$\left.\begin{array}{ll}\text { Area of a quadrilateral }\end{array}\right]$ Circumscribed, inscribed, and exscribed circles; their radii] $]$ Difficulty: 4 [ Angle between a tangent and a chord $\quad]$

In an acute triangle $P Q R(P Q>Q R)$, the altitudes $P T$ and $R S$ are drawn; $Q N$ is the diameter of the circumcircle of triangle $P Q R$. It is known that the acute angle between the altitudes $P T$ and $R S$ is $\alpha, P R=a$. Find the area of quadrilateral NSQT.
Triangle $P Q R$ is acute-angled, so its angle at vertex $Q$ is equal to the acute angle between the altitudes $P T$ and $R S$, i.e., $\alpha$.

Draw a tangent to the circumcircle of triangle $P Q R$ through vertex $Q$ and mark a point $A$ on it, lying on opposite sides of line $Q R$ from point $P$. According to the theorem on the angle between a tangent and a chord, $\angle A Q R=\angle R P Q=\angle Q T S$, hence $S T \| A Q$, and since $Q N \perp A Q$, then $S T \perp Q N$. Thus, the diagonals of quadrilateral $N S Q T$ are perpendicular. If $S$ is its area, then $S=\frac{1}{2} S T \cdot Q N$.

Triangle $T Q S$ is similar to triangle $P Q R$, with the similarity ratio equal to $\frac{Q T}{Q P}=\cos \alpha$, so $S T=P R \cos \alpha=a \cos \alpha$.

Let $r$ be the radius of the circumcircle of triangle $P Q R$. Then $r=\frac{P R}{2 \sin \alpha a}=\frac{\square}{2 \sin \alpha}$. Therefore,

## Answer

$\frac{1}{2} a 2 \operatorname{ctg} \alpha$.