Notes
two squares on a hexagon solution

Solution to the Two Squares on a Hexagon Puzzle

Solution by Angle Properties

With the points labelled as above, let $x = B \hat{O} F$. Then as angle $O \hat{B} F$ is a right angle and angles in a triangle add up to $180^\circ$, $B \hat{F} O = 90^\circ - x$. Since $K$ is the midpoint of $O F$ and triangle $O B F$ is a right-angled triangle, triangle $B K F$ is isosceles and so angle $K \hat{B} F = K \hat{F} B = 90^\circ - x$. This means that angle $B \hat{K} F = 180^\circ - 2 \times (90^\circ - x) = 2 x$ and so angle $B \hat{K} H = 90^\circ + 2 x$.

The length of $B K$ is equal to that of $K F$ which is equal to the length of $K H$. Hence triangle $B K H$ is also isosceles and so angle $K \hat{B} H$ is one half of $180^\circ - (90^\circ + 2 x)$ which is $45^\circ - x$. Then:

To get the requested angle, note that $E \hat{J} I = H \hat{B} C$ by alternate angles and $D \hat{E} J = 60^\circ$ as the exterior angle of a regular hexagon. So using angles in a triangle and angles on a straight line, angle $D \hat{I} J = 180^\circ - (180^\circ - 45^\circ - 60^\circ) = 105^\circ$.

Solution by Circle Theorems

Construct a circle centred on $K$ and passing through $O$. As $K H$ and $K O$ are sides of a square, $H$ is also on this circle. As $K$ is the midpoint of $O F$, $F$ is also on this circle. Finally, as triangle $O B F$ is a right-angled triangle and $O F$ is a diagonal, $B$ is also on this circle.

Using angles in the same segment are equal, angles $F \hat{B} H$ and $F \hat{O} H$ are equal. Since $F \hat{O} H$ is the angle between a side and a diagonal of a square, it is $45^\circ$. Hence $F \hat{B} H = 45^\circ$. The requested angle then is $105^\circ$ as above.

Solution by Transformation

With the points labelled as above, the point $F$ moves on a straight line through $B$ and $C$. The point $H$ is obtained from $F$ by a fixed transformation of a rotation of $45^\circ$ anticlockwise about $O$ and scaling by $\frac{1}{\sqrt{2}}$. It therefore also moves on a straight line.

To see where this straight line is, consider two special cases. In the first, $F$ is at $B$. In this configuration, $H$ is at a point so that angle $G \hat{B} H = 45^\circ$.

In the second, $F$ is taken to the left so that the length of $F B$ is the same as that of $O B$. This brings $H$ to coincide with $B$ and shows that $B$ is on the line that $H$ moves along.

Taken together, these show that $H$ moves on a line that goes through $B$ at an angle of $45^\circ$ to the horizontal.

Solution by Invariance Principle

In the configuration where $F$ is at $B$, the line $B H$ is the diagonal of a square so the angle $G \hat{H} B$ is $45^\circ$. From here, the given angle can be calculated as $105^\circ$ as above.