$\textbf{Claim:} \ $ In triangle $ABC$, denote $D,E,F$ as the altitudes from $A,B,C$ to $BC,CA,AB$ respectively. $AD,BE,CF$ intersect $(ABC)$ at $P,Q,R$ second time. $S$ is the antipode of $A$ on $(ABC)$. Let the tangent at $P$ to $(ABC)$ and $QR$ meet at $T$. If $O$ is the circumcenter of $(ABC)$ and $OD$ intersects $HS$ at $K$, then $TK\parallel BC$. Let the tangent at $A$ intersect $BC$ at $U$. $QR\cap AH=L$ Denote $M$ as the midpoint of $BC$. Pascal at $PPAQRC$ gives that $T,H,AQ\cap CP$ are collinear and pascal at $AAQBCP$ gives that $U,AQ\cap CP,H$ are collinear thus, $T,H,U$ are collinear. \[\frac{HL}{LA}=\frac{HT}{TU}\overset{?}{=}\frac{HK}{KM}=\frac{HD}{OM}=\frac{2.\cos{B}.\cos{C}}{\cos{A}}\]angle \[\frac{HL}{LA}=\frac{HR.\frac{\cos{C}}{\sin{90+B-C}}}{AR.\frac{\cos{A}}{\sin{90+B-C}}}=\frac{2HF}{AH}.\frac{\cos{C}}{\cos{A}}=\frac{2\cos{B}\cos{C}}{\cos{A}}\]Which gives the desired result. Let $AI\cap (ABC)=\{A,M\}, BI\cap (ABC)=\{B,R\}, CI\cap (ABC)=\{C,S\}$ The tangent at $A$ to $(ABC)$ meets $BC$ at $T$. The midpoint of the arc $BAC$ is $N$. $OP\cap IN=K$. Midpoints of $AI,XY$ are $P,G$ respectively.

By replacing

we get that the perpendicular from $T$ to $AD$ (which is the perpendicular bisector of $AD$), $OP$ and $IN$ are concurrent. $\angle XAD=\angle XBD$ yields $X=(ABD)\cap BI$ and $\angle DAY=\angle DCY$ yields $Y=(ACD)\cap CI$. Since $XD=XA$ and $YD=YA,$ the line $XY$ is the perpendicular bisector of $AD$ which passes through $T$. Also $IX.IB=IA.ID=IY.IC$ hence $(BCXY)$ is cyclic. And $IB.IR=IC.IS$ so $XY\parallel RS$. $IG$ bisects $SR$ which gives that $I,G,N$ are collinear since $ISNR$ is parallelogram. By the claim, $IN,OP,XY$ are collinear on $G$ as desired.$\blacksquare$

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Solved without v4913, CyclicISLscelesTrapezoid, ccarolyn4, ld414think, crazyeyemoody907, or kathy Let $a=x^2$, $b=y^2$, $c=z^2$. Then, $AX\cap(ABC)=xy$, so if $d=X$, then $\overline d=\frac{y^2-xz-x^2-xy}{-xy^2z-x^3y}=\frac{x^2+xy+xz-y^2}{xy(x^2+yz)}$. Similarly, if $e=Y$, then $\overline e=\frac{x^2+xy+xz-z^2}{xz(x^2+yz)}$, so the conjugate of the midpoint of $XY$ is $$\frac{x^2z+xz^2+x^2y+xy^2-y^2z-yz^2+2xyz}{2xyz(x^2+yz)}=\frac{(y+z)(x^2-yz+xy+xz)}{2xyz(x^2+yz)}.$$ The midpoint of $AI$ is $\frac{x^2-xy-yz-zx}2$, and its conjugate is $$\frac{yz-xz-xy-x^2}{2x^2yz}.$$The quotient of these two is $-\frac{x(y+z)}{x^2+yz}$, which is real. Therefore, the line through the midpoints passes through the circumcenter of $ABC$.

Hmm, I feel the solutions in this thread are rather complicated. Rephrasing in terms of the triangle formed by the midpoints of the minor arcs, it suffices to show the following. Quote: Let $ABC$ be a triangle and $H$ its orthocenter. Let $BB'$, $CC'$ be diameters of $(ABC)$, let $AH$ intersect $(ABC)$ at $D$, and let $DB' \cap CH = F$, $DC' \cap BH = E$. Then show the line connecting the midpoints of $DH, EF$ passes through the circumcenter of $(ABC)$. To prove this, let $X = AD \cap BC$, and $Y$ be the midpoint of $EF$. By angle chasing one may show htat $DEHF$ and $DC'AB'$ are homothetic, so $EF$ and $BC$ are parallel. Hence $HY$ bisects $BC$ and $DY$ bisects $B'C'$. Thus $XY$ passes through the midpoint of the midpoints of $BC$ and $B'C'$, which is precisely the circumcenter of $(ABC)$.

a_507_bc wrote: Let $I$ be the incenter of a triangle $ABC$. The points $X, Y$ lie on the prolongations of the lines $IB, IC$ after $I$ so that $\angle IAX=\angle IBA$ and $\angle IAY=\angle ICA$. Show that the line through the midpoints of $IA$ and $XY$ passes through the circumcenter of $ABC$. Let $D$, $T$ in order are insections of lines $XY$, $BC$ with line $AI$ Cuz $\angle DBI = \angle XBA = \angle IAX $ so $D$, $B$, $A$, $X$ are concyclic. Similar, $D$, $A$, $Y$, $C$ are concyclic Therefore $ \overline{IX}. \overline{IB} = \overline{ID}. \overline{IA} = \overline{IY}. \overline{IC}$. Lead to $B$, $C$, $X$, $Y$ are concylic Further, we have that $\angle YAD = \angle YCD = \angle YXB $ so $A$, $T$, $X$, $Y$ are concyclic Next, let $U$, $R$, $S$ in order are midpoints of arcs $BC$, $CA$, $AB$ which dont pass through $A$, $B$, $C$ $V$ is midpoint of arc $BAC$ and $N$, $M$ is order are midpoints of segments $XY$, $AI$ Follow Sawayama theorem, we have that $I$, $N$, $V$ are colinear So, use Menelaus theorem for $\triangle IVU$, we need to prove that $\frac{MI}{MU} = \frac{NI}{NV}$ $(*)$ (Since $X, Y$ lie on the prolongations of the lines $IB, IC$, we see that $T$ lie on prolongation of the line $IU$) Cuz line $IT$ pass through $U$; $B$, $C$, $X$, $Y$ are concylic and $U$ is circumcenter of $\triangle BIC$, so $IT \perp XY$ Lead to $TN \| AV$ and $\frac{NI}{NV} = \frac{TI}{TA}$ (Follow Thales theorem ) $(1)$ Again, since $D$, $A$, $X$, $B$ are concyclic and $B$, $C$, $X$, $Y$ are concylic, we have : $IM. ID = \frac{1}{2}. IA. ID = \frac{1}{2} IB. IX = IT. IU$. Which means $\frac{IT}{IM} = \frac{ID}{IU}$ $(2)$ $ AT = AX. \sin \angle AXT = BD. \frac{AI}{BI}. \sin \angle DIC $ (Cuz $AT \perp XY$ and $A$, $T$, $X$, $Y$ are concyclic From $(1)$, $(2)$, we have $(*) \Leftrightarrow \frac{AT}{MU} = \frac{ID}{IU} \Leftrightarrow \frac{BD. AI. \sin \angle DIC}{MU. IB} = \frac{ID}{IU} \Leftrightarrow \frac{ \sin \angle DIC. \frac{\sin \angle IBA}{\sin \angle IAB}}{MU} = \frac{\sin \angle IBD}{\sin \angle DIB}. \frac{1}{IU} $ (Sin theorem for triangles$ IAB$, $IBD$) $\hspace{0.6cm} \Leftrightarrow \sin \angle DIC . \sin \angle DIB = \frac{MU}{UV} $ (Cuz $IU = UC = UV. \sin \angle UVC = UV . \sin \angle IAB$) $\hspace{0.6cm} \Leftrightarrow \sin \angle DIB = \frac{MU}{\sin \angle SVU . UV} = \frac{MU}{SU} = \sin \angle USM$ (Cuz $SR \perp AU$ at $M$) Therefore, cuz $\angle DIB = \angle USM$, so equality $(*)$ obvious right, problem proved

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Synthetic solution https://dgrozev.wordpress.com/2024/07/10/step-by-step-solution-of-a-geometric-problem-from-239-open-mathematical-olympiad-2024/

I completely loved this problem, so many properties! Here goes my solution, completely synthetic: Firstly, let $D,E,F$ be the touchpoints of the incircle of $ABC$ with $BC, AC, AB$ respectively. Also let $L=AI\cap BC$. As stated in some of the solutions above, $L$ is symmetric to $A$ w.r.t. $XY$. Also $AIXY$ and $BCXY$ are concyclic by angle-chase (this was also in some of the above solutions, so I won't reprove it). Since $XY\perp AI$ and $EF \perp AI$, $EF\parallel XY$. By abgle-chase again, $LY\parallel DF$ and $LX\parallel DE$. Now $XE, YF, LD$ are concurrent at the center of homothety that sends one triangle to the other. Let this point be $G$, so $G\in LD\equiv BC$. Define by $O_1$ the cirucmecnter of $AXYI$. By angle-chase again $(AXYI)$ and $(ABC)$ are tangent at $A$, so $O, O_1, A$ are collinear. Now if $M$ is the midpoint of $AI$ and $N$ is the midpoint of $XY$, if we prove the symmetric point to $O_1$ wrt to $N$ and the symmetric point to $A$ wrt $M$ also lie on a line with $O$, clearly $O, M, N$ would be collinear. Now the symmetric of $O_1$ wrt $N$ is $O_2$ - the circumcenter of $XYL$, and the symmetric of $A$ wrt $M$ is $I$. So it's enough to prove that $O, I, O_2$ are collinear. However, $I$ is the circumcenter of $DEF$ and $O_2$ is the circumcenter of $XYL$, so $O_2, I, G$ are collinear. (reminder that $G$ was the center of homothety sending $XYL$ to $EFD$). Furthermore, let $R$ be the orthocenter of $DEF$. Since $I$ is the orthocenter of $XYL$, $G, I, R$ are collinear. Finally, inversion at $I$ (claim proved below) shows us the Euler line of the touchpoint triangle coincides with $OI$, so $O, R, I$ are collinear. So $G, O_2, R, I$ are collinear and we are done. $\square$ For completeness, here is the proof $O, I, R$ are collinear (which is apparently also in the Wikipedia article for Inversive geometry) -- the center of the nine-point circle for $DEF$ lies on it Euler line, so on $RI$. Also inversion at $I$ preserving the circumcircle of $DEF$ sends its midpoint triangle to $ABC$, so the centers of $(ABC)$, the nine-point circle of $(DEF)$ and $I$ are collinear, hence we're done.