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

The given equations of the half-lines are $x - \sqrt{3}y = \alpha$ for $y \ge 0$ and $x + \sqrt{3}y = \alpha$ for $y \le 0$.

The point of intersection $P$ is obtained by setting $y = 0$, which gives $x = \alpha$. Thus, $P \equiv (\alpha, 0)$.

The angle bisector of these half-lines is the x-axis ($y = 0$).

The slope of the first half-line is $\dfrac{1}{\sqrt{3}}$, so it makes an angle of $30^\circ$ with the positive x-axis. The second half-line has a slope of $-\dfrac{1}{\sqrt{3}}$, making an angle of $-30^\circ$ with the positive x-axis.

Since $A$ and $B$ are at a distance $\alpha$ from $P$, their coordinates can be written using the parametric form of a straight line:
$A \equiv (\alpha + \alpha \cos 30^\circ, 0 + \alpha \sin 30^\circ) \equiv \left(\alpha + \dfrac{\alpha\sqrt{3}}{2}, \dfrac{\alpha}{2}\right)$
$B \equiv (\alpha + \alpha \cos(-30^\circ), 0 + \alpha \sin(-30^\circ)) \equiv \left(\alpha + \dfrac{\alpha\sqrt{3}}{2}, -\dfrac{\alpha}{2}\right)$

The line segment $AB$ is vertical and intersects the angle bisector (x-axis) at $Q$. Thus, the coordinates of $Q$ are $\left(\alpha + \dfrac{\alpha\sqrt{3}}{2}, 0\right)$.

The distance $PQ$ is the difference in their x-coordinates:
$PQ = \left(\alpha + \dfrac{\alpha\sqrt{3}}{2}\right) - \alpha = \dfrac{\alpha\sqrt{3}}{2}$

Given $PQ = \dfrac{9}{2}$, we have:
$\dfrac{\alpha\sqrt{3}}{2} = \dfrac{9}{2} \Rightarrow \alpha\sqrt{3} = 9 \Rightarrow \alpha = 3\sqrt{3}$

In $\triangle PAB$, $PA = PB = \alpha$ and $\angle APB = 30^\circ - (-30^\circ) = 60^\circ$. Therefore, $\triangle PAB$ is an equilateral triangle with side length $\alpha$.

The circumradius $R$ of an equilateral triangle of side $\alpha$ is given by $R = \dfrac{\alpha}{\sqrt{3}}$.

We need to find the value of $\dfrac{\alpha^2}{R}$:
$\dfrac{\alpha^2}{R} = \dfrac{\alpha^2}{\dfrac{\alpha}{\sqrt{3}}} = \alpha\sqrt{3}$

Since we already found $\alpha\sqrt{3} = 9$, we get:
$\dfrac{\alpha^2}{R} = 9$

Answer: $9$