Symmetrical Ratio

Let $OB = 1$ and label $C$ and $D$ as follows:

By the properties of an equilateral triangle, $\angle AOD = 60°$ and by the properties of a square, $\angle AOC = 45°$ .

That means $\angle COD = \angle AOD - \angle BOC = 60° - 45° = 15°$ .

From $\triangle BOD$ , $OD = OB \sin 30° = 1 \cdot \frac{1}{2} = \frac{1}{2}$ .

From $\triangle COD$ , $OC = \cfrac{OD}{\cos 15°} = \cfrac{\frac{1}{2}}{\frac{1 + \sqrt{3}}{2\sqrt{2}}} = \frac{\sqrt{2}(\sqrt{3} - 1)}{2}$ .

From $\triangle AOC$ , $OA = \cfrac{OC}{\cos 45°} = \cfrac{\frac{\sqrt{2}(\sqrt{3} - 1)}{2}}{\frac{\sqrt{2}}{2}} = \sqrt{3} - 1$ .

Therefore, the ratio $\cfrac{OA}{OB} = \cfrac{\sqrt{3} - 1}{1} = \sqrt{3} - 1$ .