Beautiful problem. Let $O, M, N$ be the center of $(XYZ)$, and the midpoints of $\overline{AB}, \overline{AC}$, respectively.. Observe that $APXZ$ and $AQCZ$ are cyclic, and the perpendicular bisectors $p_1, p_2$ of $\overline{XZ}, \overline{ZY}$ bisect $\overline{AP}, \overline{AQ}$, respectively. Moreover $$\widehat{PZX} + \widehat{BPZ} = \widehat{PAB} + \widehat{BAQ} = \frac{\pi}{2}\to XZ\perp{BP}\hspace{0.1 cm} (\text{analogously}\hspace{0.1 cm} ZY\perp{CQ}).$$So, $p_1\parallel BP$ and $p_2\parallel CQ\to$ $M\in p_1$ and $N\in p_2 \implies \widehat{MON} = \frac{\pi}{2} - \widehat{A}$ and hence there's a fixed circle passing through $M, N$ and $O$.

Other way of proving $M \in p_1$ is reflecting $X$ across $M$ (call it $X'$) and writing $AX'/AZ=BX/AZ=PX/PZ$ which implies $XPZ \sim X'AZ \implies \angle X'ZX=90$. Nevertheless I could not see the fixed circle. Very hard and nice problem.

Let $l_1$ and $l_2$ be perpendicular bisectors of $XZ,YZ$ and $O'$ be center of $XYZ$. Claim $: l_1$ and $l_2$ bisect $AB,AC$. Proof $:$ Note that $AP,AQ$ are diameters of $APXZ$ and $AQYZ$ so $l_1$ and $l_2$ bisect $AP,AQ$ so we need to prove $l_1 || PB$ and $l_2 || QC$. Note that $\angle PZX = \angle PAX = \angle PAB = \angle PQB \implies XZ || BQ$ and Note that $l_1 \perp XZ$ and $PB || BQ$ so $PB || l_1$. we prove the other one with same approach. Now if $M,N$ be midpoints of $AB,AC$ then $\angle MO'N = \frac{\angle XO'Y}{2} = \angle ZXY + \angle ZYX = \angle PZX + \angle QZY = \angle 90 - \angle A$ so $\angle MO'N$ is fixed so $O'$ lies on $MO'N$ which is fixed.

Let the midpoints of $AP, AQ, AZ, XZ, YZ$ be $M, N, F, D, E$ respectively. In addition, define the circumcenters of $ABC$ and $XYZ$ as $O$ and $O_1$ respectively, and let the circle with diameter $AO$ be $\omega$. Because $APXZ$ is cyclic with diameter $AP$ and $AQYZ$ is cyclic with diameter $AQ$, we know $MD$ and $NE$ are the perpendicular bisectors of $XZ$ and $YZ$ respectively. It follows that $O_1 = MD \cap NE$. Since $MN$ is the $A$-midline of $APQ$, we have $F \in MN$ and $AF \perp MN$. Thus, $DMFZ$ is cyclic with diameter $MZ$, so $$\angle O_1MN = \angle DMF = 180^{\circ} - \angle DZF = 180^{\circ} - \angle XZA = \angle XPA.$$Analogously, we find $\angle O_1NM = \angle YQA$, implying $$\angle MO_1N = 180^{\circ} - \angle O_1MN - \angle O_1NM$$$$= 180^{\circ} - \angle XPA - \angle YQA$$$$= 180^{\circ} - (90^{\circ} - \angle XAP) - (90^{\circ} - \angle YAQ)$$$$= \angle PAQ - \angle XAY = 90^{\circ} - \angle BAC$$which is fixed. Since $\angle AMO = \angle ANO = 90^{\circ}$ and $\angle MAN = 90^{\circ}$, we know $MN$ is a diameter of $\omega$. Now, define $R$ as the projection of $M$ onto $NO_1$ and $S$ as the projection of $N$ onto $MO_1$. It's clear that $R$ and $S$ both lie on $\omega$, so $$\angle MAS = \angle MNS = 90^{\circ} - \angle O_1MN = 90^{\circ} - \angle XPA = \angle MAX$$which means $S \in AB$. But $S \in \omega$ gives $\angle ASO = 90^{\circ}$, so $S$ is actually the midpoint of $AB$. Similarly, we deduce that $R$ is the midpoint of $AC$. Thus, $\angle SO_1R = \angle MO_1N$ is fixed, which finishes since $R$ and $S$ are both fixed points. $\blacksquare$ Remarks: What an amazing problem! It might be too hard for its placement, however. My GGB can be found here. Discovering $\angle MO_1N = 90^{\circ} - \angle A$ is easy, but noticing that the midpoints of $AB$ and $AC$ lie on $O_1M$ and $O_1N$ respectively requires a lot more insight. In fact, it actually took me an hour to even construct the two midpoints!

sorry Let $O_1$ be the circumcenter of $XYZ$, and let $O'$ be the reflection of $O$ over $\overline{BC}$. Notice that $AZXP$ and $AZYQ$ are cyclic, so \[\angle XZY=180^\circ-\angle PZX-\angle QZY=180^\circ-\angle PAX-\angle QAY=180^\circ-(\angle PAQ-\angle BAC)=\angle BAC+90^\circ.\]Thus, $\angle XO_1Y=180^\circ-2\angle BAC$. We have $\angle OBO'=\tfrac{360^\circ-2\angle BOC}{2}=180^\circ-2\angle BAC$, so $\angle XO_1Y=\angle OBO'$. We also have $O_1X=O_1Y$ and $BO=BO'$, so $\triangle XYO_1 \sim \triangle OO'B$. Let $A=a$, $B=b$, and $C=c$ be points on the unit circle, and let $P=p$, $Q=-p$. Then, $O'=b+c$. We have \[X=\frac{\left(\frac{1}{a}-\frac{1}{b}\right)p+(a-b)\frac{1}{p}+\frac{b}{a}-\frac{a}{b}}{2\left(\frac{1}{a}-\frac{1}{b}\right)}=\frac{(b-a)p^2+ab(a-b)+(b^2-a^2)p}{2(b-a)p}=\frac{p^2+(a+b)p+ab}{2p},\]and similarly, \[Y=\frac{-p^2+(a+b)p-ab}{2p}.\]Since $\tfrac{\overrightarrow{XO_1}}{\overrightarrow{XY}}=\tfrac{\overrightarrow{OB}}{\overrightarrow{OO'}}=\tfrac{b}{b+c}$, we have \begin{align*} O_1 &= X+\frac{b}{b+c}(Y-X) \\ &= \frac{p^2+(a+b)p+ab}{2p}+\frac{b}{b+c} \cdot \frac{-2p^2-a(b+c)+(c-b)p}{2p} \\ &= \frac{p}{2}+\frac{a+b}{2}+\frac{ab}{2p}-\frac{bp}{b+c}-\frac{ab}{2p}+\frac{b(c-b)}{2(b+c)} \\ &= \left(\frac{b}{b+c}+\frac{1}{2}\right)p+\frac{a+b}{2}+\frac{b(c-b)}{2(b+c)}. \end{align*}Thus, $O_1$ lies on the fixed circle centered at $\tfrac{a+b}{2}+\tfrac{b(c-b)}{2(b+c)}$ with radius $|\tfrac{b}{b+c}+\tfrac{1}{2}|$. $\square$

Let Mb, Mc be midpoints of AC, AB and R, S = AO \cap (AB), (AC). Then ZMcX ~ ZOR and ZMbY ~ ZOS by spiral similarity, and OZ = OR = OS so ZMb = MbY, ZMc = McX. If P’, Q’ are the midpoints of AP, AQ then ZX || McQ’ and ZY || MbP’, so P’Mc \cap Q’Mb is the circumcenter of XYZ, and P’, Q’, Mb, Mc are on (AO) the angle between these two lines is clearly fixed, meaning the circumcenter travels along a circle through Mb, Mc.

Let $E,F$ denote the midpoints of $AB,AC$ and let $O'$ denote the circumcenter of $XYZ$. The main claim is that $E,F$ lie on the perpendicular bisectors of $XY,XZ$ respectively. Note that $AXYP$ and $AXZQ$ are cyclic, so the midpoint of $AX$ lies on the perpendicular bisector of $XY$. Hence, by midpoint theorem, we only need to show $BP\perp XY$. We use complex numbers : $$x-y = \frac 12 (a+ \frac {p^2}a)- \frac 12 (a+b+p- \frac {ab}p)= \frac {(p^2-ab)(p-a)}{2ap}$$ Then we have : $$\overline {\left( \frac {x-y}{b-p} \right) } = \overline {\left( \frac {(p^2-ab)(p-a)}{2ap(b-p)}\right)} = \frac {(ab-p^2)(a-p) \frac 1{ap} \frac 1{abp^2}}{\frac 1{ap} \frac 1{bp} (b-p)}= - \left( \frac {x-y}{b-p} \right)$$ This proves $XY\perp BP$, so the main claim is done. Finally note we have : $$\angle EO'F = \pi - \angle XYZ = \angle PXY+ \angle QXZ = \angle PAQ - A = \frac {\pi}2 -A$$ which is fixed, so we're done.

here's a hybrid solution containing both synthetic and (debatable) bash elements, found with mxlcv

Alright here we go. I can’t find similar triangles as I am predominantly an angle/length/cross-ratio chase nerd, so here we go. Russian geo once again. Let $O’$ be the circumcentre of $(XYZ)$. So first let’s try to eliminate $Z$ because it’s not too well-controlled w.r.t. $X$ and $Y$. Construct $X’=\overline{AB}\cap\overline{QY}$, and $Y’=\overline{AC}\cap\overline{PX}$. Note $XYX’Y’$ is clearly cyclic because of the bunch of perpendicular lines. I claim now to eliminate $Z$ that $Z\in(XYX’Y’)$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10) + blue; defaultpen(dps); /* default pen style */ pen dotstyle = blue; /* point style */ real xmin = -30.58936737805624, xmax = 40.29058823433117, ymin = -25.492178516239495, ymax = 13.491797070574005; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ffvvqq = rgb(1,0.3333333333333333,0); pen ccqqqq = rgb(0.8,0,0); draw((-6.505325726427008,7.594783551431004)--(-8.538272802918998,-5.20556409459472)--(8.586879107882611,-5.124988505997758)--cycle, linewidth(0.5) + ccqqqq); /* draw figures */ draw(circle((0,0), 10), linewidth(0.5) + qqwuqq); draw((-0.5950341408752691,9.982281020447823)--(0.5950341408752691,-9.982281020447823), linewidth(0.5) + ffvvqq); draw((0.5950341408752691,-9.982281020447823)--(-6.505325726427008,7.594783551431004), linewidth(0.5) + ffvvqq); draw((-6.505325726427008,7.594783551431004)--(-0.5950341408752691,9.982281020447823), linewidth(0.5) + ffvvqq); draw((-6.505325726427008,7.594783551431004)--(-8.538272802918998,-5.20556409459472), linewidth(0.5) + ccqqqq); draw((-8.538272802918998,-5.20556409459472)--(8.586879107882611,-5.124988505997758), linewidth(0.5) + ccqqqq); draw((8.586879107882611,-5.124988505997758)--(-6.505325726427008,7.594783551431004), linewidth(0.5) + ccqqqq); draw((-11.368255310310504,11.693280525028069)--(8.586879107882611,-5.124988505997758), linewidth(0.5) + ffvvqq); draw((-8.538272802918998,-5.20556409459472)--(-11.593796970042014,-24.4445162701222), linewidth(0.5) + ffvvqq); draw((-0.5950341408752691,9.982281020447823)--(-11.368255310310504,11.693280525028069), linewidth(0.5) + ffvvqq); draw((-11.593796970042014,-24.4445162701222)--(0.5950341408752691,-9.982281020447823), linewidth(0.5) + ffvvqq); draw((-5.990061649881782,10.839117545933437)--(-0.5950341408752691,9.982281020447823), linewidth(0.5) + ffvvqq); draw((-8.538272802918998,-5.20556409459472)--(0,0), linewidth(0.5) + ffvvqq); draw((0,0)--(8.586879107882611,-5.124988505997758), linewidth(0.5) + ffvvqq); draw((-5.990061649881782,10.839117545933437)--(-6.505325726427008,7.594783551431004), linewidth(0.5) + ffvvqq); draw((0.5950341408752691,-9.982281020447823)--(6.3077139228046635,-3.2040987517377357), linewidth(0.5) + ffvvqq); draw((-0.5950341408752691,9.982281020447823)--(8.586879107882611,-5.124988505997758), linewidth(0.5) + ffvvqq); draw(circle((-11.481026140176267,-6.375617872547062), 18.06925030436963), linewidth(0.5) + qqwuqq); draw((-11.368255310310504,11.693280525028069)--(-11.593796970042014,-24.4445162701222), linewidth(0.5) + ffvvqq); draw((-6.505325726427008,7.594783551431004)--(-0.4741479247651867,7.954296241399017), linewidth(0.5) + ffvvqq); draw(circle((-3.5501799336511386,8.788532285939413), 3.187149619597635), linewidth(0.5) + qqwuqq); draw((-0.4741479247651867,7.954296241399017)--(6.3077139228046635,-3.2040987517377357), linewidth(0.5) + ffvvqq); draw(circle((-2.955145792775873,-1.1937487345084101), 9.478506069117573), linewidth(0.5) + qqwuqq); draw((-5.990061649881782,10.839117545933437)--(-0.4741479247651867,7.954296241399017), linewidth(0.5) + ffvvqq); /* dots and labels */ dot((-6.505325726427008,7.594783551431004),dotstyle); label("$A$", (-6.279757601620247,8.175800399644892), NE * labelscalefactor); dot((-8.538272802918998,-5.20556409459472),dotstyle); label("$B$", (-8.32863131854082,-4.6711915551004655), NE * labelscalefactor); dot((8.586879107882611,-5.124988505997758),dotstyle); label("$C$", (8.782232966012078,-4.560441624456109), NE * labelscalefactor); dot((0,0),dotstyle); label("$O$", (0.19911334107453973,0.5340551851842915), NE * labelscalefactor); dot((-0.5950341408752691,9.982281020447823),dotstyle); label("$P$", (-0.3546363121472369,10.556923908498558), NE * labelscalefactor); dot((0.5950341408752691,-9.982281020447823),dotstyle); label("$Q$", (0.8082379596184941,-9.433438572807797), NE * labelscalefactor); dot((-5.990061649881782,10.839117545933437),dotstyle); label("$X$", (-5.781382913720648,11.387548388331233), NE * labelscalefactor); dot((6.3077139228046635,-3.2040987517377357),dotstyle); label("$Y$", (6.511859387802794,-2.677692803502048), NE * labelscalefactor); dot((-11.593796970042014,-24.4445162701222),dotstyle); label("$X'$", (-11.374254411260592,-23.886304521896328), NE * labelscalefactor); dot((-11.368255310310504,11.693280525028069),dotstyle); label("$Y'$", (-11.152754549971881,12.162797902841728), NE * labelscalefactor); dot((-11.481026140176258,-6.375617872547066),dotstyle); label("$O'$", (-11.263504480616238,-5.834065826866209), NE * labelscalefactor); dot((-0.4741479247651867,7.954296241399017),dotstyle); label("$Z$", (-0.24388638150288158,8.508050191577961), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We angle-chase: \begin{align*} \measuredangle XZY&=\measuredangle XZA+\measuredangle AZY\\ &=\measuredangle XPA+\measuredangle AQY\\ &=90^\circ-\measuredangle PAX+90^\circ-\measuredangle YAQ\\ &=\measuredangle BAP+\measuredangle QAC\\ &=\measuredangle BAC-\measuredangle PAQ\\ &=90^\circ+\measuredangle BAC\\ \measuredangle XX’Y&=\measuredangle AX’Y\\ &=90^\circ-\measuredangle YAX’\\ &=90^\circ+\measuredangle BAC=\measuredangle XZY \end{align*}(see I told you I was an angle-chasing nerd) So that eliminates $Z$. Note now that $O’$ is actually the midpoint of $X’Y’$ due to the right angles so we can essentially delete $X$ and $Y$ from the diagram; I choose not to do so however because it’s annoying. Now dilate everything from $A$; let $O’\to K$, and $R=\overline{BP}\cap\overline{CQ}$, $R’=\overline{BQ}\cap\overline{CP}$. I actually claim that (surprisingly) $K=R$. It suffices to show that $AX’RY’$ is a parallelogram. Meanwhile I am going to try show how this finishes. Define $S$ to be the intersection of $\overline{AC}$ with the line perpendicular to $\overline{AB}$ at $B$, and $T$ with $B\iff C$. I claim now that $K=R\in(BCST)$, which finishes. However it’s actually quite clear that $P$, $Q$, $R$, $R’$ form an orthocentric system (USAMO 2024/5 vibes anyone?) from which it’s pretty clear due to the fact that the only circle that is orthogonal to $(ABC)$ is $(BCRR’)=(BCST)$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10) + blue; defaultpen(dps); /* default pen style */ pen dotstyle = blue; /* point style */ real xmin = -35.84985463389345, xmax = 52.76035980263646, ymin = -45, ymax = 16.513505768837344; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ffvvqq = rgb(1,0.3333333333333333,0); pen ccqqqq = rgb(0.8,0,0); draw((-6.505325726427008,7.594783551431004)--(-8.538272802918998,-5.20556409459472)--(8.586879107882611,-5.124988505997758)--cycle, linewidth(0.5) + ccqqqq); /* draw figures */ draw(circle((0,0), 10), linewidth(0.5) + qqwuqq); draw((2.50481026772986,9.681215085033234)--(-2.50481026772986,-9.681215085033234), linewidth(0.5) + ffvvqq); draw((-2.50481026772986,-9.681215085033234)--(-6.505325726427008,7.594783551431004), linewidth(0.5) + ffvvqq); draw((-6.505325726427008,7.594783551431004)--(2.50481026772986,9.681215085033234), linewidth(0.5) + ffvvqq); draw((-6.505325726427008,7.594783551431004)--(-8.538272802918998,-5.20556409459472), linewidth(0.5) + ccqqqq); draw((-8.538272802918998,-5.20556409459472)--(8.586879107882611,-5.124988505997758), linewidth(0.5) + ccqqqq); draw((8.586879107882611,-5.124988505997758)--(-6.505325726427008,7.594783551431004), linewidth(0.5) + ccqqqq); draw((-11.647867456922755,11.928938787584304)--(8.586879107882611,-5.124988505997758), linewidth(0.5) + ffvvqq); draw((-8.538272802918998,-5.20556409459472)--(-10.81510214329833,-19.541503984292923), linewidth(0.5) + ffvvqq); draw((2.50481026772986,9.681215085033234)--(-11.647867456922755,11.928938787584304), linewidth(0.5) + ffvvqq); draw((-10.81510214329833,-19.541503984292923)--(-2.50481026772986,-9.681215085033234), linewidth(0.5) + ffvvqq); draw((-5.960434633485641,11.025662544556566)--(2.50481026772986,9.681215085033234), linewidth(0.5) + ffvvqq); draw(circle((0.09108898903745177,-19.35961997709105), 16.577188629599004), linewidth(0.5) + qqwuqq); draw((-8.538272802918998,-5.20556409459472)--(0,0), linewidth(0.5) + ffvvqq); draw((0,0)--(8.586879107882611,-5.124988505997758), linewidth(0.5) + ffvvqq); draw((-10.81510214329833,-19.541503984292923)--(-12.498926964267309,-30.14362114348741), linewidth(0.5) + ffvvqq); draw((8.586879107882611,-5.124988505997758)--(12.681104942342206,-8.575618810694674), linewidth(0.5) + ffvvqq); draw((-5.960434633485641,11.025662544556566)--(-6.505325726427008,7.594783551431004), linewidth(0.5) + ffvvqq); draw((-2.50481026772986,-9.681215085033234)--(4.346919174826464,-1.5515329060246381), linewidth(0.5) + ffvvqq); draw((2.50481026772986,9.681215085033234)--(8.586879107882611,-5.124988505997758), linewidth(0.5) + ffvvqq); draw((-15.957643873794087,-15.207348748139617)--(16.13982185186898,-23.51189120604247), linewidth(0.5) + ffvvqq); draw((-15.957643873794087,-15.207348748139617)--(8.586879107882611,-5.124988505997758), linewidth(0.5) + ffvvqq); draw((8.586879107882611,-5.124988505997758)--(16.13982185186898,-23.51189120604247), linewidth(0.5) + ffvvqq); draw((-15.957643873794087,-15.207348748139617)--(2.50481026772986,9.681215085033234), linewidth(0.5) + ffvvqq); draw((-8.538272802918998,-5.20556409459472)--(16.13982185186898,-23.51189120604247), linewidth(0.5) + ffvvqq); draw(circle((-6.726416803032106,-2.763066831553197), 15.494376581568185), linewidth(0.5) + qqwuqq); draw((-11.647867456922755,11.928938787584304)--(-15.957643873794087,-15.207348748139617), linewidth(0.5) + ffvvqq); /* dots and labels */ dot((-6.505325726427008,7.594783551431004),dotstyle); label("$A$", (-6.220814181678762,8.275524895441126), NE * labelscalefactor); dot((-8.538272802918998,-5.20556409459472),dotstyle); label("$B$", (-8.228389352506392,-4.53142015983871), NE * labelscalefactor); dot((8.586879107882611,-5.124988505997758),dotstyle); label("$C$", (8.870612964542739,-4.46219342981017), NE * labelscalefactor); dot((0,0),dotstyle); label("$O$", (0.28649844100390387,0.6605845923017646), NE * labelscalefactor); dot((2.50481026772986,9.681215085033234),dotstyle); label("$P$", (2.7786607220313075,10.352326796297316), NE * labelscalefactor); dot((-2.50481026772986,-9.681215085033234),dotstyle); label("$Q$", (-2.2056638400234996,-8.961930881665248), NE * labelscalefactor); dot((-5.960434633485641,11.025662544556566),dotstyle); label("$X$", (-5.667000341450449,11.73686139686811), NE * labelscalefactor); dot((4.346919174826464,-1.5515329060246381),dotstyle); label("$Y$", (4.64778243280186,-0.8624034683261078), NE * labelscalefactor); dot((-10.81510214329833,-19.541503984292923),dotstyle); label("$X'$", (-10.512871443448178,-18.861353275746417), NE * labelscalefactor); dot((-11.647867456922755,11.928938787584304),dotstyle); label("$Y'$", (-11.343592203790646,12.636808887239125), NE * labelscalefactor); dot((12.681104942342206,-8.575618810694674),dotstyle); label("$S$", (12.954990036226539,-7.854303201208613), NE * labelscalefactor); dot((-12.498926964267309,-30.14362114348741),dotstyle); label("$T$", (-12.243539694161653,-29.453042970112985), NE * labelscalefactor); dot((-15.957643873794087,-15.207348748139617),dotstyle); label("$R$", (-15.704876195588604,-14.50006928394842), NE * labelscalefactor); dot((16.13982185186898,-23.51189120604247),dotstyle); label("$R'$", (16.41632653765349,-22.807276887373177), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] So now that I’ve reduced it to showing that $K=R$, notice that there is a surprising components in the setup; $PCRY’$ is cyclic. Note that with an angle chase similar to above, we can immediately get $\measuredangle PY’C=90^\circ+\measuredangle BAC=\measuredangle PRC$, so combining this with the fact that $ABCP$ is cyclic, $\overline{AB}\parallel\overline{RY’}$ by Reim’s so we actually indeed have $AY’RX’$ a parallelogram by similar logic. Thus we are done!