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Romanian Mathematical Olympiad - Final Round

Geometry Difficulty 5.3 AIME, harder Prove it Romania

The diagonals of the parallelogram $ABCD$ intersect at $O$ and $M$ is the midpoint of the side $AB$ . Let $P$ be a point of the segment $OC$ and $Q$ be the intersection of the lines $MP$ and $BC$ . The parallel to $MP$ through $O$ intersects the line $CD$ at the point $N$ . Prove that the points $A, N$ are $Q$ collinear if and only if $P$ is the midpoint of the segment $OC$ .

Solution

If $A, N, Q$ are collinear, then the fundamental theorem for similarity yields $\frac{CR}{MB} = \frac{QC}{QB} = \frac{CN}{AB}$ and from $AB = 2 \cdot MB$ follows $CN = 2 \cdot CR$ . Now $RP \parallel ON$ shows that $RP$ is a midline in $\triangle CON$ , therefore $P$ is the midpoint of $OC$ .

For the converse, if $P$ is the midpoint of the segment $OC$ , then $RP$ is a midline of $\triangle CON$ , therefore $CN = 2 \cdot CR$ . Let the lines $AQ$ and $CD$ meet at $N'$ . Then, as above, $CN' = 2 \cdot CR = CN$ , so $N$ coincides with $N'$ , that is, $A, N, Q$ are collinear.

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