# Solution to the Angle Formed by Two Squares II Puzzle +-- {.image} [[AngleFormedbyTwoSquaresII.png:pic]] > These squares are the same size. What’s the angle? =-- ## Solution by [[Isosceles Triangles]] +-- {.image} [[AngleFormedbyTwoSquaresIIIsosceles.png:pic]] =-- As the squares are the same size, triangle $O B C$ is [[isosceles]] with $O B$ and $O C$ of the same length. Also triangle $O A C$ is isosceles, with $O A$ and $O C$ the same length. Let $a$ be angle $O \hat{C} A$ and let $b$ be angle $O \hat{C} B$. The requested angle is $b - a$. As $O B C$ is isosceles, and the [[angles in a triangle]] add up to $180^\circ$, angle $B \hat{O} C$ is $180^\circ - 2 b$. Similarly, angle $A \hat{O} C$ is $180^\circ - 2 a$. Therefore, angle $A \hat{O} B$ is $$ 180^\circ - 2 a - (180^\circ - 2 b) = 2 b - 2 a = 2(b - a) $$ It is also the corner of a [[square]], so is $90^\circ$. Hence $b - a = 45^\circ$. ## Solution by [[Angle at the Circumference is Half the Angle at the Centre]] +-- {.image} [[AngleFormedbyTwoSquaresIICircle.png:pic]] =-- As the squares are the same size, the circle centred at $O$ that passes through $A$ also passes through $B$ and $C$. Then the requested angle is half of angle $O \hat{O} B$, which is $90^\circ$, since the [[angle at the circumference is half the angle at the centre]]. ## Solution by [[Invariance Principle]] +-- {.image} [[AngleFormedbyTwoSquaresIISpecial.png:pic]] =-- By drawing the squares so that they are adjacent without overlapping, the requested angle can be seen to be $45^\circ$.