Thai Mathematical Olympiad

Solution. Here we use directed angles measured in the counterclockwise direction. Since $A{C}_1$ is the common chord of the circles $O$ and $O_2$, we see that $O{O}_2\perp A{C}_1$. Therefore, $O{O}_2\perp A{O}_1$ is equivalent to that $O_1$ lies on $A{C}_1$. Assume that $O_1$ lies on $A{C}_1$. (Claim that $O_2$ lies on $AC$.) Since $O$ and $O_1$ are circumcenters of $ABC$ and $A{B}_{1}C$, it is easy to see that $$\angle O_1{C}_1{O}_1=\angle O_{1}CA=\angle C_{1}AO=\angle C_{1}AC+\angle CAO=\angle A{C}_1{O}_1+\angle OCA=\angle OC{O}_1.$$ Therefore, $O,O_1,C_1,C$ are cyclic. Consider $△A{B}_2{C}_1$ with circumcenter $O_2$. Since $OA=O{C}_1$ and $O_{2}A=O_2{C}_1$, $△AO{O}_2\cong △C_{1}O{O}_2$ and $\angle C_1{O}_{2}O=\frac{1}{2}\angle C_1{O}_{2}A=\angle C_1{B}_{2}A$. $$\begin{array}{rl}\angle B_{2}A{C}_1& =\frac{1}{2}(\pi -\angle A_1{B}_1)=\frac{\pi }{2}-\angle AC{B}_1=\frac{\pi }{2}-\angle ACB-\angle BC{B}_1\\ & =\frac{\pi }{2}-\angle ACB-\angle BAC=\angle CBA-\frac{\pi }{2}.\end{array}$$

$$\begin{array}{rl}\angle A{C}_1{B}_2& =\angle B{C}_1{B}_2+\angle A{C}_{1}B=\angle B_{2}A{C}_1+\angle ACB=\angle CBA+\angle ACB-\frac{\pi }{2}\\ & =\frac{\pi }{2}-\angle BAC.\end{array}$$ \therefore \angle C_1B_2A = \pi - \angle B_2AC_1 - \angle AC_1B_2 = \pi + \angle BAC - \angle CBA. $$\begin{array}{rl}\angle C_1{O}_{1}O& =\angle C{O}_{1}O+\angle C_1{O}_{1}C=\angle C{C}_{1}O+\angle C_{1}OC=\frac{1}{2}(\pi -\angle C_{1}OC)+\angle C_{1}OC\\ & =\frac{\pi }{2}+\angle C_{1}AC=\frac{\pi }{2}+\angle BAC-\angle B_{2}A{C}_1\\ & =\pi +\angle BAC-\angle CBA.\end{array}$$ $$Thus,$\angle C_1O_2O = C_1O_1O$,andthen,$O, O_2, O_1, C_1, C$lieonthesamecircle.Let$O'_2$betheintersectionpointofACandtheperpendicularbisectorofAC.Then$\angle O'_2OO_1 = \angle O_1AO'_2 = \angle O'_2CO_1$.Therefore,$O'_2$alsoliesonthecirclepassingthrough$O, O_2, O_1, C_1, C$.Since$O, O_2, O'_2$lieontheperpendicularbisectorofAC,where$O_2$and$O'_2$aredifferentfromO(since$\angle CBA > 90^\circ$,Oisoutside$\triangle ABC$),thisimpliesthat$O_2 = O'_2$.(Becausealinecanintersectacircleinatmost2points.)Therefore,$O_2$ lies on AC.