Octagonal Rose

The entire octagonal rose can be tiled by different sizes of the following tile, all with the same ratio of shaded to unshaded parts, where $x = AD = DB = BC$ and $\angle ADB = 90°$ :

The area of $\triangle ADB$ is $A_{\triangle ADB} = \frac{1}{2} \cdot x \cdot x = \frac{1}{2}x^2$ and the area of $\triangle DBC$ is $A_{\triangle DBC} = \frac{1}{2} \cdot x \cdot x \sin 135° = \frac{\sqrt{2}}{4}x^2$ , so the ratio of shaded to the whole is $\cfrac{A_{\triangle ADB}}{A_{\triangle ADB} + A_{\triangle DBC}} = \cfrac{\frac{1}{2}x^2}{\frac{1}{2}x^2 + \frac{\sqrt{2}}{4}x^2} = 2 - \sqrt{2}$ .

This ratio is maintained for the whole regular octagon, which has an area of $A_{\text{oct}} = 2(1 + \sqrt{2})$ , so the area of the shaded region is $(2 - \sqrt{2})A_{\text{oct}} = (2 - \sqrt{2}) \cdot 2(1 + \sqrt{2}) = \sqrt{8}$ , which makes $a = \boxed{8}$ .