By Pascal's theorem on $CABQPR$, $CR, BQ$ meet on $EF$. Let $G$ be the end-point of the $A-$symmedian chord. By Pascal's theorem on $AYPQRC$, $CR, PY, EF$ concur. Similarly, $BQ, PY, EF$ concur, so $BQ, CR, PY, EF$ concur at a point. It suffices to show that the concurrency point is the midpoint of $EF$, or equivalently, that if this point is $X$, then $A, X, G$ are collinear. It's well known that $PG, BC, EF$ are concurrent (use harmonic ratios after constructing the Miquel point of $EFBC$). By Pascal's theorem on $PGABCR$, we have $AG \cap CR \in EF$, so $A, X, G$ are collinear, and we are done.

From here : https://artofproblemsolving.com/community/c6h1864641p12623614 We conclude that $BQ$ and $CR $ pass through the midpoint of $EF$ (call it $M$). Let $Y'=PM\cap (ABC) $. From pascal's theorem on $AY'PRQB$ We conclude that $AY', EF, RQ$ are concurrent, hence $Y'=Y$.

Arkmmq wrote: From here : https://artofproblemsolving.com/community/c6h1864641p12623614 We conclude that $BQ$ and $CR $ pass through the midpoint of $EF$ (call it $M$). Let $Y'=PM\cap (ABC) $. From pascal's theorem on $AY'PRQB$ We conclude that $AY', EF, RQ$ are concurrent, hence $Y'=Y$. This problem was proposed by me, and I didn't know this source until the problem was released and the contestants told me.

I found a weird solution Here’s the outline. Let $Z$ be the other endpoint of the $A$-symmedian chord in $\odot(ABC)$. By Pascal’s twice, one can check that $BQ, CR,$ and $EF,$ concur at the midpoint $X$ of $EF$. Since $PZ, EF,$ and $BC,$ concur, taking powers tells us that $PEFZ$ is cyclic. Let $AP \cap EF=T$ and again by powers, we see that $PZTX$ are concyclic. Now an inversion at $X$ swapping $\odot(ABC)$, swaps $\odot(BXC)$ with $QR$, line $EF$ maps to itself, and line through $A$ and $PX \cap \odot(ABC)$ maps to $\odot(XZP)$. They all pass through $T$, concluding the proof.

USJL wrote: Let point $H$ be the orthocenter of a scalene triangle $ABC$. Line $AH$ intersects with the circumcircle $\Omega$ of triangle $ABC$ again at point $P$. Line $BH, CH$ meets with $AC,AB$ at point $E$ and $F$, respectively. Let $PE, PF$ meet $\Omega$ again at point $Q,R$, respectively. Point $Y$ lies on $\Omega$ so that lines $AY,QR$ and $EF$ are concurrent. Prove that $PY$ bisects $EF$. By 2019 ELMO Shortlist G1, we find that $BQ \cap CR$ is the midpoint of $EF.$ Now, Pascal on $AYPQRC$ gives $YP \cap RC$ lies on $EF,$ which is sufficient. $\square$

Let $\{M\}=PY \cap FE$, $\{K\}=AM \cap (ABC)$ and $I,J$ be the intersections of $FE$ with the circle $(ABC)$. As $AY,QR,IJ$ are concurrent there is an involution swapping the pairs $(A,Y),(Q,R),(I,J)$.Projecting this through $P$ on the line $FE$ we obtain an involution swapping $(PA \cap FE,M),(E,F),(I,J)$.Finally ,projecting through $A$ on $(ABC)$ we obtain an involution swapping $(P,K),(B,C),(I,J)$,so $PK,BC,IJ$ are concurrent.As $-1=(FE\cap BC,PA\cap BC,B,C)\overset{P}{=}(K,A,B,C)$ $KABC$ is harmonic hence $AK$ is symmedian and $M$ is therefore the midpoint of $FE$.

Let $M'=RC \cap BQ$. By Pascal theorem $RCABQP$, $E,F,M'$ are collinear. Let $X$ be a point $\Omega$ such that $BX \parallel EF$. Obviously, $PX \parallel BE$. Let $D$ is the intersection of $BE$ and $\Omega$. Then, $$-1=(H,D;E,P_{\infty_{BE}}) \overset{P}{=} (A,D;Q,X)\overset{B}{=} (F,E;M',P_{\infty_{EF}})$$Hence, $M'$ is the midpoint of $EF$. By Pascal theorem $AYPQRC$, $YP$ intersect $EF$ at $M'$. Which mean $PY$ bisects $EF$. [asy][asy] size(300); pair A,B,C,P,Q,R,E,F,D,X,H,M,N,X_0,O; A=dir(110); B=dir(205); C=dir(335); H=orthocenter(A,B,C); E=foot(B,A,C); F=foot(C,A,B); D=2*E-H; M=foot(H,C,B); P=2*M-H; O=circumcenter(A,B,C); N=midpoint(E--F); Q=extension(B,N,P,E); R=extension(C,N,P,F); X_0=foot(B,A,O); X=2*X_0-B; draw(unitcircle,deepcyan); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$P$", P, dir(240)); dot("$E$", E, dir(30)); dot("$F$", F, dir(210)); dot("$H$", H, dir(-45)); dot("$D$", D, dir(D)); dot("$R$", R, dir(R)); dot("$Q$", Q, dir(Q)); dot("$M$", N, dir(90)); dot("$X$", X, dir(X)); draw(A--B--C--cycle,blue); draw(P--R,heavycyan); draw(P--Q,heavycyan); draw(E--F,purple); draw(A--P,lightblue); draw(P--X,heavymagenta+dashed); draw(B--D,heavymagenta+dashed); draw(B--Q,heavyblue); draw(B--X,purple); draw(C--R,heavyblue); [/asy][/asy]

Let $YP \cap RC = G$ and the concurrency point to be $D$, by pascal on $AYPQRC$ we see that $\overline{D-G-E}$ are collinear, thus $G \in EF$ but from ELMO shortlist 2019 G1 we know that $RC$ intersects $EF$ at it's midpoint, thus $PY$ bisects $EF$.

Let $S$ be the midpoint of $EF$ and $G$ be the point in $\Omega$ so that $ABGC$ is harmonic. We will use panthom points, i.e., redefine $Y'$ as the second intersection of $PS$ with $\Omega,$ we wish to show that $AY',RQ$ and $EF$ are concurrent. In order to do so, we will use Pascal's Theorem multiple times. [asy][asy] import graph; import geometry; import olympiad; size(300); pair A,B,C,H,D,E,F,P,S,Q,R,T,G,L,Y; A=dir(120); C=dir(-25.87); B=dir(205.87); H=orthocenter(A,B,C); D=extension(A,H,B,C); E=extension(B,H,A,C); F=extension(C,H,A,B); P=2*D-H; S=E*0.5+F*0.5; Q=extension(B,S,P,E); R=extension(C,S,P,F); T=extension(E,F,B,C); G=extension(A,S,T,P); L=extension(Q,R,E,F); Y=extension(A,L,P,S); draw(unitcircle,blue); draw(T--G,dotted); draw(B--A--C--T,gray); draw(P--Y); draw(T--E); draw(A--G,dotted+orange); draw(B--Q,dotted+magenta); draw(C--R,dotted+magenta); draw(P--Q,green); draw(P--R,green); dot("$A$",A,dir(A)); dot("$B$",B,dir(240)); dot("$C$",C,dir(C)); dot("$P$",P,dir(P)); dot("$E$",E,dir(45)); dot("$F$",F,dir(F)); dot("$Q$",Q,dir(Q)); dot("$R$",R,dir(R)); dot("$G$",G,dir(G)); dot("$S$",S,dir(S)); dot("$T$",T,dir(T)); dot("$Y'$",Y,dir(Y)); [/asy][/asy] It is well known that $PG,EF$ and $BC$ concurr. Then define $T$ as the intersection point. Now we apply Pascal's Theorem multiple times: $\bullet$ Apply Pascal's at $AGPQBC$ to find that $BQ\cap AG, T=PG\cap BC,E=PQ\cap BC$ are colinear, so that $BQ\cap AG$ lies on $EF.$ $\bullet$ Similarly, apply Pascal's at $AGPRCB$ to conclude that $CR \cap AG$ lies on $EF.$ $\bullet$ Finally, apply Pascal's at $ABQPRC$ to find that $BQ\cap CR$ lies on $EF.$ From above, it follows that the four lines $BQ,CR,AG$ and $EF$ must concurr. Of course, $AG$ and $EF$ concurr at $S,$ due to symmedians; so that the four lines concurr at $S.$ For finishing, apply Pascal's at $AY'PQRC$ yelding $AY'\cap RQ, S=PY'\cap CR$ and $E=PQ\cap AB$ are concurrent. So we are done.

Let $M$ be the midpoint of $EF$ , $Y'=PM \cap \Omega$ and $X'=AY' \cap RQ$. Its enough to check $\dfrac{X'Q}{X'R} = \dfrac{XQ}{XR}$. Ceva's theorem in $\triangle PQR$ implies $\dfrac{XQ}{XR} = \dfrac{FP \cdot EQ}{FR \cdot EP}$. And by a well known lemma called "Ratio Lemma" have $\dfrac{X'Q}{X'R} = \dfrac{AQ}{AR} \cdot \dfrac{Y'Q}{Y'R}$. Now with some calculations we will be done: $ 1) \ \dfrac{XQ}{XR} = \dfrac{EQ}{FR} \cdot \dfrac{FP}{EP} = \dfrac{\dfrac{EA \cdot EC}{EP}}{\dfrac{FA \cdot FB}{FP}} \cdot \dfrac{FP}{EP} = \frac{c}{b} \cdot \frac{\cos(C)}{\cos(B)} \cdot (\dfrac{FP}{EP})^2 $ $ 2)\ \dfrac{AQ}{AR} = \dfrac{\sin(\angle APQ)}{\sin(\angle APR)} \dfrac{Y'Q}{Y'R} = \dfrac{\dfrac{HE \cdot \sin(EHP)}{PE}}{\dfrac{HF \cdot \sin(FHP)}{PF}} = \dfrac{\cos(C)}{\cos(B)} \cdot \dfrac{\sin(C)}{\sin(B)} \cdot \dfrac{PF}{PE} $ $ 3)\ \dfrac{Y'Q}{Y'R} = \dfrac{\sin(Y'PQ)}{\sin(Y'PR)} = \dfrac{\sin(MPE)}{\sin(MPF)} = \dfrac{FP}{EP} $ $ (1),(2),(3) \Longrightarrow \dfrac{XQ}{XR} = \frac{c}{b} \cdot \frac{\cos(C)}{\cos(B)} \cdot (\dfrac{FP}{EP})^2 = \dfrac{AQ}{AR} \cdot \dfrac{Y'Q}{Y'R} = \dfrac{X'Q}{X'R} $ and we are done.

By Pascal on $BACRPQ$, $\overline{EF}$, $\overline{BQ}$, $\overline{CR}$ concur. By Pascal on $BAYPRQ$, $\overline{EF}$, $\overline{PY}$, $\overline{BQ}$ concur, so all four concur at some point $M'$. Let $X$ be the point on $(ABC)$ such that $\overline{PX} \parallel \overline{EF}$. Then $$(E,F;M',\infty_{\overline{EF}})\stackrel{P}{=}(Q,R;Y,X)\stackrel{M'}{=}(B,C;P,\overline{MX} \cap (ABC)).$$It thus suffices to show that $\overline{M'X} \cap (ABC)$ is the $A$-queue point $T$, since $BPCT$ is well-known to be harmonic (proof: spiral sim means this is equivalent to $FHET$ harmonic, finish with three tangents. there are like three other ways to do this). It is sufficient to prove that if $M$ is the midpoint of $\overline{EF}$, then $\overline{QM}$ and the line through $P$ parallel to $\overline{EF}$ concur on $(ABC)$, which is equivalent to $\measuredangle QAH=\measuredangle QMF$. Let $N$ be the midpoint of $\overline{BC}$, so this is equivalent to $\measuredangle QAP=\measuredangle QNB \iff \widehat{PQ}=\widehat{BQ}+\widehat{CA'}$ where $A'=\overline{QH} \cap (ABC)$ is the $A$-antipode, or $\widehat{CA'}=\widehat{PB}$ which is evident. This finishes the problem. $\blacksquare$

Desargue's involution kills it instantly let $EF$ intersect $\Omega$ again at $E_1, F_1$, $AP$ intersect $EF$ at $H'$, $YP$ intersect $EF$ at $M'$, $M$ be the midpoint of $EF$ Since the concurrency shows $(A,Y),(Q,R),(E_1,F_1)$ are pairs of involution Project it through $P$ onto $EF$ and get $(E,F),(E_1,F_1),(H',M')$ are pairs of involution. Now since $BCE_1F_1,BCEF,BCH'M$ are concyclic, we know $M=M'$