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China Girls' Mathematical Olympiad

Geometry Difficulty 5.4 AIME, harder Prove it China

Let $ABCD$ be a convex quadrilateral. Let $O$ be the intersection of $AC$ and $BD$ . Let $O$ and $M$ be the intersections of the circumcircle of $\triangle OAD$ with the circumcircle of $\triangle OBC$ . Let $T$ and $S$ be the intersections of $OM$ with the circumcircle of $\triangle OAB$ and $\triangle OCD$ respectively. Prove that $M$ is the midpoint of $TS$ .

Solution

Since $\angle BTO = \angle BAO$ and $\angle BCO = \angle BMO$ , $\triangle BTM$ and $\triangle BAC$ are similar. Hence,

$\frac{TM}{AC} = \frac{BM}{BC} \qquad \textcircled{1}$

Similarly,
$\triangle CMS \sim \triangle CBD.$
Hence,
$\frac{MS}{BD} = \frac{CM}{BC} \qquad \textcircled{2}$

Dividing ① by ②, we have
$\frac{TM}{MS} = \frac{BM}{CM} \cdot \frac{AC}{BD} \qquad \textcircled{3}$

Since $\angle MBD = \angle MCA$ and $\angle MDB = \angle MAC$ , $\triangle MBD$ and $\triangle MCA$ are similar. Hence,
$\frac{BM}{CM} = \frac{BD}{AC} \qquad \textcircled{4}$

Combining ④ and ③ yields $TM = MS$ , as desired.

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