Let A,B be points on the two half-lines x-√3|y|=α, α>0 at a distance of α from their point of intersection P. The line segment AB meets the angle…

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

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

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

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

Since $A$A and $B$B are at a distance $\alpha$α from $P$P, their coordinates can be written using the parametric form of a straight line:

$$A \equiv (\alpha + \alpha \cos 30^\circ, 0 + \alpha \sin 30^\circ) \equiv \left(\alpha + \dfrac{\alpha\sqrt{3}}{2}, \dfrac{\alpha}{2}\right)$$A≡(α+αcos30∘,0+αsin30∘)≡(α+2α3​​,2α​)

$$B \equiv (\alpha + \alpha \cos(-30^\circ), 0 + \alpha \sin(-30^\circ)) \equiv \left(\alpha + \dfrac{\alpha\sqrt{3}}{2}, -\dfrac{\alpha}{2}\right)$$B≡(α+αcos(−30∘),0+αsin(−30∘))≡(α+2α3​​,−2α​)

The line segment $AB$AB is vertical and intersects the angle bisector (x-axis) at $Q$Q. Thus, the coordinates of $Q$Q are $\left(\alpha + \dfrac{\alpha\sqrt{3}}{2}, 0\right)$(α+2α3​​,0).

The distance $PQ$PQ is the difference in their x-coordinates:

$$PQ = \left(\alpha + \dfrac{\alpha\sqrt{3}}{2}\right) - \alpha = \dfrac{\alpha\sqrt{3}}{2}$$PQ=(α+2α3​​)−α=2α3​​

Given $PQ = \dfrac{9}{2}$PQ=29​, we have:

$$\dfrac{\alpha\sqrt{3}}{2} = \dfrac{9}{2} \Rightarrow \alpha\sqrt{3} = 9 \Rightarrow \alpha = 3\sqrt{3}$$2α3​​=29​⇒α3​=9⇒α=33​

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

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

We need to find the value of $\dfrac{\alpha^2}{R}$Rα2​:

$$\dfrac{\alpha^2}{R} = \dfrac{\alpha^2}{\dfrac{\alpha}{\sqrt{3}}} = \alpha\sqrt{3}$$Rα2​=3​α​α2​=α3​

Since we already found $\alpha\sqrt{3} = 9$α3​=9, we get:

$$\dfrac{\alpha^2}{R} = 9$$Rα2​=9

Answer: $9$9