Question

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 $PQ = \frac{9}{2}$ and R is the radius of the circumcircle of $\Delta PAB$, then $\frac{\alpha^2}{R}$ is equal to ________

Answer 1

Correct Answer: 9

Step 1: Understanding the Concept:
By graphing the half-lines, we can establish the exact geometric nature of the triangle PAB. Observing the slopes of the lines determines the angle between them, which in turn identifies special properties (like an equilateral triangle) allowing swift computation of the circumradius.

Step 2: Key Formula or Approach:
Line 1: $x - \sqrt{3}y = \alpha \implies y = \frac{1}{\sqrt{3}}(x - \alpha)$ (Slope $\tan(30^\circ)$).
Line 2: $x + \sqrt{3}y = \alpha \implies y = -\frac{1}{\sqrt{3}}(x - \alpha)$ (Slope $\tan(-30^\circ)$).
Radius of circumcircle of an equilateral triangle of side $s$: $R = \frac{s}{\sqrt{3}}$.

Step 3: Detailed Explanation:
The intersection P of the two half-lines occurs where $y = 0$.
$x - \sqrt{3}(0) = \alpha \implies P = (\alpha, 0)$.
The two lines extend outward from P. Their slopes are $1/\sqrt{3}$ and $-1/\sqrt{3}$, meaning they make angles of $+30^\circ$ and $-30^\circ$ with the positive x-axis.
The angle between the two half-lines is $30^\circ - (-30^\circ) = 60^\circ$.
Since A and B are at distance $\alpha$ from P, $PA = PB = \alpha$.
In $\Delta PAB$, the sides $PA$ and $PB$ are equal, and the included angle is $60^\circ$. This forces $\Delta PAB$ to be an equilateral triangle with side length $\alpha$.
The angle bisector of the two half-lines lies directly on the x-axis. Q is the intersection of the angle bisector and segment AB. Because $\Delta PAB$ is equilateral, the angle bisector is also the median and the altitude.
Thus, PQ is the altitude of the equilateral triangle PAB.
Length of altitude $PQ = PA \cos(30^\circ) = \alpha \frac{\sqrt{3}}{2}$.
We are given $PQ = \frac{9}{2}$:
$\alpha \frac{\sqrt{3}}{2} = \frac{9}{2} \implies \alpha\sqrt{3} = 9 \implies \alpha = \frac{9}{\sqrt{3}} = 3\sqrt{3}$.
Let R be the radius of the circumcircle of equilateral $\Delta PAB$.
$R = \frac{\text{side}}{\sqrt{3}} = \frac{\alpha}{\sqrt{3}} = \frac{3\sqrt{3}}{\sqrt{3}} = 3$.
We need the value of $\frac{\alpha^2}{R}$:
$\frac{\alpha^2}{R} = \frac{(3\sqrt{3})^2}{3} = \frac{27}{3} = 9$.

Step 4: Final Answer:
The value is 9.