# Solution to the [[Four Squares XII]] Puzzle +-- {.image} [[FourSquaresXII.png:pic]] > The three smaller squares are the same size. What’s the angle? =-- ## Solution by Properties of [[Isosceles Triangles]], [[Angles in a Triangle]], and [[Angles at a Point on a Straight Line]] +-- {.image} [[FourSquaresXIIAnnotated.png:pic]] =-- With the points labelled as above, consider triangles $B C I$ and $I H J$. Both are [[right-angled triangles|right-angled]]. The sides $B C$ and $I H$ are sides of the purple squares and so have the same length. The sides $C I$ and $H J$ are the difference between the side of the outer square and one purple square, so have the same length. Therefore, triangles $B C I$ and $I H J$ are [[congruent]] so triangle $B I J$ is [[isosceles]]. Angles $B \hat{I} C$ and $H \hat{I} J$ add up to $90^\circ$ since the [[angles in a triangle]] add up to $180^\circ$, so since [[angles at a point on a straight line]] add up to $90^\circ$, angle $J \hat{I} C$ is $90^\circ$. Therefore, angle $B \hat{J} I$ is half of $180^\circ - 90^\circ = 90^\circ$, so is $45^\circ$. ## Solution by [[Invariance Principle]] +-- {.image} [[FourSquaresXIIInvariance.png:pic]] =-- The size of the outer square can vary, so drawing it as small as possible produces the configuration as above, from which the angle is clearly $45^\circ$.