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Romanian Mathematical Olympiad

Geometry Difficulty 6.0 National Olympiad Prove it Romania

Let $P$ and $Q$ be the midpoints of the diagonals $BD$ and $AC$ of the quadrilateral $ABCD$ . Consider the points $M \in (BC)$ , $N \in (CD)$ , $R \in (PQ)$ and $S \in (AC)$ such that $\frac{BM}{MC} = \frac{DN}{NC} = \frac{PR}{RQ} = \frac{AS}{SC} = k$ . Prove that the centroid of the triangle $AMN$ lies on the segment $[RS]$ .

Andrei Bud

Solution

Let $G$ be the centroid of the triangle $AMN$ . Then we have
$\overrightarrow{GR} = \frac{\overrightarrow{GP} + k\overrightarrow{GQ}}{1+k} = \frac{\overrightarrow{GB} + \overrightarrow{GD} + k(\overrightarrow{GA} + \overrightarrow{GC})}{2(1+k)}$
and
$\overrightarrow{GS} = \frac{\overrightarrow{GA} + k\overrightarrow{GC}}{1+k}.$
On the other hand,
$0 = \overrightarrow{GA} + \overrightarrow{GM} + \overrightarrow{GN} = \overrightarrow{GA} + \frac{\overrightarrow{GB} + k\overrightarrow{GC}}{1+k} + \frac{\overrightarrow{GD} + k\overrightarrow{GC}}{1+k},$
from where, we deduce that $(1+k)\overrightarrow{GA} + \overrightarrow{GB} + 2k\overrightarrow{GC} + \overrightarrow{GD} = 0$ , whence we get $\overrightarrow{GS} + 2\overrightarrow{GR} = 0$ , and thus $G, R$ and $S$ are collinear points.

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