Tile Design

Choose a triangle and note that the angle $\alpha$ it makes is the the same angle about which the top square is rotated at the centre. Let one side of the triangle by $y$ units as shown. Thus, we obtain all the sides accordingly as shown.

Notice that the common area forms a regular octagon with each side = $y\sec\alpha$ .

We know that the octagon subtends $135^{o}$ at the sides and $45^{o}$ at the centre. Therefore via trigonometric manipulation, we find $R$ = $\frac{y\sec\alpha}{2\cos67.5^{o}}$ , where $\cos67.5^{o}$ = $\frac{\sqrt(2 - \sqrt2)}{2}$

$R$ is the length of the line joining the centre of the octagon to one of its vertices.

We know that area of any polygon = $\frac{nR^{2}}{2}$ $sin(\frac{360^{o}}{n})$

For the octagon, put $n = 8$ and at the same time insert the value of $R$ . What we get is:

Common area of the squares = Area of octagon = $2y^{2}(1 + \sqrt2)\sec^{2}\alpha$

Total remaining area = 8 x $\frac{1}{2}$ x $y$ x $y\tan\alpha$ = $4y^{2}\tan\alpha$

Therefore, $\frac{\text{Common area of the squares}}{\text{Total remaining area}}$ = $\frac{1 + \sqrt2}{2\cos^{2}\alpha\tan\alpha}$