Denote the circumcenter of $APO$ as $Q$. Let $Q'$ be the intersection of $XY$ and the perpendicular bisector of $OP$. The perpendicular bisector of $OP$ intersect $BC$ at $T$. By appling the butterfly thm. , then we can find out the fact that $MQ'=MT$, and now our goal is to prove the circumcenter of $A'PO$ lies on $BC$, where $A'$ is the symmetric point of $A$ wrt $OP$. I prove this by bashing with some complex number, and this can be done in 30 minutes. Although I finished this problem using less than 90 minutes during the test, I still didn't know how to do the other problems.

Lemma (well-known) : Given a $ \triangle ABC $ and a point $ K $ such that $ \measuredangle ABK = \measuredangle KCA. $ If $ L $ is the reflection of $ K $ in the midpoint of $ BC, $ then $ AK $ and $ AL $ are isogonal WRT $ \angle A. $ Back to the main problem : Let $ T $ be the reflection of $ O $ in $ BC $ and $ U $ be the intersection of $ BC $ with the perpendicular bisector of $ OP, $ then $ U $ is the center of $ \odot (OPT) $ and $ A, T $ are symmetric WRT the midpoint of $ OH, $ so by Lemma we get $$ \measuredangle OUP = 2\measuredangle OTP = 2\measuredangle PAO. $$i.e. $ U $ and the circumcenter $ V $ of $ \triangle APO $ are symmetric WRT $ OP. $ Finally, from Butterfly theorem for $ BCYX $ and the line $ UV $ we conclude that $ V $ lies on $ XY. $ $ \qquad \blacksquare $ Remark : This problem is a generalization of the Lemma in NPC center is orthocenter (post #2).

Here is my solution during the contest. Let the perpendicular bisector of $OP$ meets $XY$ and $BC$ at $U, V$. By Butterfly theorem, $MU = MV$. That means we have to show that $V$ is the circumcenter of $\triangle A'PO$, where $A'$ is the reflection of $A$ in $OP$. Notice that $$ \measuredangle A'AO = \measuredangle POA + 90^{\circ} = \measuredangle AHP + 90^{\circ} $$Which show that $AA', A\infty_{\perp PH}$ are isogonal WRT $\angle BAC$. Hence $A'$ is the Anti-Steiner point of $P$. Let $P'$ be the reflection of $P$ in $BC$. We have $\measuredangle A'P'P = \measuredangle A'OP$, so $A', P', O, P$ are concyclic, which show that the center of $\odot(A'PO)$ is on $BC$, as desired. $\blacksquare$

Long(sad) solution somehow looks like moving point without moving point. Lemma: Let $\triangle AOP$ and $O'$ be the reflection of its circumcenter $S$ with side $PO$. $\omega=\odot(O,OA)$ and $A'$ be $A$-antipode. $A'O',AO' \cap \omega = Y,X$. $Z$ is the reflection of $X$ across $O'$. Then $A,Y,Z,P$ are concyclic. Moreover, $\measuredangle AZP = \measuredangle POA$. Proof: Let $N_E$ be the image of point $X$ with center $O'$ radius $-\sqrt{O'X \cdot O'A}$ for any point $N$. $O'$ is the midpoint of $AZ_E$ so $Z_EA' \parallel OO'$. By angle-chaseing and $\odot(APXP_E)$, It suffices to prove that $\odot(XA'Z_EP_E)$ equivalent to $\angle A'P_EZ_E = 90^{\circ} \iff \measuredangle A'P_EO' = \measuredangle PAS$. Let $\Omega = \odot(O,OS)$ and $\Omega \cap AO',AS = \{O,T\},\{S,W\}$. Easy to see that $O'X = AT$. By power of point, $O'P_E = AW$. Let $N_*$ be the reflection of point $N$ over $O$ for any point $N$ and $W'$ be the reflection of $W$ across $AO$. Consider $\square ASOW'$ be an isosceles trapezoid with $AS=SO=OW'$ and $SO \parallel AW'$ and $\square O'_*SYO$ is a parallelogram. We get that $\square AW' \parallel O'_*{P_E}_*$. Hence, $\square A{P_E}_*O'_*W'$ is an parallelogram and $\measuredangle A'P_EP = \measuredangle A{P_E}_*O'_* = \measuredangle PAS$ Which can be obtained by angle chaseing also we get the second conclusion easily. Finally, back to the main problem. Let $D$ be the foot of the altitude from $A$ to $BC$, $A'$ be the antipode of $A$ wrt $\odot(ABC)$. Since $MO \perp MS$, by Butterfly theorem, $XY$ passes through $S$, the circumcenter of $\triangle APO$ iff $BC$ passes through $O'$, the reflection of $S$ across $OP$ iff $\angle ADO' = 90^{\circ}$. ignore $X,Y$ defined from the problem . Let $Y = A'O' \cap \odot(ABC), AO' \cap \odot(ABC) = X$ and $Z$ be the reflection of $X$ across $O'$. $\angle ADO' = 90^{\circ}$ iff $\odot(ADYO')$ iff(spiral similarity) $\odot(AHYZ)$. By the lemma, $\odot(AYZP)$ and $\measuredangle PHA = \measuredangle AOP = \measuredangle PZA$ implies $\odot(AZYPH)$.

Really nice problem Solution with rg230403, arwen713, BOBTHEGR8, pluto1708 Let $O'$ be $O$ reflected across $BC$, and let $P'$ be $P$ reflected across the midpoint of $OH$. Let $N_1,N$ be the midpoints of $BC$ and $XY$, and let $T$ be the circumcentre of $AOP$. Proof Outline: $\angle OTM = \angle OAP = \angle HAP' = \angle OO'P = \angle ON_1M = \angle ONM \implies \angle ONT = \frac{\pi}{2}$ [asy][asy] size(13cm); defaultpen(fontsize(11pt)); import olympiad; pair A = dir(120), B = dir(206), C = dir(-26), O = (0,0); pair H = orthocenter(A,B,C); pair P = extension(H, H+dir(70), O, dir(140)); pair M = (O+P)/2; pair X = 2*foot(O,B,M) - B; pair Y = 2*foot(O,C,M) - C; pair N1 = (B+C)/2; pair N = (X+Y)/2; pair Op = 2*foot(O,B,C); pair Pp = O+H-P; pair T = circumcenter(A,P,O); draw(A--B--C--cycle, royalblue); draw(unitcircle, fuchsia); draw(circumcircle(A,P,O), dotted+fuchsia); draw(circumcircle(T,M,O), dashed+fuchsia); draw(B--X, purple); draw(C--Y, purple); draw(X--Y, purple); draw(A--H--Op--O--cycle, brown); draw(Pp--A--P--Op, orange); draw(M--N1, orange); draw(M--N--O--T--M, dotted + green); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$H$", H, dir(H)); dot("$O$", O, dir(-A)); dot("$P'$", Pp, dir(N1)); dot("$O'$", Op, dir(N1)); dot("$M$", M, dir(80)); dot("$T$", T, dir(T)); dot("$P$", P, dir(Y)); dot("$N$", N, dir(N)); dot("$N_1$", N1, dir(-A)); dot("$Y$", Y, dir(Y)); dot("$X$", X, dir(X)); [/asy][/asy] $\angle OTM = \cfrac{\angle OTP}{2} = \angle OAP$ By the angle condition $\angle AHP = \angle POA$, we have that the points at infinity along $HP$ and $OP$ are isogonal in $\angle OAH$, and by isogonal lemma, we can conclude that $\angle OAP = \angle HAP'$ Note that $AOO'H$ is a parallelogram, $HPOP'$ is a parallelogram, so $APO'P'$ is also a parallelogram. Since $HA || OO'$ and $AP' || OP$, $\angle HAP' = \angle OO'P$. Homothety $\frac12$ from $O$ implies that $\angle OO'P = \angle ON_1M$ Now, $MBN_1C \sim MYNX$, so $\angle(BC, MN_1) = \angle (YX, MN)$, which implies that $\angle ON_1M = \frac{\pi}{2} - \angle(BC, MN_1) = \frac{\pi}{2} - \angle (YX, MN) = \angle ONM$ Now, since $\angle OTM = \angle ONM$, we have $MONT$ concyclic. $\angle ONT = \angle OMT = \frac{\pi}{2}$, implies that $T$ lies on $XY$

This is my complete solution with some complex bashing. Let me first define some point it may appear later. The perpendicular bisector of $\overline{OP}$ intersect $\overline{XY}$ at $Q'$. Let $Q$ be the circumcenter of $\triangle APO$, $A'$ be the reflection point of $A$ with respect to $OP$, and $T$ be the intersection point of $BC$ and the perpendicular bisector of $\overline{OP}$. It is obvious to see that our goal is to prove $Q=Q'$ ($Q'$ is the unique point that lies on both $XY$ and the perpendicular bisector of $\overline{OP}$), in other words, we need to prove $\overline{MQ'}=\overline{MQ}$. Applying the butterfly theorem, then we'll get $\overline{MQ'}=\overline{TM}$ as the result. By reflecting the whole figure with the axis $OP$, then the problem is equivalent to prove $T$ is the circumcenter of $\triangle A'PO$, and it is again equivalent to prove the circumcenter of $\triangle A'PO$ lies on $BC$. Let $(ABC)$ be the unit circle on complex plane. In the following discussion, the lowercase letters denote the complex number that corresponding to the point with the same letters (but in the uppercase) in the complex plane. $$\begin{aligned} \frac{a-a'}{p}\in i\mathbb{R}&\iff \frac{a-a'}{p}=-\frac{\overline{a}-\overline{a'}}{\overline{p}}=\frac{a-a'}{aa'\overline{p}}\\ &\iff a'=\frac{p}{a\overline{p}} \end{aligned}$$Let $T'$ denote the circumcenter of $\triangle A'PO$. $$t'=\frac{\left|\begin{array}{cccc} a' & a'\overline{a'} & 1 \\ p & p\overline{p} & 1\\ o & o\overline{o} & 1 \end{array}\right| }{\left|\begin{array}{cccc} a' & \overline{a'} & 1 \\ p & \overline{p} & 1\\ o & \overline{o} & 1 \end{array}\right| }=\frac{a'p\overline{p}-p}{a'\overline{p}-p\overline{a'}}=\frac{p^2-ap}{p-a^2\overline{p}}$$ Our goal is to prove $T', B, C$ collinear. $$\begin{aligned} &\iff \frac{t'-b}{t'-c} \in \mathbb{R} \iff \frac{t'-b}{t'-c}=\frac{\overline{t'}-\overline{b}}{\overline{t'}-\overline{c}}\\ &\iff (t'-b)(\overline{t'}-\frac{1}{c})=(t'-c)(\overline{t'}-\frac{1}{b})\\ &\iff (c-b)\overline{t'}+t'=b+c \text{ (By some simplifying.)}\\ &\iff bc\frac{a^2\overline{p}^2-a\overline{p}}{a^2\overline{p}-p}+\frac{p^2-ap}{p-a^2\overline{p}}=b+c\\ &\iff bc(a^2\overline{p}^2-a\overline{p})-(p^2-ap)=(b+c)(a^2\overline{p}-p)\\ &\iff (a^2bc\overline{p}-abc-a^2b-a^2c)\overline{p}=(p-a-b-c)p\\ &\iff a^2bc(\overline{p}-\overline{h})\overline{p}=(p-h)p\text{, where } h=a+b+c. \end{aligned}$$ Notice the condition that $\measuredangle AHP=\measuredangle POA$, it is equivalent to $$\begin{aligned} &\frac{a-h}{p-h}/\frac{p-o}{a-o}\in \mathbb{R}\iff \frac{a(a-h)}{p(p-h)}\in \mathbb{R}\\ &\iff -\frac{a(b+c)}{p(p-h)}=-\frac{\overline{a}(\overline{b}+\overline{c})}{\overline{p}(\overline{p}-\overline{h})}\iff \frac{a^2bc}{p(p-h)}=\frac{1}{\overline{p}(\overline{p}-\overline{h})}. \end{aligned}$$The equation is same as the required one, so the proof is completed.

Let $\overline{OP}$ and $\overline{HP}$ meet $\overline{AH}$ and $\overline{AO}$ at $K$ and $L$ respectively. By the given angle condition, it follows that $K,L,O,H$ are concyclic. Let the perpendicular bisector of segment $OP$ meet $\overline{XY}$ and $\overline{BC}$ at $Q$ and $R$. By the butterfly theorem, $MQ=MR$, so $Q$ and $R$ are reflections of each other over line $OP$. This means that if we reflect $A$ over line $OP$, say $A'$, we just have to show that the circumcenter of $\triangle A'OP$ is $R$. Let its circumcircle be $\omega$, and let $O'$ be the reflection of $O$ over $BC$. It is known that $AOO'H$ is a parallelogram, and easy angle chasing shows that quadrilaterals $AKPL$ and $O'HPO$ have the same corresponding angles. Since $\triangle PLK\sim\triangle POH$, they are in fact similar and oppositely oriented. Thus \[\measuredangle PO'O=-\measuredangle PAO=\measuredangle PA'O\]and so $O'$ lies on $\omega$. Finally, the center of $\omega$ is the intersection of perpendicular bisectors of segments $OP$ and $OO'$, which is precisely $R$. [asy][asy] import olympiad; defaultpen(fontsize(10pt)); size(10cm); pair A = dir(120); pair B = dir(215); pair C = dir(325); pair O = circumcenter(A,B,C); pair H = orthocenter(A,B,C); real s = .75; pair L = s*O+(1-s)*A; pair K = 2*foot(circumcenter(O,L,H),A,H)-H; pair P = extension(L,H,K,O); pair M = midpoint(O--P); pair X = 2*foot(O,B,M)-B; pair Y = 2*foot(O,C,M)-C; pair Q = circumcenter(A,P,O); pair R = 2*M-Q; pair A1 = 2*foot(A,O,P)-A; pair O1 = 2*foot(O,B,C)-O; draw(A--B--C--cycle, black+1); draw(circumcircle(A,B,C)); draw(A--O); draw(A--foot(A,B,C), dotted); draw(O--K); draw(H--L); draw(B--X, dashed); draw(C--Y, dashed); draw(X--Y); draw(Q--R); draw(A--P); draw(O1--P); draw(circumcircle(O,H,L)); draw(circumcircle(O,P,O1), dotted); draw(A--K--P--L--cycle, lightblue+1); draw(H--O1--O--P--cycle, lightblue+1); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$A'$", A1, dir(A1)); dot("$L$", L, dir(45)); dot("$K$", K, dir(180)); dot("$H$", H, dir(180)); dot("$O$", O, dir(45)); dot("$P$", P, dir(90)); dot("$O'$", O1, S); dot("$M$", M, dir(280)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$R$", R, S); dot("$Q$", Q, N); [/asy][/asy]

Nice problem!, Solved with Jupiter_is_Big, jelena_ivanchic Let $O_1$ be the circumcenter of $\triangle APO$ and let $O_1M$ meet $BC$ at $D$. The main claim is the following: Claim: $PDOO_1$ is a rhombus Proof: Let $P'$ be the point such that $POP'H$ is a parallelogram. By parallelogram isgonality lemma, we have that $\angle P'AH = \angle PAO$

To finish, observe that $\angle MDO = \angle MNO = \angle HAP' = \angle PAO = \frac{\angle PO_1O}{2} = \angle PO_1M$. Since $PO_1OD$ was already a kite, this means it is a rhombus. $\square$ Now, let $O_1D$ intersect the circle at $E,F$. Obviously, $M$ is the midpoint of $EF$ and since $PDOO_1$ was a rhombus, it is also the midpoint of $DO_1$. So, by butterfly theorem, we have that $O_1$ lies on line $XY$, as desired. $\blacksquare$

Refreshing! The synthetic solutions on this thread are all amazing; I am very appreciative of AoPS For me, this is the sort of problem that you start the computations (complex) without even trying to make any synthetic observations... I know that may be quite sad and maybe it is. The angle condition is equivalent to equal arguments which are very easy to express as they involve basic elements such as $o = 0, h = a+b+c, a$ and the variability of $p$ can be fixed with the following trick, which makes this solution slightly cleaner than #7: Take $(ABC)$ as the unit circle, $o=0, h =a+b+c$ and as usual $a \cdot \overline{a} = b \cdot \overline{b} = c \cdot \overline{c} = 1$, and very importantly take $Im(p) = 0$, that is $p = \overline{p}$ as this parametrization has one more degree of freedom, namely rotation at $O$, so simply rotate the image by $\arg(p)$. Now, the computations are easy, I think this is a very instructive problem so I will take care to prove all the formulas that I blindly used.

In particular, defining $r = a(ab+bc+ca)-a-b-c$ and $q = a^2bc-1$ gives $p = \frac{q}{r}$.

We know that $p = \frac{q}{r}$ and that the collinearity is equivalent to: $$qp^3-rp^2-4qp+4r = 0$$We do not even have to expand and match coefficients this is true simply due to the relation that $p = \frac{q}{r}$!! Magic. $\blacksquare$ $\blacksquare$

Nice and easy Problem!

Let $N$ be the midpoint of $XY$, $T$ be the center of $(AOP)$, $O'$ be the reflection of $O$ in $BC$, let $P'$ be the reflection of $P$ in midpoint of $OH$. Key Claim : $MONT$ is cyclic

Now $90^\circ = \measuredangle OMT = \measuredangle ONT$ hence done

Funny problem. But i do really like this tbh, lets begin with the fact that the (generalized) locus of $P$ will be a circumrectangular hyperbola $\mathcal H$ of $\triangle AHO$ with center being the midpoint of $OH$ (call it $N_9$). Let $Q$ be the orthocenter of $\triangle AHP$ and $A'$ the circumcenter of $(BHC)$, let $O'$ the center of $(APO)$ and $O_1$ the center of $(A'PO)$ and let $O'O_1$ hit $(ABC)$ at $K,L$. Claim: $O', O_1$ are symetric over $M$. Proof: First note that by checking radiuses we get $O,A'$ are symetric in $BC$ and therefore by taking homothety with scale factor 2 at the antipode of $A$ in $(ABC)$ we get by lenghts that $AHA'O$ is a parallelogram therefore $A,N_9,A'$ are collinear and $AN_9=N_9A'$ which means that $A'$ lies in $\mathcal H$, now trivially $Q$ lies in $\mathcal H$ and with the angles and LoS we will generalize the condition of $P$ to $(AHP)$ being symetric to $(AOP)$ w.r.t. $AP$, therefore $APOQ$ is cyclic. Now as $PQ \perp AH \parallel OA'$ we have that $O$ is the orthocenter of $\triangle A'PQ$ using $\mathcal H$ and therefore $(APO), (A'PO)$ are symetric over $OP$ which implies the claim. Finishing: Note that $O,A'$ symetric w.r.t. $BC$ implies $O_1$ lies on $BC$, now note that $KL \perp OM$ and $KO=OL$ therefore $KM=ML$ and by converse of Butterfly theorem on $BCXY$ we get $X,Y,O'$ colinear as desired.