Circumcircle $(J)$ of $\triangle PQR$ is obviously the image of $(O) \equiv \odot(ABC)$ under the inversion with center $I$ and power $IP \cdot IA=IQ \cdot IB=IR \cdot IC.$ Thus, by inversion property $I$ is also a center of similitude of $(O) \sim (J)$ $\Longrightarrow$ $I,J,O$ are collinear. Since $B,C,Q,R$ are concyclic, then $QR$ is antiparallel to $BC$ WRT $IB,IC$ $\Longrightarrow$ $QR$ is perpendicular to the I-circumdiameter $IA$ of $\triangle BIC.$ Likewise, $IB,IC$ are perpendicular to $RP,PQ$ $\Longrightarrow$ $I$ is orthocenter of $\triangle PQR$ $\Longrightarrow$ $IJ$ is Euler line of $\triangle PQR,$ passing through $O.$

Hi ; Nice one .

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Let incircle of the triangle $ABC$, touches sides $BC,CA,AB$ at points $D,E,F$, respectively and $M\in ID,N\in IE,L\in IF$ $ \Rightarrow $ $IF\cdot IL=IE \cdot IN=ID\cdot IM$. So, we see not difficult $M,R,N$ - collinear; $N,P,L$ - collinear; $L,Q,M$ - collinear; and $MR=NR$, $NP=LP$, $LQ=MQ$. Therefore, $I$ - circumcenter of the triangle $MNL$, and orthocenter of the triangle $PQR$. (*) Let $IA,IB,IC$ intersect circumcircle of the triangle $ABC$ at points $A_1,B_1,C_1$. Then, we have $ \triangle A_1B_1C_1$ and $ \triangle PQR$ - gomotetic similar. (center $I$) So, points $I$, $O$ and $O'$ (circumcenter of the $ \triangle PQR$) are collinear.(**) Since (*) and (**), we get $OI$ - Euler's line of the triangle $PQR$.

Remark: the centers of the circles $\odot ABPQ,\odot BCQR,\odot CARP$ lie on a circle centered $O$! Best regards, sunken rock

Very nice to see many different solutions to this nice problem

Dear Mathlinkers, see also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=565405 Sincerely Jean-Louis

Denote by $O'$ circumcenter of $\odot (PRQ)$. From $IP\cdot IA=IQ \cdot IB=IR\cdot IC$ we have that $ABQP, BCRQ, CRPA$ are cyclic. Denote by $O_1,O_2,O_3$ circumcenters of $\odot (ABQP),\odot (BCRQ),\odot (CRPA)$. We have $\angle RPI+\angle PRI + \angle IRQ=\frac{\alpha + \beta+ \gamma}{2}=\frac{\pi}{2}$ $\implies$ $I$ is orthocenter of $\Delta PRQ$. Since $O_3$ is on perpendicular bisector of $CR$ and $O_2$ (as well) $\implies$ $O_3O_2 \perp CR$. But we have $CR \perp PQ$ $\implies$ $PQ \parallel O_3O_2$. Similarly $RQ \parallel O_1O_3$ and $RP \parallel O_1O_2$. From this we have that $\Delta O_1O_2O_3$ and $\Delta PRQ$ are homothetic$(\star)$. Now since $OO_3 \perp CA$ and $IA \perp O_1O_3$ we have that $\angle OO_3O_1=\angle IAC$. Similarly $\angle OO_1O_3=\angle IAB$. From this (and similar relations) we have that $O$ is circumcenter of $\Delta O_1O_2O_3$. Also since $O'O_1 \perp PQ || O_2O_3$ $\implies$ $O'$ is orthocenter of $\Delta O_1O_2O_3$. Let $H$ be center of homothety $\mathcal{H}$ taking $\Delta PRQ$ to $\Delta O_1O_2O_3$. We have that $\mathcal{H}: O' \mapsto O$ $\implies$ $O,H,O'$ are collinear $(1)$ and we also have that $\mathcal{H} : I \mapsto O'$ $\implies$ $O',H,I$ are collinear $(2)$. From $(1),(2)$ we have that $O,I,O',H$ are collinear $\implies$ $O,I,O'$ are collinear $\implies$ $O$ is on Euler line of $\Delta PRQ$ $\blacksquare$.