# Solution to the [[Three Squares III]] Puzzle +-- {.image} [[ThreeSquaresIII.png:pic]] > All three squares are the same size. What’s the angle? =-- ## Solution by [[Angle at the Circumference is Half the Angle at the Centre]] +-- {.image} [[ThreeSquaresIIILabelled.png:pic]] =-- With the points labelled as above, the line segments $O A$, $O B$, and $O C$ are all diagonals of [[congruent]] squares, so are the same length. Therefore the circle centre $O$ that passes through $A$ also passes through $B$ and $C$. Then since the [[angle at the circumference is half the angle at the centre]], angle $A \hat{C} B$ is half angle $A \hat{O} B$. But this is $90^\circ$ as it is formed from the diagonals of two adjacent squares. Hence angle $A \hat{C} B = 45^\circ$. ## Solution by [[Isosceles Triangles]] and [[Angles in a Triangle]] With the same labelling as above, triangles $O A C$, $O B A$, and $O C B$ are [[isosceles]]. Since $O A$ and $O B$ are the diagonals of squares, angles $O \hat{A} B$ and $O \hat{B} A$ are both $45^\circ$. Since the angles in triangle $A B C$ add up to $180^\circ$, this means that the sum of angles $O \hat{A} B$, $O \hat{C} A$, $O \hat{C} B$, and $O \hat{B} C$ is $90^\circ$. Then since triangles $O A C$ and $O B C$ are isosceles, angles $O \hat{A} C$ and $O \hat{C} A$ are equal to each other, as are angles $O \hat{B} C$ and $O \hat{C} B$. Therefore, angle $A \hat{C} B$ is half of $90^\circ$ so is $45^\circ$. ## Solution by [[Agg Invariance Principle]] and [[Angles in a Square]] The upper square is tilted, but the angle of the tilt is not specified. A special case is where it aligns with the right-hand lower square, as in the diagram below. +-- {.image} [[ThreeSquaresIIISpecial.png:pic]] =-- In this case, the requested angle is the angle between a diagonal and side in a square, hence is $45^\circ$.