4. In triangle $ABC$, the angle bisector of $\measuredangle CAB$ intersects side $BC$ at point $N$, and the angle bisector of $\measuredangle CBA$ intersects side $AC$ at point $P$, such that $PN=1$. The angle bisectors $AN$ and $BP$ intersect at point $Q$. Vertex $C$ lies on the circle passing through points $P$, $Q$, and $N$. Determine the area of triangle $NPQ$.
Solution. We have

$$
\measuredangle P Q N=\measuredangle A Q B=180^{\circ}-\frac{\alpha+\beta}{2}=90^{\circ}+\frac{\gamma}{2}
$$

Furthermore, from the cyclic quadrilateral $P Q N C$ it follows that $\measuredangle P Q N=180^{\circ}-\gamma$, so from these two relations we get $\gamma=60^{\circ}$. Now,

$$
\measuredangle Q N P=\measuredangle Q C P=\frac{\gamma}{2}=30^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_3c43981f47282d5136d0g-3.jpg?height=459&width=457&top_left_y=1727&top_left_x=990)
and similarly $\measuredangle N P Q=30^{\circ}$, so triangle $N P Q$ is isosceles, with base angles of $30^{\circ}$. By the perpendicular bisector $Q C$ of the base $N P$, it can be divided into two congruent triangles that can form an equilateral triangle with height $\frac{1}{2} P N=\frac{1}{2}$, which means the side length is $\frac{\sqrt{3}}{3}$. Finally, its area is equal to $\frac{\sqrt{3}}{12}$.