Let $$A$$ and $$B$$ be points on the two half-lines $$x - \sqrt{3}|y| = \alpha$$, $$\alpha > 0$$, at distance of $$\alpha$$ from the point of intersection $$P$$. The line $$AB$$ meets the angle bisector of the given half-lines at the point $$Q$$. If $$PQ = \frac{9}{2}$$ and $$R$$ is the radius of the circumcircle  of $$\triangle PAB$$, then $$\frac{\alpha^2}{R}$$ is equal to :

$$L_1: x - \sqrt{3}y = \alpha \quad (y > 0)$$

$$L_2: x + \sqrt{3}y = \alpha \quad (y < 0)$$

For $$L_1$$: $$y = \frac{1}{\sqrt{3}}x - \frac{\alpha}{\sqrt{3}} \implies \tan\theta_1 = \frac{1}{\sqrt{3}} \implies \theta_1 = 30^\circ$$

For $$L_2$$: $$y = -\frac{1}{\sqrt{3}}x + \frac{\alpha}{\sqrt{3}} \implies \tan\theta_2 = -\frac{1}{\sqrt{3}} \implies \theta_2 = -30^\circ$$

The total angle between the two half-lines at vertex $$P$$ is: $$\angle APB = 30^\circ - (-30^\circ) = 60^\circ$$

Given that $$PA = PB = \alpha$$ and the vertical angle $$\angle APB = 60^\circ$$, $$\triangle PAB$$ must be an equilateral triangle. Therefore, the side length of the triangle is $$AB = \alpha$$.

Since $$Q$$ lies on the angle bisector (the x-axis) and $$\triangle PAB$$ is equilateral, $$PQ$$ is the median/altitude of $$\triangle PAB$$.

$$PQ = PA \cos(30^\circ) = \alpha \frac{\sqrt{3}}{2}$$

$$\alpha \frac{\sqrt{3}}{2} = \frac{9}{2} \implies \alpha = \frac{9}{\sqrt{3}} = 3\sqrt{3}$$

Circumradius, $$R = \frac{a}{2\sin(60^\circ)} = \frac{\alpha}{\sqrt{3}}$$

$$R = \frac{3\sqrt{3}}{\sqrt{3}} = 3$$

$$\frac{\alpha^2}{R} = \frac{(3\sqrt{3})^2}{3} = \frac{27}{3} = 9$$