Berov s.l.

Two circles $\sigma_{1}$ and $\sigma_{2}$ intersect at points $A$ and $B$. Let $P Q$ and $R S$ be the segments of the common external tangents to these circles (points $P$ and $R$ lie on $\sigma_{1}$, points $Q$ and $S$ lie on $\sigma_{2}$). It turns out that $R B \| P Q$. Ray $R B$ intersects $\sigma_{2}$ again at point $W$. Find the ratio $R B / B W$.
Let $X$ be the point of intersection of lines $AB$ and $PQ$. Then $XP^2 = XA \cdot XB = XQ^2$, i.e., $X$ is the midpoint of $PQ$. Lines $AB$ and $PR$ are parallel, as both these lines are perpendicular to the line connecting the centers of circles $\sigma_1$ and $\sigma_2$. From the condition, we now get that quadrilateral $PXBR$ is a parallelogram, hence $BR = XP = \frac{1}{2} PQ = \frac{1}{2} RS$ (the last equality is due to the symmetry of $PQ$ and $RS$). Further, since $RS$ is a tangent segment to $\sigma_2$, we have $RB \cdot RW = RS^2 = (2RB)^2$, from which $RW = 4RB$. Therefore, $RB / BW = 1 / 3$.