I have a very different and much shorter solution. Let $A' = PH\cap\omega$. Then using Reim's theorem with $\overline{PHA'}$ and $\overline{BBC}$ gives $CA'\parallel BH$ or $A'$ is the $A$-antipode in $\omega$. This means $\odot(PCH)$ is also tangent to $BC$ which means that it suffices to show that $O_1\in AB$. Now we use a common trick. We will reflect $O$ across $AB$ to get $O'$ and show that $P,X,O,O'$ are concyclic instead. Notice that $O'X$ and $OO'$ are perpendicular bisectors of $BH, BA$ respectively. Thus $$\angle XO'O = \angle ABH = 45^{\circ}$$On the other hand, we can easily chase $\angle XPO$ Since $OX$ is the perpendicular bisector of $BP$, we get $$\angle XPO = \angle XBO = 90^{\circ} - \angle OBC = 45^{\circ}$$so we are done.

Notice that since $\angle BPC + \angle BHC = 180$, we know from the condition of the problem that $H$ is the $P-$HM point of $\triangle BPC.$ This means that $HP$ bisects $BC$, and so $P \in HA'.$ We have $\angle PYC = 2 (180 - \angle PHC) = 2 \angle A'HC = 2 \angle PCB = \angle POB.$ Analogously $\angle PXB = \angle POC.$ This means that $PXBO$ and $POCY$ are similar kites. Hence, we've that $\angle PO_1O = 2 \angle PXO = 2 \angle POY = \angle POC.$ Similarly, $\angle PO_2O = \angle POB$, so we get that $\triangle PO_1O \sim \triangle POC$ and $\triangle PO_2O \sim \triangle POB.$ We are now set up to begin the complex bash. WLOG let $B = 1, C = i, A = a$ for some complex number $a$ on the unit circle. If we let $E, F$ be the feet of $B, C$ onto $AC, AB$ respectively, then we know that $\triangle PEF \sim \triangle PCB.$ We have $F = \frac{a+i+1-ai}{2}$ and $E = \frac{a+i+1+ai}{2}$. This means that $P = \frac{EB - CF}{E + B - F - C} = \frac{ai - 1 - a}{ai - 1 + i}$. Hence, we have from $\triangle PO_1O \sim \triangle POC$ that $O_2 = P + \frac{P^2}{1-P} = \frac{P}{1-P} = \frac{ai-1-a}{a+i}.$ To show that $A, O_2, C$ are collinear, we need to show that: $$\frac{a - \frac{ai-1-a}{a+i}}{a - i} \in \mathbb{R}.$$ This simplifies to $\frac{a^2+a+1}{a^2+1}$, and this is clearly equal to its complex conjugate, hence it's real. Therefore, $O_2 \in AC$. Analogously, $O_1 \in AB$ and we're done. $\square$

I am only going to prove $O_1 \in AB$ since the other case is anologous. Once you get $POO_1 \sim PCO$, write: \[ \frac{PC}{CO} = \frac{PO}{OO_1} = \frac{AO}{OO_1}. \]Let $P'$ be the point where the parallel to $AB$ through $P$ cuts $\omega$ and let $A'$ be the point where $AH$ cuts $\omega$. Notice that $HO \perp CP'$, thus $HP'=HC$. Furthermore $CHP' \sim COP$. Thus we write: \[ \frac{PC}{CO} = \frac{P'C}{CH} = \frac{P'C}{CA'}. \]Finally $\angle AOO_1 = \angle P'CA' \implies AOO_1 \sim P'CA' \implies \angle O_1AO= \angle A'P'C = \angle BAO \implies O_1 \in AB$.

Bonus problem: In the same configuration, let $B'$ and $C'$ be the $B$ and $C$-antipodes respectively. Show that the circumcenter of $AO_1O_2$ lies on $B'C'$. (This was actually included in the original proposal for this problem, but it was deemed too difficult for the National Olympiad)

juckter wrote: Bonus problem: In the same configuration, let $B'$ and $C'$ be the $B$ and $C$-antipodes respectively. Show that the circumcenter of $AO_1O_2$ lies on $B'C'$. (This was actually included in the original proposal for this problem, but it was deemed too difficult for the National Olympiad) We can use MarkBcc168's trick. Reflect $A$ across $B'C'$ to get $K$. We may easily check that $\angle O_1C'K = \angle O_2B'K$. Also: \[ \frac{O_2B'}{O_1C'} = \frac{O_2P}{O_1P} = \frac{CP}{BP} = \frac{BA'}{CA'} = \frac{B'K}{C'K}. \]$\implies O_1C'K \sim O_2B'K \implies O_1KO_2 \sim C'KB' \implies \angle O_1KO_2=135^{\circ} \implies AO_1KO_2 \text{ cyclic}$.

juckter wrote: Bonus problem: In the same configuration, let $B'$ and $C'$ be the $B$ and $C$-antipodes respectively. Show that the circumcenter of $AO_1O_2$ lies on $B'C'$. (This was actually included in the original proposal for this problem, but it was deemed too difficult for the National Olympiad) We might also let $O_3$ be said circumcenter. Check that $O_3$ lies on the circumcircles of $PC'O_1$ and $PB'O_2$. Furthermore $O_3$ lies on $B'C'$.

Solution from Twitch Solves ISL: We let $\overline{BK}$ and $\overline{CL}$ be the altitudes of $\triangle ABC$. The circle with diameter $\overline{BC}$ will be denoted by $\gamma$; and we'll denote the center by $M$. Claim: The circle $(PHC)$ is also tangent to line $BC$. Proof. We are given $\measuredangle PBC = \measuredangle PHB$. Since $\measuredangle BHC = -\measuredangle BAC = -\measuredangle BPC$, it follows $\measuredangle PCB = \measuredangle PHC$ too. $\blacksquare$ Claim: Points $P$, $H$, $M$ are collinear. Actually, $P$ is the inverse of $H$ with respect to $\gamma$. Proof. Line $PH$ bisects $\overline{BC}$ by radical axis on $(BPH)$ and $(CPH)$. Also, $MH \cdot MP = MB^2 = MC^2$. $\blacksquare$ [asy][asy]size(10cm); pair B = dir(180); pair C = dir(0); pair K = dir(65); pair L = dir(90)*K; filldraw(unitcircle, invisible, blue); filldraw(B--C--K--L--cycle, invisible, blue); pair A = extension(B, L, C, K); pair H = extension(B, K, C, L); draw(B--K, blue); draw(C--L, blue); pair O = circumcenter(A, B, C); pair P = 1/conj(H); pair M = origin; filldraw(circumcircle(P, B, H), invisible, orange); pair X = circumcenter(P, H, B); pair O_1 = circumcenter(P, X, O); draw(L--A--K, dotted+deepcyan); draw(circumcircle(X, O, L), dashed+deepgreen); draw(P--M, blue); draw(CP(O_1, P), dotted+grey); dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$K$", K, dir(K)); dot("$L$", L, dir(L)); dot("$A$", A, dir(A)); dot("$H$", H, dir(110)); dot("$O$", O, dir(60)); dot("$P$", P, dir(P)); dot("$M$", M, dir(-90)); dot("$X$", X, dir(X)); dot("$O_1$", O_1, dir(O_1)); /* TSQ Source: B = dir 180 R225 C = dir 0 R315 K = dir 65 L = dir(90)*K unitcircle 0.1 lightcyan / blue B--C--K--L--cycle 0.1 lightcyan / blue A = extension B L C K H = extension B K C L R110 B--K blue C--L blue O = circumcenter A B C R60 P = 1/conj(H) M = origin R-90 circumcircle P B H 0.1 yellow / orange X = circumcenter P H B O_1 = circumcenter P X O L--A--K dotted deepcyan circumcircle X O L dashed deepgreen P--M blue CP O_1 P dotted grey */ [/asy][/asy] We now actually use the condition that $\measuredangle BAC = 45^{\circ}$, which is equivalent to $\measuredangle BHC = 135^{\circ}$ and $\measuredangle BOC = 90^{\circ}$. This means that $O$ is the arc midpoint of $(BC)$. Claim: $LOKH$ is a parallelogram. Proof. Since arcs $KL$ and $OB$ of $\gamma$ measure $90^{\circ}$, the arcs $BL$ and $OK$ are equal, hence $\overline{LO} \parallel \overline{BK}$. Similarly $\overline{KO} \parallel \overline{CL}$. $\blacksquare$ Claim: We have $\measuredangle XPO = 45^{\circ}$. Proof. Since $\triangle HLK \sim \triangle HBC$, and $\overline{HO}$ bisects $\overline{LK}$ by previous claim, it follows \[ \measuredangle BHM = \measuredangle OHL. \]Now \begin{align*} \measuredangle XPO &= \measuredangle XPH + \measuredangle MPO = (90^{\circ} - \measuredangle HBP) + \measuredangle HOM \\ &= 90^{\circ} - \measuredangle MBP + \measuredangle MBH + \measuredangle HOM \\ &= 90^{\circ} - \measuredangle BHM + \measuredangle MBH + \measuredangle HOM \\ &= 90^{\circ} - \measuredangle OHL + \measuredangle MBH + \measuredangle HOM \\ &= 90^{\circ} + \measuredangle(LH, HO) + \measuredangle(MB, BH) + \measuredangle(HO,MO) \\ &= 90^{\circ} + \measuredangle(LH, BH) + \measuredangle(MB, MO) = 45^{\circ}. \end{align*}$\blacksquare$ Claim: We have $X$, $O_1$, $O$, $L$, $H$ are concyclic, in the circle with diameter $\overline{XO}$. Proof. First, since $\triangle LBH$ is a $45^{\circ}$-$45^{\circ}$-$90^{\circ}$ triangle and $XB=XH$, it follows that $\overline{XL}$ is the perpendicular bisector of $\overline{BH}$. Hence, $\measuredangle XLO = 90^{\circ}$. On the other hand, since $\measuredangle XPO = 45^{\circ}$, we have $\measuredangle XO_1O = 90^{\circ}$. This implies concyclic. $\blacksquare$ Finally, $\measuredangle BLO = 135^{\circ}$ and \[ \measuredangle OLO_1 = \measuredangle OXO_1 = 90^{\circ} - \measuredangle XPO = 45^{\circ} \]this implies $O_1$ lies on line $BL$, ergo lies on line $AB$.

Solved with geometry6, mueller.25 Let $P'$ be the $H$ humpty point in $\triangle HBC$. So, $\angle BP'C = \angle BP'H + \angle CP'H = \angle HBC + \angle HCB = \angle BAC$ and so $P' \in (ABC)$ and since $(P'BH)$ is tangent to $BC$, we have $P' = P$, which gives $(PHC)$ is tangent to $BC$ too. So, it suffices to show $O_1 \in AB$. Let $BX \cap (ABC) = Q$. Since $\angle XQP = \angle BQP = \frac{\angle BOP}{2} = \angle XOP$, we have $PQOX$ is cyclic. So, $\angle XPO = \angle XQO = \angle BQO = 90 - \angle OBC = 45^\circ$. Define $O'$ to be the reflection of $O$ across line $AB$. It suffices to show that $O' \in (PXO)$, or that $\angle XO'O = 45^\circ$. But note that $O'$ is the center of $(AHB)$, so $\angle XO'O = \angle HBA = 45^\circ$, and so we are done. $\blacksquare$

Complex cuz Let $A', C'$ be the antipode of $A, C,$ and $M$ the midpoint of $BC$. Note that \[\angle BPA' = 90 - \angle C = \angle HBC = \angle BPH\]so $P, H, A'$ are collinear. It is well known that $M$ lies on this line. Next, note that $OX$ perpendicularly bisects $PB$ so \[\angle PXO = 90 + \angle PHB = 90 + \overarc{PB} + 45 = 135 + \overarc{PB} = 180 - (45 + \overarc{PB}) = 180 - \angle PC'O\]so $(PXOC')$ concyclic. Now, we set complex numbers with the circumcenter as the origin, $C = -1, B = i, C' = 1, A' = a$. We have $M = \frac{i-1}{2}, A = -a$, and \[ap\overline{m} + m = a + p \implies p\left(a\frac{1+i}{2} + 1\right) = \frac{i-1}{2} -a \implies p = \frac{i-1-2a}{a(i+1)+2}\]$O_1$ is the circumcenter of $OPC'$. Since $C' = 1$, then $o_1 = \frac{p}{p+1} = \frac{i-1-2a}{i+ai + 1 - a}$. Now, we want to show that $O_1\in AB$, or $-ai\overline{o_1} + o_1 = -a + i$. However, this resolves to \[-ai\overline{o_1} + o_1 = \frac{-ai(-i-1-\frac{2}{a})(-a)}{(-a)(-i - \frac{1}{a}i + 1 - \frac{1}{a})} + \frac{i-1-2a}{i+ai+1-a} = \frac{ai(-ai-a-2) + i-1-2a}{i+ai + 1 -a} = i-a.\]We conclude $O_1\in AB$. Similarly, because $\angle HPC = \angle A'PC = 90 - \angle B = \angle HCB$, we have $(PHC)$ is tangent to $BC$ as well, so by symmetry, $O_2\in AC$.

Like everyone else, we can show $P$ is the $A$ - Queue point, also $(PHC)$ tangent to $BC$. We just want to show $O_1\in AB$. Notice that the antipode $C'$ of $C$ is on $(POX)$ due to some angles. We will now use moving points. Animate $A$ on $(BAC)$ projectively. Let $\ell$ be the bisector of $C'O$. Clearly $A\to AB\to \ell \cap AB$ is projective. Also $A\to A' \to A'M \cap (ABC)=P\to P'\to P'O\cap \ell$ is projective, where $A'$ is the antipode of $A$, $M$ is the midpoint of $BC$, and $P'$ is the midpoint of $C'P$. We want to show those two bisectors intersect on $AB$, so it's enough to check for 3 cases of $A$. It's easy to verify what happens when $A$ is the antipode of $B$ or $C$, and also when $A$ is the midpoint of major arc $BC$.

Let $A'$ be the $A-$ antipode. Observe that $$\measuredangle BPH = \measuredangle CBH = \measuredangle BAA' = \measuredangle BPA'$$ so $P$ is actually the $A-$ orthic Miquel point. Now let $O'$ be the reflection of $O$ over $AB$. As $OX$ is the perpendicular bisector of $BH$, so we have $$\measuredangle XO'O = \measuredangle HBX = 90^\circ - \measuredangle BAC$$ and $$\measuredangle XPO = \measuredangle BPO + \measuredangle XPB = (90^\circ + \measuredangle BA'P) - (90^\circ + \measuredangle PHB = \measuredangle BA'H + \measuredangle A'HB = \measuredangle A'BH = \measuredangle BAC$$ so from $\angle BAC = 45^\circ$ it follows follows that $O'$ lies on $(PXO)$. Since $AB$ is the perpendicular bisector of $OO'$, we are done.

Rename $P$ to $Q$ since it is the queue point. Let $O'$ be the reflection of $O$ over $AB$ and $F$ the foot from $C$ to $AB$. Let $H$ be the orthocenter of $ABC$. Let $AB$ hit $(BHQ)$ again at $J$. Claim: $FXJH$ is cyclic Proof. First, note that $\angle JHF = 90^{\circ}$. Then, $\angle JXH = 2\angle JBH=2\angle ABH = 90^{\circ}$ as desired. $\blacksquare$ Note by reflection that $\angle XQO=\angle XBO=45 ^{\circ}$, so it suffices to show $\angle XO'O=45^{\circ}$. But by the claim, as $\angle XFB=135 ^{\circ}$ and $\angle BFO=180^{\circ} - \angle OCB = 45 ^{\circ}$ we get that $F,X,O'$ collinear, so \[\angle XO'O=\angle FO'O=\angle FOO'=90^{\circ} - \angle FA = 45 ^{\circ}\]as desired.

Mexico 2019/4 Label the points as the following. We divide the problem into four claims. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(11.11960960330805cm); 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 = -20.76039899260374, xmax = 32.35921061070431, ymin = -9.566482834569173, ymax = 14.026720905139266; /* image dimensions */ pen ffccww = rgb(1.,0.8,0.4); pen ffwwqq = rgb(1.,0.4,0.); pen qqzzcc = rgb(0.,0.6,0.8); pen afeeee = rgb(0.6862745098039216,0.9333333333333333,0.9333333333333333); pen ffcctt = rgb(1.,0.8,0.2); pen zzffff = rgb(0.6,1.,1.); pen ffzztt = rgb(1.,0.6,0.2); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); /* draw figures */ draw(circle((0.5058327053282743,-1.7350287966840203), 4.489852339053012), linewidth(1.) + ffccww); draw(circle((0.5058327053282743,2.754823542368992), 6.349610070941334), linewidth(1.) + ffwwqq); draw((-0.7078110158086872,8.98736852585242)--(-3.9840196337247384,-1.7350287966840203), linewidth(1.) + qqzzcc); draw((-0.7078110158086872,8.98736852585242)--(4.995685044381286,-1.7350287966840203), linewidth(1.) + qqzzcc); draw((-3.9840196337247384,-1.7350287966840203)--(4.995685044381286,-1.7350287966840203), linewidth(1.) + qqzzcc); draw((-3.9840196337247384,-1.7350287966840203)--(3.015283336501507,1.988065555626175), linewidth(1.) + afeeee); draw((4.995685044381286,-1.7350287966840203)--(-3.217261646981921,0.7744218344892114), linewidth(1.) + afeeee); draw(circle((-3.9840196337247384,2.215903286798792), 3.950932083482812), linewidth(1.) + ffcctt); draw(circle((-2.0085535919833317,4.7302895841103965), 3.197593553624637), linewidth(1.) + zzffff); draw((0.5058327053282743,-1.7350287966840203)--(-4.919015326353041,6.054607154390085), linewidth(1.) + ffzztt); draw((-0.7078110158086872,-1.7350287966840203)--(-0.7078110158086872,8.98736852585242), linewidth(1.) + afeeee); draw(circle((-1.7390934641982323,2.485363414583891), 2.2610400852463703), linewidth(1.) + dotted + rvwvcq); draw((-3.217261646981921,0.7744218344892114)--(0.5058327053282743,2.754823542368992), linewidth(1.) + eqeqeq); draw((0.5058327053282743,2.754823542368992)--(-2.0085535919833317,4.7302895841103965), linewidth(1.) + eqeqeq); draw((-3.9840196337247384,2.215903286798792)--(-3.217261646981921,0.7744218344892114), linewidth(1.) + eqeqeq); draw((-3.9840196337247384,2.215903286798792)--(-2.0085535919833317,4.7302895841103965), linewidth(1.) + eqeqeq); /* dots and labels */ dot((0.5058327053282743,-1.7350287966840203),dotstyle); label("$M$", (0.6340298423685562,-1.3996046169777905), NE * labelscalefactor); dot((-3.9840196337247384,-1.7350287966840203),dotstyle); label("$B$", (-3.833322345031858,-1.3996046169777905), NE * labelscalefactor); dot((4.995685044381286,-1.7350287966840203),linewidth(4.pt) + dotstyle); label("$C$", (5.136283218733036,-1.4694069949059219), NE * labelscalefactor); dot((0.5058327053282743,2.754823542368992),linewidth(4.pt) + dotstyle); label("$O$", (0.6340298423685562,3.0328463814585582), NE * labelscalefactor); dot((-0.7078110158086872,8.98736852585242),dotstyle); label("$A$", (-0.5526105824096789,9.349961583954457), NE * labelscalefactor); dot((3.015283336501507,1.988065555626175),linewidth(4.pt) + dotstyle); label("$H_1$", (3.1469154477812893,2.265020224249112), NE * labelscalefactor); dot((-3.217261646981921,0.7744218344892114),linewidth(4.pt) + dotstyle); label("$H_2$", (-3.065496187822412,1.043478610506811), NE * labelscalefactor); dot((-0.7078110158086893,0.007663847746394663),linewidth(4.pt) + dotstyle); label("$H$", (-0.5526105824096789,0.2756524532973649), NE * labelscalefactor); dot((-3.9840196337247384,2.215903286798792),linewidth(4.pt) + dotstyle); label("$X$", (-3.833322345031858,2.509328546997572), NE * labelscalefactor); dot((-4.919015326353041,6.054607154390085),linewidth(4.pt) + dotstyle); label("$P$", (-4.775654447061633,6.348459333044803), NE * labelscalefactor); dot((-2.0085535919833317,4.7302895841103965),linewidth(4.pt) + dotstyle); label("$O_1$", (-1.878855763044177,5.022214152410305), NE * labelscalefactor); dot((-0.7078110158086872,-1.7350287966840203),linewidth(4.pt) + dotstyle); label("$H_3$", (-0.5526105824096789,-1.4694069949059219), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim 1: $P$ is the $A$-queue point of $\triangle ABC$. Proof: Easy angle chase gives us that $BC$ is tangent to $(PHC)$, then by PoP we have that $MC^2=MB^2=MH \cdot MP$. $\blacksquare$ Claim 2: $HOH_1H_2$ is a parallelogram. Proof: $\angle BMH_1=\angle OMH_2$ hence $\angle OH_1H_2=\angle BH_1H_2$ hence $OH_2 \| HH_1$. And similarly $OH_1 \| HH_2$. $\blacksquare$ Claim 3: $\angle XO_1O=90$. Proof: Long angle chase. $\blacksquare$ Finally we how our last claim. Claim 4: $OH_2O_1X$ is cyclic. Proof: By our previous claim $\angle XO_1O=90$, and we also have that $\angle XH_2O=90$ since $XH_2$ is perpendicular to $BH$, hence it’s cyclic. $\blacksquare$ Now since $\angle O_1H_2O= \angle H_2BH=45$ and $\angle O_1XO=45$, hence we have that $O_1$ must lie on $AB$. So we are done.

truly rizztastic problem

Solved with stillwater_45. Very interesting problem. After I realized that $P$ with the $A-$Queue Point of $\triangle ABC$ I thought this could turn into a orthocenter-configuration problem. Guess my wish was not to be granted. Denote by $B_1$ and $C_1$ the feet of the altitudes from $B$ and $C$. We can start off by noting that $O$ lies on $(BC)$ as well since we have $\measuredangle COB = 2\measuredangle BAC = \frac{\pi}{2}$. Now, we prove the following claim, invoking some symmetry into the problem. Claim : Circle $(PHC)$ is also tangent to $\overline{BC}$ and in particular, $P$ is the $A-$Queue Point of $\triangle ABC$. Proof : Let $P'$ denote the intersection of the circles tangent to $\overline{BC}$ passing through points $B$ and $H$ and $C$ and $H$ respectively. Note that, \[\measuredangle CPB = \measuredangle CPH + \measuredangle HPB = \measuredangle BCH + \measuredangle HBC = \frac{\pi}{4} \]which implies that $P'$ lies on $(ABC)$ and indeed $P'\equiv P$. Thus, the first part of the claim is proved. For the second part simply note that, since the radical axis $\overline{PH}$ of circles $(PHB)$ and $(PHC)$ must bisect the common tangent $\overline{BC}$, $\overline{PH}$ passes through the midpoint of side $BC$. It is well known that this implies that $P$ is the $A-$Queue Point of $\triangle ABC$. Also note that, \[\measuredangle OPX = \measuredangle BPX + \measuredangle OPB = \frac{\pi}{2}+\measuredangle PHB + \frac{\pi}{2} + \measuredangle BCP = \measuredangle PHC_1 + \measuredangle C_1HB + \measuredangle BAP = \frac{\pi}{4}\]A similar calculation yields that $\measuredangle YPO = \frac{\pi}{4}$ as well. With the symmetry now established and the above observations, we can prove the following results. Claim : Quadrilaterals $XO_1OC_1$ and $YO_2OB_1$ are cyclic. Proof : We prove the first one, the other follows similarly due to symmetry. First note that, \[\measuredangle OO_1X = 2\measuredangle OPX = \frac{\pi}{2}\]so it suffices to show that $\measuredangle XC_1O=\frac{\pi}{2}$ as well, which we shall do as follows. Note that $C_1$ lies on the perpendicular bisector of segment $BH$ (since $\triangle BC_1H$ is a 45-45-90 right triangle). Also, $X$ lies on this perpendicular bisector as it is the center of $(BHP)$. Thus, $\overline{XC_1}$ is the perpendicular bisector of segment $BH$ implying that $XC_1 \perp BH$. Further note that, $O$ and $C_1$ (due the 45-45-90 right triangle $\triangle AC_1B$) lie on the perpendicular bisector of $AC$. Thus, $OC_1 \perp AC \perp BH$, implying that $OC_1 \parallel BH$. Thus, $XC_1 \perp OC_1$ and indeed $\measuredangle XC_1O = \frac{\pi}{2}$ as desired. Now we are essentially done since, \[\measuredangle OC_1O_1 = \measuredangle OXO_1 = \frac{\pi}{4} = \measuredangle OC_1A\]which implies that $O_1$ lies on $AB$. A similar argument shows that $O_2$ also lies on $AC$ which finishes the proof of the problem.

Nice! Let $A'$ be the $A$-antipode and $C'$ be the $C$-antipode. Claim: $P$ is the orthocenter Miquel point, i.e. it lies on $\overline{HA'}$. Proof: The actual Miquel point $P'$ satisfies $\measuredangle BP'A' = \measuredangle BAA' = \measuredangle CBH$, i.e. $\overline{CB}$ is tangent to $(P'BH)$. So $P = P'$. $\blacksquare$ Claim: $\overline{BO}$ and $\overline{CP}$ intersect at the intersection point $Q$ of $(PXO)$ and $(PBH)$. Proof: $Q$ lies on $\overline{BO}$ as \[\angle PQO + \angle PQB = 180^\circ - \frac 12 \angle PXB + \angle PQB = 180^\circ - \angle PQB + \angle PQB = 180^\circ.\]$Q$ lies on $\overline{CP}$ because \[\measuredangle QPH = \measuredangle OBH = \angle C - 45^\circ = \measuredangle A'PC. \ \blacksquare\]Now note that $C'$ lies on $(POQX)$ because $CO \cdot CC' = 2BC^2$ as $\angle BOC = 90^\circ$. In particular, $\angle COQ = 90^\circ$, so it follows that $O_1$ is the midpoint of $\overline{C'Q}$. To finish, let $N$ be the midpoint of $\overline{AB}$. Note that $\overline{QH} \parallel \overline{AB}$ by the same angle chase as before, so a homothety with ratio $\frac 12$ at $C'$ takes $\overline{QH}$ to $\overline{O_1N}$. It follows that $O_1$ lies on $\overline{AB}$, and the other part follows similarly.