XXXI Brazilian Math Olympiad

One can guess, by drawing a good diagram, that the common point is $O$ and that the lines $A_2{A}_3$, $B_1{B}_3$ and $C_1{C}_2$ are the perpendicular bisectors of the triangle $ABC$. So it is sufficient to prove that $A_2{A}_3$ is the perpendicular bisector of $BC$. For the sake of simplicity, let $\angle A=\angle BAC$, $\angle B=\angle ABC$ and $\angle C=\angle ACB$. Then, considering that the perpendicular bisector of $BC$ also bisects the angle $\angle BOC$, it suffices to prove that the angle between either the lines $O{A}_2$, $O{A}_3$ and either the lines $OB$, $OC$ is $\angle A$ or $180^{\circ }-\angle A$.

By the definition of $A_2$ and $A_3$, $O,A,A_2,C$ lies on the same circle, as well as $O,A,A_3,B$. Now, if $A$ and $A_2$ are opposite vertices in the quadrilateral $AO{A}_{2}C$ then $\angle A_{2}OC=\angle A_{2}AC$ and, since $A_2$ lies on $AB$, $\angle A_{2}AC=\angle A$ or $\angle A_{2}AC=180^{\circ }-\angle A$. Either way, we are done. If $A$ and $A_2$ are neighbors in the quadrilateral $A{A}_{2}OC$, then $\angle A_{2}OC=180^{\circ }-\angle A_{2}AC$ and we are done again. Finally, if $A$ and $A_2$ are neighbors in the quadrilateral $A{A}_{2}CO$, then $\angle A_{2}OC=\angle A_{2}AC$ and we are done. The result follows analogously to $O,A,A_3,B$, so all cases are covered.

---

Alternative solution.

Another solution can be found by applying an inversion with respect to the circumcircle of $ABC$. It can be shown that it maps $A_2$ to $A_3$, $B_1$ to $B_3$ and $C_1$ to $C_2$, proving directly that all lines $A_2{A}_3$, $B_1{B}_3$ and $C_1{C}_2$ pass through the center of inversion $O$.