Let D be the A-Humpty point , you can show that Y lies on (AEF) by angle chasing and using that M is the center of DXY , now it is well known that DKX=90 where K is instrection of A symmedian with ABC now let AK insrects AEF at Y' , you can see that Y'DKX is cyclic with center M those Y=Y' and with some easy angle chasing you can finish the problem

Let $Z$ be the intersection of $(BFS)$ and $(CES)$. Radical axises of $(BFS),(CES),(BCEF)$ concur hence $A,Z,S$ are collinear. Since $Y$ is unique, the two claims below are sufficient. Note that $AS$ is $A-$symmedian. Claim: $Q,Z,M,X$ are concyclic. Proof: Invert at $A$ with radius $\sqrt{AF.AB}$. Note that $Z$ swaps with $S$. $X^*$ lies on $EF$ and $M^*$ is $A-$humpty. $Q^*$ lies on $BC$ and $EF$. $AX^*$ is symmedian in $\triangle AEF$ since $EF$ and $BC$ are antiparallel and $M^*$ is the center of spiral homothety sending $BC$ to $EF$, thus $M^*EX^*F\sim M^*BSC$ which shows $\measuredangle M^*X^*Q^*=\measuredangle M^*X^*F=\measuredangle M^*SC$ so $Q^*,S,M^*,X^*$ are concyclic.$\square$ Claim: $MX=MZ$. Proof: Let $AS\cap (ABC)=L$ and $K$ be $A-$humpty. It's wellknown that $K$ is the reflection of $L$ with respect to $BC$ (which can be proved by cross ratio and using $EF\cap BC$). Also since $M$ is the midpoint of $BC$ and $LX\parallel BC$, we observe that $MK=ML=MX$. We have $AK.AM=AF.AB=AZ.AS$ thus, $S,Z,K,M$ are concyclic. \[\measuredangle AZK=\measuredangle SMA=\measuredangle LXA\]Hence $Z,K,L,X$ are concyclic. Since $M$ is the circumcenter of $(KLX)$, we conclude that $MZ=MK=ML=MX$ as desired.$\blacksquare$

This problem was proposed by me and none other than Rawlat_vanak Here is the modified solution we had submitted : Redefine $Y$ as the second intersection of $(BSF)$ and $(CSE)$. Note that $Y$ also lies on $(AEF)$ by Miquel. By radical axis on $(BSE)$, $(CSF)$, $(BEFC)$ we obtain $Y$ lies on $AS$. It is sufficient to show the following two claims. Claim 1 : $MX=MY$. Proof : Introduce the $A$-humpty point, call it $H_A$. Define the second intersection of $AS$ with $\omega$ as $K$. Observe that $AY\cdot AS=AE\cdot AC=AF\cdot AB=AH_A \cdot AM$ hence $YH_AMS$ is cyclic. Since $KX\parallel SM$, we infer that $YH_AKX$ is also cyclic. But since $MH_A=MK=MX$, we must have $MX=MY$. $\blacksquare$ Claim 2 : $QM$ bisects $\angle YQX$. Proof : Define $O$ to be the circumcenter of $ABC$ and $A'$ to be the $A$-antipode. It is well known that $Q,H,M,A'$ are collinear. \begin{align*} \angle YQM=\angle YQH&=\angle YAH=\angle SAH\\ &=\angle MAO=\angle XAA'\\ &=\angle XQA'=\angle XQM \hspace{1cm}\blacksquare \end{align*}

This is the solution I got while test-solving! Turns out participants who solved had similar solution. ( None of us thought about inversion). Great problem, Malay and Ketti Peri! Might have typos. We begin with finding nice properties and lemmas. Define $O$ to be circumcenter of $ABC$, $H$ be orthocentre of $ABC$ , $P=BB\cap CC$,$T=AP\cap (ABC)$. Lemma: $AOPX$ is cyclic Proof: This is true by PoP. Note that $BOCP$ is cyclic. So $$OM\cdot MP=BM\cdot MC=AM\cdot MX.$$ Lemma: $PM$ is angle bisector of $XPA$. Proof: Note that $(A,T;B,C)=-1\implies$ projecting from $A$, we get $TX||BC$. So $\widehat{BT}=\widehat{XC}$. Since $MP$ is perpendicular bisector of $BC$. It also implies $MP$ is perpendicular bisector of $TX$. So $$\angle TPM=\angle XPM\implies \angle APO=\angle XPO.$$ Now, define the following. $R=$ parallel line through $P$ to $BC\cap AX$. $D=$ parallel line through $P$ to $CB\cap (MXP)$. Note that $(A,X;M,R)=-1$ and hence $(A,S,T,P)=-1$. Re-Define $Q$ as $AD\cap (MXP)$. So $(Q,M;X,P)=-1$. But note that $$OM\cdot OP=OX^2\implies OQ^2=OM\cdot OP\implies Q\in (ABC).$$So by cyclicity, $\angle MQD=90\implies \angle MQA=90.$ Define $Y$ as $AT\cap (ABC)$. Note that $MX=MY$ by the angle conditions we got. Now, we have all the definitions of points satisfying the question's condition. Claim: $Y\in (AH)$ Proof: Consider circles $(AHQA_{HM}),(A_{HM}YTX)$ and $(QXYMP)$, where $A_{HM}$ is the humpty point. Note that $MA_{HM}=MX=MY=MT$. So $$\measuredangle QYP=\measuredangle QDP$$$$\measuredangle A_{HM}YP=\measuredangle A_{HM}XT=\measuredangle A_{HM}RP.$$So $\measuredangle QYA_{HM}=\measuredangle A$. Now, by miquel theorem applying on $S,E,F$, we get $BFSY, CESY$ to be cyclic. So done.

Many congratulations to proposers !! Another day with double inversion Consider $H$ as orthocenter of $\triangle ABC$ and $D$ as feet of perpendicular from $A$ to $BC$. Define $R = EF \cap BC$ and $H_a$ be $A$-humpty point. Lemma : If $N = AM \cap EF$ then $NS \perp BC$ We present two proof for this

Back to the main problem. We use $\sqrt{AH.AD}$ inversion from $A$, thus $$Q \leftrightarrow R, M \leftrightarrow H_a, X \leftrightarrow N$$Thus $(QMN) \leftrightarrow (RNH_a)$. Let feet of symmedian in inverted diagram be $G$. As $\angle RH_aN = 90 = \angle RGN$ we get $G \in (QMN)$. Let $G$ is imagine of point $G'$. As $\angle NSM = 90$ we also have $\angle XG'H_a = 90$.. But it's known that $MX=MH_a=MY$ hence $\angle XYH_a = 90$. Therefore $G' \equiv Y$. As $G \in BC \Rightarrow Y \in (AEF)$. Now $\angle AEF = \angle AYF = \angle ABC$ and $\angle AFE = \angle AYE = \angle ACB$ which give us $BFYS$ and $CEYS$ are cyclic.

Nice problem!

The main claim is $Z \equiv Y$ In order to prove this, we need: $\angle ZQM=\angle MQX$ and $ZMXQ$ cyclic. Part 1: $\angle ZQM=\angle MQX$ Proof: $\angle ZQM=\angle ZQH=\angle HAS=\angle XAA’=\angle XQM$. Note that we used $\overline{Q-H-M}$ collinear.

Part 2: $ZMQX$ cyclic. Note that because of the similarity between $\triangle AFE$ and $\triangle ACB$, we have $\angle AMZ=\angle APX$ so $ZMXP$ is cyclic. Since $QBH’C$ is harmonic, we have $\overline{Q-H’-P}$ collinear. Also, the similarity between $\triangle AFE$ and $\triangle ACB$ gives $\angle AMH =\angle APA’=\angle XPQ$ by isogonality and the above collinearity. Combining these two results, we are done. $\square$ Therefore $Y\equiv Z$ and we finish by angle chasing $\blacksquare$