Round 3

Geometry Difficulty 6.3 National Olympiad Prove it Mongolia

Let $ABCD$ be a trapezoid with obtuse angle at vertex $D$ . The diagonals $AC$ and $BD$ meet at $O$ and the line through $O$ parallel to $AB$ meets the circumcircle of $BCO$ again at $P$ . The circumcircles of $ADO$ and $BCO$ intersect again at $Q$ . Prove that the line $PQ$ bisects the segment $BC$ .

Solution

Since $OP \parallel AB$ and $P$ , $C$ , $O$ , $Q$ , $B$ are cyclic, we have $\angle POC = \angle PQC = \angle PBC = \alpha$ and $\angle POB = \angle PCB = \angle PQB$ . It follows that $\triangle OAB \sim \triangle BPC$ . Hence
$\frac{BP}{PC} = \frac{OA}{OB}.$
Because $ABCD$ is a trapezoid, we have
$\frac{OC}{OA} = \frac{OD}{OB} \iff 1 + \frac{OC}{OA} = \frac{OC}{OB} + 1 \iff \frac{AC}{BD} = \frac{OA}{OB}$
From this, we deduce that
$\frac{BP}{PC} = \frac{AC}{BD}.$
On the other hand, we have $\frac{QC}{QB} = \frac{AC}{DB}$ since $\triangle AQC \sim \triangle DQB$ . Consequently,
$\frac{QC}{QB} = \frac{BP}{PC}. \text{ Now by the law of sines}$
$\frac{CM}{MB} = \frac{CM}{MP} \cdot \frac{MP}{MB} = \frac{\sin \angle CPM}{\sin \angle MCP} \cdot \frac{\sin \angle CBP}{\sin \angle MPB} = \frac{CQ}{PB} \cdot \frac{CP}{BQ} = 1.$

Thus $CM = MB$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.