66th Belarusian Mathematical Olympiad

There is nothing to prove if some two of four points $A_1$, $B_1$, $C_1$, $G$ coincide. So, we can assume that all these points are pairwise distinct.

Let $\Gamma$ and $\Gamma (A_1)$ denote the circumcircles of the triangles $ABC$ and $A{A}_{1}B$, respectively. Let $M_A$ be the intersection point of the lines $A{A}_1$ and $BC$. Since $\angle A_{1}AB=\angle A_{1}BC$, it follows that $\Gamma (A_1)$ touches $BC$ at $B$. By the Power of a Point Theorem, we have $M_A{B}^2=M_A{A}_1\cdot M_{A}A$, Similarly, $M_A{C}^2=M_A{A}_1\cdot M_{A}A$, hence $M_A{B}^2=M_A{C}^2$, so $M_{A}B=M_{A}C$, i.e., $M_A$ is the midpoint of the side $BC$. Thus, $A_1$ lies on the median of the triangle $ABC$ from the vertex $A$. By the same arguments, $B_1$ and $C_1$ lie on the medians of the triangle $ABC$ from the vertices $B$ and $C$, respectively.



Let $H$ be the orthocenter of the triangle $ABC$. It is well known that the points which are symmetric to $H$ with respect to the sides of the triangle $ABC$ lie on the circumcircle of $ABC$. If $H_A$ is symmetric to $H$ with respect to the side $BC$, then $H_A$ belongs to $\Gamma$ and we have $$\begin{array}{rl}\angle BHC& =\angle B{H}_{A}C=180^{\circ }-\angle BAC=180^{\circ }-(\angle A_{1}AB+\angle A_{1}AC)=\\ & =180^{\circ }-(\angle A_{1}BC+\angle A_{1}CB)=\angle B{A}_{1}C,\end{array}$$ hence the quadrilateral $B{A}_{1}HC$ is cyclic and $A_1$ belongs to the circle passing through the points $B$, $H$, $C$.

Further, $$\begin{array}{rl}\angle H{A}_1{M}_A& =\angle H{A}_{1}B+\angle B{A}_1{M}_A=\angle BCH+(\angle A_{1}BA+\angle A_{1}AB)=\\ & =(90^{\circ }-\angle ABC)+(\angle A_{1}BA+\angle A_{1}BC)=90^{\circ }-\angle ABC+\angle ABC=90^{\circ }.\end{array}$$ Therefore, $A_1$ is the projection of $H$ on the median $A{M}_A$. Similarly, $B_1$ and $C_1$ are the projections of $H$ on the median of the triangle $ABC$ from the vertices $B$ and $C$, respectively.

Since $G$ is the intersection point of the medians of $ABC$, we have $\angle H{A}_{1}G=\angle H{B}_{1}G=\angle H{C}_{1}G=90^{\circ }$. Thus $A_1,B_1,C_1$ lie on the circle with the diameter $GH$, so all points $A_1,B_1,C_1,G$ lie on the same circle.