26th Hellenic Mathematical Olympiad

Let $H$ be the orthocenter of the triangle $AB\Gamma$. Then $\overrightarrow{AH}=2\cdot \overrightarrow{O{A}_1}$ and from $\overrightarrow{O{A}_2}=\lambda \cdot \overrightarrow{O{A}_1}$, we find: $\overrightarrow{AH}=\frac{2}{\lambda }\cdot \overrightarrow{O{A}_2}$. If $A{A}_2$ meets $OH$ at $C$ (from the similarity of the triangles $CHA$ and $CO{A}_2$), we have: $\overrightarrow{HC}=\frac{2}{\lambda }\cdot \overrightarrow{CO}$. It means that $A{A}_2$ passes through $C$ which divides $OH$ in ratio $\frac{2}{\lambda }$.

Similarly, we have $\overrightarrow{BH}=2\cdot \overrightarrow{O{B}_1}$ and $\overrightarrow{BH}=\frac{2}{\lambda }\cdot \overrightarrow{O{B}_2}$. Let now $C^{\prime }$ be the point of intersection of the lines $B{B}_2$ and $OH$. Then we have $\overrightarrow{H{C}^{\prime }}=\frac{2}{\lambda }\cdot \overrightarrow{C^{\prime }O}$, which means that $B{B}_2$ passes through $C^{\prime }$ which divides $OH$ in ratio $\frac{2}{\lambda }$.

Similarly, if $C^{\prime \prime }$ is the point of intersection of the lines $\Gamma {\Gamma }_2$ and $OH$, then we have that $\Gamma {\Gamma }_2$ passes through $C^{\prime \prime }$ which divides $OH$ in ratio $\frac{2}{\lambda }$.



Since the points $C$, $C^{\prime }$, $C^{\prime \prime }$ coincide, the lines $A{A}_2,B{B}_2,\Gamma {\Gamma }_2$ are concurrent.