Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.0 National Olympiad, round 2 Prove it Romania

The medians $AD$ , $BE$ and $CF$ of triangle $ABC$ intersect at $G$ . Let $P$ be a point lying in the interior of the triangle, not belonging to any of its medians. The line through $P$ parallel to $AD$ intersects the side $BC$ at $A_1$ . Similarly one defines the points $B_1$ and $C_1$ . Prove that
$\overrightarrow{A_1D} + \overrightarrow{B_1E} + \overrightarrow{C_1F} = \frac{3}{2} \overrightarrow{PG}.$

Solution

Let us draw through point $P$ parallels to the triangle's sides. If the parallels to $AB$ and $AC$ intersect the side $BC$ at points $A_B$ and $A_C$ , then the triangles $ABC$ and $PA_B A_C$ are similar, and since $PA_1$ is parallel to the median $AD$ , it follows that $A_1$ is the midpoint of the line segment $A_B A_C$ . With the notations in the figure, we have
$2\overrightarrow{A_1D} = \overrightarrow{A_1B} + \overrightarrow{A_1C} = \overrightarrow{A_B}B + \overrightarrow{A_C}C = \overrightarrow{PC_B} + \overrightarrow{PB_C}.$
and the similar equalities. We deduce that the sum $2(\overrightarrow{A_1D} + \overrightarrow{B_1E} + \overrightarrow{C_1F})$ equals
$\vec{v} = \overrightarrow{PC_B} + \overrightarrow{PB_C} + \overrightarrow{PA_C} + \overrightarrow{PC_A} + \overrightarrow{PB_A} + \overrightarrow{PA_B}.$
But $\overrightarrow{PC_A} + \overrightarrow{PB_A} = \overrightarrow{PA}$ and adding up all similar equalities yields
$\vec{v} = \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} = 3\overrightarrow{PG},$
hence the conclusion.

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