Mongolian Mathematical Olympiad

Let $R$ be the line $l$ tangent to $\omega_1$, in the $\omega_1$ $R'$ be the opposite point of $R$ for diameter. Hence, it's enough to show that $R'$, $P$, $C$ are collinear. Now assume $R'P \cap BC = C'$ and $R'P \cap AB = Q'$. The triangles $O_1AR$ and $CAB$ are similar, hence $\frac{BC}{RO_1} = \frac{BA}{RA}$. The triangles

$R'Q'R$ and $C'Q'B$ are similar, hence $\frac{BC'}{RR'} = \frac{BQ'}{RQ'}$.

Calculating length of $BC$, $BC'$ and observe that $RR' = 2 \cdot RO_1$, $BC = BC'$ if and only if
$$BA \cdot RQ' = 2BQ' \cdot RA. \qquad (1)$$
Let if $RP \cap \omega_2 = T$, $R'P \cap \omega_2 = T'$, then the points $T, T'$ be the diameters end point of $\omega_2$. Furthermore, $TT' \parallel RR' \perp AB$, we get that $TT'$ passes through the point $M$, which is midpoint of $AB$. Thus diameter with $Q'T$ circles on the points $P, Q', M, T$.
For above circles we get $RQ' \cdot RM = RP \cdot RT$ and for $\omega_2$ we have $RP \cdot RT = RA \cdot RB$. From here $RQ' \cdot RM = RA \cdot RB$. In the last equation if we put $RM = RA + \frac{AB}{2}$ and $RB = RQ' + Q'B$ then $RQ' \cdot \frac{AB}{2} = RA \cdot Q'B$. Considering with (1) and then $C \equiv C'$.