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

he lines \[ x-\sqrt{3}|y|=\alpha \] represent two half-lines making angles \[ 30^\circ \text{ and } -30^\circ \] So angle between them is \[ 60^\circ \] Given \[ PA=PB=\alpha \] Hence $\triangle PAB$ is equilateral. Altitude of equilateral triangle: \[ PQ=\frac{\sqrt{3}}{2}\alpha \] Given \[ PQ=\frac{9}{2} \] \[ \frac{\sqrt{3}}{2}\alpha=\frac{9}{2} \] \[ \alpha=3\sqrt{3} \] Circumradius of equilateral triangle: \[ R=\frac{\alpha}{\sqrt{3}}=3 \] Now \[ \frac{\alpha^2}{R} = \frac{(3\sqrt{3})^2}{3} = \frac{27}{3} = 9 \] Final Answer: \[ \boxed{9} \]