@Complex2Liu: I think $\angle DSP = \angle EAC$ in your proof is not correct.

Solution with the base idea of my friend Nguyen Canh Hoang. Let $(PKL)$ intersects $AB,AC$ at $G,H$. Since $\angle AGL=\angle KPL=\angle 90^\circ-\frac{1}{2}\angle A=\angle ALG$ so $GL\parallel BC$. Similarly, $GL\parallel BC$. Let $BC$ intersects $(PKL)$ at $I,J$ and we have $D$ is the midpoint of $IJ$. Since $K(IJDH)=-1$ so $HJLI$ is hamoric quaterial $\implies AI,AJ$ is the tangent line of $(PKL)$ $\implies P(IJPA)=-1$. Let the tangent line at $P$ intersects $BC$ at $T$. Hence $(TEIJ)=-1$. But $TP^2=TI.TJ=TE.TD$ so $TP$ is the tangent line of $(PDE)$ i.e $(TPE)$ tangent to $(PKL)$. Hence $XY\parallel KL\implies \angle MDP=\angle XYP=\angle KLP=\angle 180^\circ-\angle XGP\implies D,M,G,P$ are concyclic. Hence $\angle GMP=\angle GDP$. Similarly, $\angle PNA=\angle GDP$ i.e $P,M,A,N$ are concyclic. $\blacksquare$

Thanks to baopbc.

@Complex2Liu you have been on the point to reach the goal : in stead of taking $S,T$ as midpoints define them as the intersection of $DY,PX$ and $XD,PY$ and by angle chase prove $\widehat{PSN} =\widehat{PAN}$ ,consider the triangle $SPY$ and so on... In this configuration it s interesting to prove an additional question : $(KLP)$ and $(DEP)$ are tangent R HAS

@baopbc, I don't think that it is well-known that $\omega $ tangents to $\odot (DEF)$ . Can you please prove it? Thanks.

Can anyone prove the well-known lemma used by baopbc?

PROF65 wrote: @Complex2Liu you have been on the point to reach the goal : in stead of taking $S,T$ as midpoints define them as the intersection of $DY,PX$ and $XD,PY$ and by angle chase prove $\widehat{PSN} =\widehat{PAN}$ ,consider the triangle $SPY$ and so on... In this configuration it s interesting to prove an additional question : $(KLP)$ and $(DEP)$ are tangent R HAS Can you show how to continue after definining S and T? Please help me...

Any help?

It is indeed the hardest problem of that TST.Only 2 silver medalists solved this problem in that test.

Can someone revive this problem with his solution?

After lots of tries... In fact, this question is equivalent to the following problem (with some reverse constructions) Equivalent problem: Let $ABC$ be a triangle, $I$ be its incenter, the perpendicular line from $I$ to $AI$ intersects $AB$ and $AC$ at $D$ and $E$ respectively. $BI$ and $CI$ intersects $(ABC)$ at $M_B$ and $M_C$ respectively.Let $X$ be an arbitrary point on the line $DE$ (solving for the line segment case doesn't matter), $Y$ be the second intersection of $AX$ and $(ABC)$, $YM_B$ and $YM_C$ intersects $CI$ and $BI$ at $L$ and $K$ respectively. Prove that $X,I,Y,K,L$ are concyclic. Proof: Let $T_A$ be the A-mixtillinear tangency point of the triangle $ABC$. We will prove that $T_A$ lies on this circle as well, We first prove the following three claims: Claim 1: $T_A,Y,X,I$ are concyclic Proof: It is well known that $T_AI$ intersects $(ABC)$ again at the midpoint of the arc $BAC$, which means that if we denote this intersection point with $M$, $DE \parallel MA$. As $YT_AAM$ is cyclic, we have $\angle(T_AYX)=\angle(T_AMA)=\angle(T_AIX)$, which proves the claim. Claim 2: $Y,K,I,L$ are cyclic Proof:Angle chasing. Claim 3:$Y,T_A,K,L$ are cyclic Proof:Using $BT_AI \sim IT_AC$, it suffices to show that $\angle(BT_AK)=\angle(IT_AL)$, which is equivalent to $\frac{BK}{KI}=\frac{IL}{LC}$. However, from angle chasing, we know that $BM_CI \sim IM_BC$ and $\angle(BM_CK)=\angle(IM_BL)$, which gives us $M_CBKI \sim M_BILC$, which gives us the desired ratio. (Similarities are positive directed) Using Claim 3 with Claim 2, we have $Y,T_A,K,I,L$ are cyclic. Using this with Claim 1, we get $Y,T_A,K,I,L,X$ are cyclic, which gives the desired cyclicity. Finally, observe that if $YM_C$ and $YM_B$ intersect $AB$ and $AC$ at $P$ and $Q$ respectively. we have $P,I,Q$ are collinear from Pascal's theorem on the quadrilateral $BACM_CYM_B$, which ends the problem.

First we make a simple observation. Claim. $D$ is the incentre of $\triangle AMN$. Proof. $\angle KPL = \angle XPY = 180^{\circ} - \angle XDY = 180^{\circ} - \angle MDN$, so $\angle MDN = 90^{\circ} + \frac{1}{2} \angle MAN$. Also $AD$ bisects $\angle MAN$. Therefore $D$ is the incentre of $\triangle AMN$. Then we prove an equivalent problem. Equivalent problem. Given a triangle $AMN$, its incentre $D$, and its circumcircle. $P$ is a point on the circumcircle of $\triangle AMN$. Let $B$ be a point on $AM$ and $C$ be a point on $AN$ such that $BC$ is perpendicular to $AD$ and passes through $D$. Let $AP$ intersects $BC$ at $E$. Let $MD$ and $ND$ intersect the circumcircle of triangle $PDE$ at $X$ and $Y$ respectively. Let $PX$ meets $AM$ at $K$, and $PY$ meets $AN$ at $L$. Prove that $K, D, L$ are collinear. Proof. Since $\angle APX = \angle EPX = \angle BDM = \frac{1}{2} \angle ANM$, we have the intersection of $PX$ and $ND$ is on the circumcircle of $\triangle AMN$. By the same argument, $PY$ meets $MD$ on the circumcircle of $\triangle AMN$. By Pascal's Theorem, $K,D,L$ are collinear.