Team Selection Test for IMO

Let us denote the intersections of $a_B$, $b_C$ and $c_A$ with $b_A$, $c_B$ and $a_C$ respectively by $A^{\prime }$, $B^{\prime }$ and $C^{\prime }$. The triangle $A{A}_1{C}_2$ is similar to $ABC$, since they have a common angle and their sides are at a ratio $1:3$. Let $O_A$ be the center of the circumscribed circle around the triangle $A{A}_1{C}_2$. Then: $$\angle O_A{A}_{1}A=\frac{1}{2}(180^{\circ }-\angle A{O}_A{A}_1)=90^{\circ }-\angle A{C}_2{A}_1=90^{\circ }-\gamma$$ Since $a_B$ is perpendicular to $O_A{A}_1$ it follows that the angle between $a_B$ and $AB$ equals $$180^{\circ }-90^{\circ }-(90^{\circ }-\gamma )=\gamma$$ Analogously, the angle between $b_A$ and $AB$ equals $\gamma$, so that the triangle $A_1{A}_2{A}^{\prime }$ is isosceles with base $A_1{A}_2$, that is, the perpendicular to $AB$ through $C^{\prime }$ passes through the midpoint of $A_1{A}_2$, which is also the midpoint of $AB$. Thus, the perpendicular passes through the center of the circumscribed circle around the triangle $ABC$. From symmetry reasons, all the three perpendiculars pass through the center of the circumscribed circle, that is, they pass through a common point.