Competition problem — Geometry — Olympiad Maths Prep

Geometry Difficulty 6.7 National olympiad Prove it Greece

A triangle $AB\Gamma$ is given and let $O$ its circumcenter and $A_1, B_1, \Gamma_1$ the middles of its sides $B\Gamma, A\Gamma$ and $AB$ , respectively. We consider the points $A_2, B_2, \Gamma_2$ such that $\overrightarrow{OA_2} = \lambda \cdot \overrightarrow{OA_1}$ , $\overrightarrow{OB_2} = \lambda \cdot \overrightarrow{OB_1}$ and $\overrightarrow{\Gamma_2} = \lambda \cdot \overrightarrow{\Gamma_1}$ , with $\lambda > 0$ . Prove that the lines $AA_2, BB_2, \Gamma\Gamma_2$ are concurrent.

Solution

Let $H$ be the orthocenter of the triangle $AB\Gamma$ . Then $\overrightarrow{AH} = 2 \cdot \overrightarrow{OA_1}$ and from $\overrightarrow{OA_2} = \lambda \cdot \overrightarrow{OA_1}$ , we find: $\overrightarrow{AH} = \frac{2}{\lambda} \cdot \overrightarrow{OA_2}$ .
If $AA_2$ meets $OH$ at $C$ (from the similarity of the triangles $CHA$ and $COA_2$ ), we have: $\overrightarrow{HC} = \frac{2}{\lambda} \cdot \overrightarrow{CO}$ . It means that $AA_2$ passes through $C$ which divides $OH$ in ratio $\frac{2}{\lambda}$ .

Similarly, we have $\overrightarrow{BH} = 2 \cdot \overrightarrow{OB_1}$ and $\overrightarrow{BH} = \frac{2}{\lambda} \cdot \overrightarrow{OB_2}$ .
Let now $C'$ be the point of intersection of the lines $BB_2$ and $OH$ . Then we have $\overrightarrow{HC'} = \frac{2}{\lambda} \cdot \overrightarrow{C'O}$ , which means that $BB_2$ passes through $C'$ which divides $OH$ in ratio $\frac{2}{\lambda}$ .

Similarly, if $C''$ is the point of intersection of the lines $\Gamma\Gamma_2$ and $OH$ , then we have that $\Gamma\Gamma_2$ passes through $C''$ which divides $OH$ in ratio $\frac{2}{\lambda}$ .

Since the points $C$ , $C'$ , $C''$ coincide, the lines $AA_2, BB_2, \Gamma\Gamma_2$ are concurrent.

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