Problem 1520

(b) In a scalene triangle $ABC$, $\angle BAC = 120^{\circ}$. The bisectors of angles $A, B, C$ meet the opposite sides in $P, Q, R$ respectively. Prove that the circle on $QR$ as diameter passes through the point $P$.

1. Let $I$ be the incenter of $\Delta ABC$. The angle bisectors of $\angle A, \angle B, \angle C$ meet the opposite sides at $P, Q, R$ respectively.
2. Since $\angle BAC = 120^{\circ}$, we have $\angle BIC = 180^{\circ} - \frac{\angle BAC}{2} = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
3. Consider the quadrilateral $BICR$. Since $I$ is the incenter, $\angle BIR = 90^{\circ} + \frac{\angle BAC}{2} = 90^{\circ} + 60^{\circ} = 150^{\circ}$.
4. Since $\angle BIC = 120^{\circ}$, we have $\angle BIR = 30^{\circ}$ (since $\angle BIR = 180^{\circ} - \angle BIC$).
5. Now, consider the quadrilateral $AIMR'$, where $M$ is the incenter of $\Delta APB$ and $R'$ is a point on $AB$ such that $PR'$ is the bisector of $\angle APB$.
6. Since $M$ is the incenter of $\Delta APB$, we have $\angle IMR' = 120^{\circ}$.
7. Therefore, $AIMR'$ is cyclic, and $\angle MIR' = \angle MAB = 30^{\circ}$.
8. Since $\angle BIR = 30^{\circ}$, we have $\angle MIR = \angle MIR' = 30^{\circ}$.
9. Thus, $R \equiv R'$, and $PR$ bisects $\angle APB$.
10. Similarly, $PQ$ bisects $\angle APC$.
11. Therefore, $\angle QPR = 90^{\circ}$.
12. Since $\angle QPR = 90^{\circ}$, the circle on $QR$ as diameter passes through the point $P$.

$\blacksquare$