Assume $AB < AC$. Let $Z' = (EFK)\cap (ABC)$, then we want to show $Z'\in XY$. In fact if $J=EF\cap BC$, then $Z' = JK\cap (ABC)$, so projecting $(D,J;B,C) = -1$ through $K$ we have $(A,Z;B,C) = -1$ or that $AZ'$ is the $A$-symmedian. So, it suffices to show that $XY$ is also the $A$-symmedian, which is true by HMMT 2018 Team 4. This proves a). Now by Pascal $BS\cap CT = Q$ lies on $EF$ by Pascal on $BACTKS$. Now move $K$ on $(ABC)$, we have that for any point $K\in (ABC)$, projecting through $E$ to $S\in (ABC)$, then projecting through $B$ to $Q\in EF$ is a projective map. We have $A\to J$, $B\to E$, and $C \to F$. Therefore, $(A,K;B,C) = (J,Q;E,F)$. To show that $Q\in XY$ it suffices to show that $QE = QF$, for which it suffices to show that $\frac{JE}{JF} = \frac{AB}{AC}\cdot \frac{KC}{KB}$. But \[ \frac{AB}{AC}\cdot \frac{KC}{KB} = \frac{AB}{AC}\cdot \frac{HC}{HB} = \frac{AB}{AC}\cdot \frac{CE}{BF} = \frac{AB}{BF} \cdot \frac{EC}{CA}, \]which equals $\frac{FJ}{EJ}$ by Menelaus on $\triangle AEF$ and line $BCJ$, which solves b). $\blacksquare$

Let $Q = EF \cap BC.$ (a) It suffices to prove that $Q \in KZ$, since that would imply that $QK \cdot QZ = QB \cdot QC = QF \cdot QE$ by PoP, and hence PoP would finish. Let $Z'$ be the $A-$HM point of $\triangle ABC.$ It's well-known/easy to see that $Q, H, Z'$ are collinear. Hence, since $K, H$ are symmetric w.r.t $BC$, in order to prove that $Q \in KZ$, we just need to show that $Z, Z'$ are symmetric w.r.t $BC.$ However, it's well-known that the point which is the reflection of the $A-$HM point over $BC$ is just the point where the $A-$symmedian meets $(\triangle ABC)$ for a second time, and so it would therefore suffice to show that $XY$ is the $A-$symmedian of $\triangle ABC$. By Pascal on $EDFPMN$ (cyclic hexagon on the nine-point circle of $\triangle ABC$), we see that $A, X, Y$ are collinear. Hence, we just need to show that $AX$ is the $A-$symmedian. To show this, let $W = AA \cap MP,$ where $AA$ denotes the tangent to $(\triangle ABC)$ at $A.$ It's easy to see by angle-chasing that $AEXW$ is an isosceles trapezoid with $AE || XW$, and so $PA = PE \Rightarrow P$ is the midpoint of $WX.$ Hence, if we let $P_{\infty}$ denote the point at infinity along $WX$, projecting the harmonic bundle $(W, P, X, P_{\infty})$ from $A$ onto $(\triangle ABC)$ implies that $ABZC$ is harmonic, where we used that $Z = AX \cap (\triangle ABC).$ This clearly implies the desired, and so we're done. $\square$ (b) By Pascal on $BSKTCA$ we know that $BS \cap CT \in EF.$ Since $EF$ is anti-parallel to $BC$, it suffices to show that $BS \cap CT$ is the midpoint of $EF$, since the median in $\triangle AEF$ is the $A-$symmedian of $\triangle ABC$, which we know from part (a) is $XY$. So let's show this. If we let $M$ be the midpoint of $EF$, it suffices by symmetry to show that $CM \cap KF \in (\triangle ABC).$ Notice that $\triangle FHD \sim \triangle FEC$, and so hence it's clear that $\triangle FHK \sim \triangle MEC$. This implies that $\angle ECM = \angle HKF$ and so now the result is clear . $\square$

It was posted here.

For part (a), by Sharygin 2016/12, we know that it suffices to show $XY$ is the $A$-symmedian. This is just standard projective stuff. For part (b), we make the following claim: if $L$ is the midpoint of $EF$ then lines $BL$ and $KE$ meet on $\odot(O)$. Proving this will imply that $BS \cap CT=L$, which lies on the $A$-symmedian, so we would be done. Let $X$ be the spiral center for $EF \mapsto CB$. Then $\triangle XEL \sim \triangle XCM \sim \triangle XKB$ since $KXBC$ is harmonic. Hence $X$ is the spiral center for $EL \mapsto KB$, proving the claim.

Copying from here for storage. 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draw((-8.700395581102367,1.6266480077497738)--(-6.31182053170465,-0.01789907082801534), linewidth(2) + wrwrwr); draw((-8.69334964667858,-4.83304059113157)--(-6.605232996287319,-2.6227446999341564), linewidth(2) + wrwrwr); draw((-6.8454883720930155,-4.755674418604645)--(-8.69334964667858,-4.83304059113157), linewidth(2) + wrwrwr); draw((-5.778152691316285,4.7198762769486144)--(-7.837158267576747,-2.027355351689403), linewidth(2) + wrwrwr); draw((-7.837158267576747,-2.027355351689403)--(-8.69334964667858,-4.83304059113157), linewidth(2) + wrwrwr); draw(circle((-5.621754572459052,0.9843673611662406), 3.7387815169479595), linewidth(2) + wrwrwr); draw((-5.778152691316285,4.7198762769486144)--(-0.7644361470609574,-12.53120527816307), linewidth(2) + wrwrwr); draw((-5.778152691316285,4.7198762769486144)--(-5.30262885711987,-6.637827609050367), linewidth(2) + wrwrwr); /* dots and labels */ dot((-5.778152691316285,4.7198762769486144),dotstyle); label("$A$", (-5.668215771704715,4.9860494348393605), NE * labelscalefactor); dot((-6.8454883720930155,-4.755674418604645),dotstyle); label("$B$", (-6.742002630645836,-4.485301320948982), NE * labelscalefactor); dot((7.134232558139544,-4.17037209302325),dotstyle); label("$C$", (7.2447595319718525,-3.907108396903763), NE * labelscalefactor); dot((-0.012026025833968617,-0.7275143400315739),linewidth(4pt) + dotstyle); label("$O$", (0.08618047236437369,-0.5205498417817688), NE * labelscalefactor); dot((-5.465356453601818,-2.751141554616133),linewidth(4pt) + dotstyle); label("$H$", (-5.365352811490553,-2.530458577748481), NE * labelscalefactor); dot((-5.383992655360844,-4.69448458183325),linewidth(4pt) + dotstyle); label("$D$", (-5.282753822341236,-4.485301320948982), NE * labelscalefactor); dot((-2.0763306387369003,2.1711514028780616),linewidth(4pt) + dotstyle); label("$E$", (-1.9787942563685528,2.3979477748274296), NE * labelscalefactor); dot((-6.605232996287319,-2.6227446999341564),linewidth(4pt) + dotstyle); label("$F$", (-6.494205663197885,-2.3927935958329525), NE * labelscalefactor); dot((-5.30262885711987,-6.637827609050367),linewidth(4pt) + dotstyle); label("$K$", (-5.2001548331919185,-6.412611067766377), NE * labelscalefactor); dot((0.14437209302326437,-4.4630232558139475),linewidth(4pt) + dotstyle); label("$M$", (0.2513784506630078,-4.237504353501031), NE * labelscalefactor); dot((0.6780399334116298,0.27475209196268224),linewidth(4pt) + dotstyle); label("$N$", (0.7745053819420159,0.4981710243931401), NE * labelscalefactor); dot((-6.31182053170465,-0.01789907082801534),linewidth(4pt) + dotstyle); label("$P$", (-6.1913427029837225,0.19530806417897797), NE * labelscalefactor); dot((-3.9232454823069314,-1.6624461494058054),linewidth(4pt) + dotstyle); label("$X$", (-3.823505014036634,-1.429138722424255), NE * labelscalefactor); dot((-0.7644361470609574,-12.53120527816307),linewidth(4pt) + dotstyle); label("$Y$", (-0.6572104299794799,-12.304672293750984), NE * labelscalefactor); dot((-1.9601787228614225,-8.416921490483075),linewidth(4pt) + dotstyle); label("$Z$", (-1.8411292744530243,-8.202255832668243), NE * labelscalefactor); dot((-8.700395581102367,1.6266480077497738),linewidth(4pt) + dotstyle); label("$U$", (-8.393982413632179,1.599490879717366), NE * labelscalefactor); dot((-8.69334964667858,-4.83304059113157),linewidth(4pt) + dotstyle); label("$T$", (-9.109840319592927,-4.540367313715193), NE * labelscalefactor); dot((-7.837158267576747,-2.027355351689403),linewidth(4pt) + dotstyle); label("$G$", (-7.733190500437641,-1.814600671787734), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] (a) We make a very important structural claim first. Claim: $XY$ is the $A$-symmedian. Proof of Claim 1: It suffices to show that $AX$ is the isogonal of $AM$ in $\angle A$. To do this, let $U$ be on $MP$ such that $AU$ tangent to $(ABC)$. Then, we have \[\angle AUX=\angle(AU,AC)=\pi-\angle CAU=\angle ABC\]and \[\angle EXU = \angle(ED,MP)=\angle(ED,AC)=\angle CED=\angle ABC.\]Thus, $\angle AUX=\angle EXU$, so $AUXE$ is an iscoleces trapezoid. But we have that $AEDB$ cyclic with center $P$, so $PA=PE$, implying that $P$ is the midpoinit of $UX$. Thus, $(UX;P\infty_{AC})=-1$. Projecting through $A$, we get that \[(AU,AX;AB,AC)=-1,\]which is certainly enough to imply that $AX$ is the $\angle A$-symmedian. It similarly follows that $AY$ is the $\angle A$-symmedian, so we are done. $\blacksquare$ Now, we have that $Z$ is the harmonic conjugate of $A$ wrt $BC$ in $(ABC)$. Let $T=EF\cap BC$ and $G=AT\cap (ABC)$. It is well known that if we define an inversion $\phi$ at $T$ with power $TB\cdot TC$, then $\phi(G)=A$. Also, we have that $\phi$ swaps $(E,F)$. Now, we know $(BC;TD)=-1$, so projecting from $A$, we get $(GK;BC)=-1$. Inverting this statement, we get $(A\phi(K);CB)=-1$, which readiliy implies $\phi(K)=Z$. Thus, $\phi(K)=Z$ and $\phi(E)=F$, so $EFKZ$ cyclic, as desired. (b) We claim that they are concurrent on the midpoint of $EF$ which we'll denote $L$. Note that $AL$ is the $A$ symmedian of $ABC$ since $AEF$ and $ABC$ are inversely similar, so this in fact suffices. So it suffices to show that $T,L,C$ collinear in the following diagram: [asy][asy] unitsize(0.3inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); 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 = -13.84, xmax = 16.96, ymin = -10.51, ymax = 9.99; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-5.22,6.21)--(-7.28,-3.47), linewidth(2) + wrwrwr); draw((-7.28,-3.47)--(4.6,-2.97), linewidth(2) + wrwrwr); draw((4.6,-2.97)--(-5.22,6.21), linewidth(2) + wrwrwr); draw(circle((-1.4905531357814978,0.35714250616837606), 6.9400803133775195), linewidth(2) + wrwrwr); draw((-6.66349327180283,-0.573016927694852)--(-1.4902815641864529,2.723358936785299), linewidth(2) + wrwrwr); draw(circle((-3.2047234321092506,-0.29357125308418885), 3.4700401566887606), linewidth(2) + wrwrwr); draw((-5.22,6.21)--(-4.715017339545034,-5.788388012409966), linewidth(2) + wrwrwr); draw((-4.715017339545034,-5.788388012409966)--(-7.954668182394771,2.882995118708958), linewidth(2) + wrwrwr); draw((-6.66349327180283,-0.573016927694852)--(-4.918893728437006,-0.944285012336753), linewidth(2) + wrwrwr); draw((-4.918893728437006,-0.944285012336753)--(4.6,-2.97), linewidth(2) + wrwrwr); draw((-4.918893728437006,-0.944285012336753)--(-1.4902815641864529,2.723358936785299), linewidth(2) + wrwrwr); draw((-4.918893728437006,-0.944285012336753)--(-7.28,-3.47), linewidth(2) + wrwrwr); draw((-7.954668182394771,2.882995118708958)--(-4.076887417994643,1.0751710045452236), linewidth(2) + wrwrwr); draw((-4.076887417994643,1.0751710045452236)--(4.6,-2.97), linewidth(2) + wrwrwr); /* dots and labels */ dot((-5.22,6.21),dotstyle); label("$A$", (-5.14,6.41), NE * labelscalefactor); dot((-7.28,-3.47),dotstyle); label("$B$", (-7.2,-3.27), NE * labelscalefactor); dot((4.6,-2.97),dotstyle); label("$C$", (4.68,-2.77), NE * labelscalefactor); dot((-4.81695553399102,-3.36633651237336),linewidth(4pt) + dotstyle); label("$D$", (-4.74,-3.21), NE * labelscalefactor); dot((-1.4902815641864529,2.723358936785299),linewidth(4pt) + dotstyle); label("$E$", (-1.42,2.89), NE * labelscalefactor); dot((-6.66349327180283,-0.573016927694852),linewidth(4pt) + dotstyle); label("$F$", (-6.58,-0.41), NE * labelscalefactor); dot((-4.715017339545034,-5.788388012409966),linewidth(4pt) + dotstyle); label("$K$", (-4.64,-5.63), NE * labelscalefactor); dot((-7.954668182394771,2.882995118708958),linewidth(4pt) + dotstyle); label("$T$", (-7.88,3.05), NE * labelscalefactor); dot((-4.076887417994643,1.0751710045452236),linewidth(4pt) + dotstyle); label("$L$", (-4,1.23), NE * labelscalefactor); dot((-4.918893728437006,-0.944285012336753),linewidth(4pt) + dotstyle); label("$H$", (-4.84,-0.79), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We have $\angle FHD=\angle FEC=\pi-B$ and $\angle DFH=\angle CFE$, so $\triangle DHF\sim \triangle CEF$. But then by SAS, we have $\triangle KHF\sim\triangle CEL$, so $\angle FKH=\angle LCE$, so $\angle TKA=\angle LCA$, implying that $LC$ passes through $T$, as desired.

VMO 2019 P6 wrote: Given an acute triangle $ABC$ and $(O)$ be its circumcircle, and $H$ is its orthocenter. Let $M, N, P$ be midpoints of $BC, CA, AB$, respectively. $D, E, F$ are the feet of the altitudes from $A, B$ and $C$, respectively. Let $K$ symmetry with $H$ through $BC$. $DE$ intersects $MP$ at $X$, $DF$ intersects $MN$ at $Y$. a) $XY$ intersects smaller arc $BC$ of $(O)$ at $Z$. Prove that $K, Z, E, F$ are concyclic. b) $KE, KF$ intersect $(O)$ at $S, T$ ($S, T\ne K$), respectively. Prove that $BS, CT, XY$ are concurent. $(a)$ By Pascal's Theorem on $DFPMNE$ we get that $\overline{A-X-Y}$. So, redifine $Z$ as $AX\cap\odot(ABC)$. Claim:- $AZ$ is the $A-\text{Symmedian}$ of $\triangle ABC$. By Three Tangents Lemma we get that $MF=ME$. Hence, $MM\|EF$ WRT $\odot(X_5)$. So, by Pascal's Theorem on $DEFPMM$ we get that $BX\|FE$. Let $BX\cap AC=L$. Now as $MP\|AC$ we get that $BX=XL$. Hence, $AZ$ bisects $FE$. Now as $FE$ and $BC$ are antiparallel, we get that $AZ$ is the $A-\text{Symmedian}$ of $\triangle ABC$. Let $KZ\cap BC=X_A'$ also it's well known that $K\in\odot(ABC)$. Then $$-1=(AZ;BC)\overset{K}{=}(DX_A';BC)$$Now if $EF\cap BC=X_A$ then $(X_AD;BC)=-1\implies\boxed{X_A'\equiv X_A}$. Hence, $EF,BC,KZ$ are concurrent at a point $X_A$.So, $$X_AF\cdot X_AE=X_AB\cdot X_AC=X_AK\cdot X_AZ\implies E,F,K,X\text{ are concyclic}.\blacksquare$$ $(b)$ From ELMO Shortlist 2019 G1 we get that $CT$ bisects $EF$ and $BS$ bisects $EF$. Hence, $CT,BS,XY$ are concurrent at the midpoint of $XY$. $\blacksquare$

$\text{I think this one is just about angle chasing and some basic ratios; the problem would be interesting if it didn't have part a.}$

as neither a geo main nor exactly an anti-geo main, this would have been impossible without peeking at the "Muricaaaa" handout