India TST 2019 Day 3 P2 wrote: Let $ABC$ be an acute-angled scalene triangle with circumcircle $\Gamma$ and circumcenter $O$. Suppose $AB < AC$. Let $H$ be the orthocenter and $I$ be the incenter of triangle $ABC$. Let $F$ be the midpoint of the arc $BC$ of the circumcircle of triangle $BHC$, containing $H$. Let $X$ be a point on the arc $AB$ of $\Gamma$ not containing $C$, such that $\angle AXH = \angle AFH$. Let $K$ be the circumcenter of triangle $XIA$. Prove that the lines $AO$ and $KI$ meet on $\Gamma$. Proposed by Anant Mudgal Solution: WLOG, assume, $AB <AC$. Let $F'$ be midpoint of arc $BC$ and $D$ be projection of $A$ over $\overline{BC}$. Let $\stackrel{\longrightarrow}{F'D}$ $\cap$ $\odot (ABC)$ $=$ $X'$ and $F_1$ $\in$ $\stackrel{\longrightarrow}{F'D}$ such $DF'$ $=$ $DF_1$. Also, Let $H'$ be reflection of $H$ over $\overline{BC}$ $\implies$ $\angle X'F'H'$ $= $ $\angle HFD$ $=$ $\angle DF_1H$ $\implies$ $\angle AX'H$ $=$ $\angle AFH$ $\implies$ $X \equiv X'$. Let $A_1$ be the $A-$antipode in $\odot (ABC)$. Let $AI$ $\cap$ $\overline{BC}$ $=$ $E$. By Shooting Lemma, $AXDE$ is cyclic. Let $XD$ $\cap$ $\odot (XIA)$ $=$ $G$ $\implies$ By Reim's Theorem, $GI$ $||$ $DE$ and $A_1$ $\in$ $XE$. Let $A_1I$ $\cap$ $\odot (ABC)$ $=$ $J$. Applying $\sqrt{F'E \cdot F'A}$ inversion, $J$ $\in$ $I_DF'$ where, $I_D$ is the $A-$incircle touch point. Let $I'$ be foot from $I$ to $AH$ $\implies$ $AJI'I$ is cyclic. By Reim's Theorem $\implies$ $F'$ $\in$ $JI'$. Also, Since, $II'DI_D$ is a rectangle. Hence, If $I'F'$ $\cap$ $ID$ $=$ $L$ $\implies$ $\Delta I'LD$ is isosceles. Also, notice, $\sqrt{F'E \cdot F'A}$ inversion swaps $X$ and $D$ $\implies$ $\odot (XID)$ is tangent to $AF'$. $$\angle AIK=\frac{1}{2} (180^{\circ}-\angle AKI)=90^{\circ}-\frac{1}{2}\angle AKI=90^{\circ}-\angle AXI=\angle XAI+\angle AIX-90^{\circ}$$$$=\angle AIX-\angle F'EA_1=\angle IDX-\angle ADX=\angle IDA=90^{\circ}-\angle F'I_DC=90^{\circ}-\angle JI_DB=90^{\circ}-\angle JAF'$$$$=90^{\circ}-\angle JA_1F'=\angle F'IA_1 \implies \angle AIK=\angle F'IA_1 \implies K \in IA_1 $$

Kayak wrote: Let $ABC$ be an acute-angled scalene triangle with circumcircle $\Gamma$ and circumcenter $O$. Suppose $AB < AC$. Let $H$ be the orthocenter and $I$ be the incenter of triangle $ABC$. Let $F$ be the midpoint of the arc $BC$ of the circumcircle of triangle $BHC$, containing $H$. Let $X$ be a point on the arc $AB$ of $\Gamma$ not containing $C$, such that $\angle AXH = \angle AFH$. Let $K$ be the circumcenter of triangle $XIA$. Prove that the lines $AO$ and $KI$ meet on $\Gamma$. Proposed by Anant Mudgal Solution. Let $Y$ be the reflection of $A$ in $BC$ and let $A'$ be the $A-$ antipode in $\odot(ABC)$. Let $AI$ meet $\odot(ABC)$ at $M$. Let $YM$ meet $\odot(ABC)$ again at $Z$. Let $J$ be the midpoint of $AZ$. Let $E$ be the feet from $A$ on $BC$, $AH$ meet $\odot(ABC)$ at $H'$ and $AI$ meet $BC$ at $D$. Let $I_a$ be the A-excenter of $\triangle ABC$. We have $F$ is the reflection of $M$ in $BC$. Therefore, $\angle AXH=\angle HFA = \angle YMA = \angle ZAY = \angle ZXH'$. Also $\angle XAH = \angle XZH'$, so $X$ is the center of spiral similarity sending $H'Z$ to $HA$. Therefore $X$ is the center of spiral similarity sending $HH'$ to $AZ$. This gives $\angle XEH = \angle XJA$ or $X,E,J,A$ are concyclic. We also have $\angle AZM= \angle ACM = ADB$, and as $DA = DY$ we get $D$ is the circumcenter of $\triangle AYZ$. This gives $\angle AJD = 90^{\circ} = \angle DEA$, so $A,J,D,E,X$ are concyclic or $\angle AXD = 90^{\circ}$ which gives $X,D,A'$ are collinear. We also have $\angle XED = \angle EAD = \angle MAA' = \angle MXD$ giving $M,E,X$ are collinear. We have $\angle KIA = \angle IXD$, therefore we need to show that $A'I$ is tangent to $\odot(XID)$. Therefore we need to show $180^{\circ}=\angle ADX + \angle XIA' = \angle AEX + \angle XIA'$. From shooting lemma, $MI^2 = ME\cdot MX$, or $\angle AIX = \angle IEX$. So we need to show $\angle MIA' = \angle IEA$. Now as $\triangle IA'I_a$ is isosceles, $\sqrt{bc}$ inversion and flip over angle bisector gives us that $\angle AEI = \angle AI_aA' = \angle A'IM$ and we are done. $~\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -9.274342948502314, xmax = 28.233233326026244, ymin = -8.177485964177107, ymax = 11.83680268723112; /* image dimensions */ pair A = (0.,7.428260407433981), B = (-2.,0.), C = (6.,0.), O = (2.,2.9064040779614393), H = (0.,1.6154522515111018), I = (1.0720099711944011,2.3542947625003086), D = (1.5694172532186919,0.), M = (2.,-2.0380053573898533), F = (2.,2.0380053573898533), X = (-2.943533520656443,2.9994685423772225), Y = (0.,-7.428260407433981), Z = (5.225139830055946,6.654157770638907), J = (2.612569915027973,7.041209089036444), K = (-0.6191905853877252,4.6472118467586165); draw(A--B--C--cycle, linewidth(2.)); /* draw figures */ draw(circle(O, 4.944409435351292), linewidth(1.)); draw(A--M, linewidth(1.)); draw(M--F, linewidth(1.)); draw(A--X, linewidth(1.)); draw(X--H, linewidth(1.)); draw(A--F, linewidth(1)); draw(F--H, linewidth(1.)); draw(Y--Z, linewidth(1.)); draw(A--Y, linewidth(1.)); draw(A--(4.,-1.615452251511102), linewidth(1.)); draw((0.,-1.615452251511102)--M, linewidth(1.)); draw(A--Z, linewidth(1.)); draw(X--(0.,-1.615452251511102), linewidth(1.)); draw(X--Z, linewidth(1.)); draw(X--(0.,0.), linewidth(1.)); draw(X--J, linewidth(1.)); draw(X--D, linewidth(1.)); draw(D--(4.,-1.615452251511102), linewidth(1.)); draw(X--I, linewidth(1.)); draw(I--(4.,-1.615452251511102), linewidth(1.)); draw(K--I, linewidth(1.)); draw(circle((0.7847086266093466,3.714130203716991), 3.7961204931927455), dotted); /* dots and labels */ dot(A,dotstyle); label("$A$", (-0.2971197745987578,7.5), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (-2.387705993179038,-0.4), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (6.05,-0.35), NE * labelscalefactor); dot(O,linewidth(4.pt) + dotstyle); label("$O$", (2.1,3), NE * labelscalefactor); dot(H,linewidth(4.pt) + dotstyle); label("$H$", (0.1,1.7), NE * labelscalefactor); dot(I,linewidth(4.pt) + dotstyle); label("$I$", (1.2093320593782086,2.3), NE * labelscalefactor); dot((4.,-1.615452251511102),linewidth(4.pt) + dotstyle); label("$A'$", (4.,-2), NE * labelscalefactor); dot(D,linewidth(4.pt) + dotstyle); label("$D$", (1.25,-0.5), NE * labelscalefactor); dot(M,linewidth(4.pt) + dotstyle); label("$M$", (1.9779299338562528,-2.5), NE * labelscalefactor); dot(F,linewidth(4.pt) + dotstyle); label("$F$", (2.131649508751862,1.5683350842044723), NE * labelscalefactor); dot(X,linewidth(4.pt) + dotstyle); label("$X$", (-3.3715112725109346,3.228506493077044), NE * labelscalefactor); dot(Y,dotstyle); label("$Y$", (0.1,-8), NE * labelscalefactor); dot(Z,linewidth(4.pt) + dotstyle); label("$Z$", (5.359760581559647,6.6), NE * labelscalefactor); dot((0.,-1.615452251511102),linewidth(4.pt) + dotstyle); label("$H'$", (-0.543071094431732,-2.028702968352767), NE * labelscalefactor); dot((0.,0.),linewidth(4.pt) + dotstyle); label("$E$", (0.05,-0.5), NE * labelscalefactor); dot(J,linewidth(4.pt) + dotstyle); label("$J$", (2.746527808334297,7.1), NE * labelscalefactor); dot(K,linewidth(4.pt) + dotstyle); label("$K$", (-0.4815832644734884,4.5), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Nice synthetic solutions above! Here's one with complex numbers. Let $M, M'$ be the midpoints of arcs $BC$ not containing $A$, containing $A$ respectively. Let $A'$ be the point diametrically opposite $A$ on $\Gamma.$ Let $Y, D$ be the points on $\Gamma$ such that $MY \perp IA', MD \perp AF.$ We will use complex numbers to show that $YI, DH$ concur on $\Gamma.$ Clearly, this is equivalent simply to $\angle ADH = \angle AYI,$ or $$\frac{\frac{a-y}{i-y}}{\frac{a-d}{h-d}} \in \mathbb{R}.$$ So let's simply show this. First of all, let $a, b, c$ be complex numbers of magnitude $1$ such that $A = a^2, B = b^2, C = c^2, I = -ab-bc-ca, M = -bc, M' = bc, H = a^2+b^2+c^2.$ Since $MBFC$ is a rhombus, we've that $F = b^2+c^2+bc.$

Hence, we obtain that: $$\frac{a-d}{h-d} = \frac{a^2 + \frac{a^2bc(b^2+bc+c^2-a^2)}{a^2b^2 + a^2bc+a^2c^2-b^2c^2}}{a^2+b^2+c^2 + \frac{a^2bc(b^2+bc+c^2-a^2)}{a^2b^2 + a^2bc+a^2c^2-b^2c^2}} = \frac{a^2(a^2+bc)}{(a^4+2a^2bc+a^2b^2+a^2c^2-b^2c^2)},$$ and $$\frac{a-y}{i-y} = \frac{a^2 + \frac{a^2(ab+bc+ca-a^2)}{ab+ac+a^2-bc}}{-ab-bc-ca + \frac{a^2(ab+bc+ca-a^2)}{ab+ac+a^2-bc}} = \frac{a^2(2ab+2ac)}{a^2(ab+bc+ca-a^2) - (ab+bc+ca)(ab+ac+a^2-bc)} = \frac{2a^3(b+c)}{-a^4-a^2b^2-a^2c^2+b^2c^2-2a^2bc}.$$ As a result, we obtain that: $$\frac{\frac{a-y}{i-y}}{\frac{a-d}{h-d}} = - \frac{2a^3(b+c)}{a^2(a^2+bc)},$$ so we just need to show that $$\frac{a^3(b+c)}{a^2(a^2+bc)} \in \mathbb{R}.$$ Since its complex conjugate is $$\frac{\frac{1}{a^3b}+\frac{1}{a^3c}}{\frac{1}{a^4} + \frac{1}{a^2bc}} \cdot \frac{a^6bc}{a^6bc} = \frac{a^3b+a^3c}{a^4+a^2bc},$$ which is equal to itself, we've shown that $\frac{a^3(b+c)}{a^2(a^2+bc)} \in \mathbb{R}.$ As a result, we have our desired $\frac{\frac{a-y}{i-y}}{\frac{a-d}{h-d}} \in \mathbb{R}.$ Hence, we have that $\angle ADH = \angle AYI$ and so $YI \cap DH \in \Gamma,$ say at $Z.$ We claim that $Z = X.$ Indeed, we have that $\angle AZH = 180 - \angle AM'D = \angle M'AF = \angle AFH,$ where we used that $AF || M'D$ and that $AM'FH$ is a parallelogram. Hence, we have that $Z = X.$ Now, observe that $\angle KIA = 90 - \angle AXI = 90 - \angle IMY = \angle MIA',$ which implies that $A' \in KI.$ As $A' \in \Gamma, AO,$ the problem is solved. $\square$

H',A' reflect A,H wrt BC, M,J,L be projection of I wrt ABC, (AI)-(O)=T, AT-BC=K, MN perp LJ, PQN perp IN with P,Q in AC,AB. Firstly, IQB=INJ=90-TIA=TAI=IDC because K is radical center of (AI), (IBC) so KIA=KDA so IDC=IAK so BQ=BD, similarly CD=CP. Let GD cuts (O) at X then XB/XC=DB/DC=QB/PC so AIPQX cyclic. Notice DH.DA'=DX.DG so HXA'G cyclic so AXH=AXD-DXH=AH'G-HA'G=H'GA'=HFA so X satisfy the problem condition. Now center (AXI) lies on IN but IN pass E antipode A by parallel lemma with U,V be projection of N wrt AC,AB which UJ/JC=VL/LB by UJ/VL=NJ/NL=BL/CJ, done!

Beautiful configuration! Let $I_A$ be the excentre opposite to $A$. Let $M$ be midpoint of arc $BC$ not containing $A$. Let $AM\cap BC = L$. Let $A'$ be antipode of $A$ w.r.t. $\Gamma$. Let $M'$ be reflection of $M$ in $L$. Let $A_{1}$ be midpoint of $BC$. Lemma 1: $M'$ is orthocentre of $\triangle AHF$ . Proof : It easily follows by angle chasing that $AI\perp HF$. $\therefore AM'\perp HF$. Also $A_{1}$ is midpoint of $MF$ and $L$ is midpoint of $MM'$. $\therefore LA_{1}\parallel M'F\implies BC\parallel M'F\implies M'F\perp AH$. So $M'$ is orthocentre of $\triangle AHF$. Lemma 2: $X,L$ and $A'$ are collinear. Proof: Let $A''$ be reflection of $A'$ in $L$. Let $A'L\cap\Gamma =X'$. $LX'.LA''=LX'.LA'=LM.LA=LM'.LA$ So $X'A''M'A$ is cyclic. Also now $MA'M'A''$ is a parallelogram. So $\angle AM'A''= 90^{\circ}$. Now $A_1$ is midpoint of $A'H$ and $L$ is midpoint of $A'A''$. So $A''H\parallel LA_{1} \implies\angle AHA''=90^{\circ}$ So $AM'HA''X'$ is cyclic $\implies\angle AX'H= 180^{\circ}-\angle AM'H=\angle AFH$. So $X=X'$ as desired. Now we want to prove that $K,I,A'$ are collinear. This is true iff $\angle A'IL=\angle AIK$ Now $\angle AIK=90^{\circ}-\angle AXI=\angle A'XA-\angle AXI=\angle IXA'$. So all we need to prove is that $\angle A'IL=\angle IXA'$ Now the $A-$excentre comes into play. $IBI_{A}C$ is cyclic. So $IL.LI_{A}=BL.LC=XL.LA'$ So $XIA'I_{A}$ is cyclic. Finally note that $A'M$ is $\perp$ bisector of $II_{A}$. So $A'I=A'I_{A}$ which means that $A'$ is midpoint of arc $II_{A}$ of $\odot (XIA'I_{A})$ $\implies \angle A'IL=\angle A'II_{A}=\angle A'I_{A}I=\angle A'XI$ as desired. Q.E.D.

This was just too beautiful. Heres my solution Proof : We'll alter the wording a bit. Let $E$ be the centre of $\odot(AIK)$. Let $J$ be the midpoint of $\widehat{BC}$ not containing $A$ . Let $\{H_A,A'\}$ be the reflections of $H,A$ in $BC$. Let $K=AH \cap BC$ and let $D$ be the foot of the $A$-angle bisector. Its a well known fact that $A' \in \odot(BHC)$. Let $JK \cap \odot(ABC)=X'$. Notice that by PoP we have $JK \cdot KX'=BK \cdot KC=HK \cdot AK=HK \cdot KA' \implies HX'A'J$ is concyclic. Now $\angle AX'J=\angle AH_AJ$ and $\angle HX'J=\angle HA'J \implies \angle A'JH_A=\angle HX'A \implies X'=X$ by reflection arguements. Now notice that $\angle JBC=\angle JAC=\angle JAB=\angle JXB \implies JB$ is tangent to $\odot(XBK) \implies JB^2=JK \cdot JX$. But its well known that $JB^2= JD \cdot JA \implies XKDA$ is cyclic. Now let $G$ be the $A$-antipode w.r.t $\omega$. Notice that $\angle XDK=\angle XAK =\angle XGH_A$. But its well known that $H_AG \parallel BC \implies \{X,D,G\}$ is a collinear triplet. Now we claim that $\{E,I,G\}$ is a collinear triplet. Notice that $\angle EIA=90^\circ-\angle AXI=\angle IXD$. So we just need to show that $EI$ is tangent to $\odot(XID) \iff \odot(XID)$ othogonal to $\odot(AXI) \iff EX$ tangent to $\odot(XID)$. But for this notice that $\angle EXI=90^\circ -\angle XAI=\angle IDX$. We are done $\blacksquare$. 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label("$H_A$", (-4.110599817501999,-4.21740763323496), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

here's my solution : angle chasing and a tiny inversion and the and (that can be avoided) [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -1.14, xmax = 18.42, ymin = -8.12, ymax = 4.92; /* image dimensions */ pen sqsqsq = rgb(0.12549019607843137,0.12549019607843137,0.12549019607843137); draw((8.8,2.6)--(10.66,-2.56)--(0.66,-2.84)--cycle, linewidth(2) + sqsqsq); /* draw figures */ draw((8.8,2.6)--(10.66,-2.56), linewidth(2) + sqsqsq); draw((10.66,-2.56)--(0.66,-2.84), linewidth(2) + sqsqsq); draw((0.66,-2.84)--(8.8,2.6), linewidth(2) + sqsqsq); draw(circle((5.625269143988581,-1.4596122853064415), 5.153578166198897), linewidth(2)); draw((10.674744952164046,-0.4289998568093245)--(5.769512800207099,-6.611171435967791), linewidth(2)); draw((9.899513598642306,1.4196588476339498)--(2.450538287977162,-5.519224570612883), linewidth(2)); draw((8.8,2.6)--(9.899513598642306,1.4196588476339498), linewidth(2)); draw((8.8,2.6)--(13.531750109264305,-2.4795909969406), linewidth(2)); draw((13.531750109264305,-2.4795909969406)--(10.66,-2.56), linewidth(2)); draw((10.674744952164046,-0.4289998568093245)--(0.4968587230860095,-0.9509134845413034), linewidth(2)); draw((0.4968587230860095,-0.9509134845413034)--(5.769512800207099,-6.611171435967791), linewidth(2)); draw((8.8,2.6)--(5.769512800207099,-6.611171435967791), linewidth(2)); draw((8.8,2.6)--(8.945823914051383,-2.6079969304065616), linewidth(2)); draw(circle((9.151554464478997,0.722918555564357), 1.9097186417217886), linewidth(2)); draw((8.8,2.6)--(10.674744952164046,-0.4289998568093245), linewidth(2)); draw(circle((11.165875054632155,0.06020450152969948), 3.47101223682382), linewidth(1.6) + linetype("4 4")); draw(circle((5.595161286332185,-3.2111263369392353), 3.4045124652170946), linewidth(1.2) + linetype("4 4")); /* dots and labels */ dot((8.8,2.6),linewidth(3pt) + dotstyle); label("$A$", (8.88,2.72), NE * labelscalefactor); dot((10.66,-2.56),linewidth(3pt) + dotstyle); label("$B$", (10.74,-2.44), NE * labelscalefactor); dot((0.66,-2.84),linewidth(3pt) + dotstyle); label("$C$", (0.74,-2.72), NE * labelscalefactor); dot((8.945823914051383,-2.6079969304065616),linewidth(3pt) + dotstyle); label("$D$", (9.02,-2.48), NE * labelscalefactor); label("$d$", (3.04,2.54), NE * labelscalefactor); dot((5.769512800207099,-6.611171435967791),linewidth(3pt) + dotstyle); label("$N$", (5.84,-6.5), NE * labelscalefactor); dot((10.674744952164046,-0.4289998568093245),linewidth(4pt) + dotstyle); label("$X$", (10.76,-0.26), NE * labelscalefactor); dot((-3.58,3.32),dotstyle); label("$F$", (-1.08,3.26), NE * labelscalefactor); dot((7.754182609049265,-0.5787638896551299),linewidth(3pt) + dotstyle); label("$I$", (7.84,-0.46), NE * labelscalefactor); dot((2.450538287977162,-5.519224570612883),linewidth(3pt) + dotstyle); label("$A'$", (2.54,-5.4), NE * labelscalefactor); dot((9.899513598642306,1.4196588476339498),linewidth(3pt) + dotstyle); label("$H$", (9.98,1.54), NE * labelscalefactor); dot((13.531750109264305,-2.4795909969406),linewidth(3pt) + dotstyle); label("$T$", (13.62,-2.36), NE * labelscalefactor); dot((0.4968587230860095,-0.9509134845413034),linewidth(3pt) + dotstyle); label("$J$", (0.58,-0.84), NE * labelscalefactor); dot((2.2159995941140034,-2.796432011364808),linewidth(3pt) + dotstyle); label("$W$", (2.3,-2.68), NE * labelscalefactor); dot((9.151554464478997,0.722918555564357),linewidth(3pt) + dotstyle); label("$K$", (9.24,0.84), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] let $N=AI \cap (ABC)$ claim(1): $N,D,X$ are collinear proof: notice that $N$ is the reflection of $F$ wrt $BC$ let $F'$ be the reflection of $F$ wrt $AH$ and let $X'=ND \cap(AHF')$ and $H'$ be the reflection of $H$ wrt $BC$ It's clear that $DN=DF'$ and $D,N,F'$ are collinear so $DF'.DX=DH.DA=DH'.DA=DX.DN$ SO $X' \in (ABC) \implies X'=X$ $\blacksquare$ now suppose that $K,I,A'$ are collinear notice that $$\angle{MAI}=90-\angle{KIA}=90-\angle{KIA}=\angle{}AXI$$so if we put $J=XI \cap (ABC)$ and $M=IA' \cap (ABC)$ it suffices to show that $NJ \parallel AM$ claim(2):$NJ \parallel AM$ proof: let $J' $ a point on $(ABC)$ such that $NJ \parallel AM$ let $W=J'N \cap BC$ and $T=AM \cap BC$ since $(AMI) , (BIC) $ are tangent at $I$ so by radical axis on $(ABC),(BIC),(AIM)$ we have $\angle{TIA}=90$ so $ATDI$ is cyclic $\angle{ANW}= \angle{TAI}=\angle{IDC}$ so $IDNW$ is cyclic now invert wrt $N$ with radius $NI$ that swaps $BC$ and $(ABC) $ $so W^*=J' , D^*=X$ so $W,I,J'$ are collinear so $J=J'$ $\blacksquare$ and we win

Nice one! Here is a nice synthetic solution that hasn't been posted yet Let $AH$ meet $BC,\odot (ABC)$ at $D,H'$ respectively, and $M =AI \cap \odot (ABC)$. We claim that $M,D,X$ are collinear. Indeed, it suffices to prove that $\angle{AX'H} = \angle{AFH}$ where $X' = MD \cap \odot(ABC)$. Let $A'$ be the reflection of $A$ with respect to $BC$, we know that $M,H'$ are the reflections of $F,H$ with respect to $BC$ respectively. $$ \Rightarrow \angle{AFH} = \angle{A'MH'} $$Clearly, $A' \in \odot (BHFC)$, so $$ X'D \cdot DM = AD \cdot DH' = A'D \cdot DH $$which yeilds that $X' \in \odot (A'MH)$. Since $X' \in \odot (ABC)$, then $$ \angle{AX'M} = \angle{AH'M} \Rightarrow \angle{AX'H} + \angle{HX'M} = \angle{A'MH'} + \angle{HA'M} $$But $\angle{HX'M} = \angle{HA'M}$ because $X' \in \odot (A'MH)$, thus $\angle{AX'H} = \angle{A'MH'} = \angle{AFH}$, as required. $\blacksquare$ The key observation is to define $P,Q = AB,AC \cap \odot (XIA)$. We claim that $BD=BP$ and $CD=CQ$. Since $X = \odot (APQ) \cap \odot (ABC)$, thus $X$ is the center of spiral similarity that sends $PQ \rightarrow BC$, then there is a spiral similarity at $X$ that sends $BP \rightarrow CQ$, hence $$ \Rightarrow \frac{BP}{CQ} = \frac{XB}{XC} = \frac{BD}{CD} $$because $X,D,M$ are collinear (proved previously) and $XD$ is the angle bisector of $\angle{BXC}$. Take $P',Q'$ on $AB,AC$ respectively such that $BP'=BD$ and $CQ' = CQ$, we will prove that $I \in \odot (AP'Q')$. The proof is simple, notice that $BI$ is the angle bisector of $\angle {P'BD}$ so $IP' = ID$, similarly $IQ'=ID$ giving that $I$ is the center of $\odot (P'DQ')$, thus by angle chaising $$ \angle{P'IQ'} + \angle{P'AQ'} = 2\angle{P'DQ'} + \angle{A} = 2(180 - (90-\frac{\angle{B}}{2})- (90-\frac{\angle{C}}{2})) +\angle{A} = \angle{A}+\angle{B}+\angle{C} = 180 $$Therefore $I \in \odot (AP'Q')$. Now suppose that $P \not\equiv P'$, WLOG assume $BP < BP'$, then clearly $CQ' < CQ$ giving that $$ \frac{BD}{CD} = \frac{BP}{CQ} < \frac{BP'}{CQ'} = \frac{BD}{CD} $$contradiction! Thus $P \equiv P', Q \equiv Q'$, as desired. $\blacksquare$ Finally, let $T = AO \cap \odot (ABC)$, it is well known that $TB=CH$, $TC=BH$ and $TB \perp BP$, $TC \perp CQ$, giving that $$ CH^2 - CD^2 = HD^2 = BH^2 - BD^2 $$$$ TB^2 - CQ^2 = TC^2 - BP^2 \Rightarrow TB^2 + BP^2 = TC^2 + CQ^2 $$$$ TP^2 = TQ^2 \Rightarrow TP = TQ $$Since $I \in \odot (APQ)$ then $IP=IQ$, combined with the previous result yields that $TI$ is the perpendicular bisector of $PQ$, hence $K \in TI$, we are done.. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); 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 = -191.6577745848904, xmax = 2019.1212555497297, ymin = -918.1201735445003, ymax = 362.1274424261491; /* image dimensions */ pen qqccqq = rgb(0,0.8,0); pen ffqqtt = rgb(1,0,0.2); pen ffdxqq = rgb(1,0.8431372549019608,0); pen ffwwqq = rgb(1,0.4,0); pen wwqqcc = rgb(0.4,0,0.8); pen qqqqcc = rgb(0,0,0.8); /* draw figures */ draw((-33.62454245658835,0.5388180213982423)--(-33.62454245658835,184.72606467649211), linewidth(0.8)); draw((-33.62454245658836,35.15420172808046)--(87.27579128238118,0.5388180213982423), linewidth(0.8)); draw((-86.35981637741066,0.5388180213982423)--(-33.62454245658836,35.15420172808046), linewidth(0.8)); draw(circle((0.45798745248525563,75.32474949560406), 114.58737543163688), linewidth(0.8) + qqccqq); draw(circle((0.45798745248525585,-74.24711345280757), 114.58737543163688), linewidth(0.8) + qqccqq); draw((-33.62454245658835,-183.6484286336956)--(-33.62454245658835,0.5388180213982423), linewidth(0.8)); draw((0.4579874524852565,-39.26262593603281)--(0.4579874524852559,40.34026197882931), linewidth(0.8)); draw((34.540517361558855,-34.07656568528396)--(-33.62454245658835,184.72606467649211), linewidth(0.8)); draw((-33.62454245658835,-34.07656568528397)--(0.4579874524852565,-39.26262593603281), linewidth(0.8)); draw((0.4579874524852565,-39.26262593603281)--(-33.62454245658835,-183.6484286336956), linewidth(0.8)); draw((-33.62454245658836,35.15420172808046)--(0.4579874524852559,40.34026197882931), linewidth(0.8)); draw((0.4579874524852559,40.34026197882931)--(-33.62454245658835,184.72606467649211), linewidth(0.8)); draw((-112.76449325016864,92.95814031665898)--(0.4579874524852565,-39.26262593603281), linewidth(0.8)); draw((-33.62454245658836,35.15420172808046)--(-112.76449325016864,92.95814031665898), linewidth(0.8)); draw((-112.76449325016864,92.95814031665898)--(-33.62454245658835,184.72606467649211), linewidth(0.8)); draw(circle((-47.1519837412555,116.3832217650892), 69.66875802626887), linewidth(0.8) + ffqqtt); draw((34.540517361558855,-34.07656568528396)--(-47.1519837412555,116.3832217650892), linewidth(0.8) + linetype("2 2") + ffdxqq); draw((-71.8442437707701,51.23701589277734)--(20.93251487709415,101.61037868020031), linewidth(0.8)); draw((-112.76449325016864,92.95814031665898)--(-86.35981637741066,0.5388180213982423), linewidth(0.8)); draw((-112.76449325016864,92.95814031665898)--(87.27579128238118,0.5388180213982423), linewidth(0.8)); draw((34.540517361558855,-34.07656568528396)--(-86.35981637741066,0.5388180213982423), linewidth(0.8)); draw((34.540517361558855,-34.07656568528396)--(87.27579128238118,0.5388180213982423), linewidth(0.8)); draw((-86.35981637741066,0.5388180213982423)--(-33.62454245658835,0.5388180213982423), linewidth(0.8) + ffwwqq); draw((-33.62454245658835,0.5388180213982423)--(87.27579128238118,0.5388180213982423), linewidth(0.8) + wwqqcc); draw((-86.35981637741066,0.5388180213982423)--(-71.8442437707701,51.23701589277734), linewidth(0.8) + ffwwqq); draw((-71.8442437707701,51.23701589277734)--(-33.62454245658835,184.72606467649211), linewidth(0.8)); draw((-33.62454245658835,184.72606467649211)--(20.93251487709415,101.61037868020031), linewidth(0.8)); draw((20.93251487709415,101.61037868020031)--(87.27579128238118,0.5388180213982423), linewidth(0.8) + wwqqcc); draw(shift((-174.21144397270015,-74.24711345280757))*xscale(178.1384984814033)*yscale(178.1384984814033)*arc((0,0),1,-80.63635218665581,80.61423623252896), linewidth(0.8) + qqqqcc); /* dots and labels */ dot((-33.62454245658835,184.72606467649211),linewidth(3pt) + dotstyle); label("$A$", (-29.070417604592244,191.35958949715675), NE * labelscalefactor); dot((-86.35981637741066,0.5388180213982423),linewidth(3pt) + dotstyle); label("$B$", (-82.24364127110485,6.275868657949452), NE * labelscalefactor); dot((87.27579128238118,0.5388180213982423),linewidth(3pt) + dotstyle); label("$C$", (91.59189763864791,6.275868657949452), NE * labelscalefactor); dot((-33.62454245658835,0.5388180213982423),linewidth(3pt) + dotstyle); label("$D$", (-29.070417604592244,6.275868657949452), NE * labelscalefactor); dot((-33.62454245658836,35.15420172808046),linewidth(3pt) + dotstyle); label("$H$", (-29.070417604592244,41.042976439899995), NE * labelscalefactor); dot((0.45798745248525563,75.32474949560406),linewidth(3pt) + dotstyle); label("$O$", (4.674128183771525,81.94545618337122), NE * labelscalefactor); dot((-33.62454245658835,-183.6484286336956),linewidth(3pt) + dotstyle); label("$A'$", (-29.070417604592244,-177.78529018767108), NE * labelscalefactor); dot((-33.62454245658835,-34.07656568528397),linewidth(3pt) + dotstyle); label("$H'$", (-29.070417604592244,-27.46867713041431), NE * labelscalefactor); dot((0.4579874524852565,-39.26262593603281),linewidth(3pt) + dotstyle); label("$M$", (4.674128183771525,-33.604049091934996), NE * labelscalefactor); dot((-13.909082061860427,55.15704000551125),linewidth(3pt) + dotstyle); label("$I$", (-9.641739726443408,61.49421631163561), NE * labelscalefactor); dot((0.4579874524852559,40.34026197882931),linewidth(3pt) + dotstyle); label("$F$", (4.674128183771525,46.1557864078339), NE * labelscalefactor); dot((34.540517361558855,-34.07656568528396),linewidth(3pt) + dotstyle); label("$T$", (38.418673972135295,-27.46867713041431), NE * labelscalefactor); dot((-112.76449325016864,92.95814031665898),linewidth(3pt) + dotstyle); label("$X'$", (-108.83025310436116,99.3290100743465), NE * labelscalefactor); dot((-71.8442437707701,51.23701589277734),linewidth(3pt) + dotstyle); label("$P$", (-67.92777336088991,57.40396833728849), NE * labelscalefactor); dot((20.93251487709415,101.61037868020031),linewidth(3pt) + dotstyle); label("$Q$", (25.125368055507142,107.50950602304074), NE * labelscalefactor); dot((-47.1519837412555,116.3832217650892),linewidth(3pt) + dotstyle); label("$K$", (-43.38628551480718,122.84793592684245), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

[asy][asy] size(10cm); defaultpen(fontsize(9pt)); defaultpen(linewidth(0.35)); dotfactor *= 1.5; pair A = dir(130), B = dir(200), C = dir(340), O = origin, A1 = 2O-A, I = incenter(A,B,C), N = dir(270), D = foot(A,B,C), H = orthocenter(A,B,C), Q = 2D-N, X = intersectionpoints(unitcircle,circumcircle(A,Q,H))[1], K = circumcenter(A,X,I), P = D+dir(A--D)*abs(B-D)*abs(C-D)/abs(A-D), L = dir(90), R = extension(A,L,I,K), F = B+C-N, M = (B+C)/2; filldraw(L--R--X--cycle, magenta+white+white+white+white); filldraw(X--I--N--cycle, purple+white+white+white+white); draw(A--B--C--A); draw(unitcircle); draw(circumcircle(A,Q,H), orange); draw(N--Q, purple); draw(N--A--P^^X--I); draw(R--A1, heavyred); draw(Q--H^^P--N, blue); draw(circumcircle(A,X,I), heavygreen); draw(L--R--X--L, magenta); draw(Q--F--N^^L--F, dotted); dot("$A$", A, dir(90)); dot("$B$", B, dir(200)); dot("$C$", C, dir(340)); dot("$D$", D, dir(225)); dot("$N$", N, dir(270)); dot("$P$", P, dir(225)); dot("$A'$", A1, dir(310)); dot("$K$", K, dir(75)); dot("$I$", I, dir(80)); dot("$H$", H, dir(30)); dot("$X$", X, dir(225)); dot("$Q$", Q, dir(210)); dot("$L$", L, dir(90)); dot("$R$", R, dir(135)); dot("$F$", F, dir(45)); dot("$M$", M, dir(270)); [/asy][/asy] Let $D = \overline{AH} \cap \overline{BC}$, $P = \overline{AH} \cap \Gamma$, $N = \overline{AI} \cap \Gamma$, and $M$ be the midpoint of $\overline{BC}$. Claim: $X$, $D$, $N$ are collinear. Proof. Let $Q$ be the reflection of $F$ over $\overline{AH}$. Recall that $F$ is the reflection of $N$ over $M$, so $Q$ is simply the reflection of $N$ over $D$. We also have $HD = DP$, so $QHNP$ is a parallelogram. By definition $\angle AQH = \angle AFH = \angle AXH$, so $AQXH$ is cyclic and we are done by Reim's. $\square$ Let $A'$ be the antipode of $A$ on $\Gamma$ and $L = \overline{NO} \cap \Gamma$. It remains to show that $\overline{A'I}$ passes through the circumcenter $K$, which is equivalent to $R = \overline{AL} \cap \overline{A'I}$ lies on $(AXI)$ from $\angle LAI = 90^\circ$. The key claim is $\triangle XRL \sim \triangle XIN$. We already have $\angle XNI = \angle XLR$. Furthermore $$\frac{LX}{NX} = \frac{DM}{NM} = \frac{AL \cdot \tfrac{AN}{2R}}{NI \sin \tfrac{A}{2}} = \frac{AL \cdot \tfrac{AN}{NI}}{NI} = \frac{AL + NA' \cdot \frac{AI}{NI}}{NI} = \frac{LR}{NI}$$from where the desired similarity follows. Now note that $\measuredangle ARX = \measuredangle NIX = \measuredangle AIX$ and we win. $\blacksquare$

I will instead solve the following problem. Quote: Let $ABC$ be an acute triangle with circumcircle $\Omega$ and altitudes $\overline{AD}$, $\overline{BE}$, $\overline{CF}$ meeting at $H$. Let $\omega$ be the circumcircle of $\triangle DEF$. Point $S \neq A$ lies on $\Omega$ such that $DS = DA$. Line $\overline{AD}$ meets $\overline{EF}$ at $Q$, and meets $\omega$ at $L \neq D$. Point $M$ is chosen such that $\overline{DM}$ is a diameter of $\omega$. Point $P$ lies on $\overline{EF}$ with $\overline{DP} \perp \overline{EF}$. Prove that lines $SH, MQ, PL$ are concurrent. Claim: $\overline{PL}$ and $\overline{MQ}$ intersect on $\omega$. Proof: Suppose that $\overline{LP}$ meets $\omega$ again at a point $T$. We aim to show that $\angle DTQ = 90^\circ$, or $DTPQ$ is cyclic. To this end, note that $-1 = (AH;QD)$ and moreover $P,T$ are inverses around $(AH)$, so $LP \cdot LT = LH^2 = LQ \cdot LD$ by a well-known lemma. Claim: $\overline{STH}$ is collinear. We now apply a negative inversion with radius $-\sqrt{HA \cdot HD}$. Redefine $S$ to be the inverse of $T$ under this inversion; note that $S$ must lie on $\Omega$ and $TH \cdot HS = AH \cdot HD$, so $ASDT$ is cyclic. Moreover, $\overline{THS}$ is collinear. We will show that $DS = DA$. Note that as $\angle QTD = 90^\circ$ and $-1 = (AH;QD)$, we have that $\overline{QT}$ bisects $\angle ATH$ and thus $\angle ATL + \angle DTH = 90^\circ$. Now $\angle DSA = 180^\circ - \angle DTA = 180^\circ - (90^\circ + \angle QTH ) = 90^\circ - \angle QTS = \angle DTS = \angle DAS,$ as desired.

Let $I_A$ be the $A$-excenter of $\triangle ABC$, $A'$ be the antipode of $A$ wrt $(O)$, $M$ is the midpoint of arc $BC$ and $D$ is the foot from $A$ to $BC$. We ultimately want to show that $K,I,A'$ are collinear. The crucial part about this problem is figuring out what the $X$ really is. Firstly note that if $F'$ is the reflection of $F$ over $AD$, then $F',D,M$ are collinear as $F$ is reflection of $M$ over $BC$. Note that $X=(AF'H)\cap (ABC)$ and therefore \begin{align*} \measuredangle F'XA=\measuredangle F'HA=\measuredangle AHF=\measuredangle DHF=\measuredangle MXA, \end{align*}hence $F',X,M$ are collinear. We conclude that $F',X,D,M$ are collinear. Now we angle chase with inversions, note that by inversion at $M$ with radius $MB$ implies $\measuredangle IXA=\measuredangle CDI$ and inversion at $A$ with radius $\sqrt{AB\cdot AC}$ implies $\measuredangle IDA=\measuredangle A'I_AA$. Also as $MI=MI_A$ and $A'M\perp II_A$, we obtain that $\measuredangle I_AIA'=\measuredangle IA'I_A$. Hence, \begin{align*} \measuredangle IXA=\measuredangle CDI=90^\circ+\measuredangle ADI=90^\circ+\measuredangle AI_AA'=90^\circ+\measuredangle A'IM=\measuredangle IA'M, \end{align*}thus $\measuredangle AIK=\measuredangle MIA'$, which means that $K,I,A'$ are collinear. [asy][asy]import olympiad; size(10cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta; pair O,A,B,C,I,M,a,D,F,Ia,H,X,K,f; O=(0,0);A=dir(125);B=dir(200);C=dir(335);I=incenter(A,B,C);path w=circumcircle(A,B,C); M=intersectionpoints(w,A--100I-99A)[1];a=-A;Ia=2M-I;D=foot(A,B,C);H=orthocenter(A,B,C);F=2midpoint(B--C)-M; X=intersectionpoints(w,M--100D-99M)[0];K=circumcenter(A,X,I);f=2foot(F,A,D)-F; draw(A--B--C--cycle,deep);draw(w,heavyblue); draw(circumcircle(A,X,I),heavyblue);draw(A--Ia,deep);draw(A--D,heavygreen);draw(A--a,magenta);draw(K--a,magenta);draw(M--f,deep);draw(circumcircle(B,H,C),heavyblue);draw(M--a--Ia,deep); draw(circumcircle(A,H,X),heavyblue+dashed);draw(f--F,deep); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$I$",I,dir(I)); dot("$O$",O,dir(90)); dot("$M$",M,dir(M)); dot("$A'$",a,dir(a)); dot("$F$",F,dir(F)); dot("$H$",H,dir(H)); dot("$D$",D,dir(D)); dot("$I_A$",Ia,dir(Ia)); dot("$X$",X,dir(X)); dot("$K$",K,dir(K)); dot("$F'$",f,dir(f)); [/asy][/asy]

Let $E$ be reflection of $F$ across $AH$. Let $AH$ meet $AB$ at $D$. Let $T,T'$ be midpoints of arcs $BC , BAC$. Let $M$ be midpoint of $BC$. Claim $: E,D,T$ are collinear. Proof $:$ Note that $\angle EFT = \angle 90$ and $D$ is center of $EFT$ since lies on perpendicular bisectors of $FT,TE$ so $D$ lies on $ET$. Claim $: X$ lies on $ET$. Proof $:$ Let $TE$ meet $AHE$ at $X'$. Note that $DX'.DE = DX'.DT$ and also $DX'.DE = DH.DA$ so $DX'.DT$ equals power of $D$ wrt $\Gamma$ so $X'$ lies on $\Gamma$ so $X'$ is intersection of $AHE$ and $\Gamma$ which is $X$ so $X'$ is $X$. Let $AO$ meet $\Gamma$ at $A'$ and Let $A'I$ meet $AIX$ at $S$. we need to prove $K$ lies on $A'I$ or in fact $\angle IAS = \angle 90$ or instead proving $S,A,T'$ are collinear. Let $AT'$ and $A'I$ meet at $S'$. Claim $: XST'$ and $XIT$ are similar. Proof $:$ Note that $\angle XTI = \angle XTA = \angle XT'A = \angle XT'S$ and $\frac{TI}{T'S} = \frac{TI}{T'A' . \frac{\sin{T'A'S}}{\sin{T'SA'}}} = \frac{TI}{TA} . \frac{\sin{TA'I}}{\sin{TIA'}} = \frac{TI^2}{TA.TA'} = \frac{TB}{TA} . \frac{TB}{TA'.TM} . TM = \frac{\sin{\frac{A}{2}}}{\sin{ABT}} . \frac{TM}{TA'.\sin{\frac{A}{2}}} = \frac{TM}{TA'.\sin{ABT}} = \frac{TM}{MD} = \frac{TX}{T'X}$. Now we have $\angle XS'T' = \angle XS'A = \angle XIT$ so $AS'XI$ is cyclic so $S'$ is $S$. we're Done.

Let $SH$ intersect $\omega$ at $U$. Claim: $AUDS$ is cyclic. Let $SH$ intersect $(ABC)$ at $V$. By homothety, $HV=2HU$, since the nine-point circle is a homothety with center $H$ and ratio 1/2 of the circumcircle. Furthermore, let $H'$ be the reflection of $H$ across $BC$. Hence, $$HU\cdot HS=\frac{1}{2}HV\cdot HS=\frac{1}{2}HA\cdot HH'=HA\cdot HD,$$which shows the claim. Now, let $UM$ intersect $AD$ at $Q'$. Due to $AUDS$ cyclic, as well as the fact that $\triangle DAS$ is isosceles, $$\angle AUH+2\angle DUH=SDA+2\angle DAS=180,$$which means that $UD$ is an external bisector of $\angle HUA$. Then, since $UM\perp UD$, $UM$ bisects $\angle HUA$. Hence, by the right angle harmonic lemma, $(AH;Q'D)=-1$. Thus, clearly $(AH;QD)=-1$, so $Q=Q',$ which means that $U,Q,M$ are collinear. Now, it remains to show that $P,U,L$ are collinear. Note that due to $\angle DPQ=\angle DUQ=90$, we have that $UQDP$ is cyclic. Claim 2: Lines $DP$ and $DM$ are reflections over $AD$. Since $H$ is the incenter of $\triangle DEF$, it suffices to show that $\angle PDF=\angle N_9DE$. We have $$\angle N_9DE=90-\frac{\angle DN_9E}{2}=90-\frac{arc~ DE}{2}$$$$=90-\frac{(360-2\angle EFD)}{2}=\angle EFD-90=(180-2\angle C)-90=90-2\angle C$$and $$\angle PDF=\angle EFD-90=90-2\angle C,$$hence shown. Finally, we have $$\angle LUM=\angle LDM$$and $$\angle PUD=\angle PQD=90-\angle PDQ.$$By Claim 2, $\angle PDQ=\angle LDM$, so $$\angle LUM+\angle PUD=90,$$and since $\angle DUM=90$, $P,U,L$ are collinear, as desired.

Wow, that was amazing. Decided to lessen my inhibitions and solved almost instantly. We'll solve the following translation from OTIS (it looks almost completely different, so this solution is likely extremely out of place): Problem from OTIS wrote: Let $\triangle ABC$ be an acute triangle with circumcircle $\Gamma$ and altitudes $AD$, $BE$, $CF$ meeting at $H$. Let $\omega$ be the circumcircle of $\triangle DEF$. Point $S\neq A$ lies on $\Gamma$ such that $DS = DA$. Line $AD$ meets $EF$ at $Q$, and meets $\omega$ at $L\neq D$. Point $M$ is chosen such that $DM$ is a diameter of $\omega$. Point $P$ lies on $EF$ with $DP\perp EF$. Prove that lines $SH$, $MQ$, $PL$ are concurrent. There isn't really a good place to begin without immediately realizing that $\triangle DEF$ should be our reference triangle. First, let's show that $PL$ and $MQ$ concur on $(DEF)$. Notice that $\triangle DPQ\sim \triangle DLM$, and so $X=PL\cap QM$ should lie on $(DPQ)$ and $(DLM)$, meaning $\angle DXM=90^{\circ}\implies X\in (DEF)$. Now let's phantom point this into oblivion. Let $S'$ be the point such that $DS'=DA$ and $S'\in XH$. As $(D,Q;H,A)=-1$ it follows that $XQ$ bisects $\angle AXH$, therefore $XDAS'$ should be cyclic. This means the midpoint of $HS'$ lies on $\triangle DEF$, hence by homothety $S'\in (ABC)\implies S'\equiv S$. So $X\in SH$, done.

We will solve the following problem: Equivalent Problem wrote: Let $\triangle ABC$ be an acute triangle with circumcircle $\Gamma$ and altitudes $AD$, $BE$, $CF$ meeting at $H$. Let $\omega$ be the circumcircle of $\triangle DEF$. Point $S$ lies on $\Gamma$ such that $DS = DA$ and $S \neq A$. Line $AD$ meets $EF$ at $Q$, and meets $\omega$ at $L\neq D$. Let $M$ be the antipode of $D$ on $\omega$ and $P$ denote the feet of perpendicular of $D$ on $EF$. Prove that lines $SH$, $MQ$, $PL$ are concurrent. Here is the proof (projective+angle chase+homothety): Let $X = \omega \cap SH$. Claim: $X, Q, M$ are collinear. Proof: Let $S_0 = XS \cap \omega$. Notice that, a homothety centered at $H$ with a factor of $2$ maps $\omega$ to $\Gamma$. Since $XLS_0D$ is cyclic, it follows that $XASD$ is cyclic. Since $DA = DS$, we must have $\angle AXS = 180^\circ - \angle DAS$ and $\angle DXS = \angle DAS$. Hence, $DX$ is the external angle-bisector of $\angle AXS$. Since $\angle MXD = 90^\circ$ ($M$ is the antipode of $D$ in $\omega$), $MX$ is the angle bisector of $\angle AXH$. Let $Q' = XM \cap AD$. By Right Angles and Bisectors Lemma (EGMO Lemma 9.18), $(AH; Q'D) = -1$. Notice that, by EGMO Lemma 9.12, $(AH; QD) = -1$. Therefore $Q' = Q$. Claim: $X, P, L$ are collinear. Proof: Note that $DP, DM$ are isogonal with respect to $\angle FDE$. Hence; $\angle PDH = \angle MDH$ which implies that $\angle PDQ = \angle MDL$ (as $H$ is the incenter of $\triangle DEF$). Hence; $$\angle PXM = \angle PXQ = \angle PDQ = \angle QDM = \angle LDM = \angle LXM$$Hence, $X, P, L$ are collinear. Therefore, lines $SH$, $MQ$, $PL$ are concurrent.

I am probably the only one who couldn't relate the original wording with the OTIS one.Very beautiful problem. India TST 2019 #8 wrote: Let $ABC$ be an acute-angled scalene triangle with circumcircle $\Gamma$ and circumcenter $O$. Suppose $AB < AC$. Let $H$ be the orthocenter and $I$ be the incenter of triangle $ABC$. Let $F$ be the midpoint of the arc $BC$ of the circumcircle of triangle $BHC$, containing $H$. Let $X$ be a point on the arc $AB$ of $\Gamma$ not containing $C$, such that $\angle AXH = \angle AFH$. Let $K$ be the circumcenter of triangle $XIA$. Prove that the lines $AO$ and $KI$ meet on $\Gamma$. Proposed by Anant Mudgal Call the midpoint of $\overarc{BC}$ as $D$.Let the foot of the altitude from $A$ to $BC$ be $E$.Let the incircle touch $BC$ at $J$,$G$ be the $\text{A-sharkydevil point}$, $A'$ be the antipode of $A$ in $\Gamma$.It is well known that $G,J,D$ and $G,I,A'$ are collinear.Let $F'$ be the reflection of $F$ over $AE$.Let $AI$ meet $BC$ at $M$.We also have $A,I,M,D$ collinear.Let $AH$ meet $\Gamma$ at $N$.Its well known that $EH=EN$ By the problem conditions we have $AF'XH$ cyclic.Now we have $F'$ and $D$ as the reflection of $F$ over $AH$,$BC$ respectively. Since $AH\perp BC$ we have $D,E F'$ collinear.Let $DF'$ meet $\Gamma$ at $X'$ Claim : $X' \equiv X$ Proof : By Shooting lemma,we have $EX'AM$ cyclic.Thus we get that $\angle EX'A=\angle DME$.Again , $$\angle EHF'=\angle EHF=\angle END=\angle AND=\angle ABD\stackrel{\triangle ABD \sim \triangle BMD}=\angle BMD=\angle DME$$This gives $\angle EHF'=\angle EX'A$, which gives $EX'F'A$ cyclic and hence $X' \equiv X$ and the claim is proved $\square$ Now let $L \in AH$ be such a point such that $IL \perp AH$.Note that $IL \parallel BC$ also we have $\angle ILA=\angle IGA=90$ so $ILGA$ is cyclic.Again by shooting lemma $DJ.DG=DM.DA$ so $AGJM$ is cyclic but $IL \parallel MJ$ since $M,J \in BC$ , thus by converse of reim we have $L \in \overline{GJD}$. Now we will finally prove that $\angle GIA=90-\angle AXI$ which should finish the problem.For this proof note that $DI^2=DM.DA=DJ.DG=DE.DX$ (which could be obtained as $M,A$ are inverses wrt $\odot(\triangle BIC)$. Clearly we have $DA(DI)$ tangent to $\odot(\triangle IXE)$,so $$\angle AXI=\angle AXD-\angle DXI=\angle DME-\angle DIE=\angle IEM$$Similarly as $DA(DI)$ is tangent to $\odot(\triangle IGJ)$ as well, we have $\angle GIA=\angle GJI=\angle LJI$ but by definition $LEJI$ is a rectangle, thus we have $\angle GIA=\angle LJI =\angle IEL=\angle IEA$.Finally we get $$\angle AXI+\angle GIA=\angle IEM+\angle IEA=90 \implies \angle GIA=90-\angle AXI \text{ } \blacksquare$$

OTIS Rewording wrote: Let $ABC$ be an acute triangle with circumcircle $\Gamma$ and altitudes $AD$, $BE$, $CF$ meeting at $H$. Let $\omega$ be the circumcircle of $\triangle DEF$. Point $S \neq A$ lies on $\Gamma$ such that $DS = DA$. Line $AD$ meets $EF$ at $Q$, and meets $\omega$ at $L \neq D$. Point $M$ is chosen such that $DM$ is a diameter of $\omega$. Point $P$ lies on $EF$ with $DP \perp EF$. Prove that lines $SH$, $MQ$, $PL$ are concurrent. Let $T = SH \cap \omega$, where $T$ is not on segment $SH$. We claim this is the desired concurrency point. First note the reflections of $H$ across $T$ and $D$ land on $\Gamma$, so Reim's tells us $ASDT$ is cyclic. We then show $T$ also lies on the other two segments: Since $D$ is the midpoint of arc $STA$ on $(ASDT)$, $TD$ is the external angle bisector of $\angle ATH$. It follows from Apollonius that \[(AH; TM \cap AD, D) = -1 = (AH;QD) \implies Q \in TM.\] Isogonals gives us the spiral similarity $\triangle DLM \sim \triangle DPQ$, so $D$ is the Miquel point of $PQLM$. Consequently, it lies on the circle formed by $L$, $M$, and $LP \cap MQ$, which is $\omega$, so $P \in TL$ as desired. $\blacksquare$

Let $R = PL \cap \omega$. Since $L$ is the midpoint of arc $EF$ not containing $D$, so we have $LR \cdot LP = LF^2 = LQ \cdot LD$, hence $DPRQ$ is cyclic. Therefore we get $\angle DRQ = \angle DPQ = 90^{\circ}$, so $R, Q, M$ are collinear. Now we'll prove that $R, H, S$ are collinear, finishing the problem. Note that $(A, H; Q, D) = -1$ and $\angle DRQ = 90^{\circ}$, so we get $\angle ARQ = \angle HRQ$. Assume ray $RH$ intersects $\Gamma$ at $S'$. We'll prove that $S' = S$. Since $H$ is exsimicenter of $\omega$ and $\Gamma$, thus $AH \cdot HD = RH \cdot HS'$, hence $ARDS'$ is cyclic. Since $RQ$ is bisector of $\angle ARH$ and $\angle DRQ = 90^{\circ}$, so $DR$ is external angle bisector of $\angle ARS'$, which means $AD = DS'$, finishing the problem. $\blacksquare$

We will solve the revised OTIS formulation. In particular, this problem is much more difficult than the solution seems: it's incredibly easy to find a red herring and go down a deep rabbit hole. We will define the point $G = \overline{MQ} \cap \omega$. In particular, because $\overline{EF}, \overline{MG}, \overline{AH}$ meet at $Q$, radical axis implies $(AMGH)$ is also cyclic. Claim. Suppose that $S' = \overline{GH} \cap \Gamma$ (on minor arc $\widehat{BC}$). Then $DS'=DA$. Proof. Negative inversion at $H$ implies $AGDS'$ cyclic, so by Fact 5 it suffices to show that $\overline{GD}$ externally bisects $\angle AQS'$. This follows by harmonic bisector lemma with the bundle $(AH;QD) = -1$ (say by looking at complete quadrilateral $BFEC$). $\blacksquare$ So $S=S'$. Thus $GH \cdot HS = 2 GH \cdot HS = LH \cdot HH_A$ where $H_A$ is the reflection of $H$ about $\overline{BC}$. It then follows that $GLSH_A$ is cyclic. We have the information we need now to finish; it remains to show that $G$ lies on $\overline{LP}$. To see this, notice that $QGPD$ is cyclic along $(QD)$, so $$\angle QGP + \angle MGL = \angle FQD + 90^\circ + \angle EDH + \angle EFM = 90^\circ + E + \frac D2 + F-90^\circ + \frac D2 = 180^\circ.$$

Here's a solution to the problem in OTIS. Problem: Let $ABC$ be an acute triangle with circumcircle $\tau$ and altitudes $\overline{AD}, \overline{BE}, \overline{CF}$ meeting at $H$. Let $\omega$ be the circumcircle of $\triangle DEF$. Point $S\ne A$ lies on $\tau$ such that $DS = DA$. Line $\overline{AD}$ meets $\overline{EF}$ at $Q$, and meets $\omega$ at $L \ne D$. Point $M$ is chosen such that $\overline{DM}$ is a diameter of $\omega$. Point $P$ lies on $\overline{EF}$ with $\overline{DP} \perp \overline{EF}$. Prove that lines $\overline{SH}, \overline{MQ}, \overline{PL}$ are concurrent. Solution: $\omega$ is the nine-point circle. We perform $\sqrt{-HA\cdot HD}$ inversion, which is inversion about $H$ with radius $\sqrt{HA\cdot HD}$ and then reflection about $H$. Let $A',H',S'$ be the reflections of line $A,H,S$ in line $\overline{BC}$, respectively. Let $M,T$ be the intersections of line $\overline{QM}$ with the nine-point circle $\omega$. Under this inversion: $A,B,C\longleftrightarrow D,E,F$, respectively, hence $\text{circumcircle}\longleftrightarrow\text{nine-point circle}$. $L,Q\longleftrightarrow H',A'$, respectively. Note that $DA'=DA=DS=DS'$, which means that $(ASA'S')$ is a cyclic trapezoid. Reflect circle $(ASCH'B)$ about line $\overline{BC}$, to get $(A'S'CHB)$. Under $\sqrt{-HA\cdot HD}$ inversion, circle $(A'S'CHB)$ goes to line $\overline{EQF}$. Hence, image of $S'$ lies on line $EF$. Let $P',T'$ be the images of $S',S$ under this inversion. Claim: $P'\equiv P$ and $T'\equiv T$. Proof. Under this inversion, circle $(ASA'S')$ goes $(P'T'QD)$. Note that since $A,H,Q,D,H'$ are all on the same line, the diameter of $(P'T'QD)$ is $\overline{QD}$, as the diameter of $(ASA'S')$ was $\overline{AA'}$. Hence, $\measuredangle QP'D=90^\circ$. However, note that $P'$ also lies on line $\overline{EF}$, which means that $90^\circ=\measuredangle QP'D=\measuredangle EP'D$, which means $P'$ is the foot of perpendicular from $D$ to line $\overline{EF}$. So, $P'\equiv P$. Note that since $S$ lies on $(ABC)$, it means that $T'$ lies on $(DEF)$. Also, $\angle DT'Q=90^\circ$. Hence, $T'$ when extended meets the antipode of $D$ in $(DEF)$, which is $M$. So, $T',Q,M$ are collinear. So, $T'$ is the intersection of $MQ$ with $\omega$ other than $M$. It follows that $T'\equiv T$. $\blacksquare$ Hence, $T,H,S$ are collinear. Finally, note that under this inversion, line $\overline{PT}$ goes to $(HSS')$. However, since $\overline{HH'}$ and $\overline{SS'}$ have same perpendicular bisector of $\overline{BC}$, it follows that $(HSS'H')$ is a cyclic trapezoid, which means $H'\in(HSS')$. Now, pre-image of $H'$ under the inversion is $L$, which means that $P,L,T$ are collinear. Hence, $\overline{SH},\overline{MQ},\overline{PL},\omega$ all meet at $T$, as desired. $\blacksquare$

Let $D=AH \cap BC$, $E$ the projection of $I$ in $BC$, $M$ the midpoint of the small arc $BC$ in $\Gamma$, $N=AH \cap \Gamma$ and $L= AH \cap (BHC)$. I define $X$ as $MD \cap \Gamma$ and I will prove that $\angle AXH = \angle AFH$. By power of point we get that $ DX \times DM = DB \times DC = DH \times DL => XHML$ is cyclic. It is trivial that $AHF$ is symmetric to $NLM$ with respect to $BC$, hence $\angle AFH = \angle NML = \angle ANM - \angle NLM = \angle AXM - \angle HLM = \angle AXM - \angle HXM = \angle AXH$. Because there is a unique point $X$ satisfying this conditions, we conclude that $X= MD \cap \Gamma$. Let $S = ME \cap \Gamma$. It is well known that $\angle ASI = 90$, hence $AO,SI$ meet on $\Gamma$, so it remains to prove that $S,K,I$ are collinear, which is equivalent to $\angle SIA = 90 - \angle IXA.<=> \angle IXA = 90 - \angle SIA = \angle IAS$. Using that $M$ is the midpoint of arc $BC$ we get that: $\angle BXD = \angle BXM = \angle CXM = \angle CBM => MB$ is tangent to $(XBD) => MD \times MX = MB^2 = MI^2$. With the same way we can prove that $ MD \times MX = MI^2 = ME \times MS = MP \times MA$, where $P = AM \cap BC$ and so $MI$ is tangent to $(IDX)$. Hence, we get that $\angle AXI = \angle AXM - \angle IXM = \angle ABM - \angle IXD = \angle B + \frac {\angle A} {2} -\angle MID = \angle APC - \angle PID = \angle IPC - \angle PID = \angle IDP$. Inverting through $M$ with radius $MI^2$ and using the above equalities we get that $P$ is the inverse of $A$, $E$ is the inverse of $S$ and $I$ is the inverse of itself, hence $\angle IAS = \angle MAS = \angle MEP$.Now, let $R = ME \cap ID$, which is well known to be the midpoint of $ID$. Using that $IED$ is a right angle triangle, we get that $\angle AXI = \angle IDP = \angle RDE = \angle RED = \angle MEP = \angle IAS$, which completes the proof.

Let $AO \cap \Gamma=A'$, reflections of $H,A$ over $BC$ be $H_A, A_1$ respectively, let $M$ midpoint of minor arc $BC$ on $\Gamma$, also let $I_A$ the $A$-excenter of $\triangle ABC$, let $A_1M \cap \Gamma=V$ and $(ADL) \cap AV=U$ where $AI \cap BC=L$ and $AH \cap BC=D$, finally re-define $X$ as $A'L \cap \Gamma=X$, our first aim is to prove $X$ is also the $X$ from the problem statement. Claim 1: $U$ is midpoint of $AV$. Proof: Fitst note that from reflection we have $A_1, L,F$ colinear and also $AL=A_1L$, in addition to this, it's clear that $L$ lies inside $\triangle AA_1V$ so notice $\angle ALA_1=2\angle ALB=2\angle ACM=2\angle AVA_1$ proving that $L$ is circumcenter of $\triangle AA_1V$, therefore $AL=LV$ but also we have $AO=OV$ so $LO$ is perpendicular bisector fo $AV$ and since $U$ is projection from $L$ to $AV$ it's now clear that $AU=UV$ must be true. Claim 2: $\angle AXH=\angle AFH$ Proof: Note that $AXDL$ is cyclic with diameter $AL$ therefore we have that $X$ is miquel point of $H_ADUV$, so by homologue points (reflection) we have $\triangle AXU \sim \triangle HXD$ which gives that $\angle AXH=\angle UXD=\angle DAU=\angle H_AMA_1=\angle AFH$ and since $AB<AC$ we have that this is the $X$ from the problem statement. Finish: From isogonallity of altitude-diameter and also parallels we have that all we need to prove is $\angle MIA'=\angle IXA'$, notice that from shooting lemma we have that $X,D,M$ are colinear, from I-E Lemma we have that $IA'=I_AA'$ and also from PoP we have that $XL \cdot LA'=BL \cdot LC=IL \cdot I_AL$ which means that $XIA'I_A$ is cyclic, and with the other thing we get $A'$ midpoint of minor arc $II_A$ on this cyclic, so by shooting lemma $A'I^2=A'L \cdot A'X$ which is enough to get the desired angle equality thus we are done .