Belarus2022

Let's first consider the case when the radii of ${\omega }_2$ and ${\omega }_3$ are different (without loss of generality assume that the radius of ${\omega }_3$ is greater than the radius of ${\omega }_2$). Consider three homotheties: $l_1$ centered at $Q$ and mapping ${\omega }_3$ to ${\omega }_1$; $l_2$ centered $P$ and mapping ${\omega }_1$ to ${\omega }_2$; and $l_3=l_2\circ l_1$ mapping ${\omega }_3$ to ${\omega }_2$. Since $l_1$ maps $A$ to $C$, $l_2$ maps $C$ to $B$, then $l_3$ maps $A$ to $B$, so its center (denoted by $I$) lies on the line $AB$. Denote $\angle ACB=\angle PCQ=\alpha$, this angle is fixed (by a fixed value we denote any value which doesn't depend on the position of $C$). Then the angle $\angle AOB=2\alpha$ is also fixed, which means that all triangles $AOB$ are similar to each other and the ratio $\frac{OA}{AB}=\frac{1}{2\sin \alpha }$ is fixed. Let $\frac{IA}{IB}=t>1$ be the homothety coefficient of $l_3$, this number is fixed. Then the ratio $$\frac{AB}{IA}=\frac{IB-IA}{IA}=\frac{1}{t}-1\text{ is also fixed.}$$ In the triangle $IAO$ we can find the ratio $$\frac{OA}{IA}=\frac{OA}{AB}\cdot \frac{AB}{IA}=\frac{1}{2\sin \alpha }\cdot \left(\frac{1}{t}-1\right)$$ and the angle $$\angle OAI=180^{\circ }-\angle OAB=90^{\circ }+\alpha .$$ Hence all triangles $IAO$ are similar to each other, in particular, the angle $AIO$ and the ratio $\frac{IO}{IA}$ are fixed. Therefore for any position of the point $C$ on the circle ${\omega }_1$, the spiral similarity $\ell$ with center $I$, angle $AIO$ and coefficient $\frac{IO}{IA}$ maps $A$ to $O$. Since the point $A$ varies over all positions on ${\omega }_3$, the locus of $O$ is the circle $\omega =\ell ({\omega }_3)$. Note that the homothety ${\ell }_3$ and the spiral similarity $\ell$ have a common center $I$. Therefore $\ell (B)=\ell \circ {\ell }_3(A)={\ell }_3\circ \ell (A)={\ell }_3(O_3)=O_2$. Hence $\ell$ maps the triangle $AOB$ to the triangle $O_3{O}^{\prime }{O}_2$ and $$\angle O_3{O}^{\prime }{O}_2=\angle AOB=2\angle ACB=\angle Q{O}_{1}P=\angle O_3{O}_1{O}_2,$$ from which we conclude that $O^{\prime }$ lies on the circumcircle of the triangle $O_1{O}_2{O}_3$.

Let now the radii of the circles ${\omega }_2$ and ${\omega }_3$ be equal. In this case ${\ell }_3$ is a translation by the vector $\overrightarrow{O_3{O}_2}$, the point $I$ is not defined and the quadrilateral $AB{O}_2{O}_3$ is a parallelogram. Let $D$ be a such point that triangles $ABD$ and $O_3{O}_2{O}_1$ are equal and have the same orientation. The quadrilateral $O_{3}AD{O}_1$ is a parallelogram, while the triangles $QA{O}_3$ and $QC{O}_1$ are similar. Hence the segments $D{O}_1$ and $O_{1}C$ are parallel and $$D{O}_1+O_{1}C=O_{3}A\left(1+\frac{Q{O}_1}{Q{O}_3}\right)=O_{3}Q\left(1+\frac{Q{O}_1}{Q{O}_3}\right)=O_1{O}_3=DA=DB.$$