Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.9 National olympiad Prove it

Let $ABCD$ be a quadrilateral with $AC = BD$ . Diagonals $AC$ and $BD$ meet at $P$ . Let $\omega_{1}$ and $O_{1}$ denote the circumcircle and circumcenter of triangle $ABP$ . Let $\omega_{2}$ and $O_{2}$ denote the circumcircle and circumcenter of triangle $CDP$ . Segment $BC$ meets $\omega_{1}$ and $\omega_{2}$ again at $S$ and $T$ (other than $B$ and $C$ ), respectively. Let $M$ and $N$ be the midpoints of minor arcs $\widehat{SP}$ (not including $B$ ) and $\overparen{TP}$ (not including $C$ ). Prove that $\overline{MN} \| \overline{O_{1}O_{2}}$ .

Solution

Let $Q$ be the second intersection point of $\omega_{1}, \omega_{2}$ . Suffice to show $\overline{Q P} \perp \overline{M N}$ . Now $Q$ is the center of a spiral congruence which sends $\overline{A C} \mapsto \overline{B D}$ . So $\triangle Q A B$ and $\triangle Q C D$ are similar isosceles. Now,
$\measuredangle Q P A=\measuredangle Q B A=\measuredangle D C Q=\measuredangle D P Q$
and so $\overline{Q P}$ bisects $\angle B P C$ . !
Now, let $I=\overline{B M} \cap \overline{C N} \cap \overline{P Q}$ be the incenter of $\triangle P B C$ . Then $I M \cdot I B=I P \cdot I Q=$ $I N \cdot I C$ , so $B M N C$ is cyclic, meaning $\overline{M N}$ is antiparallel to $\overline{B C}$ through $\angle B I C$ . Since $\overline{Q P I}$ passes through the circumcenter of $\triangle B I C$ , it follows now $\overline{Q P I} \perp \overline{M N}$ as desired.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.