Bump, really hard problem.

/Bump This problem is proposed by 2021 TWN5.

The tangent point may be the Feuerbach point.

/Bump. Any solution?

bobaboby1 wrote: The tangent point may be the Feuerbach point. Yes, it is...

I found a generalization for this problem. However, my friend told me it has been found by @KevinCheng here. But I still would like to state my proof below. Generalization: Given a $\triangle ABC$ and an arbitrary point $P$. The projections of $B$, $C$ to $AP$ are $U$, $V$, respectively; the projections of $C$, $A$ to $BP$ are $W$, $X$, respectively; and the projections of $A$, $B$ to $CP$ are $Y$, $Z$, respectively. Show that the circumcircle of the triangle formed by $UX$, $VY$, $WZ$ is tangent to the nine-point-circle of $\triangle ABC$ at the Poncelet point $O'$ of $ABCP$. When $P$ is the Gergonne point of $\triangle ABC$, the previous problem can be proved by Feuerbach's theorem. Proof. Let $\mathcal{H}$ be the rectangular hyperbola passing through $ABCP$. Let $H,\triangle DEF$ be the orthocenter and the orthic triangle of $\triangle ABC$ respectively. Then $O'$ is the center of $\mathcal{H}$. And by Brocard's theorem we can get the polar triangle of the triangle formed by $UX,VY,WZ$ is the pedal triangle $\triangle P_1P_2P_3$ of $P$ WRT $\triangle ABC$.

we can find a transformation which is an inversion with center $O'$ composing a reflection with an axis passing through $O'$. Under this transformation, the nine-point-circle of $\triangle ABC$ becomes to $\mathbf{S}_{\triangle DEF}^{O'}$ since $\triangle DEF$ is a self-polar triangle WRT $\mathcal{H}$ by Brocard's theorem and the circumcircle of the triangle formed by $UX$, $VY$, $WZ$ becomes to $\mathbf{S}_{\triangle P_1P_2P_3}^{O'}$. So we only need to prove that $\mathbf{S}_{\triangle DEF}^{O'}//\mathbf{S}_{\triangle P_1P_2P_3}^{O'}$. However, it follows from here. Q.E.D. $\quad\Box$