Dadgarnia wrote: Given a triangle $ABC$ with orthocenter $H$. Reflection of nine-point circle about $AH$ intersects circumcircle at points $X$ and $Y$. Prove that $AH$ is the external bisector of $\angle XHY$. Proposed by Mohammad Javad Shabani Hopefully correct.

Iran TST #3 2019 P4 wrote: Given a triangle $ABC$ with orthocenter $H$. Reflection of nine-point circle about $AH$ intersects circumcircle at points $X$ and $Y$. Prove that $AH$ is the external bisector of $\angle XHY$. Solution: Denote, $\omega_{N}$ as the nine-point circle WRT $\Delta ABC$, $\omega_{N}'$ as its reflection over $AH$ and $X', Y'$ as the reflections of $X, Y$ over $AH$. Hence, $X', Y' $ $\in$ $\omega_{N}$. and $X, X' , Y, Y'$ are concyclic points. Also reflection of $H$ over $Y'$ be $H_{1}'$ lies on $\odot (ABC)$ Now, Let $HH_1' \cap \odot (ABC)=T_1$ and $HY \cap \odot (ABC)=T_2$. Let $M'$ be the midpoint of $HT_2$ $\implies$ $M'$ $\in$ $\omega_{N}$ $$HY \cdot HT_2 = HH_1' \cdot HT_1 \implies HY \cdot 2 HM' = 2 HY' \cdot HT_1 \implies HM'=HT_1$$Also, $$HY \cdot HM' = HY' \cdot HT_1 \implies Y, Y', M', T_1 \text{ concyclic}$$$HD$ bisects $\angle YHY'$ $\implies$ $HD$ bisects $\angle T_1HM'$, Hence, $T_1$ is the reflection of $M'$ over $AH$ $\implies$ $T_1 $ $\equiv$ $ \omega_{N}' $ $\cap $ $\odot (ABC)$ $ \equiv X$

Let $\Gamma$ be the circumcircle and $\omega_9$ be the $9$-point circle, and let $\Gamma'$, $\omega_9'$ be their reflection in $AH$. Considering negative inversion with center $H$ taking $\Gamma$ to $\omega_9$, by symmetry it also takes $\Gamma’$ to $\omega_9'$, so $X$ goes to one of the intersections of $\omega_9$ and $\Gamma'$, say $Y'$, which is symmetric to $Y$ in $AH$. Now the conclusion is obvious.

Dadgarnia wrote: Given an acute-angled triangle $ABC$ with orthocenter $H$. Reflection of nine-point circle about $AH$ intersects circumcircle at points $X$ and $Y$. Prove that $AH$ is the external bisector of $\angle XHY$. Proposed by Mohammad Javad Shabani Redefine $X$ and $Y$ by reflecting $\odot(ABC)$ in $A$-altitude and intersecting with the nine-point circle; noting that it still suffices to show $AH$ bisects angle $XHY$. Suppose $G$ lies on $AD$ with $\frac{AG}{DG}=2$, and $N$ is the center of the nine-point circle, $O$ the circumcenter and $M$ the centroid of $\triangle ABC$. Observe that $(HM;NO)=-1$ so $\angle HGM=90^{\circ}$ yields that $GH$ bisects angle $OGN$; so $G$ lies on the perpendicular bisector of $XY$. Let $XY$ meet $AH$ at $Z$ and $\odot(DEF)$ meet $AH$ at $K$. Reflect $H$ in $BC$ to get $H_A$. Note that $r^2=ZX \cdot ZY=ZA \cdot ZH_A=ZK \cdot ZD$. So it is sufficient to show that $r^2=ZG \cdot ZH$, or equivalently, $(AG;HD)=(H_AH;GK)$. This is a direct computation, notably using $HD=DH_A, AG=2DG, AH=2AK$, so I'll leave it at that.

Let $D,E,F$ denote foot of altitude from $A,B,C$ to $BC$, $CA$, $AB$. Let $P_A = EF\cap BC$ and similarly $P_B = DF\cap AC$, $P_C=DE\cap AB$. Let $G$ be the centroid, $N$ be the nine-point center, $O$ be the circumcenter of $\triangle ABC$. Finally, let $M$ be the midpoint of $BC$. We first prove the following lemma. Lemma: $\odot(ABC)$, $\odot(DEF)$ and $\odot(HG)$ are coaxal with radical axis $\overline{P_AP_BP_C}$. Proof: It suffices to show that $P_A$ lie on the radical axis. First, note that $P_AB\cdot P_AC = P_AE\cdot P_AF$ so the first two circles are done. To see the third circle, recall that the projection $K_A$ of $H$ on to $AM$ lie on both $\odot(HG)$ and $\odot(BHC)$. Thus this power is also equal to $PK_A\cdot PH$, done. $\blacksquare$ Back to the main problem. Let $P$ be the projection of $G$ onto $AH$ and let $O'$ be the reflection of $O$ across $AH$. Notice that $P$ is the centroid of $\triangle HOO'$. Therefore if $N'$ is the reflection of $N$ across $AH$, then $P\in ON'$. Therefore $PX=PY$. However, if $T=\overline{P_AP_BP_C}\cap AH$, then by the lemma, $TX\cdot TY = TH\cdot TP$ so $X,Y,H,P$ are concyclic. Hence we are done.

whoa these solutions are so overcomplicated

Solved with eisirrational, goodbear, Th3Numb3rThr33. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ size(7cm); defaultpen(fontsize(10pt)); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-2.7610213408027877,5.933356659746497)--(-5.28,1.09)--(2.46,1.09)--cycle, rvwvcq); /* draw figures */ draw((-2.7610213408027877,5.933356659746497)--(-5.28,1.09), rvwvcq); draw((-5.28,1.09)--(2.46,1.09), rvwvcq); draw((2.46,1.09)--(-2.7610213408027877,5.933356659746497), rvwvcq); draw(circle((-1.41,2.15397929378948), 4.0135958861864465), wrwrwr); draw(circle((-2.085510670401394,2.979688682978509), 2.0067979430932237), wrwrwr); draw(circle((-4.112042681605576,2.15397929378948), 4.0135958861864465), wrwrwr); draw(circle((-3.4365320112041817,2.979688682978509), 2.0067979430932237), wrwrwr); draw((-4.5578028381004945,4.64401729150342)--(-0.12037089975989268,2.5729164294431177), wrwrwr); draw((-5.4016717818456845,2.572916429443112)--(-0.9642398435050779,4.644017291503414), wrwrwr); draw(circle((-2.3842613779589934,3.5116783298732486), 2.45081088682496), linetype("2 2") + wrwrwr); draw(circle((-3.1377813036465847,3.5116783298732486), 2.4508108868249603), linetype("2 2") + wrwrwr); draw((-2.7610213408027877,5.933356659746497)--(-2.7610213408027877,-1.6253980721675363), wrwrwr); /* dots and labels */ dot("$A$",(-2.7610213408027877,5.933356659746497),N); dot("$B$",(-5.28,1.09),SW); dot("$C$",(2.46,1.09),SE); dot("$H$",(-2.7610213408027877,3.8053980721675362),dir(300)); dot("$D$",(-2.7610213408027877,1.09),dir(-60)); dot("$Y'$",(-0.9642398435050779,4.644017291503414),NE); dot("$X'$",(-0.12037089975989268,2.5729164294431177),SE); dot("$X$",(-4.5578028381004945,4.64401729150342),NW); dot("$Y$",(-5.4016717818456845,2.572916429443112),W); dot("$H_A$",(-2.7610213408027877,-1.6253980721675363),S); /* end of picture */ [/asy][/asy] Let $\Gamma$ be the circumcircle, $\omega$ the nine-point circle, and $\Gamma'$ and $\omega'$ their respective reflections across $\overline{AH}$. Also let $\Psi$ denote negative inversion at $H$ with radius $\sqrt{AH\cdot HD}$. Note that: $\Psi$ swaps $\Gamma$ and $\omega$; $\Psi$ swaps $\Gamma'$ and $\omega'$. Let $\Psi$ send $X$ and $Y$ to $X'$ and $Y'$, respectively. Then $X'$ and $Y'$ are the intersections of $\Gamma'$ and $\omega$. Thus $X$ and $Y$ are the reflections of $Y'$ and $X'$ across $\overline{AH}$, so $XYX'Y'$ is an isosceles trapezoid and $\overline{AH}$ bisects $\angle XHY$. End proof.

The problem is just a shadow of the following: Claim: Let $\ell$ be a line through the exsimilicenter $H$ of circles $\omega, \gamma$. Let the reflection $\gamma'$ of $\gamma$ across $\ell$ intersect $\omega$ at $X,Y$; then $\ell$ is the external angle bisector of $\angle XHY$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); 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 = -2.643470659725064, xmax = 3.651868867601753, ymin = -2.191103221604837, ymax = 1.9984781363025583; /* image dimensions */ /* draw figures */ draw(circle((0,0), 1), linewidth(2)); draw((xmin, -3.4494843194342395*xmin-1.743302900012535)--(xmax, -3.4494843194342395*xmax-1.743302900012535), linewidth(2)); /* line */ draw(circle((-0.30322843739700656,0), 0.4), linewidth(2)); draw(circle((-0.6761890095104617,-0.10812067473744177), 0.4), linewidth(2)); draw(circle((-0.9324014302836379,-0.2703016868436044), 1), linewidth(2)); draw((-0.9901987764884226,0.13966525352008985)--(0.06186946814305705,-0.16341211512693504), linewidth(2)); draw((-0.17043406540790262,0.3773137351966124)--(-0.8972773630680387,-0.44146725102284434), linewidth(2)); /* dots and labels */ label("$\omega$", (-0.19772028586812995,1.0552739562501081), NE * labelscalefactor); dot((-0.505380728995011,0),dotstyle); label("$H$", (-0.48287503797701015,0.05723232386902697), NE * labelscalefactor); dot((-0.7096344581992715,0.7045700360760642),dotstyle); label("$\gamma$", (-0.5102937641413255,0.4301269997037166), NE * labelscalefactor); dot((-0.9901987764884226,0.13966525352008985),linewidth(4pt) + dotstyle); label("$X$", (-1.0860870135919491,0.13948850236197322), NE * labelscalefactor); dot((-0.8972773630680387,-0.44146725102284434),linewidth(4pt) + dotstyle); label("$Y$", (-0.9928633446332767,-0.5569471422116383), NE * labelscalefactor); dot((-0.17043406540790262,0.3773137351966124),linewidth(4pt) + dotstyle); label("$X'$", (-0.1483665787723622,0.41915950923799045), NE * labelscalefactor); dot((0.06186946814305705,-0.16341211512693504),linewidth(4pt) + dotstyle); label("$Y'$", (0.10388570193933952,-0.20598744730840102), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Proof: Let the reflections of $\omega, X,Y$ across $\ell$ be $\omega', X', Y'$. Consider the inversion at $H$ swapping $\omega$ and $\gamma$; then this inversion must also swap $\omega'$ and $\gamma'$. It follows that $X$ is swapped with $Y'$, and the desired result is immediate.

Let $X'$ and $Y'$ be the reflection of $X$ and $Y$ about $AH$. By $\sqrt{-HA.HD}$ inversion, $X$ is sent to $Y'$ and $Y$ is sent to $X'$ and oobviously $XX'Y'Y$ is an isoceles trapezoid so $H$ is the intersection of diagonals and $AH$ is the perpendicular bisector of parallel sides so the conclusion follows.