Notes
regular hexagon and rectangle solution

Solution to the Regular Hexagon and Rectangle Puzzle

Regular hexagon and rectangle

Here’s a regular hexagon and a rectangle. What’s the angle?

Solution by Angle at the Circumference is Half the Angle at the Centre

Regular hexagon and rectangle labelled

The circle in the above diagram is the circumcircle through the vertices of the hexagon and the point labelled OO is its centre. This is also the centre of the rectangle and so the vertices of the rectangle also lie on the circle. The angle AD^BA \hat{D} B is then half of angle AO^BA \hat{O} B since the angle at the circumference is half the angle at the centre. Triangle AOBA O B is an equilateral triangle as it is formed from the regular hexagon so angle AO^BA \hat{O} B is 60 60^\circ meaning that angle AD^BA \hat{D} B is 30 30^\circ.

Solution by Invariance Principle

Regular hexagon and rectangle extreme

Under the assumption that the angle doesn’t depend on the size of the rectangle (providing it shares a diagonal with the hexagon) there are two configurations where the angle can be easily deduced. These are both shown in the above diagram. In one, the rectangle is viewed as actually being the line EAE A so the angle in question is AE^BA \hat{E} B. In the other, the rectangle is EGACE G A C so the angle in question is AC^BA \hat{C} B.

Triangle ABCA B C is isosceles and angle CB^AC \hat{B} A is the interior angle of a regular hexagon so is 120 120^\circ, meaning that angle AC^BA \hat{C} B is 180 120 2=30 \frac{180^\circ - 120^\circ}{2} = 30^\circ. Then angle BE^CB \hat{E} C is also 30 30^\circ, while angle AE^CA \hat{E} C is 120 2\frac{120^\circ}{2}, meaning that angle AE^BA \hat{E} B is 120 230 =30 \frac{120^\circ}{2} - 30^\circ = 30^\circ.