Let $Z$ be the antipode of $A$ on $\omega$ on the Euler line (hence delete point $D$). We now apply complex numbers; with $k \in {\mathbb R}$ we have the relations \begin{align*} z &= k(a+b+c) \\ s &= \frac{1}{2}(a+z) \\ x &= \frac{1}{2}(a+b+z-ab\overline z) \\ y &= \frac{1}{2} (a+c+z-ac \overline z). \end{align*}We now determine the coordinates of the circumcenter $W$ of $\triangle XSY$. We know that in general if $u$ and $v$ lie on a circle centered at $0$, then the circumcenter is given by $\frac{uv}{u+v}$ (the midpoint of $0$ and $\frac{2uv}{u+v}$). So if we shift accordingly we get \[ w = s + \frac{(x-s)(y-s)}{x+y-2s}. \]Then \begin{align*} w - \frac{a+b+c}{2} &= \frac{z-b-c}{2} + \frac{(b-ab \overline z)(c-ac \overline z)} {b+c-ab \overline z - ac \overline z} \\ &= \frac{z-b-c}{2} + \frac{bc(1-a\overline z)}{2(b+c)} \\ &= \frac{(b+c)(z-b-c)+bc(1-a\overline z)}{2(b+c)} \\ &= \frac{(b+c)(k(a+b+c)-b-c)+bc-k(ab+bc+ca)}{2(b+c)} \\ &= \frac{k\left( b^2+c^2+bc \right) - (b^2+bc+c^2)}{2(b+c)} \\ &= (k-1) \cdot \frac{b^2+bc+c^2}{b+c} \end{align*}Thus \[ \frac{w - \frac{a+b+c}{2}}{b-c} = (k-1) \cdot \frac{b^2+bc+c^2}{b^2-c^2} \]is pure imaginary. So the line through $W$ and the nine-point center is perpendicular to $BC$ as desired. Remark: The only synthetic part is replacing the point $D$ with the antipode $Z$. After this the entire calculation is routine. There is a nice trick about a circumcenter here, but it is not strictly necessary; with enough pain the circumcenter formula will work equally well.

Note that the result is immediate when $S$ is the midpoint of $AH$ and when $S$ is the midpoint of $AO$. Move $S$ uniformly over the $A$-midline of $\triangle AOH$. By spiral similarity at $D$ we see that $X, Y$ move uniformly on lines $AB$ and $AC$. As $\triangle XLY$ has a fixed shape ($L$ is the circumcenter of $XSY$), we see that $L$ moves uniformly over a fixed line. Hence, this line is the perpendicular bisector of $PM$, as desired. $\blacksquare$

My solution (basically the idea is the same as above with more detail, but posting it anyways) : Note that $S$ is on the midline of $\triangle AHO$. If $T$ is the circumcenter of $XYS$, we know that $\angle XTY = 360^\circ -4A $ or $4A$. Also by a simple angle chase, we know that the configuration $XYTDS$ has a fixed shape. So as $S$ varies on a line, $T$ also varies on a line. For the point $S$ being the midpoints of $AH, AO$, we know that that line is perpendicular to the line $BC$ and passes through the nine point center. So by using the "functional relation" we just derived, the locus of $T$ is the perpendicular bisector of $PM$, as required. EDIT: The complex bash in post #2 can be a bit reduced if we directly notice the fact that $XYT$ has a fixed shape. Then by similarity using determinants, we have that if the displacement vector $TN$ is $x$ ($N$ is the nine point center of $ABC$), then $\frac{x}{\overline{x}}=bc$, so it is perpendicular to $BC$.

Let $\omega$ passes through $A,D$ cuts the Euler line at $X$ , then $AX$ is the diameter $P,N$ be the projections of $B,C$ on $CA,AB$ ; $U,V$ are the midpoints of $CA,AB$ We have $(APN) , (AUV) , (AXY)$ are concurrent $D$ . Let $O_3,O_1,S$ be the centers of these circles, then $\overline{O_1,S,O_3}$ because it's parrallel with $OH$ Let $I_1,I_2,I_3$ be the centers of $(VO_1U),(XSY),(PO_3N)$ , then we have $\triangle VO_1U \cap I_1 \sim \triangle XSY \cap I_2 \sim \triangle NO_3P \cap O_3$ so the rotation - homothetic centre $K$ transform ${O_1,S,O_3}$ to $\overline{I_1,I_2,I_3}$ That means $I_1,I_2,I_3$ are collinear. But $(I_1) \equiv (UO_1V)$ then $I_1 \in $ perpendicular bisector of $UV$ $I_3$ is the NPC so $I_3 \in$ perpendicular bisector of $ UV $ , it followed that the circumcenter of $\triangle XSY$ lies on the perpendicular bisector of $UV$ so it is equidistant from $P$ and $M$

I think the solutions those show that locus of the circumcenter is a straight line are incomplete. If A = 60. Then in both special cases we get the same point. So we can not conclude that this line is the perpendicular bisector of PM. In fact, in this case the circumcenter of XSY is always the nine point centre Nof ABC. As AH =AO so and HN =NO so AN bisects HAO hence XAY. And D=N. So NX=NY = 2*sinXAN *radius of w = 2*sin30 *SN =SN. So N is the circumcentre of XSY.

The problem is even more easier to bash if you set the triangle coordinates to be $a,b,\overline{b}$ (all in the unit circle), and apply the formula for circumcenter of $0,x,y$ is $\frac{xy(\overline{y}-\overline{x})}{x\overline{y}-y\overline{x}}$. As obviously $SX = SY$, $x\overline{x} = y \overline{y}$ (after shifting), so a lot of terms cancel and the bash ends very nicely. BTW I am a bit surprised at the position of this problem in the shortlist, especially because G4 looks lot harder than this.

Supravat wrote: I think the solutions those show that locus of the circumcenter is a straight line are incomplete. If A = 60. Then in both special cases we get the same point. So we can not conclude that this line is the perpendicular bisector of PM. In fact, in this case the circumcenter of XSY is always the nine point centre Nof ABC. As AH =AO so and HN =NO so AN bisects HAO hence XAY. And D=N. So NX=NY = 2*sinXAN *radius of w = 2*sin30 *SN =SN. So N is the circumcentre of XSY. I think that this case can be handled by continuity as well, as for angles $60^\circ - \epsilon$ and $60^\circ + \epsilon$ the statement is true, hence it is also true for $60^\circ$ from continuity.

Supravat wrote: I think the solutions those show that locus of the circumcenter is a straight line are incomplete. If A = 60. Then in both special cases we get the same point. So we can not conclude that this line is the perpendicular bisector of PM. In fact, in this case the circumcenter of XSY is always the nine point centre Nof ABC. As AH =AO so and HN =NO so AN bisects HAO hence XAY. And D=N. So NX=NY = 2*sinXAN *radius of w = 2*sin30 *SN =SN. So N is the circumcentre of XSY. I believed the $\angle A = 60^{\circ}$ case was problematic as well, but it is not. The circumcenter $O$ of $\triangle XSY$ moves with fixed velocity as $S$ varies with fixed velocity; it follows that if $O$ is ever on the same location twice, then $O$ is always fixed at this point.

My solution is the same as that of anantmudgal09

Lamp909 wrote: My solution is the same as that of anantmudgal09 Thank you for informing us.

Here's an alternative solution not based on showing that the circumcenter of $XSY$ varies linearly with $S$. [asy][asy] size(10cm); pair A=dir(70); pair B=dir(220); pair C=dir(-40); pair O=(0, 0); pair H=A+B+C; pair D = foot(A, O, H); pair D1=2*D-A; pair O1 = B+C; pair H1=2*foot(A, B, C)-H; pair E=circumcenter(B, O, C); draw(unitcircle); draw(A--B--C--cycle); draw(A--D1, orange+dashed); draw(circumcircle(O, H, D1), red); draw(D1--H1, dotted); draw(A--H1); draw(O--O1); draw(D1--O1, heavygreen); draw(D--2*H-O, heavygreen); string[] names = {"$A$", "$B$", "$C$", "$O$", "$H$", "$D$", "$D'$", "$O'$", "$H'$", "$E$"}; pair[] pts = {A, B, C, O, H, D, D1, O1, H1, E}; pair[] labels = {A, B, C, dir(90), dir(45), D, D1, O1, H1, dir(225)}; for(int i=0; i<names.length; ++i){ dot(names[i], pts[i], dir(labels[i])); } [/asy][/asy] Let $O$ and $H$ be the circumcenter and orthocenter of $\triangle ABC$. Let $E$ be the circumcenter of $\triangle BOC$, let $AH$ meet $(ABC)$ again at $H'$, let $O'$ be the reflection of $O$ across $BC$, and let $D'$ be the reflection of $A$ across $OH$. We first prove that $D', E, H'$ are collinear. Since $OHH'O'$ is an isosceles trapezoid, it is cyclic. $AOO'H$ is a parallelogram, so $D'OHO'$ is an isosceles trapezoid as well. Hence $D', O, H, H', O'$ all lie on a circle. The inverse of $O'$ about $(ABC)$ is $E$, so after inverting about $(ABC)$ we find that $D', E, H'$ are collinear. [asy][asy] size(10cm); pair A=dir(70); pair B=dir(220); pair C=dir(-40); pair O=(0, 0); pair H=A+B+C; pair Q=1.5*H-0.5*O; pair X1=2*foot(Q, A, B)-A; pair Y1=2*foot(Q, A, C)-A; pair D = foot(A, O, H); pair D1=2*D-A; pair P=foot(A, B, C); pair M=midpoint(B--C); pair H1=2*foot(A, B, C)-H; pair E=circumcenter(B, O, C); pair T=circumcenter(Q, X1, Y1); draw(unitcircle); draw(X1--A--Y1); draw(B--C); draw(D1--H1, dotted); draw(A--H1); draw(O--T, red); draw(circumcircle(Q, X1, Y1), orange); draw(circumcircle(A, X1, Y1), blue+dashed); string[] names = {"$A$", "$B$", "$C$", "$O$", "$D'$", "$H'$", "$E$", "$Q$", "$X_1$", "$Y_1$", "$T$", "$P$", "$M$"}; pair[] pts = {A, B, C, O, D1, H1, E, Q, X1, Y1, T, P, M}; pair[] labels = {A, B, C, dir(90), D1, H1, dir(225), dir(90), X1, Y1, T, P, dir(135)}; for(int i=0; i<names.length; ++i){ dot(names[i], pts[i], dir(labels[i])); } [/asy][/asy] Returning to the problem, take a homothety with ratio $2$ at $A$ which sends points $D, X, Y, S$ to $D', X_1, Y_1, Q$. This homothety also takes the perpendicular bisector of $PM$ to the perpendicular bisector of $BC$. Let $T$ be the circumcenter of $\triangle X_1QY_1$. Since $AD'X_1Y_1$ is cyclic, $D'$ is the center of the spiral similarity sending $\overline{X_1Y_1}$ to $\overline{BC}$, and this spiral similarity also sends $Q$ to $O$ and $T$ to $E$. As a result, \[\measuredangle D'ET = \measuredangle D'BX_1 = \measuredangle D'BA = \measuredangle D'H'A. \]Therefore $ET\parallel AH'$, so $ET\perp BC$. Since $E$ lies on the perpendicular bisector of $BC$, so does $T$, as desired. (Note that when $\angle A=60^{\circ}$, $D'=E$. Then $T=D'=E$ as well and the statement still holds.)

We firstly state some well known lemma that will be used later on. Let $\omega$ be a circle with center $O$ and let $\overline{AB}$. Let $O'$ be the reflection of $O$ in $\overline{AB}$ and let $X$ be the circumcenter of $\triangle{OAB}$. Then $X,O'$ are inverses of each other with respect to $\omega$. Proof: Notice that we have that $\overline{OX}=\frac{OA}{2\sin \theta}$ where $\theta=\angle OAB$ by sine rule in $\triangle{OAB}$ and also we have that $\overline{OO'}=2 \cdot OA \sin \theta$ and with this we are done . In a triangle $\triangle{ABC}$ let $H$ be the orthocenter and let $O$ be the circumcenter. Let $X$ be the midpoint of $\overline{AH}$ and let $M$ be the midpoint of $\overline{BC}$. Then $\overline{XM} \parallel \overline{AO}$. This can be easily bashed using complex numbers. [asy][asy] size(10cm); pair A=(8.227184338605278,8.917180830035385); pair B=(5.836037486558229,-5.602600636257744); pair C=(22.852265578227577,-5.038029693332168); pair O=(14.151548373898125,0.48476445751051883); pair H=(8.61239065559483,-2.6929784145755735); pair D=(13.333865353835646,0.015670324291781175); pair S=(14.5510849593405,6.629546963587926); pair X=(7.827568156749471,6.490588073520484); pair Y=(17.132485773397857,0.41976777226100737); pair P=(8.705768203458195,-5.507387609724802); pair M=(14.344151532392903,-5.320315164794956); pair T=(11.291157898982467,1.632964751029067); pair N9=(11.381969514746476,-1.1041069785325273); pair E=(8.419787497100053,3.1121012077299053); pair Q=(10.408968970806821,0.28080888219357325); pair G=(8.39272486321513,3.927771862972893); pair L=(12.480026965073659,3.455177922890749); draw(A--B--C--A,purple); draw(A--P,orange); draw(circumcircle(A,X,Y),red); draw(circumcircle(A,B,C),orange); draw(circumcircle(S,G,D),blue+dashed); draw(circumcircle(T,Q,D),cyan+dashed); draw(E--M,magenta); draw(O--H,green); draw(T--Q--N9--D--T,lightblue); draw(T--N9,lightblue); draw(Q--D,magenta); draw(S--A,green); draw(S--Y,green); draw(S--X,green); draw(X--Y,green); draw(G--D,lightblue); draw(S--T,red); draw(E--D,magenta+dotted); draw(A--O,orange); draw(A--D,cyan); draw(S--G,green); draw(S--D,green); dot("$A$",A,NW); dot("$B$",B,SW); dot("$C$",C,SE); dot("$D$",D,SE); dot("$M$",M,SE); dot("$H$",H,W); dot("$P$",P, SE); dot("$S$",S,NE); dot("$X$",X,NW); dot("$Y$",Y,SW); dot("$G$",G,W); dot("$E$",E,SW); dot("$T$",T,N); dot("$N_9$",N9,SE); dot("$Q$",Q,SW); dot("$L$",L,N); dot("$O$",O,NE); [/asy][/asy] We begin with a few claims. Claim 1: Define $G$ as the intersection of $\odot(AXY)$ with $\overline{AH}$ and define $Q$ as the intersection of $\overline{AH}$ with the perpendicular bisector of $\overline{XY}$. Then show that $(SQDEG)$ is cyclic where $E$ is the midpoint of $\overline{AH}$ Proof: For this firstly notice that \begin{align*} &\angle PEQ \\ &=\angle OAQ \\ &=\angle BAC-2 \angle XAG \\ &=\angle XSQ-\angle XSG \\ &=\angle GSQ \end{align*}and hence $(SGEQ)$ is cyclic. Now notice that $$\angle DEH=2 \angle DAH =\angle DSG $$and with this we are done $\qquad \square$ Claim 2: $Q$ is the reflection of $S$ in $\overline{XY}$. Proof: We present a computational proof. Notice that $$SQ=SG \cdot \frac{\sin \angle SGQ}{\sin \angle SQG}=SG \cdot \frac{\sin \angle SEQ}{\sin \angle SEG}=SG \cdot \frac{\sin \angle HOA}{\sin \angle AHO}=SG \cdot \frac{AH}{AO}=2\cdot SG \cdot \cos \angle BAC$$and notice that this is exactly twice the distance from $S$ to $\overline{XY}$ and we are done $\qquad \square$ Claim 3: $(TQDN_9)$ is cyclic where $N_9$ is the nine point center. Proof: Notice by the second lemma we have that $T,Q$ are inverses with respect to $\odot(AXY)$ and hence inverting gives that $T \in \overline{GD}$. Now notice that \begin{align*} & \angle SQE \\ &=\angle SGA \\ &=\angle SEA+ \angle GSE \\ &=\angle DHE+\angle GDE \\ &=\angle GDH \\ &=\angle TDN_9 \end{align*}and this finishes the proof $\qquad \square$. Now back to the main problem. Notice that by reims we have that $\overline{TN_9} \parallel \overline{EG}$ and with this we are done $\qquad \blacksquare$

v_Enhance wrote: We know that in general if $u$ and $v$ lie on a circle centered at $0$, then the circumcenter is given by $\frac{uv}{u+v}$ (the midpoint of $0$ and $\frac{2uv}{u+v}$). So if we shift accordingly we get \[ w = s + \frac{(x-s)(y-s)}{x+y-2s}. \] I don't understand this line. Can you explain more

Same for me. Can somebody also explain why you have the right to say z=k(a+b+c)?

KrysTalk wrote: v_Enhance wrote: We know that in general if $u$ and $v$ lie on a circle centered at $0$, then the circumcenter is given by $\frac{uv}{u+v}$ (the midpoint of $0$ and $\frac{2uv}{u+v}$). So if we shift accordingly we get \[ w = s + \frac{(x-s)(y-s)}{x+y-2s}. \] I don't understand this line. Can you explain more What don't you understand?

but why can you let z=k(a+b+c) (is it because z is on the euler line?) and why is uv/u+v the center of a circle passing through 0,u and v ?

Let $O$ and $H$ be the circumcenter and orthocenter of $\triangle ABC$. Let $AH$ and $AD$ meet $(ABC)$ again at $J$ and $E$ respectively. Let $I$ and $G$ be the circumcenter of $(XSY)$ and $(BOC)$ respectively. CLAIM 1. $J,G,E$ are collinear. Proof. Firstly notice that $$\measuredangle HJE=\measuredangle AJE=\measuredangle ACE=\measuredangle DOE=\measuredangle HOE$$Hence $E$ lies on $(HOJ)$. Let $K$ be the reflection of $O$ in $BC$, then since $H$ and $J$ are reflections of each other in $BC$. Hence $K$ also belongs to $(HOJ)$. Let $B_1$ be the reflection of $O$ over $B$. Let $B_2$ be the midpoint of $OB$. Then $\angle OKB_2=\angle OB_1G=90^{\circ}$. Hence $$OG\times OK=OB_1\times OB_2=OB^2=OE^2$$Since $OJ=OE$, $G$ lies on $JE$ by shooting lemma. $\blacksquare$ Let $G'$ be the reflection of $G$ in $I$. Let $F$ be the second intersection of $(AXY)$ and $(ABC)$. CLAIM 2. $G'$ lies on $AH$ Proof. Firstly notice that $$\angle XSY=2\angle XAY=2\angle BAC=\angle BOC$$Together with $SX=SY$ and $BO=OC$, $\triangle XSY\sim\triangle BOC$. Since $A,D,E$ are collinear, while $I$ and $K$ are corresponding elements in the two triangle, therefore by spiral similarity lemma, $F$ is the center of spiral sim. sending $DE$ to $IG$. Now $A$ and $G'$ are the reflections of $E$ in $D$ and $G$ in $I$ respectively. Therefore $$\triangle FG'A\sim\triangle FGE$$Hence $$\measuredangle G'AF=\measuredangle GEF=\measuredangle JEF=\measuredangle JAF$$which implies $G'$ lies on $AH$ as desired. Now $I$ is the midpoint of $G'$ and $G$. While $P$ and $M$ are the projection of $G"$ and $G$ on $BC$ respectively. This implies $I$ lies on the perpendicular bisector of $PM$ as desired.

Another finish with complex numbers: set without loss of generality that $\Re(m) = 0$, and by circumcenter formula the center $O$ of $\omega$ is expressed by \[o = \frac{1}{2} \left(a+e+\frac{bc(1-a\overline{e})}{b+c}\right) = \frac{1}{2} \cdot \frac{ab+bc+ca+k(b^2+bc+c^2)}{b+c}\]Notice that $b^2+bc+c^2$ is real, so $\Im(ab+bc+ca+k(b^2+bc+c^2))$ is constant. Since $b+c=2m$ is imaginary, we have $\Re(o)$ is a constant not depending on $k$. This means that the locus of $O$ is the line passing through the nine-point center perpendicular to $BC$. ($E = e = k(a+b+c)$ is the antipode of $A$ in $\omega$, which lies on Euler line)

Here are some problems related with perpendicularity from vertex to the Euler Line https://artofproblemsolving.com/community/c6h2237565p17170407 (Lemma) https://artofproblemsolving.com/community/c6h1888731p12879517 https://artofproblemsolving.com/community/c6h2226600p16967110

Another finish with complex numbers Throw the configuration onto the complex plane, so that we have that $h$ is a real number in other words we have that $a+b+c \in \mathbb{R}$. Then we can express $s$ as the following $s=k(a+b+c)+iq$, where $k$ is an arbitrary real number and we shall determine $q$ later on. First we calculate $d$. Since we know that $h$ is a real number then we have that $d$ is also a real number, since the Euler line is the real number line. Since we have that $AD \perp OH$, then we must have that: \[\frac{a-d}{\overline{a-d}}=-\frac{h}{h}=-1\]this implies that $d=\frac{1}{2}\left(a+\frac{1}{a}\right)$. Denote with $T$ the midpoint of $AD$, since we have that $ST \parallel OH$, we must have that: $$\frac{s-t}{\overline{s-t}}=\frac{0-h}{\overline{0-h}}$$where $t=\frac{a+d}{2}$, pluging all of this we get that: $$2iq=\frac{a-\frac{1}{a}}{2}$$this implies that $q=\frac{a-\frac{1}{a}}{4i}$, thus giving us that: $$s=k(a+b+c)+\frac{a-\frac{1}{a}}{4}$$Now we calculate $x$ and $y$. We have that $\overline{x}=\frac{a+b-x}{ab}$, and since we have that $\mid s-x\mid=\mid s-a \mid$, we must have that $(s-a)\overline{(s-a)}=(s-x)\overline{(s-x)}$, plugging this in we get the following quadratic: $$-x^2+(a+b+s-ab\overline{s})x+a^2b\overline{s}-ab-as=0$$let $z=b+s-ab\overline{s}$, then the equation has turned into: $$-x^2+(a+z)x-az=0$$this implies that: $$x_{1,2}=\frac{a+z \mp a-z}{2}$$giving us that $x=z=b+s-ab\overline{s}$. Similarly we have that $y=c+s-ac\overline{s}$. Denote with $G$ the center of $(SXY)$, then we have that: \[ g=\frac{\begin{vmatrix} x & x\overline{x} & 1 \\ s & s\overline{s} & 1 \\ y & y\overline{y} & 1 \end{vmatrix}} {\begin{vmatrix} x & \overline{x} & 1 \\ s & \overline{s} & 1 \\ y & \overline{y} & 1 \end{vmatrix}} \]Calculating this(calculation done in 30 min), we get that $\mid g-p\mid = \mid g- m\mid$, where $p=\frac{1}{2}\left(a+b+c-bc\overline{a}\right)$ and $m=\frac{1}{2}\left(b+c\right)$. This implies that $GM=GP$, which is what we needed to prove.

perhaps we could use laplace vectors/matrixes? I am not that good at advanced math, this is just my guess but i think laplace vectors or matrixes would help. Sorry if my concepts are unclear, as I said I am not that great at math

First let's switch the labels of $P$ and $D$. Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$ and let its circumcenter be $N_9$. Let $Q$ be the circumcenter of $\triangle XSY$ and let $R$ be the midpoint of $AH$. Notice that $P$ also lies on the circle with diameter $AH$. Since $\measuredangle PFX=\measuredangle PEY$ and $\measuredangle PXF=\measuredangle PYE$, it follows that $P$ is the center of spiral similarity which maps segment $FX$ to segment $EY$. Thus it also maps segment $FE$ to segment $XY$. Claim: $\triangle FN_9E\stackrel{+}{\sim}\triangle XQY$. Proof: Angle chasing gives us \[\measuredangle QXY=90^\circ-\measuredangle YSX=90^\circ-\measuredangle ERF=\measuredangle N_9FE.\]and similarly we have $\measuredangle QYX=\measuredangle N_9EF$. Therefore, the spiral similarity mentioned before maps $Q$ to $N_9$. But this means that \[\measuredangle PN_9Q=\measuredangle PFX=\measuredangle PFA=\measuredangle PHA.\]and so $N_9Q$ is parallel to $AD$. Since $N_9$ is the midpoint of $AD$ and $AD,OM\perp DM$, the line $N_9Q$ is precisely the perpendicular bisector of $DM$ and the result follows.

v_Enhance wrote: We now determine the coordinates of the circumcenter $W$ of $\triangle XSY$. We know that in general if $u$ and $v$ lie on a circle centered at $0$, then the circumcenter is given by $\frac{uv}{u+v}$ (the midpoint of $0$ and $\frac{2uv}{u+v}$). So if we shift accordingly we get \[ w = s + \frac{(x-s)(y-s)}{x+y-2s}. \] $\frac{2uv}{u+v}$, this formula is applicable for $u,v$ on the unit circle. But $x,y$ are not on the unit circle. How $w$? I didn't understand the shifting part.

Let's imagine a world without moving points. *shivers* Ok, let's stop doing that. Notice $XY$ are pedal points of a point on the Euler line. By properties of continuous spiral similarity centered at $D$, it suffices to prove the claim for two points on the Euler line. For $O$, the midpoint of the midpoints, by 1/2 homothety at $A$, is equidistant from $P$ and $M$. For $H$, the $XSY$ is the nine-point circle, which is the midpoint of $OH$ so projecting onto $BC$, it is equidistant from $P$ and $M$. Since the locus is the perpendicular bisector of $PM$, we are done.

Denote by $O,H$ circumcenter and orthocenter and let $Q=AS\cap OH\in \omega.$ Move $Q$ on $OH$ with degree $1:$ $XSY$ and it's circumcenter move with the same degree. When $Q=O$ points $X,Y$ are midpoints of $AB,AC,$ so $XYMP$ form isosceles trapezoid. When $Q=H$ circumcenter of $XSY$ is the nine-point center. In both cases this circumcenter lies on perpendicular bisector of $PM,$ so it does for every $Q.$

Let $OH$ be the Euler line, with $E = AS \cap OH$. Note that $S$ is the midpoint of $AE$, and let $R$ be the midpoint of $OE$, $T$ the midpoint of $AO$. Let $O_1$ be the circumcircle of $(SXY)$. Let $O_1'$ be the point on $SO_1$ such that $TO_1' \; || \; AH$. Let $\Omega$ be the 9-point circle of $\triangle AOE$, through $R, S, T$ and $D$, the foot of the altitude from $A$ onto $OE$. Note that $ST \; || \; EO$. Claim 1: $O_1'$ lies on $\Omega$. Proof: Let $Z = O_1'T \cap AE$. $\measuredangle TO_1'S = \measuredangle ZO_1'S = \measuredangle ZSO_1' + \measuredangle O_1'ZS = \measuredangle ESO_1' + \measuredangle HAE = \measuredangle ESY + \measuredangle YSO_1 + \measuredangle HAE = \measuredangle EAB + \measuredangle EAC + \measuredangle HAE = \measuredangle HAB + \measuredangle EAC = \measuredangle CAO + \measuredangle EAC = \measuredangle EAO = \measuredangle SAT = \measuredangle TRS$, where $\measuredangle SAT = \measuredangle TRS$ is due to $\triangle RST \sim \triangle ABC$. Hence due to angles in a circumcircle, $O_1'$ lies on $\Omega$. $\square$ Claim 2: $O_1' = O_1$.

Proof: We prove $SO_1 = SO_1'$, since $O_1, O_1'$ lie on the same side of $AE$. Remark

Let $r$ be the radius of $\Omega$. By the sine rule, \begin{align*} 2r = \frac{SO_1'}{\sin(\angle SQO_1')} = \frac{SO_1'}{\sin(\angle AHO)}. \end{align*}The radius of $\Omega$ is half the radius of $(AOE)$, and since $AE = 2 \cdot AS$, \begin{align*} 2 \cdot 2r = \frac{AE}{\sin(\angle AOE)} \Rightarrow \frac{AS}{\sin(\angle AOE)} = 2r. \end{align*}Thus \begin{align*} \frac{SO_1'}{SA} = \frac{\sin(\angle AHO)}{\sin(\angle HOA)} = \frac{AO}{AH} = \frac{AO}{AC} \cdot \frac{AC}{AF} \cdot \frac{AF}{AH} = \frac{2}{\sin{\hat{B}}} \cdot \cos{\hat{A}} \cdot \sin{\hat{B}} = 2 \cos{\hat{A}}. \end{align*}But we also have \begin{align*} \frac{SY}{\sin(\angle SXY)} = \frac{SY}{\sin(90 - \hat{A})} = 2SO_1 \Rightarrow \frac{SO_1}{SA} = 2\cos{\hat{A}}. \square \end{align*} The perpendicular from $T$ to $BC$ bisectors $PM$, hence so does that of $O_1$ since $TO_1 \perp BC$. Thus, $O_1$ lies on the perpendicular bisector of $PM$. $\blacksquare$

[asy][asy] size(7cm); defaultpen(fontsize(10pt)); pair A, B, C; A = dir(110); B = dir(-162); C = dir(-18); pair O, H, D, P, M; O = circumcenter(A, B, C); H = orthocenter(A, B, C); D = foot(A, H, O); P = foot(A, B, C); M = (B + C)/2; pair K, S, X, Y, N, S1; K = 0.57 * O + 0.43 * H; S = (A + K)/2; X = foot(K, A, B); Y = foot(K, A, C); N = (X + Y)/2; S1 = 2 * circumcenter(S, X, Y) - S; draw(A--B--C--A--P); draw(circumcircle(A, B, C)); draw(K--A, deepgreen+dashed); draw(X--K--Y, deepgreen); draw(circumcircle(A, D, K), red); draw(A--D, blue); draw((4.1*O-3.1*H)--(3.1*H-2.1*O), blue); draw(X--Y, purple); draw(S--S1, purple+dotted); draw(circumcircle(S, X, Y), purple+dotted); dot("$A$", A, dir(110)); dot("$B$", B, dir(-162)); dot("$C$", C, dir(-18)); dot("$O$", O, dir(80)); dot("$H$", H, dir(190)); dot("$D$", D, dir(200)); dot("$P$", P, dir(260)); dot("$M$", M, dir(260)); dot("$K$", K, dir(-100)); dot("$S$", S, dir(60)); dot("$X$", X, dir(185)); dot("$Y$", Y, dir(5)); dot("$N$", N, dir(-95)); dot("$S'$", S1, dir(-95)); [/asy][/asy] Define point $K$ to be the dilation of $S$ by a factor of $2$ about $A$. Here, $\omega = (AK)$ and $K$ is allowed to vary linearly on the Euler line. Let $l$ be the perpendicular bisector of $AD$. Note that as $K$ varies linearly, $S$, $X$, and $Y$ vary linearly on $l$, $AB$, and $AC$, respectively. Hence, the midpoint $N$ of $XY$ also varies linearly. Let $S'$ be the antipode of $S$ in $(SXY)$. Note that \[\frac{SN}{NS'} = \left(\frac{SN}{NY}\right)^2 = (\tan{\angle NSY})^2 = \tan^2{A},\]which is constant. Since $N$ and $S$ both vary linearly, so does $S'$. It follows that the center of $(XSY)$, midpoint of $SS'$, also varies linearly. Thus, it suffices to prove the problem statement for two such points $K$ on the Euler line. We choose $H$ and $O$. If $K = H$, $S$ is the midpoint of $AH$, and $X$, $Y$ are the feet of the altitudes from $B$ and $C$. The circle $(XSY)$ in question is therefore the nine-point circle, whose center lies on the perpendicular bisector of $MP$ . If $K = O$, we find $S$ to be the midpoint of $AO$, and $X$ and $Y$ to be the midpoints of $AB$ and $AC$, respectively. The circumcenter of $(SXY)$ then lies on $SN$, which is the perpendicular bisector of $MP$. This completes our proof.

Easy and nice problem. Let $T$ denote the antipode of $A$ in $(AXY)$, then we have the following cases: $\text{CASE 1:} \quad T \equiv O$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -4.36471658380817, xmax = 14.498175842200546, ymin = -5.39223726459042, ymax = 4.219781556321238; /* image dimensions */ pen zzttff = rgb(0.6,0.2,1); draw((3.12,3.61)--(0.46,-3.87)--(11.44,-3.87)--cycle, linewidth(0.4) + zzttff); /* draw figures */ draw((3.12,3.61)--(0.46,-3.87), linewidth(0.4) + zzttff); draw((0.46,-3.87)--(11.44,-3.87), linewidth(0.4) + zzttff); draw((11.44,-3.87)--(3.12,3.61), linewidth(0.4) + zzttff); draw(circle((4.535,1.0003208556149736), 2.968610826066321), linewidth(0.8) + blue); draw((3.12,3.61)--(5.95,-1.6093582887700522), linewidth(0.8)); draw((5.95,-1.6093582887700522)--(5.95,-3.87), linewidth(0.8)); draw((3.12,3.61)--(3.12,-3.87), linewidth(0.8)); draw((1.79,-0.13)--(4.535,1.0003208556149739), linewidth(0.8)); draw((4.535,1.0003208556149739)--(7.28,-0.13), linewidth(0.8)); draw((1.79,-0.13)--(7.28,-0.13), linewidth(0.8)); /* dots and labels */ dot((3.12,3.61),dotstyle); label("$A$", (3.177662346473665,3.7475147356406078), NE * labelscalefactor); dot((0.46,-3.87),dotstyle); label("$B$", (0.5107438296889277,-3.725413191599959), NE * labelscalefactor); dot((11.44,-3.87),dotstyle); label("$C$", (11.497892510817715,-3.725413191599959), NE * labelscalefactor); dot((5.95,-1.6093582887700522),linewidth(4pt) + dotstyle); label("$O$", (6.039043671773957,-1.7668949058361674), NE * labelscalefactor); dot((3.12,-0.9112834224598962),linewidth(4pt) + dotstyle); label("$H$", (3.177662346473665,-0.7945808632583985), NE * labelscalefactor); dot((2.0687040121969074,-0.6519606994429188),linewidth(4pt) + dotstyle); label("$D$", (2.1220071002463734,-0.5445572523098293), NE * labelscalefactor); dot((1.79,-0.13),linewidth(4pt) + dotstyle); label("$X$", (1.8442030880812965,-0.016729629196183367), NE * labelscalefactor); dot((7.28,-0.13),linewidth(4pt) + dotstyle); label("$Y$", (7.330832328341564,-0.016729629196183367), NE * labelscalefactor); dot((4.535,1.0003208556149739),linewidth(4pt) + dotstyle); label("$S$", (4.594462808515557,1.1083766200723777), NE * labelscalefactor); dot((3.12,-3.87),linewidth(4pt) + dotstyle); label("$P$", (3.177662346473665,-3.7531935928164666), NE * labelscalefactor); dot((5.95,-3.87),linewidth(4pt) + dotstyle); label("$M$", (6.011263270557449,-3.7531935928164666), NE * labelscalefactor); dot((4.535,-2.8979750100503),linewidth(4pt) + dotstyle); label("$O''$", (4.594462808515557,-2.780879550238698), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $O''$ be the circumcenter of $\triangle XSY$. Just notice that perpendicular bisectors of $PM$ and $XY$ are the same to get that $O''$ lies on perpendicular bisector of $PM$. $\text{CASE 2:} \quad T \equiv H$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -1.7419188729163546, xmax = 12.001886155190885, ymin = -4.429974238189027, ymax = 2.573496365235269; /* image dimensions */ pen zzttff = rgb(0.6,0.2,1); pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0); draw((2.78,2.05)--(0.76,-3.09)--(10.56,-3.09)--cycle, linewidth(0.8) + zzttff); /* draw figures */ draw((2.78,2.05)--(0.76,-3.09), linewidth(0.8) + zzttff); draw((0.76,-3.09)--(10.56,-3.09), linewidth(0.8) + zzttff); draw((10.56,-3.09)--(2.78,2.05), linewidth(0.8) + zzttff); draw(circle((2.78,1.0087548638132295), 1.0412451361867703), linewidth(0.8) + qqwuqq); draw(circle((4.22,-1.0406225680933903), 2.504705144005615), linewidth(0.8) + ccqqqq); /* dots and labels */ dot((2.78,2.05),dotstyle); label("$A$", (2.8224817660205854,2.148430230345355), NE * labelscalefactor); dot((0.76,-3.09),dotstyle); label("$B$", (0.7983573141638491,-2.9928458773707467), NE * labelscalefactor); dot((10.56,-3.09),dotstyle); label("$C$", (10.605240283409735,-2.9928458773707467), NE * labelscalefactor); dot((5.66,-2.0487548638132296),linewidth(4pt) + dotstyle); label("$O$", (5.696738487657151,-1.9706630291830964), NE * labelscalefactor); dot((2.78,-0.03249027237354091),linewidth(4pt) + dotstyle); label("$H$", (2.8224817660205854,0.05346142267363664), NE * labelscalefactor); dot((1.8016049241530079,0.6524761678588333),linewidth(4pt) + dotstyle); label("$D$", (1.8407814068700683,0.7315431140456422), NE * labelscalefactor); dot((2.78,1.0087548638132295),linewidth(4pt) + dotstyle); label("$S$", (2.8224817660205854,1.0857648931205706), NE * labelscalefactor); dot((2.071079344262295,0.24611278688524568),linewidth(4pt) + dotstyle); label("$X$", (2.1140382078707276,0.3267182236742956), NE * labelscalefactor); dot((3.7377807425127645,1.4172245480057057),linewidth(4pt) + dotstyle); label("$Y$", (3.7738202583932514,1.5007104057512006), NE * labelscalefactor); dot((4.22,-1.0406225680933852),linewidth(4pt) + dotstyle); label("$N_{9}$", (4.259610126838869,-0.9586008032547298), NE * labelscalefactor); dot((2.78,-3.09),linewidth(4pt) + dotstyle); label("$P$", (2.8224817660205854,-3.013087121889314), NE * labelscalefactor); dot((5.66,-3.09),linewidth(4pt) + dotstyle); label("$M$", (5.696738487657151,-3.013087121889314), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] all lie on nine point circle $\text{CASE 3 :} \quad T \not\equiv H,O$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -9.249767422915646, xmax = 8.321611580789027, ymin = -3.820030982745853, ymax = 5.133867596461554; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen zzttff = rgb(0.6,0.2,1); draw((-1.6544806223452169,4.69393616337911)--(-5.5089610303002345,-2.345013343607709)--(5.06,-1.64)--cycle, linewidth(2) + zzttqq); draw((-2.353213494872263,3.417928793244585)--(-1.0334166145108656,2.758924130777999)--(-1.1611468347842868,4.228562161516616)--cycle, linewidth(0.8) + zzttff); draw((-2.7738546959360164,2.6497650301767286)--(-0.8934889087838629,0.6612468565858854)--(-0.18054456324227472,3.303535709144152)--cycle, linewidth(0.8) + zzttff); draw((-3.5817208263227256,1.1744614098857005)--(-0.624749532151955,-3.367465542523285)--(1.7027596888273915,1.5269680816895552)--cycle, linewidth(0.8) + zzttff); /* draw figures */ draw((-1.6544806223452169,4.69393616337911)--(-5.5089610303002345,-2.345013343607709), linewidth(0.4) + zzttqq); draw((-5.5089610303002345,-2.345013343607709)--(5.06,-1.64), linewidth(0.4) + zzttqq); draw((5.06,-1.64)--(-1.6544806223452169,4.69393616337911), linewidth(0.4) + zzttqq); draw(circle((-1.8629437636790365,3.978774801703462), 0.744924596538825), linewidth(0.4) + linetype("4 4")); draw((-2.353213494872263,3.417928793244585)--(-1.0334166145108656,2.758924130777999), linewidth(0.8) + zzttff); draw((-1.0334166145108656,2.758924130777999)--(-1.1611468347842868,4.228562161516616), linewidth(0.8) + zzttff); draw((-1.1611468347842868,4.228562161516616)--(-2.353213494872263,3.417928793244585), linewidth(0.8) + zzttff); draw((-2.7738546959360164,2.6497650301767286)--(-0.8934889087838629,0.6612468565858854), linewidth(0.8) + zzttff); draw((-0.8934889087838629,0.6612468565858854)--(-0.18054456324227472,3.303535709144152), linewidth(0.8) + zzttff); draw((-0.18054456324227472,3.303535709144152)--(-2.7738546959360164,2.6497650301767286), linewidth(0.8) + zzttff); draw((-3.5817208263227256,1.1744614098857005)--(-0.624749532151955,-3.367465542523285), linewidth(0.8) + zzttff); draw((-0.624749532151955,-3.367465542523285)--(1.7027596888273915,1.5269680816895552), linewidth(0.8) + zzttff); draw((1.7027596888273915,1.5269680816895552)--(-3.5817208263227256,1.1744614098857005), linewidth(0.8) + zzttff); draw((-1.0334166145108656,2.758924130777999)--(-0.624749532151955,-3.367465542523285), linewidth(0.4) + linetype("4 4")); /* dots and labels */ dot((-1.6544806223452169,4.69393616337911),dotstyle); label("$A$", (-1.6027239831590467,4.823327761344535), NE * labelscalefactor); dot((-5.5089610303002345,-2.345013343607709),dotstyle); label("$B$", (-5.678559319069924,-2.6296282814639436), NE * labelscalefactor); dot((5.06,-1.64),dotstyle); label("$C$", (5.112699951246494,-1.5168605389612888), NE * labelscalefactor); dot((-0.3164638350777267,-0.613570893400371),linewidth(4pt) + dotstyle); label("$O$", (-0.2699905241151726,-0.5076060748309741), NE * labelscalefactor); dot((-1.470513982489999,1.9360646065721419),linewidth(4pt) + dotstyle); label("$H$", (-1.4215757460074523,2.041408405087898), NE * labelscalefactor); dot((-2.5378389363536256,4.294098699353272),linewidth(4pt) + dotstyle); label("$D$", (-2.4825868493239347,4.396335488058632), NE * labelscalefactor); dot((-1.8629437636790365,3.978774801703462),dotstyle); label("$S$", (-1.8097505399037264,4.111673972534698), NE * labelscalefactor); dot((-2.353213494872263,3.417928793244585),linewidth(4pt) + dotstyle); label("$X$", (-2.30143861217234,3.5164726218937425), NE * labelscalefactor); dot((-1.1611468347842868,4.228562161516616),linewidth(4pt) + dotstyle); label("$Y$", (-1.111035910890433,4.33163968907592), NE * labelscalefactor); dot((-1.2040956347666916,-2.0578528992097027),linewidth(4pt) + dotstyle); label("$P$", (-1.1498533902800605,-1.9567919720437337), NE * labelscalefactor); dot((-0.22448051515011747,-1.9925066718038544),linewidth(4pt) + dotstyle); label("$M$", (-0.16647724574283285,-1.8920961730610213), NE * labelscalefactor); dot((-1.0334166145108656,2.758924130777999),linewidth(4pt) + dotstyle); label("$E$", (-0.9816443129250083,2.8565754722700754), NE * labelscalefactor); dot((-0.8934889087838629,0.6612468565858854),linewidth(4pt) + dotstyle); label("$N_9$", (-0.8393135551630412,0.7604315852301906), NE * labelscalefactor); dot((-2.071406905012856,3.263613440027813),linewidth(4pt) + dotstyle); label("$T$", (-2.0167770966484055,3.361202704335233), NE * labelscalefactor); dot((-2.7738546959360164,2.6497650301767286),linewidth(4pt) + dotstyle); label("$F$", (-2.7284308854582413,2.753062193897735), NE * labelscalefactor); dot((-0.18054456324227472,3.303535709144152),linewidth(4pt) + dotstyle); label("$G$", (-0.12765976635320544,3.4129593435214027), NE * labelscalefactor); dot((-0.7142880749584045,-2.0251797855067784),linewidth(4pt) + dotstyle); label("$I$", (-0.6581653180114466,-1.9179744926541062), NE * labelscalefactor); dot((-3.5817208263227256,1.1744614098857005),linewidth(4pt) + dotstyle); label("$Q$", (-3.5306587928438744,1.2779979770918906), NE * labelscalefactor); dot((1.7027596888273915,1.5269680816895552),linewidth(4pt) + dotstyle); label("$N$", (1.7485184041454525,1.627355291598538), NE * labelscalefactor); dot((-0.624749532151955,-3.367465542523285),linewidth(4pt) + dotstyle); label("$O''$", (-0.5675911994356494,-3.263647111494526), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] $F$ and $G$ denote the feet of perpendiculars from $B$ and $C$ to opposite side. Let $Q$ and $N$ be midpoints of $AB$ and $AC$.$E$ is circumcenter of $\triangle XSY$. Notice $\measuredangle XEY = \measuredangle FN_9G= \measuredangle QO''N$ and $\frac{XE}{EY}=\frac{FN_9}{N_9G}=\frac{FO''}{O''N}=1$. By gliding principle we're done.

Complex numbers is most definitely the nicest way to go about this ... solution from the OTIS walkthrough. We will instead define $z = k(a+b+c)$ to be the $A$-antipode in $(AXY)$; the condition implies that $Z$ lies on $\overline{OH}$, hence $k$ is real. Hence, we have the following formulas: \begin{align*} z &= k(a+b+c) \\ s &= \frac 12(a+z) \\ x &= \frac 12(a+b+z-ab\overline z) \\ y &= \frac 12(a+c+z-ac\overline z) w = s + \frac{(x-s)(y-s)}{x+y-2s}. \end{align*}Here, the last line follows from the shifted circumcenter formula, observing that $x, y$ lie on a circle centered at $s$. It then suffices to show that $\overline{WN_9} \perp \overline{BC}$, i.e. $\frac{w-\frac{a+b+c}2}{b-c}$ is pure imaginary. The rest is pure computation: \begin{align*} w - \frac{a+b+c}2 &= s+\frac{(b-ab\overline z)(c-zc\overline z)}{2(b+c-ab\overline z - ac\overline z)} -\frac{a+b+c}2\\ &= \frac 12\left(a+z+\frac{bc(1-a\overline z)}{b+c}\right) \\ &= \frac z2 - \frac b2 - \frac c2 + \frac{bc(1-a\overline z)}{2(b+c)} \\ &= \frac{b^2k+c^2k+bck-b^2-c^2-2bc+bc}{b+c} \\ &= \frac{(k-1)(b^2+c^2+bc)}{b+c}. \end{align*}This is a constant multiple of $\frac{b^2+c^2+bc}{b+c}$, and note that $\frac{b^2+c^2+bc}{b^2-c^2}$ clearly equals its own conjugate. This finishes the problem.

We will use moving points. Let $W$ be the circumcenter of $\triangle XSY$. Fix $\triangle ABC$(thus fixing $D$) and move $X$ on line $\overline{AB}$. Then since $D$ and $A$ are fixed and $Y$ lies on line $\overline{AC}$ and on circle $(ADX)$ we have the map of moving points $X \to Y$ is linear. Since $\angle BOC = 2\angle A = \angle XSY$, so $\triangle XSY \sim \triangle BOC$ is fixed, which implies that $X \to W$ is a linear map. We now check whether the problem is true for $deg(W) + 1 = 1 + 1 = 2$ cases. If $X$ is the projection of $H$ onto $AB$ we have that $\angle HXA = 90^\circ = \angle HSA \implies S$ is the midpoint of $HA$. It follows that $W$ is the center of the nine-point circle which passes through $P$ and $M$. If $X$ is the projection of $O$ onto $AB$, similarly we have $S$ being the midpoint of $AO$. Then note that $S$ lies on the perpendicular bisector of $PM$ by homothety since $OM \parallel AP$. Clearly the perpendicular bisector of $PM$ also perpendicularly bisects $XY$, so it passes through $W$ as desired.