Question

P is the point of intersection of the two half-lines \[ x-\sqrt{3}y=\alpha, \quad \alpha>0 \] Points \(A\) and \(B\) lie on these lines at a distance \(\alpha\) from \(P\). If the length of perpendicular from \(P\) on \(AB\) is \(\frac{\alpha}{2}\) and the radius of the circumcircle of \(\triangle PAB\) is \(R\), then find \[ \frac{\alpha^2}{R} \]

Hint:

In an equilateral triangle: \[ R=\frac{a}{\sqrt3}, \quad \text{Altitude}=\frac{\sqrt3}{2}a \]

Answer 1

Correct Answer: 9

Concept: The lines \[ x-\sqrt{3}y=\alpha \] represent two symmetric half-lines forming an angle \(60^\circ\) between them.
Step 1: Point of intersection For the lines \[ x-\sqrt3 y=\alpha \] intersection occurs at \[ P(\alpha,0) \]
Step 2: Points \(A\) and \(B\) Since \(A\) and \(B\) lie on the two rays and are at distance \(\alpha\) from \(P\), \[ PA=PB=\alpha \] Thus triangle \(PAB\) is symmetric with vertex angle \[ \angle APB=60^\circ \]
Step 3: Altitude The perpendicular from \(P\) to \(AB\) is given as \[ \frac{\alpha}{2} \] In an equilateral triangle of side \(\alpha\), altitude \[ =\frac{\sqrt3}{2}\alpha \] From the figure geometry, \[ PQ=\frac{\sqrt3\alpha}{2}=\frac{9}{2} \] \[ \alpha=\frac{9}{\sqrt3}=3\sqrt3 \]
Step 4: Circumradius For an equilateral triangle: \[ R=\frac{a}{\sqrt3} \] Thus \[ R=\frac{3\sqrt3}{\sqrt3}=3 \]
Step 5: Required value \[ \frac{\alpha^2}{R} = \frac{(3\sqrt3)^2}{3} \] \[ =\frac{27}{3} \] \[ =9 \]