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Iranian Mathematical Olympiad

Geometry Difficulty 5.7 AIME, harder Prove it Iran

Cyclic quadrilateral $ABCD$ with circumcenter $O$ is given. Point $P$ is the intersection of diagonals $AC$ and $BD$ . Let $M$ and $N$ be midpoints of the sides $AD$ and $BC$ , respectively. Suppose that $\omega_1$ , $\omega_2$ and $\omega_3$ are circumcircles of triangles $ADP$ , $BCP$ and $OMN$ , respectively. Let $E$ and $F$ be intersection points of $\omega_1$ and $\omega_3$ , which is not on the arc $APD$ of $\omega_1$ , and the intersection point of $\omega_2$ and $\omega_3$ , which is not on the arc $BPC$ of $\omega_2$ , respectively. Prove that $OE = OF$ .

Solution

Let $E'$ be the second intersection of $NP$ and $\omega_1$ . Since $\triangle BPC \sim \triangle APD$ we have $\angle MPD = \angle NPC = \angle APE'$ . So $APDE'$ is a harmonic quadrilateral. Then
$\begin{aligned} \angle E'MA &= \angle PMA = \angle PNB \implies \angle E'MO = \angle E'MA + 90^\circ \\ &= \angle PNB + 90^\circ = 180^\circ - \angle PNO, \end{aligned}$
hence $E'NOM$ is cyclic and $E' \equiv E$ . Similarly, we can show that $M, P$ and $F$ are collinear. Finally
$\angle ENO = 90^\circ - \angle PNB = 90^\circ - \angle PMA = \angle FMO \implies OE = OF.$

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