Isn't $O$ should be circumcenter?

demmy wrote: Isn't $O$ should be circumcenter? Yes, $O$ is circumcenter and $H$ is orthocenter Let $D$ be $A$-humpty point. Then $$\angle DAO =\angle DAC - 90 + \angle B = \angle DCB - (\angle HCB) = \angle DCH = \angle DXO$$ hence $A,X,D,O$ cyclic.Let $M$ be midpoint of $BC$. Do inversion about circle $\omega$ center $M$ with radius $MB$,We know that $MD.MA = MB^2$ hence $(ADO)$ is orthogonal to $\omega$ from $(AOD) \leftrightarrow (AOD)$ and $(ABC) \leftrightarrow (HBC)$ we have $X \leftrightarrow Y$ which give us $Y,X,M$ are collinear.

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Let $M$ be the midpoint of $BC$ and let $Z$ be the reflection of $Y$ about $M$. Since triangles $ABC$ and $BHC$ have equal circumradius $Z$ lies on the circumcircle of $BHC$. We conclude with an angle chase: $$\angle HXZ+\angle HXY=\angle HCZ+\angle YAO=90^{\circ}-\angle ABC+\angle YBC+\angle YCB+90^{\circ}-\angle ACB=180^{\circ}$$

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A really interesting extension of the Humpty Point configuration which I'm surprised I didn't know before. We denote by $H_A$ and $Q_A$ the $A-$Humpty and $A$-Queue points of $\triangle ABC$ respectively. We first deal with the following claim. Claim : $X$ lies on $(AH_AO)$. Proof : Simply note that, $$\measuredangle OXH_A= \measuredangle HCH_A = \measuredangle BCH_A + \measuredangle HCB + \measuredangle BCH_A + \measuredangle ACO = \measuredangle CAM + \measuredangle OAC = \measuredangle OAH_A$$ using the well known facts that $(AH_AB)$ and $(AH_AC)$ are tangent to line $BC$ and that $H$ and $O$ are isogonal conjugates. Now, we can also note the following. Claim : Quadrilateral $Q_AHXY$ is cyclic. Proof : It is well known that $Q_AH$ passes through the $A-$anitpodal point $A'$. Further $A$ , $O$ and $A'$ are collinear by definition. Thus, $$\measuredangle YQ_AH = \measuredangle YQ_AA' = \measuredangle YAA' = \measuredangle YAO = \measuredangle YXO = \measuredangle YXH$$ which finishes the proof of the claim. Now, we are simply left to note that by Radical Center on circles $(Q_AHXY)$ , $(AH_AOY)$ and $(AH_AHQ_A)$ (it is well known that the last is a circle), it follows that $Q_AH$ , $AH_A$ and $YX$ concur. Since it is well known that $Q_AH$ and $AH_A$ pass through the midpoint of $BC$, it follows that $XY$ bisects the segment $BC$ as desired.

To begin with, let $T$ be the reflection of $H$ with respect to the midpoint $M$ - it is well known that $T$ lies on $\Omega$ (due to $\angle BTC = \angle BHC = 180^{\circ} - \angle BAC$ and is the antipode of $A$, in particular it lies on $AO$. Now, instead of chasing the collinearity of $X$, $M$, $Y$, let $S$ be the reflection of $X$ with respect to $M$ and aim to show that $Y$, $X$, $S$ are collinear. The advantage is that $S$ lies on $\Omega$, as $\angle BSC = \angle BXC = \angle BHC = 180^{\circ} - \angle BAC$. Moreover, $XM = MS$ and $HM = MT$ imply that $HXTS$ is a parallelogram; in particular, $OH \parallel TS$. Therefore $\angle AYX = 180^{\circ} - \angle AOX = 180^{\circ} - \angle ATS = 180^{\circ} - \angle ACS = \angle AYS$ and we are done!

Here is an alternative great solution from JBMO training. Let $XY$ intersect $BC$ at $M$. Having no other ideas and seen many similarity problems, we boldly claim that $MB^2 = MX \cdot MY$, as then we would analogously have $MC^2 = MX \cdot MY$ and would conclude $MB = MC$. The equality $MB^2 = MX \cdot MY$ is equivalent to (by Power of a Point) $\angle CBX = \angle BYX$. Without loss of generality treat $O$ to be between $H$ and $X$. We compute $\angle CBX = 180^{\circ} - \angle XHC = 180^{\circ} - \angle OHC$ from the circumcircle of $BXHC$. On the other, $\angle BYX = \angle AYB - \angle AYX = 180^{\circ} - \angle ACB - (180^{\circ}- \angle AOX) = 180^{\circ} - \angle ACB - \angle AOH$. It remains to argue $\angle OHC- \angle AOH = \angle ACB$. For this, let $CH \cap AO = T$. Then $\angle TAC = \angle OAC = 90^{\circ} - \angle ABC$ and $\angle ACT = \angle ACH = 90^{\circ} - \angle BAC$, hence $\angle OHC- \angle AOH = \angle OTH = \angle ACB$, as desired.