Iran TST 2012

[Phlembac Adib Hasan]

Points $A$ and $B$ are on a circle $\omega$ with center $O$ such that $\dfrac{\pi}{3}< \angle AOB <\dfrac{2\pi}{3}$. Let $C$ be the circumcenter of the triangle $AOB$. Let $\ell$ be a line passing through $C$ such that the angle between $\ell$ and the segment $OC$ is $\dfrac{\pi}{3}$. $\ell$ cuts tangents in $A$ and $B$ to $\omega$ in $M$ and $N$ respectively. Suppose circumcircles of triangles $CAM$ and $CBN$, cut $\omega$ again in $Q$ and $R$ respectively and themselves in $P$ (other than $C$). Prove that $OP\perp QR$.

Proposed by Mehdi E'tesami Fard, Ali Khezeli

Comment: ভালো প্রবলেম

[Tahmid Hasan]

WLOG $\angle MCO=60^{\circ}$.
By symmetry we conclude $MA,NB,CO$ are concurrent at a point $D$.
Let $P'$ be a point such that $P',M$ are on opposite sides of $CO$ and $\triangle P'CO$ equilateral.
It immediately follows that $P' \in \odot OADB$.
$\angle OAP'=\angle ODP'=30^{\circ}$.
So $\angle MAP'=120^{\circ}=\angle MCP' \Rightarrow P' \in CAM$.
Again $\angle P'CN+\angle P'BN=180^{\circ}-\angle OCP'-\angle DCN+\angle P'BN=180^{\circ} \Rightarrow P' \in \odot CBN$.
Hence $P=P'$.
Let $O_1,O_2$ be the circumcentres of $\triangle CAM,\triangle CBN$ respectively.
We have $CO=PO \Rightarrow$ $O \in$ the perpendicular bisector of $CP$.
Also $O_1,O_2$ both lie on the perpendicular bisector of $CP$ implying $O,O_1,O_2$ are collinear.
So the common chords of these three circles $AQ,CP,BR$ are parallel hence $ACPQ,BPCR$ are isosceles trapezoids.
Now $RP=BC=AC=PQ \Rightarrow P \in$ the perpendicular bisector of $RQ$.
But $O \in$ perpendicular bisector of $RQ \Rightarrow RQ \perp OP$.
Edit: Fixed typo, thanks for noticing Mahi

[*Mahi*]

$\angle OAP=\angle ODP=30^{\circ}$.
So $\angle MAP=120^{\circ}=\angle MCP \Rightarrow P' \in CAM$.

Shouldn't those be $P'$?

[*Mahi*]

I did the first part in a little different way.
At first, $\angle APC + \angle BPC = \angle AMC+ \angle CND = 180 ^\circ - \angle ADB$, so $P \in \bigcirc OADB$
$180^\circ - \angle POC = 180^\circ - \angle DAP = \angle MAP = \angle MCP = \angle MCO + \angle OCP = 60^\circ + \angle OCP$, which implies $\angle OCP + \angle POC = 120^\circ = 180^\circ - \angle OPC$, or $\angle OPC = 60^\circ$, and the rest is same as Tahmid.

[Thanks Adib, edited now]

[Phlembac Adib Hasan]

$P \in \bigcirc OADB$
...
$180^\circ - \angle POC = 180^\circ - \color {red}{\angle OAP}$

সম্ভবত typo.
$P \in \bigcirc OADB$ আমি আরও সহজে দেখাইসি। Suppose $AM\cap BN=X$. Then $P$ is the Miquel point of $\triangle XMN$ wrt $A,B,C$. বাকিটা মাহি ভাই ও তাহমিদ ভাইয়ের মতো করসি।