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We have $\bigtriangleup AEF\sim\bigtriangleup ABC$ so we also have $\bigtriangleup AE'F'\sim^{+}\bigtriangleup ABC$ By spiral similarity we have $\angle E'YF'=180^{o}-\angle BAC=180^{o}-\angle E'AF'\implies Y\in (AEF),(ABC)$. Similarly we have $X$ lies on $(AEF)$ Let $N=\overline{AX}\cap\overline{BC}$ . Since we have $\angle NBX=\angle BF'F=\angle FAX=\angle BAN\implies NB^2=NX.NA$ and $\angle NXC=\angle AXE'=\angle AEE'=\angle ACB\implies NC^2=NX.NA $ Hence $N$ is the midpoint of side $\overline{BC}$. We have $(AEF)\cap(ABC)=Y$ this is well known that $\overline{YH}$ passes through the midpoint of $\overline{BC}$. Hence $\overline{AX},\overline{YH}$ intersect at $N$ is the midpount of $\overline{BC}$

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This problem is proposed by Viktor Ahmeti, Albania.

It is easy to se that $AEFHE'F'$ are concyclic, call that circle $\omega$.Let $M$ be the midpoint of $BC$. Let $X'$ be the $A$-humpty point of $\triangle ABC$. We claim that $X'=X$ .We know that $1.$$X'$ lies on $AM$ and $\omega$ $2.$$BC$ is tangent to $(AX'C)$ $3.$$HX' \perp AM$ . Let $X'C$ intersect $\omega$ second time at $E''$. $\angle BCX' = \angle CAX' = \angle EAX'= \angle EE''X'$. Thus $EE'' \parallel BC$, which means $EE'' \perp AD$, $E''=E'$. So $E', X',C$ are collinear. Similarly $B,X', F'$ are collinear. $BF' \cap CE' = X' \rightarrow X=X'$ Now let $\omega$ and $(ABC)$ intersect at points $A$ and $Y'$. We claim that $Y'=Y$ Let $AA'$ be the diameter of $(ABC)$.Because $AH$ is the diameter of $\omega$, $\angle AY'H = 90^{\circ}$,so $Y' H M A'$ are collinear. $\angle BY'H= \angle BYA'=\angle BCA'= 90-\angle C= \angle CAH = \angle EAH=\angle E'AH= \angle E'Y'H$. Thus $Y',E',B$ are collinear. Similarly $Y',F',C$ are collinear. $BE' \cap CF' = Y' \rightarrow Y=Y'$ We finish the problem by saying $Y,H,M$ and $A,X,M$ are collinear. Thus lines $AX, YH$, and $BC$ are concurrent at $M$.

Redefine $X,Y$: let $X=AM\cap (AH)$ and $Y=MH\cap (AH)$, then note that we get bunch of cyclic quads by PoP, such as $CMFY,BMEY,CMXE,BFXM$. Also note that $ME,MF$ are tangents from $M$ to $(AH)$. Now, we claim that $E'$ lies on $BY$ and $CX$. Indeed, $$\measuredangle E'YH=\measuredangle HAE=\measuredangle HEM=\measuredangle BYH$$and $$\measuredangle AXE'=\measuredangle EFA=\measuredangle ECM=\measuredangle MEC=\measuredangle MXC$$Similarly, $F'$ lies on $BX$ and $CY$. We are done.

First Note that $AEF'HFE'$ is cyclic ( $E'$ and $F'$ are reflections across $AH$). Claim1 : $Y,X$ are on this circle as well. Proof : First Note that because of spiral similarity between $AEF$ and $AE'F'$ we have $\angle E'AB = \angle F'AC$. $\frac {AE'}{AF'} = \frac {AE}{AF} = \frac {AB}{AC}$ so we have $AE'B$ and $AF'C$ are similar so $Y$ lies on $ABC$ so $\angle E'YF' = \angle BAC = \angle E'AF'$ so $Y$ lies on $AEF'HFE'$. we have same approach for $X$. Claim2 : $YH$ passes through $M$ midpoint of $BC$. Proof : $\angle AYH = \angle AFH = \angle 90$ so we have $YH$ passes through $M$. Let $AX$ meet $BC$ at $N$. Claim3 : $N$ is $M$. Proof : Note that $EE' || BC || FF'$. $\angle NBX = \angle XF'F = \angle XAB$ so $NB^2 = NX.NA$. $\angle NCX = \angle XE'E = \angle XAC$ so $NC^2 = NX.NA$ so $NB = NC$ so $N$ is midpoint of $BC$ so $N$ is $M$. Now we have $AX$ and $YH$ meet at $BC$.

The official solution uses a very clever idea. After we get that $$Y = \odot(AEF) \cap \odot(ABC) \ne A$$we can analogously conclude $$ X = \odot(HBC) \cap \odot(HEF) \ne H $$By looking at $\triangle HBC$ as the reference triangle instead of $\triangle ABC$.

Claim 1. $A,Y,E,'F,H,X,F',E$ is cyclic. See above posts. Claim 2. $B,H,X,C$ is cylic. Since $EE'HX$ is cylic, we have $ \angle E'EH=\angle E'XH=180-\angle HXC $ But $EE' \parallel BC \Rightarrow \angle E'EH=\angle CBH$. Therefore $\angle CBH=180- \angle HXC$ $\square$ Inverse the diagram with center $A$, radius $\sqrt{AH \cdot AD}$. Let this tranformation be $\phi$. We have: $(1)$ $\phi(D)=H,\phi(E)=C,\phi(F)=B$ $(2)$ Since $\omega =(AYE'FHXF'E)\in H,E,F$ and $\phi(H,E,F)=\phi(D,C,B) \Rightarrow \overline{BC}$, The circle transforms to line $BC$. Therefore, $\phi(BC)=\omega$ $(3)$ From $(2)$ We know that $\phi(X) \in BC$. But Observe that $\phi((BHXC)) $ transforms to nine point circle, meaning $\phi(X)$ is on nine-point circle. This gives that $\phi(X)$ is Midpoint of $BC$ $(3)$$ \phi(AX)=AX$ $(4)$ $\phi(YH)$ containst $X$ since $\phi(X) \in YH$, $X \in \phi (YH)$ Since $X \in \phi(AX),\phi(YH),\phi(BC)$, reversing the transformation we know that these lines concur at $\phi(X) \Rightarrow$ midpoint of $BC$ $\blacksquare$ [asy][asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(26.05cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.61, xmax = 23.44, ymin = -7.72, ymax = 5.13; /* image dimensions */ draw((7.6,0.86)--(5.06,-4.84)--(14.94,-5.92)--cycle, blue); draw((5.06,-4.84)--(7.23,-2.49)--(8.67,-1.93)--(14.94,-5.92)--cycle, red); /* draw figures */ draw((7.6,0.86)--(5.06,-4.84), blue); draw((5.06,-4.84)--(14.94,-5.92), blue); draw((14.94,-5.92)--(7.6,0.86), blue); draw(circle((7.6,0.86), 4.47)); draw((7.6,0.86)--(6.95,-5.05)); draw((xmin, 0.57*xmin-5.63)--(xmax, 0.57*xmax-5.63)); /* line */ draw((xmin, -0.11*xmin-4.29)--(xmax, -0.11*xmax-4.29)); /* line */ draw(circle((8.71,-3.1), 2.62), red); draw(circle((11.15,-11.83), 9.87)); draw(circle((7.42,-0.81), 1.68), red); draw((5.06,-4.84)--(7.23,-2.49), red); draw((7.23,-2.49)--(8.67,-1.93), red); draw((8.67,-1.93)--(14.94,-5.92), red); draw((14.94,-5.92)--(5.06,-4.84), red); draw((7.23,-2.49)--(14.94,-5.92)); draw((5.06,-4.84)--(8.67,-1.93)); draw((xmin, -2.6*xmin + 20.62)--(xmax, -2.6*xmax + 20.62)); /* line */ draw((xmin, -1.05*xmin + 5.08)--(xmax, -1.05*xmax + 5.08)); /* line */ draw(circle((4.79,-1.82), 3.88)); draw((xmin, -0.54*xmin + 2.2)--(xmax, -0.54*xmax + 2.2)); /* line */ draw((xmin, 5.79*xmin-34.12)--(xmax, 5.79*xmax-34.12)); /* line */ draw((xmin, -0.92*xmin + 7.88)--(xmax, -0.92*xmax + 7.88)); /* line */ draw(circle((10.18,-3.71), 5.25)); draw(circle((9.82,-7.05), 5.25)); /* dots and labels */ dot((7.6,0.86),dotstyle); label("$A$", (7.66,0.95), NE * labelscalefactor); dot((5.06,-4.84),dotstyle); label("$B$", (5.12,-4.76), NE * labelscalefactor); dot((14.94,-5.92),dotstyle); label("$C$", (15,-5.83), NE * labelscalefactor); dot((6.95,-5.05),dotstyle); label("$D$", (7.01,-4.96), NE * labelscalefactor); dot((9.07,-0.5),dotstyle); label("$E$", (9.14,-0.41), NE * labelscalefactor); dot((6.3,-2.07),dotstyle); label("$F$", (6.35,-1.98), NE * labelscalefactor); dot((7.23,-2.49),dotstyle); label("$H$", (7.29,-2.39), NE * labelscalefactor); dot((8.24,-2.28),dotstyle); label("$F'$", (8.3,-2.2), NE * labelscalefactor); dot((5.87,-0.15),dotstyle); label("$E'$", (5.92,-0.06), NE * labelscalefactor); dot((8.67,-1.93),dotstyle); label("$X$", (8.73,-1.84), NE * labelscalefactor); dot((5.74,-0.92),dotstyle); label("$Y$", (5.79,-0.83), NE * labelscalefactor); dot((8.85,-5.25),dotstyle); label("$F''$", (8.91,-5.16), NE * labelscalefactor); dot((-1.03,-4.17),dotstyle); label("$E''$", (-0.98,-4.09), NE * labelscalefactor); dot((1.99,-4.5),dotstyle); label("$Y'$", (2.04,-4.41), NE * labelscalefactor); dot((10,-5.38),dotstyle); label("$X'$", (10.06,-5.29), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Diagram has random lines and circles but meeeh

Lol what. Its easy to see that $E',F'$ lie in $(AEF)$ Let $CF' \cap (AEF)=X'$ and $CE' \cap (AEF)=Y'$, now its known that $AD$ is the angle bisector of $\angle EDF$ hence $D,E,F'$ and $D,F,E'$ are colinear points, now by Reim's theorem twice $DF'X'C$ and $DY'E'C$ are cyclic (becuase $EE'$ and $FF'$ are perpendicular to $AD$ so they are parallel to $BC$...) so by radax we get that $B,F',X'$ and $B,E',Y'$ are colinear hence $X=X'$ and $Y=Y'$. Note that $AE',HF'$ meet at a point in $BC$ becuase of the reflection over $AD$, now we finish by pascal

Nice problem ! Since the circle with diameter $\overline{AH}$ is fixed after reflection along $(AD)$, it follows that $E',F'$ lie on it. Let's show that both $Y$ and $X$ lie on it ! Define $Y'$ the second intersection of the circumcircle of $(ABC)$ and the circle with diameter $\overline{AH}$. Define $F''$ to be the intersection of $(DE)$ and $(CY')$. First note that it is well-known that $(Y'H)$ intersect $(BC)$ at the midpoint of $\overline{BC}$, say $G$, since reflecting $H$ about $G$ gives you the antipode of $A$. Now notice $C,G,D$ are colinear, hence by inverse Pascal on $AEEF''Y'H$, we deduce that $F''$ lies the circle with diameter $\overline{AH}$. Since $\measuredangle FDA = \measuredangle ADE$, it follows that $F''=F'$. Proceeding similarly with $E'$, we get that $Y'=Y$. Inverse Pascal yet again on $E'XF'EAF$ gives that $X$ lies on the circle with diameter $\overline{AH}$. Finally, Pascal on $AXF'YHE$ finishes.

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Step 1. Deleting points $E'$ and $F'$ from the diagram cause why not. Reflect the problem statement over the line $AD$. Let $B'$ be a reflection of $B$ and $C'$ be a reflection of $C$. Redefine the points $X$ and $Y$ as $X=B'F \cap C'E$ and $Y=C'F \cap B'E$. We need to show that $AX, HY$ and $BC$ are concurrent. Step 2. Showing that $AY, HX$ and $BC$ are concurrent. Let $Z = EF \cap BC$ and $Z'$ be the reflection of $Z$ over $AD$. Consider the degenerate quadrilateral $EFC'B'$. By DDIT, there exists an involution swapping the pairs of lines $(AE, AC'), (AF, AB'), (AY, AZ)$. Notice that the first two pairs are symmetric wrt the angle bisector of $C'AC$(which clearly coincides with $AD$), thus, the involution is actually a reflection over the line $AD$, implying that $AY, AZ$ are isogonal in $\angle{C'AC}$. Since $AZ, AZ'$ are also isogonal in $\angle{C'AC}$, it implies that $Z' \in AY$. Similarly, consider the degenerate quadrilateral $FEC'B'$. By DDIT, there exists an involution swapping the pairs of lines $(HF, HC'), (HE, HB'), (HX, HZ)$. Again, the first two pairs are symmetric wrt $AD$ and thus the involution is a reflection, implying $Z' \in HX$. Since $Z' \in BC$, we got that $AY, HX$ and $BC$ are concurrent. $\ \blacksquare$ Step 3. Finishing the problem. Since $XH \cap YA = Z', HF \cap AE = C, FY \cap EX = C'$ are collinear, by the inverse Pascal's theorem, the six points $X, H, F, Y, A, E$ lie on a conic. Now Pascal's theorem on $AXFHYE$ implies that $AX \cap HY, XF \cap YE = B', FH \cap EA = C$ are collinear or equivalently, $AX$ and $HY$ intersect on $BC$ and we are done. $\ \square$

It suffices to show that $X$ is the $A$-humpty point, and $Y$ is the $A$-Queue point, because then $AX$ and $YH$ would intersect at the midpoint $M$ of $BC$. Let $B'$ and $C'$ denote the reflections of $B$ and $C$ across $AD$ respectively, and let $T$ denote the $A$-Ex point. Since $BA \cdot BF = BD \cdot BC = (\frac{1}{2} BB') \cdot (2 BM) = BB' \cdot BM$, we have that $A,F,M,B'$ are concyclic. We also show that $A,F,T,C'$ are concyclic. Using directed lengths, this is equivalent to showing that $BC' \cdot BT = BA \cdot BF = BD \cdot BC$. Recall that $(B,C;D,T)=-1$, so $\frac{BD}{CD}=-\frac{BT}{CT}$. $BC' \cdot BT = BD \cdot BC \iff (BD + CD) \cdot BT = BD \cdot (BT + TC) \iff CD \cdot BT = BD \cdot TC \iff \frac{BD}{CD}=-\frac{BT}{CT}$. So $A,F,T,C'$ are concyclic. Take inversion centred at $A$ with radius $\sqrt{AH \cdot AD}$, the image of $(AFMB')$ is the line through $B,F',$ and the $A$-Humpty point. Similarly, $C,E'$ are collinear with the $A$-Humpty point, so $X$ is the $A$-Humpty point. The image of $(AFTC')$ is the line through $B,E',$ and the $A$-Queue point. Similarly, $C,F'$ are collinear with the $A$-Queue point, so $Y$ is the $A$-Queue point. Hence proved. $\square$

Apparently $A,E,F,H,E',F'$ lies on the circle with diameter $AH.$ So use inverse Pascal on $EHFF'XE'$ and $FAEE'YF'$ we get $X,Y$ also lies on $\odot (AH).$ Now use Pascal on $HEAXF'Y$ we are done.$\Box$

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Redefine $X$ is $A$ - Humpty point of $\triangle ABC$ and $Y$ be second intersection of $\bigodot(AH)$ with $(ABC)$. We have $$(XF', XB) \equiv (XF', XA) + (XA, XB) \equiv (HF', HA) + (BA, BC) \equiv (HA, HF) + (BA, BC) \equiv (BC, BA) + (BA, BC) \equiv 0 \pmod \pi$$Then $B, F', X$ are collinear. Similarly, we have $C, E', X$ are collinear. We also have $$(YE', YB) \equiv (YE', YA) + (YA, YB) \equiv (HE', HA) + (CA, CB) \equiv (HA, HE) + (CA, CB) \equiv (CB, CA) + (CA, CB) \equiv 0 \pmod \pi$$So $E', Y, B$ are collinear. Similarly, we have $F', Y, C$ are collinear. This means $AX, YH$ intersect at midpoint of $BC$

Knock knock who’s there? Queue point and Humpty point from Muricaaaaaaa. If we can show that $Y$ is the $A$-Queue point and $X$ the $A$-Humpty point then we win since then the concurrency point is none other than the midpoint of $BC$. We focus our attention on $X$ since $Y$ basically follows from orthocenter duality. Redefine $X$ as the $A$-queue point. Then $\measuredangle AXC=\measuredangle ABC=\measuredangle AFF’=\measuredangle AXF’$ (second one follows from the fact $EE’$ perpendicular to $AH$ so it’s parallel to $BC$), thus done ($X-E’-C$); similarly $X-F’-B$. Oh well I mixed my point names up