Ratio of Square of Alpha to Circumradius of Triangle PAB

To find the value of $\frac{\alpha^2}{R}$, we analyze the geometry of the given half-lines and triangle $\Delta PAB$.

Step 1: Identify the half-lines and their point of intersection

The equation of the two half-lines is given by: $x - \sqrt{3}|y| = \alpha \quad (\alpha > 0)$

We can rewrite this as: $x = \alpha + \sqrt{3}|y|$

Since $|y| \ge 0$, we have $x \ge \alpha$. The two individual half-lines are:

1. Upper half-line ($y \ge 0$): $y = \frac{1}{\sqrt{3}}(x - \alpha)$
2. Lower half-line ($y \le 0$): $y = -\frac{1}{\sqrt{3}}(x - \alpha)$

The point of intersection $P$ occurs when $y = 0$, giving $P(\alpha, 0)$.

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Step 2: Determine the angle between the half-lines

The angle of inclination $\theta_1$ of the upper half-line with the positive $x$-axis satisfies: $\tan \theta_1 = \frac{1}{\sqrt{3}} \implies \theta_1 = 30^\circ$

Similarly, for the lower half-line: $\tan \theta_2 = -\frac{1}{\sqrt{3}} \implies \theta_2 = -30^\circ$

Thus, the angle between the two half-lines at the intersection point $P$ is: $\angle APB = \theta_1 - \theta_2 = 30^\circ - (-30^\circ) = 60^\circ$

The angle bisector of these two half-lines is the line $y = 0$ (the $x$-axis) for $x \ge \alpha$.

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Step 3: Analyze the geometry of $\Delta PAB$

Let point $A$ lie on the upper half-line and point $B$ lie on the lower half-line such that: $PA = PB = \alpha$

Since $PA = PB = \alpha$ and the vertex angle $\angle APB = 60^\circ$, $\Delta PAB$ is an equilateral triangle with side length equal to $\alpha$.

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Step 4: Relation between altitude $PQ$ and circumradius $R$

The line segment $AB$ intersects the angle bisector at point $Q$. In an equilateral triangle, the angle bisector from a vertex is also the altitude to the opposite side.

Therefore, $PQ$ is the altitude of $\Delta PAB$: $PQ = \frac{\sqrt{3}}{2}\alpha$

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

The radius $R$ of the circumcircle of an equilateral triangle with side length $\alpha$ is: $R = \frac{\alpha}{\sqrt{3}}$

Substituting $\alpha = 3\sqrt{3}$: $R = \frac{3\sqrt{3}}{\sqrt{3}} = 3$

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Step 5: Compute $\frac{\alpha^2}{R}$

Using the values obtained: $\alpha^2 = (3\sqrt{3})^2 = 27$

Thus: $\frac{\alpha^2}{R} = \frac{27}{3} = 9$

(Alternatively, notice that $\frac{\alpha^2}{R} = \frac{\alpha^2}{\alpha / \sqrt{3}} = \sqrt{3}\alpha = 2 PQ = 2 \times \frac{9}{2} = 9$.)