} [[FourSquaresIVExtremeC.png:pic]] =--