Cute! Let $K \in (ABC)$ such that $AK \parallel BC$, let $V$ be the midpoint of $EF$, let $U$ be the queue point, let $W \in (ABC)$ such that $KW \parallel NR$. Let $H_A$ denote the reflection of orthocenter in $BC$. Let $G,I$ be the midpoints of $HB,HC$ . Let $T$ be the intersection of $GI$ and $UH$. Claim 1: $PUVLNT$ is cyclic. Proof: $PLVNA_1$ cyclic is trivial. $TUVN$ follows from PoP, and the claim follows Claim 2: $UV$, the line through $K$ parallel to $NR$, line through $H_A$ parallel to $EF$ are concurrent at $W$. Proof: The latter two concur at circumcircle is trivial as $K,H_A$ are antipodes so suffices to show $W=W'$, $\measuredangle H_AUW' = \measuredangle H_AUH + \measuredangle A'UW = \measuredangle H_AUW$ as desired. let $PL \cap (ABC) = H'_A$ then by reims we get $H'_AW \parallel EF$, and it follows $H'_A = H_A$ and $LH_A,GI,EF$ concur, and then reims finishes.

Nice Problem, it seems that this configuration finally showed up. Let $Q$ the A-queue point, $M, M_A, M_B, M_C, R, S$ are midpoints of $EF, AH, BH, CH, M_BM_C, BC$ respectively. Let $PH \cap (PN)=J$ and let $H'$ reflection of $H$ over $J$ as well as $K$ reflection of $M_A$ over $AN$. Claim 1: $PQMJNSL$ is a cyclic heptagon (crazy huh?). Proof: From spiral similarity we have $\angle QSP=\angle QRB=\angle QMP$ (we use the well-known fact that $Q,H,R$ are colinear), so $QMSP$ is cyclic, now it has diameter $PN$ so $QMNSP$ is cyclic, now $-\sqrt{QH \cdot HS}$ inversion sends $(AH) \to M_BM_C$ trivially and since $AH \perp M_BM_C$ we have that $A,N,J$ colinear as well as $AQHJ$ cyclic, therefore the claim is complete (as this circle is just $(PN)$). The Finish: (This early?, das crazy), since $AJ \perp HH'$ we have $A,K,H'$ colinear by symetry and by mid bade we have $AH' \parallel M_AJ$, now $-\sqrt{AH \cdot HD}$ inversion sends $(ABC) \to (DEF)$ and therefore it will sent $L \to K$ if and only if the midpoint of $HK$ lies in $(PN)$ (and is not $R$ unless $L=Q$ which is trivial to verify) which by homothety at $K$ happens if and only if $KR \perp HH'$ which is true as $M_A,N,R$ colinear is trivial from $AH \perp BC$ so by reflection and synetry this will follow thus we are done .

Let $A'$ be the reflection of $A$ about the perpendicular bisector of $BC$, let $M$ be the midpoint of $AA'$, and let $O$ and $O'$ be the centers of $(ABC)$ and $(BCH)$. We claim that triangles $AA'N$ and $HGP$ are similar. Notice that $M$ and $D$ are the midpoints of $AA'$ and $DD'$ so it suffices to show that $AMN$ is similar to $HDP$. Consider the negative inversion centered at $H$ swapping the nine-circle and the circumcircle. Then $P$ is taken to $P'$, the intersection of $(BHC)$ and the circle centered at $A$ passing through $H$. Notice $N$ is the center of parallelogram $AHOO'$. Thus $AO'\perp HP'$ so $AN\perp HP$. Also clearly $AM\perp GH$. Thus $\angle MAN=\angle DHP$ but as $DP=HP$ it suffices to show that $AN=MN$ which follows from the fact that $N$ is the circumcenter of right triangle $AMO'$. From above it follows $A'N\perp GP$. Now $L$ must be the intersection of $A'N$ with $GP$ as it lies on both $(ABC)$ and the circle with diameter $NP$. Now consider the same inversion as above. The point $G$ is taken to $G'$, the midpoint of $AH$ which lies on the nine-point circle. Thus $L$ is taken to the point such that $L'$ lies on the nine-point circle and $L'$, $G'$, $P'$, and $H$ are concyclic. Thus $L'$ must be the intersection of segment $AP'$ with the nine-point circle. Thus as $A$, $P'$, and $L'$ are collinear, $DHLP$ must be cyclic. Now $$\angle ADL=\angle HDL=180^{\circ}-\angle HPL=180^{\circ}-\angle ANA'=\angle ANL$$

Attachments:

Let $X$ be such that $AX \parallel BC$. Let $A'$ be antipode of $A$. Let $H_A$ be the midpoint of $HA$. Let $M$ be midpoint of $BC$. Let $AD$ meet $ABC$ at $D'$ . WLOG assume $\angle B > \angle C$. Claim $: \angle ADN = \angle B - \angle C$. Proof $:$ Note that $\angle ADN = \angle H_ADN = \angle DH_AN = \angle DH_AM = \angle DAA' = \angle B-\angle C$ Note that we need to prove $\angle ALN = \angle B-\angle C$ and we have $\angle ALX = \angle B-\angle C$ so we need to prove $L,N,X$ are collinear and since $\angle PLN = 90 = \angle XLD'$ we need to prove $P,L,D'$ are collinear. Let $PH$ meet $AFE$ at $J$. Claim $: A,N,J$ are collinear. Proof $:$ Let $H'$ be reflection of $H$ w.r.t $J$. Since $\angle AJH = 90$ then $A$ lies on perpendicular bisector of $H'$. Let $R,S$ be midpoints of $BH,CH$. Note that $PH.PJ = PE.PF = PR.PS$ so $HJSR$ is cyclic and sine $H',B,C$ are reflections of $H$ w.r.t $J,S,R$ then $HH'CB$ is cyclic so $O'$ also lies on perpendicular bisector of $HH'$ and since $N$ lies on $AO'$ then $A,N,J,O'$ are collinear. Claim $:PHD'$ and $NAX$ are similar. Proof $:$ Let $HP$ meet $BC$ at $P'$ and Let $O'$ be circumcenter of $BHC$. Note that since $P$ lies on the perpendicular bisector of $DH$ then $PH = HP'$ and by homothety we have that $O'$ is the reflection of $A$ across $N$. so we can instead prove $P'HD'$ and $OAX$ are similar. Note that $O'$ lies on perpendicular bisector of $AX$ and $P'$ lies on perpendicular bisector of $HD'$ and $\angle P'HD' = \angle AHJ = \angle JAX = \angle O'AX$ so $P'HD'$ and $OAX$ are similar as wanted. Now since $\angle HD'P = \angle AXN$ then $XN$ and $PD'$ meet on $ABC$ at $L'$ and since $\angle NL'P = \angle 180 - \angle NL'D' = \angle 180 - \angle XL'D' = 90$ and $L,L'$ lie on the same side of $PN$ then we have $L'$ is $L$ so $P,L,D'$ are collinear as wanted.

Let $\omega$ be the locus of points satisfying $\frac{XH}{XD} =2$. Let $A'$ be the point on $(ABC)$ s.t $AA'||BC$ and let $H'=AH \cap (ABC)$. It's enough to show that $(H'P \cap A'N \in (ABC)$. We'll prove that $H'P , A'N , (ABC) , \omega $ are concurrent. First we'll go for $H'P,(ABC), \omega$. Fix $B,H,D,H'$ and move $C$ on $BD$ with degree 1. Then one can easily get that $A,E,F$ all move with degree 1. By inversion through $H'$ , $\omega \cap (ABC)$ moves with degree 2 and so the line connecting $H'$ and that intersection has degree 1.the line $EF$ has degree at most 2 and so for the concurrency of $EF$ and the degree-1 line through $H'$ we need to check 4 special cases for $C$.$C=B,D, \infty , B'$ where $B'$ is the reflection of $B$ through $D$ works. Also , We got that $L' = \omega \cap (ABC)$ moves with degree 2. But one can easily get that $O_{ABC}$ also moves with degree 1 since $O$ is a point on the perpendicular bisector of $BH'$ s.t $\angle OBH' = 90 - \angle BAD$ which makes a projective movement on the perpendicular bisector of $BH'$ which implies that $O,N$ have degree 1 since $N$ is just the midpoint of $HO$. Also , since $\angle A'BC$ is fixed , $A'$ is the projection of $A$ onto a fixed line which has degree 1. So $A'$ has degree 1. So for the collinearity of $L',N,A'$ we need 4 cases for $C$ the previous 4 works here too and we're done.

Let $O$ be center of $(ABC); Y, Z, M, Q, R$ be midpoint of $HB, HC, BC, YZ, EF;$ $S$ be second intersection of $(AEF)$ and $(ABC)$. We have $\angle{MSQ} = \angle{BSF} = \angle{(EF, BC)} = \angle{(EF, YZ)}$. Then $Q, R, S, P$ lie on a circle. But $\angle{PQN} = \angle{PRN} = 90^{\circ}$ so $N, Q, P, R, S, L$ lie on a circle. Suppose that $A' \in (ABC)$ such as $AA' \parallel BC;$ $X$ is second intersection of $AD$ with $(ABC)$. We have $\angle{SLN} = 180^{\circ} - \angle{SRN} = 90^{\circ} - \angle{SRF} = 90^{\circ} - \angle{SMB} = 90^{\circ} - \angle{SAH} = 90^{\circ} - \angle{SA'X} = \angle{SLA'}$. Then $L, N, A'$ are collinear. Hence $\angle{ALN} = \angle{ABC} - \angle{ACB} = \angle{HAO} = \angle{(AH, NM)} = \angle{NDA}$ or $A, N, L, D$ lie on a circle

nice problem!

Attachments:
IranMO3rdRound2024p3.pdf (473kb)