GATE CS 2015 Set 2 QGA-8 — General Aptitude

We can find the length of the angle bisector $PS$ by using the area of the triangles.

Let $PQ = r$, $PR = q$, and the length of the angle bisector $PS = x$.

Since $PS$ is the angle bisector of $\angle QPR$ and $\angle QPS = 60^\circ$, we have: $$\angle SPR = 60^\circ$$ $$\angle QPR = \angle QPS + \angle SPR = 60^\circ + 60^\circ = 120^\circ$$

The total area of $\triangle PQR$ is the sum of the areas of $\triangle PQS$ and $\triangle PSR$: $$\text{Area}(\triangle PQR) = \text{Area}(\triangle PQS) + \text{Area}(\triangle PSR)$$ $$\frac{1}{2} \cdot PQ \cdot PR \cdot \sin(120^\circ) = \frac{1}{2} \cdot PQ \cdot PS \cdot \sin(60^\circ) + \frac{1}{2} \cdot PR \cdot PS \cdot \sin(60^\circ)$$ $$\frac{1}{2} q r \sin(120^\circ) = \frac{1}{2} r x \sin(60^\circ) + \frac{1}{2} q x \sin(60^\circ)$$

Since $\sin(120^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}$, we can divide both sides by $\frac{1}{2} \sin(60^\circ)$: $$q r = r x + q x$$ $$q r = x (q + r)$$ $$x = \frac{qr}{q + r}$$

Therefore, the length of $PS$ is $\dfrac{qr}{q+r}$.