45th Mongolian Mathematical Olympiad

Let $R$ be the line $l$ tangent to ${\omega }_1$, in the ${\omega }_1$ $R^{\prime }$ be the opposite point of $R$ for diameter. Hence, it's enough to show that $R^{\prime }$, $P$, $C$ are collinear. Now assume $R^{\prime }P\cap BC=C^{\prime }$ and $R^{\prime }P\cap AB=Q^{\prime }$. The triangles $O_{1}AR$ and $CAB$ are similar, hence $\frac{BC}{R{O}_1}=\frac{BA}{RA}$. The triangles

$R^{\prime }{Q}^{\prime }R$ and $C^{\prime }{Q}^{\prime }B$ are similar, hence $\frac{B{C}^{\prime }}{R{R}^{\prime }}=\frac{B{Q}^{\prime }}{R{Q}^{\prime }}$.



Calculating length of $BC$, $B{C}^{\prime }$ and observe that $R{R}^{\prime }=2\cdot R{O}_1$, $BC=B{C}^{\prime }$ if and only if $$BA\cdot R{Q}^{\prime }=2B{Q}^{\prime }\cdot RA.(1)$$ Let if $RP\cap {\omega }_2=T$, $R^{\prime }P\cap {\omega }_2=T^{\prime }$, then the points $T,T^{\prime }$ be the diameters end point of ${\omega }_2$. Furthermore, $T{T}^{\prime }\parallel R{R}^{\prime }\perp AB$, we get that $T{T}^{\prime }$ passes through the point $M$, which is midpoint of $AB$. Thus diameter with $Q^{\prime }T$ circles on the points $P,Q^{\prime },M,T$. For above circles we get $R{Q}^{\prime }\cdot RM=RP\cdot RT$ and for ${\omega }_2$ we have $RP\cdot RT=RA\cdot RB$. From here $R{Q}^{\prime }\cdot RM=RA\cdot RB$. In the last equation if we put $RM=RA+\frac{AB}{2}$ and $RB=R{Q}^{\prime }+Q^{\prime }B$ then $R{Q}^{\prime }\cdot \frac{AB}{2}=RA\cdot Q^{\prime }B$. Considering with (1) and then $C\equiv C^{\prime }$.