Let A,B be points on the two half-lines x- 3 |y|=α , α>0 at a...

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$.<br><br>The point of intersection$P$is obtained by setting$y = 0$, which gives$x = \alpha$. Thus,$P \equiv (\alpha, 0)$.<br><br>The angle bisector of these half-lines is the x-axis ($y = 0$).<br><br>The slope of the first half-line is$\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.<br><br>Since$A$and$B$are at a distance$\alpha$from$P$, their coordinates can be written using the parametric form of a straight line:<br>$A \equiv (\alpha + \alpha \cos 30^\circ, 0 + \alpha \sin 30^\circ) \equiv \left(\alpha + \dfrac{\alpha\sqrt{3}}{2}, \dfrac{\alpha}{2}\right)$<br>$B \equiv (\alpha + \alpha \cos(-30^\circ), 0 + \alpha \sin(-30^\circ)) \equiv \left(\alpha + \dfrac{\alpha\sqrt{3}}{2}, -\dfrac{\alpha}{2}\right)$<br><br>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: $9$