South-Afrika, 2011

Geometry Difficulty 5.9 AIME, harder Prove it South Africa

Let $O$ be the intersection point of the diagonals of the convex quadrilateral $ABCD$ , with $AO = OC$ . Points $P$ and $Q$ are marked on the segments $AO$ and $CO$ , respectively, such that $PO = OQ$ . Let $N$ be the intersection of $AB$ and $DP$ , and $K$ be the intersection of $CD$ and $BQ$ .
Prove that the points $N$ , $O$ and $K$ are collinear.

Solution

Draw $NM \parallel KL \parallel AC$ as in the figure. From the similarity of the triangles $BOQ$ and $BLK$ it follows that
$\frac{LK}{OQ} = \frac{BL}{BO} = 1 + \frac{LO}{BO},$
From the similarity of the triangles $DOC$ and $DLK$ it follows that
$\frac{LK}{OC} = \frac{DL}{DO} = 1 - \frac{LO}{DO}.$
Therefore
$LK \left( \frac{1}{OQ} - \frac{1}{OC} \right) = LO \left( \frac{1}{BO} + \frac{1}{DO} \right).$
Hence
$\frac{LK}{LO} = \frac{DO + BO}{DO \cdot BO} \cdot \frac{OC \cdot OQ}{OC - OQ} = \frac{BD}{DO \cdot BO} \cdot \frac{OC \cdot OQ}{QC}.$
Similarly we obtain
$\frac{NM}{OM} = \frac{BD}{DO \cdot BO} \cdot \frac{OC \cdot OQ}{QC}.$
Therefore, since $\angle NMO = \angle KLO$ and $\frac{LK}{LO} = \frac{NM}{OM}$ , triangles $ONM$ and $OKL$ are similar. Hence $\angle NOM = \angle KOL$ . Since $MOL$ is a straight line, it follows that so is $NOK$ .

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