# Solution to the [[Seven Squares]] Puzzle +-- {.image} [[SevenSquares.png:pic]] > Seven squares. What’s the angle? =-- ## Solution by [[Angle in a Semi-Circle]] and [[Angle at the Circumference is Half the Angle at the Centre]] +-- {.image} [[SevenSquaresAnnotatedCircle.png:pic]] =-- With the points labelled as above, consider a circle centred on $O$ which passes through $A$. Since the pink squares are all the same size, $O A$, $O B$, and $O C$ all have the same length and therefore the circle also passes through $B$ and $C$. Triangle $A D C$ is a [[right-angled triangle|right-angled]], since $D$ is the corner of the purple square. Therefore, as the [[angle in a semi-circle]] is a right-angle, $D$ also lies on the same circle. Then as the [[angle at the circumference is half the angle at the centre]], angle $A \hat{D} B$ is half of angle $A \hat{O} B$, but this is the corner of a square and so is $90^\circ$. Therefore, angle $A \hat{D} B$ is $45^\circ$. ## Solution by [[Isosceles Triangles]] and [[Angles in a Quadrilateral]] +-- {.image} [[SevenSquaresAnnotatedTriangles.png:pic]] =-- In the above diagram, $E$ and $F$ are so that $O E$ and $O F$ are [[perpendicular]] to $A D$ and $D C$ respectively. Triangles $O E A$ and $C F O$ are [[congruent]] because $C O$ and $O A$ have the same length, so $E$ is at the [[midpoint]] of $A D$. This means that triangle $O E D$ is also congruent to $O E A$, so $O D$ also has the same length as $O A$. Hence triangles $O D A$ and $O D B$ are each [[isosceles]]. So the sum of angles $D \hat{B} O$ and $O \hat{A} D$ is equal to angle $A \hat{D} B$, meaning that the sum of the angles inside [[quadrilateral]] $A D B O$ is $2 A \hat{D} B + B \hat{O} A$, where the latter is taken inside the quadrilateral. That angle is $360^\circ - 90^\circ = 270^\circ$ and so, since the [[angles in a quadrilateral]] add up to $360^\circ$, angle $A \hat{D} B$ is $\frac{360^\circ - 270^\circ}{2} = 45^\circ$. ## Solution by [[Invariance Principle]] The larger square can tilt in relation to the arrangement of smaller squares, giving configurations where the answer can be more simply deduced. +-- {.image} [[SevenSquaresAnnotatedInvarianceA.png:pic]] =-- In this configuration, point $D$ aligns with $C$, meaning that angle $A \hat{D} B$ coincides with the diagonal in the upper left square, whence is $45^\circ$. +-- {.image} [[SevenSquaresAnnotatedInvarianceB.png:pic]] =-- In this configuration, $D$ is such that $B O D$ is a straight line and the corners of the large square are at $A$, $B$, $C$, and $D$, whence angle $A \hat{D} O = 45^\circ$.