[Inscribed and Circumscribed Circles] The Law of Sines

Through the center of the circle circumscribed around triangle $A B C$, lines perpendicular to sides $A C$ and $B C$ are drawn. These lines intersect the altitude $C H$ of the triangle or its extension at points $P$ and $Q$. It is known that $C P=p, C Q=q$. Find the radius of the circle circumscribed around triangle $A B C$.
## First method.

Let $R$ be the radius of the circumcircle of triangle $ABC$, $\angle BAC = \alpha$, $\angle ABC = \beta$. From the isosceles triangles $APC$ and $BQC$, we find that

$$
AC = 2p \sin \alpha, \quad BC = 2q \sin \beta
$$

Since

$$
AC = 2R \sin \beta, \quad BC = 2R \sin \alpha
$$

Multiplying these equations term by term, we get that $R^2 = pq$.

## Second method.

Assume that $ABC$ is an acute triangle. Then $\angle COP = \angle ABC$ and $\angle OQP = \angle ABC$, so $\angle COP = \angle OQP$.

Triangles $OQC$ and $POC$ are similar by two angles, hence $\frac{OC}{CP} = \frac{CQ}{OC}$, from which $R^2 = OC^2 = CP \cdot CQ = pq$.

Similarly for an obtuse triangle.

## Answer

$\sqrt{pq}$.