We have $C$ is the center of $(ZDH)$ and $(ZDH)$ tagents $\overline{AB}$ at $D$. Hence $KD^2=\overline{KZ}.\overline{KH}=\overline{KA}.\overline{KB}$ and since we have $(ED,AB)=-1$ $\implies K$ is the midpoint of side $\overline{ED}$ and since $\overline{KH}\parallel\overline{CE}(\bot CO)\implies I \ \text{is the midpoint of side} \ \overline{CD}$ We also have $\overline{ML}\parallel\overline{KO}\equiv\overline{AB}$ and $\overline{ML}$ passes through $I$ is the midpoint of $\overline{CD}$ $\implies M,L$ are the midpoint of sides $\overline{CO},\overline{CK}$ and hence $(DML)$ is the nine-point circle of $\bigtriangleup CKO$ $\implies \overline{KS},\overline{OU},\overline{CD}$ are the altitudes of $\bigtriangleup CKO$ and $P$ is the midpoint of $\overline{KO}$. We have $MLDP$ is a isosceles trapezoid and $Q$ is the center of $(DML)$ and $\overline{MM}\cap\overline{LL}=R$ $\implies \overline{RQ}$ is the perpendicular bisector of sides $\overline{ML},\overline{PD}$ $\implies \overline{RP},\overline{RD}$ are isogonal WRT $\angle XRT$ Similarly $ \overline{XL}, \overline{XS}$ are isogonal WRT $\angle RXT$ and $ \overline{TM},\overline{TU}$ are isogonal WRT $\angle RTX$ and since $\overline{RP},\overline{XL},\overline{TM}$ are concurrent at the Lemoine point of $\bigtriangleup MPL$ $\implies \overline{RD},\overline{XS},\overline{TU} $ are concurrent (1). Easy to see that $PU=PS=\frac{KO}{2}\implies \angle TPS=\angle PUS=\angle PSU\implies \overline{US}\parallel\overline{TX}$ Similarly we have $\overline{SD}\parallel\overline{XR}$ and $\overline{DU}\parallel\overline{RT}$ $\implies \bigtriangleup USD \sim \bigtriangleup TXR$ And hence we well known that $I$ is the incenter of $\bigtriangleup USD$ (since $I$ is the orthocentre of $\bigtriangleup CKO$) and $Q$ is the incenter of $\bigtriangleup TXR$ By (1) $\implies \overline{RD},\overline{XS},\overline{TU} $ are concurrent at the homethy centre WRT $\bigtriangleup USD,\bigtriangleup TXR$ which lies on $\overline{IQ}$.(as needed)

Attachments:

This problem is so anti. First we prove that $(Q)$ is the nine point circle of $\triangle COK$. First note that $HZ \parallel CC \perp OC$. From Cyclic Quad $SODI$, we have $CS\cdot CO = CI \cdot CD$. Now we length chase : $$CD^2=CH^2=2 CS \cdot CO = 2 CI \cdot CD \implies CD=2CI \implies CO=2CL$$The claim follows. Now note that $\triangle SUD \sim \triangle XTR$ (well known), so $SX, UT , DR$ concur at the center of homothety of the two triangles. Since $I,Q$ are the incenters of the corresponding triangles, $IQ$ passes through that point as well. $\square$

Nice problem [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15.560478951732144cm); 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 = -1.2541358001173941, xmax = 33.30634315161475, ymin = -9.609733272779662, ymax = 8.603476879013366; /* image dimensions */ pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen qqwuqq = rgb(0.,0.39215686274509803,0.); /* draw figures */ draw(circle((7.871123856092187,-3.90742753599081), 7.764356089536745), linewidth(0.8)); draw((xmin, 0.013020833333333318*xmin-4.00991612786701)--(xmax, 0.013020833333333318*xmax-4.00991612786701), linewidth(0.8)); /* line */ draw((xmin, -76.8*xmin + 855.4305011163551)--(xmax, -76.8*xmax + 855.4305011163551), linewidth(0.8)); /* line */ draw((xmin, -0.4568189236009429*xmin + 4.735695376626565)--(xmax, -0.4568189236009429*xmax + 4.735695376626565), linewidth(0.8) + yqqqyq); /* line */ draw((11.097338283586263,3.1549209369291864)--(7.871123856092187,-3.90742753599081), linewidth(0.8)); draw((11.097338283586263,3.1549209369291864)--(18.614030369756385,-3.7675459407608067), linewidth(0.8)); draw((xmin, 0.013020833333333318*xmin-0.4997458915860096)--(xmax, 0.013020833333333318*xmax-0.4997458915860096), linewidth(0.8)); /* line */ draw(circle((12.181382096743317,-1.2186767046049138), 2.8256505189684122), linewidth(0.8) + yqqqyq); draw((xmin, 3.201657279057725*xmin-30.741510740547024)--(xmax, 3.201657279057725*xmax-30.741510740547024), linewidth(0.8) + red); /* line */ draw((xmin, -2.9311783847347*xmin + 43.23834828686522)--(xmax, -2.9311783847347*xmax + 43.23834828686522), linewidth(0.8) + red); /* line */ draw((xmin, 0.4052203109413472*xmin-9.203647953739752)--(xmax, 0.4052203109413472*xmax-9.203647953739752), linewidth(0.8) + red); /* line */ draw((12.87091826544215,1.5215496961515154)--(7.701894601012492,-6.082683828679985), linewidth(0.8) + qqwuqq); draw((9.778792941569568,0.2685577109422571)--(15.718144329863657,-2.8343366209714276), linewidth(0.8) + qqwuqq); draw((11.143035877538429,-0.35465427859722826)--(12.181382096743317,-1.2186767046049138), linewidth(0.8) + qqwuqq); draw((12.062912343172838,7.879800369607578)--(11.188733471490595,-3.8642294941236432), linewidth(0.8) + qqwuqq); /* dots and labels */ dot((7.871123856092187,-3.90742753599081),dotstyle); label("$O$", (7.953879131858188,-3.7076796489803634), NE * labelscalefactor); dot((0.10742587549472482,-4.008517353446506),dotstyle); label("$A$", (0.3887038683628309,-3.7279616201961683), NE * labelscalefactor); dot((15.634821836689653,-3.8063377185351137),linewidth(4.pt) + dotstyle); label("$B$", (15.194542855900663,-4.275574843022907), NE * labelscalefactor); dot((11.097338283586263,3.1549209369291864),dotstyle); label("$C$", (11.178712555171224,3.3504463341198303), NE * labelscalefactor); dot((11.188733471490595,-3.8642294941236432),linewidth(4.pt) + dotstyle); label("$D$", (11.259840440034445,-3.7076796489803634), NE * labelscalefactor); dot((4.083363794406663,2.8703375233946513),linewidth(4.pt) + dotstyle); label("$H$", (4.384252197876465,2.194373974818937), NE * labelscalefactor); dot((15.47422208873247,-2.3332221015101355),linewidth(4.pt) + dotstyle); label("$Z$", (15.985539733317067,-2.348787577521418), NE * labelscalefactor); dot((9.778792941569568,0.2685577109422571),linewidth(4.pt) + dotstyle); label("$S$", (8.886849807785229,0.12561291080681095), NE * labelscalefactor); dot((11.143035877538429,-0.35465427859722826),linewidth(4.pt) + dotstyle); label("$I$", (11.219276497602834,-0.19889862864607152), NE * labelscalefactor); dot((18.614030369756385,-3.7675459407608067),linewidth(4.pt) + dotstyle); label("$K$", (18.703323876234972,-3.6062697929013376), NE * labelscalefactor); dot((9.484231069839222,-0.37625329953081155),linewidth(4.pt) + dotstyle); label("$L$", (9.637282742770024,-0.9290495924150571), NE * labelscalefactor); dot((14.855684326671325,-0.30631250191581033),linewidth(4.pt) + dotstyle); label("$M$", (14.930877230095195,-0.13805271499865607), NE * labelscalefactor); dot((13.242577112924279,-3.8374867383758087),linewidth(4.pt) + dotstyle); label("$P$", (13.389447417693995,-4.376984699101933), NE * labelscalefactor); dot((12.87091826544215,1.5215496961515154),linewidth(4.pt) + dotstyle); label("$U$", (12.801270252435643,0.8760458457916016), NE * labelscalefactor); dot((12.062912343172838,7.879800369607578),linewidth(4.pt) + dotstyle); label("$R$", (12.355066885687927,7.690788174302133), NE * labelscalefactor); dot((15.718144329863657,-2.8343366209714276),linewidth(4.pt) + dotstyle); label("$X$", (15.012005114958416,-2.9166827715639623), NE * labelscalefactor); dot((7.701894601012492,-6.082683828679985),linewidth(4.pt) + dotstyle); label("$T$", (8.217544757663656,-6.425463791898254), NE * labelscalefactor); dot((12.181382096743317,-1.2186767046049138),linewidth(4.pt) + dotstyle); label("$Q$", (12.273939000824708,-1.9634301244211199), NE * labelscalefactor); dot((11.432069748299176,-0.5951634089760371),linewidth(4.pt) + dotstyle); label("$F$", (12.213093087177292,-0.7262298802570055), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note $I$ is orthocenter of $\triangle COK$ from $CD \perp AB$ and $KS \perp CO$ Claim : $I$ is midpoint of $CD$ Let $Q = (c) \cap CO$. Now as $CQ$ is diameter in $(c)$ we get $\angle QHC = \angle QZC = 90$ so $QH$ and $QZ$ are tangents. which give us $CS.CQ = CH^2$ From $\angle ISO = \angle IDO = 90 \implies I,D,O,S$ are cyclic. $$CI.CD = CS.CO = \frac{CH^2}{2} = \frac{CD^2}{2}$$ and which give us $I$ is midpoint of $CD$ which also give us $M,L$ are midpoint of $CO$ and $CK$ hence $(LMD)$ is Nine point circle of $\triangle COK$ Now we need to show segment by meeting points of tangent from midpoint to nine point circle and feet of altitude of respective vertex are concurrent on Eular line. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12.866840638118546cm); 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 = -10.584715025928322, xmax = 43.28212561219023, ymin = -18.59182175212301, ymax = 9.795750368199968; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((6.612210623799663,5.401682968729039)--(2.6,-8.64)--(21.38,-7.46)--cycle, linewidth(2.) + zzttqq); /* draw figures */ draw((6.612210623799663,5.401682968729039)--(2.6,-8.64), linewidth(0.8) + zzttqq); draw((2.6,-8.64)--(21.38,-7.46), linewidth(0.8) + zzttqq); draw((21.38,-7.46)--(6.612210623799663,5.401682968729039), linewidth(0.8) + zzttqq); draw(circle((9.439337302146196,-3.5241557841326756), 5.195107897293627), linewidth(0.8) + yqqqyq); draw((xmin, -1.826361923233006*xmin + 24.53279529957761)--(xmax, -1.826361923233006*xmax + 24.53279529957761), linewidth(0.8) + red); /* line */ draw((xmin, 0.5635772192315726*xmin-14.807290858586567)--(xmax, 0.5635772192315726*xmax-14.807290858586567), linewidth(0.8) + red); /* line */ draw((xmin, 2.5371332915659144*xmin-13.305461646915138)--(xmax, 2.5371332915659144*xmax-13.305461646915138), linewidth(0.8) + red); /* line */ draw((xmin, 14.234722965244465*xmin-114.74166861836844)--(xmax, 14.234722965244465*xmax-114.74166861836844), linewidth(0.8) + qqwuqq); /* line */ draw((xmin, -0.24254461743139402*xmin-1.5379561107589121)--(xmax, -0.24254461743139402*xmax-1.5379561107589121), linewidth(0.8) + qqwuqq); /* line */ draw((xmin, 1.3754212337231166*xmin-14.18949688081432)--(xmax, 1.3754212337231166*xmax-14.18949688081432), linewidth(0.8) + qqwuqq); /* line */ draw((xmin, -0.05533803089070806*xmin-3.0018014449186965)--(xmax, -0.05533803089070806*xmax-3.0018014449186965), linewidth(0.8) + qqwuqq); /* line */ draw((6.612210623799663,5.401682968729039)--(7.475241630246895,-8.333674913541463), linewidth(0.8)); draw((4.32927323912288,-2.5879980322979432)--(21.38,-7.46), linewidth(0.8)); /* dots and labels */ dot((6.612210623799663,5.401682968729039),dotstyle); label("$A$", (6.738658606518252,5.717802925525509), NE * labelscalefactor); dot((2.6,-8.64),dotstyle); label("$B$", (1.9020232675322557,-9.550790987743977), NE * labelscalefactor); dot((21.38,-7.46),dotstyle); label("$C$", (21.501460588913414,-7.1482793160908065), NE * labelscalefactor); dot((4.606105311899832,-1.6191585156354806),linewidth(4.pt) + dotstyle); label("$C'$", (3.7987430083110776,-1.205224128317177), NE * labelscalefactor); dot((13.99610531189983,-1.0291585156354803),linewidth(4.pt) + dotstyle); label("$B'$", (14.135865595555655,-0.7626561888021194), NE * labelscalefactor); dot((11.99,-8.05),linewidth(4.pt) + dotstyle); label("$A'$", (11.954637893660012,-9.139835043908565), NE * labelscalefactor); dot((8.67154771206422,8.69541074276523),linewidth(4.pt) + dotstyle); label("$R$", (9.488902230647543,8.752554510771619), NE * labelscalefactor); dot((7.475241630246895,-8.333674913541463),linewidth(4.pt) + dotstyle); label("$D$", (7.781854463946603,-8.065027190800569), NE * labelscalefactor); dot((16.460706241078373,-5.530411808651826),linewidth(4.pt) + dotstyle); label("$X$", (16.380317288810595,-5.1251115925934005), NE * labelscalefactor); dot((4.32927323912288,-2.5879980322979432),linewidth(4.pt) + dotstyle); label("$F$", (3.103279103358843,-3.2283918518145827), NE * labelscalefactor); label("$j$", (-10.331819060491146,0.47021164270411236), NE * labelscalefactor,qqwuqq); dot((-0.7609762057051925,-15.236159712499292),linewidth(4.pt) + dotstyle); label("$T$", (-1.2591763004324474,-14.577098300807844), NE * labelscalefactor); dot((11.284949955387354,1.3320629093281848),linewidth(4.pt) + dotstyle); label("$E$", (10.78499405351307,2.114035418045755), NE * labelscalefactor); dot((7.81941149192228,-3.4345122796058476),linewidth(4.pt) + dotstyle); label("$M$", (8.192810407782014,-4.492871679000461), NE * labelscalefactor); dot((11.713536019507275,-3.650005463005612),linewidth(4.pt) + dotstyle); label("$O$", (11.828189910941424,-3.3864518302128177), NE * labelscalefactor); dot((7.165138584785112,-3.3983061052597394),linewidth(4.pt) + dotstyle); label("$H$", (6.485762641081075,-2.5961519382216434), NE * labelscalefactor); label("$l$", (-10.331819060491146,-2.912271895018113), NE * labelscalefactor,qqwuqq); dot((9.439337302146193,-3.5241557841326756),linewidth(4.pt) + dotstyle); label("$N$", (9.552126222006837,-3.2600038474942297), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] This results is true for any triangle. Now rename everything as shown.($D,E,F$ are feet of altitude and $A',B',C'$ are midpoint of them. With $N$ as nine point circle center) Claim: $EF \parallel A'$ tangent As $A'$ is midpoint of $BC$ from three tangent lemma we get $A'E = A'F$ $$\angle XA'E = \angle DEA' = \angle EFA'$$ which give us $EF \parallel TX$ , $DE \parallel RT$ and $DF \parallel RX$ hence $\triangle DEF$ and $\triangle RTS$ So there exits homothety replace this triangle, Let $M$ be its center As $H$ is incenter of $\triangle DEF$ and $N$ is incenter of $\triangle RXT$ $M$ also lie on $NH$ which give us our result. $\blacksquare$

To the BMO PSC: pleaaaaaaassee don’t make a problem that dies to triangle centres next time, and pleeeeaaaaaaasssssseeeeeeee don’t make a problem with such a contrived condition. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10) + blue; defaultpen(dps); /* default pen style */ pen dotstyle = blue; /* point style */ real xmin = -21.155402630929544, xmax = 20.319367274214898, ymin = -11.322117178422829, ymax = 11.489006269406696; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ffvvqq = rgb(1,0.3333333333333333,0); pen ccqqqq = rgb(0.8,0,0); draw((-3.9060310728863175,9.20559184722235)--(-14.94608371585861,-0.11021290679384937)--(0,0)--cycle, linewidth(0.5) + ccqqqq); /* draw figures */ draw(circle((0,0), 10), linewidth(0.5) + qqwuqq); draw((-3.9060310728863175,9.20559184722235)--(-3.8379400464017186,-0.028301094564701156), linewidth(0.5) + ffvvqq); draw((-14.94608371585861,-0.11021290679384937)--(5.299580161866532,8.480238800172476), linewidth(0.5) + ffvvqq); draw((-3.9060310728863175,9.20559184722235)--(-14.94608371585861,-0.11021290679384937), linewidth(0.5) + ccqqqq); draw((-14.94608371585861,-0.11021290679384937)--(0,0), linewidth(0.5) + ccqqqq); draw((0,0)--(-3.9060310728863175,9.20559184722235), linewidth(0.5) + ccqqqq); draw(circle((-3.9060310728863175,9.20559184722235), 9.234143991094815), linewidth(0.5) + qqwuqq); draw((-9.780992452373184,2.081390556003085)--(-3.9060310728863175,9.20559184722235), linewidth(0.5) + ffvvqq); draw((-3.9060310728863175,9.20559184722235)--(5.299580161866532,8.480238800172476), linewidth(0.5) + ffvvqq); draw((-9.999728129312253,-0.07373832003298214)--(-9.780992452373184,2.081390556003085), linewidth(0.5) + ffvvqq); draw((5.299580161866532,8.480238800172476)--(9.999728129312253,0.07373832003298214), linewidth(0.5) + ffvvqq); draw((9.999728129312253,0.07373832003298214)--(0,0), linewidth(0.5) + ffvvqq); draw((-9.780992452373184,2.081390556003085)--(-3.8379400464017186,-0.028301094564701156), linewidth(0.5) + ffvvqq); draw((-3.8379400464017186,-0.028301094564701156)--(5.299580161866532,8.480238800172476), linewidth(0.5) + ffvvqq); draw((-3.8379400464017186,-0.028301094564701156)--(-3.7698490199171197,-9.262194036351751), linewidth(0.5) + ffvvqq); draw((-9.780992452373184,2.081390556003085)--(-3.7698490199171197,-9.262194036351751), linewidth(0.5) + ffvvqq); draw((-3.7698490199171197,-9.262194036351751)--(5.299580161866532,8.480238800172476), linewidth(0.5) + ffvvqq); draw((-8.836037234120463,-4.682354749187292)--(8.904128260605063,-4.551538192599756), linewidth(0.5) + ffvvqq); draw((8.904128260605063,-4.551538192599756)--(-3.8379400464017186,-0.028301094564701156), linewidth(0.5) + ffvvqq); draw((-3.8379400464017186,-0.028301094564701156)--(-8.836037234120463,-4.682354749187292), linewidth(0.5) + ffvvqq); /* dots and labels */ dot((0,0),dotstyle); label("$O$", (0.13281910944537725,0.31025969341069415), NE * labelscalefactor); dot((-9.999728129312253,-0.07373832003298214),dotstyle); label("$A$", (-9.879449562968398,0.24545536543390575), NE * labelscalefactor); dot((-3.9060310728863175,9.20559184722235),dotstyle); label("$C$", (-3.787842733150308,9.544876430103045), NE * labelscalefactor); dot((-3.8379400464017186,-0.028301094564701156),dotstyle); label("$D$", (-3.72303840517352,0.31025969341069415), NE * labelscalefactor); dot((5.299580161866532,8.480238800172476),dotstyle); label("$H$", (5.414371839553614,8.799626658369977), NE * labelscalefactor); dot((-9.780992452373184,2.081390556003085),dotstyle); label("$Z$", (-9.652634415049638,2.416400352656318), NE * labelscalefactor); dot((-2.2407061452533257,5.280814678087781),dotstyle); label("$S$", (-2.102930205753815,5.591812423518951), NE * labelscalefactor); dot((-3.8719855596440174,4.588645376328823),dotstyle); label("$I$", (-3.755440569161914,4.911366979762672), NE * labelscalefactor); dot((-14.94608371585861,-0.11021290679384937),dotstyle); label("$K$", (-14.8045784892043,0.21305320144551154), NE * labelscalefactor); dot((9.999728129312253,0.07373832003298214),dotstyle); label("$B$", (10.145087781859152,0.4074661853758768), NE * labelscalefactor); dot((-3.7698490199171197,-9.262194036351751),dotstyle); label("$C'$", (-3.6258319132083376,-8.924357043281656), NE * labelscalefactor); dot((-3.8038945331594194,-4.645247565458226),dotstyle); label("$E$", (-3.658234077196732,-4.323249756929678), NE * labelscalefactor); dot((-8.836037234120463,-4.682354749187292),dotstyle); label("$M_1$", (-8.712971659386211,-4.355651920918072), NE * labelscalefactor); dot((8.904128260605063,-4.551538192599756),dotstyle); label("$M_2$", (9.043414206253752,-4.2260432649644954), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] 0: $I$ bisects $CD$ Notice that if $C'$ is the reflection of $C$ over $AB$, then $D$ is the incentre of $\triangle CZH$ by Incentre-Excentre. Now recall that if the midpoints of arcs $C'Z$ and $C'H$ without the other point on $(ABC)$ are $M_1$ and $M_2$ then $M_1M_2$ bisects $C'D$ at $E$. However $H-D-M_1$ and $Z-D-M_2$, and $D$ is the midpoint of $CC'$ so by the Butterfly theorem $ID=DE$. Hence $CI=EC'=DE=ID$, as desired. 0.5: $I$ is the orthocentre of $\triangle COK$ Notice that $KH\perp CO$ and $CD\perp KO$, thus done. 1: $Q$ is the nine-point centre of $\triangle COK$ Pretty clear. 1.5: $U$, $S$, $P$ all lie on $(LMD)$ $P$ and $U$ are self-evident; for $S$, note it's the foot of $K$ to $CO$, thus done. So now we can actually erase a bunch of points. We erase $H$, $K$, $A$, $B$. 2: $RD$, $TU$, $XS$ concur They actually concur at the isogonal conjugate of the Gergonne point, which uh surprisingly is $V=X(55)$ on ETC. The proof of this is actually quite easy; first relabel so that it's like the following diagram: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10) + blue; defaultpen(dps); /* default pen style */ pen dotstyle = blue; /* point style */ real xmin = -62.29819418912363, xmax = 70.39867134040271, ymin = -31.874452398881893, ymax = 28.46115364657525; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ffvvqq = rgb(1,0.3333333333333333,0); /* draw figures */ draw(circle((-0.12124663883273357,-0.12124663883273445), 10), linewidth(0.5) + qqwuqq); draw((-13.911185558395536,5.985559817506855)--(2.125782836508121,-29.073679105418847), linewidth(0.5) + ffvvqq); draw((-13.911185558395536,5.985559817506855)--(14.527524044643178,15.032871927611714), linewidth(0.5) + ffvvqq); draw((14.527524044643178,15.032871927611714)--(2.125782836508121,-29.073679105418847), linewidth(0.5) + ffvvqq); draw((-13.911185558395536,5.985559817506855)--(8.353023539125129,-5.4303656760152705), linewidth(0.5) + ffvvqq); draw((-3.9702686152598967,-9.350820292311031)--(14.527524044643178,15.032871927611714), linewidth(0.5) + ffvvqq); draw((1.4037544070292052,9.761787905785635)--(2.125782836508121,-29.073679105418847), linewidth(0.5) + ffvvqq); draw((-3.1528668873218875,9.408143618101366)--(1.4037544070292052,9.761787905785635), linewidth(0.5) + ffvvqq); draw((-3.1528668873218875,9.408143618101366)--(2.125782836508121,-29.073679105418847), linewidth(0.5) + ffvvqq); /* dots and labels */ dot((0,0),dotstyle); label("$O$", (0.42180865881655966,0.9887557049152115), NE * labelscalefactor); dot((-9.215021370215418,-4.280965515736053),dotstyle); label("$A$", (-8.80477027253332,-3.261690769077474), NE * labelscalefactor); dot((-3.1528668873218875,9.408143618101366),dotstyle); label("$C$", (-2.6882741270317143,10.422673488655073), NE * labelscalefactor); dot((9.50544626968545,-2.8280498538976406),dotstyle); label("$B$", (9.959395868751265,-1.8103188023482641), NE * labelscalefactor); dot((1.4037544070292052,9.761787905785635),dotstyle); label("$C'$", (1.7695111993508117,10.837351193434849), NE * labelscalefactor); dot((8.353023539125129,-5.4303656760152705),dotstyle); label("$B'$", (8.819032180606898,-4.402054457221853), NE * labelscalefactor); dot((-3.9702686152598967,-9.350820292311031),dotstyle); label("$A'$", (-3.517629536591254,-8.341492652629707), NE * labelscalefactor); dot((14.527524044643178,15.032871927611714),dotstyle); label("$D$", (14.935528326108503,16.020822503182025), NE * labelscalefactor); dot((-13.911185558395536,5.985559817506855),dotstyle); label("$E$", (-13.46989445130573,7.001582424221937), NE * labelscalefactor); dot((2.125782836508121,-29.073679105418847),dotstyle); label("$F$", (2.4951971827154087,-28.03868362966898), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that $F$ lies on the perpendicular bisectors of $AB$, and $CC'$ and $AB$ share a perpendicular bisector so $FC$ is the reflection of $FC'$ over the angle bisector of $\angle DFE$ since it is just the perpendicular bisector of $AB$. Thus we are done. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10) + blue; defaultpen(dps); /* default pen style */ pen dotstyle = blue; /* point style */ real xmin = -16.90941397775261, xmax = 9.063470187490015, ymin = -1.6558050483462503, ymax = 12.629281242537303; /* image dimensions */ pen ffvvqq = rgb(1,0.3333333333333333,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0); draw((-3.941951998904229,9.190267375780476)--(-14.84293399122863,-0.10945227737396124)--(0,0)--cycle, linewidth(0.5) + ccqqqq); /* draw figures */ draw((-3.941951998904229,9.190267375780476)--(-3.873972016282619,-0.028566795480979904), linewidth(0.5) + ffvvqq); draw((-9.392442995066427,4.54040754920326)--(-1.9709759994521148,4.595133687890239), linewidth(0.5) + ffvvqq); draw(circle((-5.67321199943157,3.415416347139067), 3.885651091337576), linewidth(0.5) + qqwuqq); draw((-6.198206273586069,7.265437829639894)--(0,0), linewidth(0.5) + ffvvqq); draw((-3.941951998904229,9.190267375780476)--(-14.84293399122863,-0.10945227737396124), linewidth(0.5) + ccqqqq); draw((-14.84293399122863,-0.10945227737396124)--(0,0), linewidth(0.5) + ccqqqq); draw((0,0)--(-3.941951998904229,9.190267375780476), linewidth(0.5) + ccqqqq); draw((-9.392442995066427,4.54040754920326)--(-7.421466995614315,-0.05472613868697855), linewidth(0.5) + ffvvqq); draw((-7.421466995614315,-0.05472613868697855)--(-1.9709759994521148,4.595133687890239), linewidth(0.5) + ffvvqq); draw((-5.67321199943157,3.4154163471390664)--(-3.907962007593424,4.580850290149748), linewidth(0.5) + ffvvqq); /* dots and labels */ dot((0,0),dotstyle); label("$O$", (0.07441730842557488,0.2109960010305776), NE * labelscalefactor); dot((-3.941951998904229,9.190267375780476),dotstyle); label("$C$", (-3.8620979478690103,9.402962037636046), NE * labelscalefactor); dot((-3.873972016282619,-0.028566795480979904),dotstyle); label("$D$", (-3.801224000606723,0.1704133695223857), NE * labelscalefactor); dot((-3.907962007593424,4.580850290149748),dotstyle); label("$I$", (-3.8215153163608186,4.7765420457021674), NE * labelscalefactor); dot((-14.84293399122863,-0.10945227737396124),dotstyle); label("$K$", (-14.758534507818455,0.08924810650600186), NE * labelscalefactor); dot((-2.266789444409748,5.284793189900176),dotstyle); label("$S$", (-2.177918740279059,5.486738097095526), NE * labelscalefactor); dot((-1.9709759994521148,4.595133687890239),dotstyle); label("$L$", (-1.8938403197217177,4.7968333614562635), NE * labelscalefactor); dot((-9.392442995066427,4.54040754920326),dotstyle); label("$M$", (-9.32046188572078,4.735959414193976), NE * labelscalefactor); dot((-7.421466995614315,-0.05472613868697855),dotstyle); label("$P$", (-7.331912941819392,0.15012205376828974), NE * labelscalefactor); dot((-6.198206273586069,7.265437829639894),dotstyle); label("$U$", (-6.114433996573644,7.47528704099693), NE * labelscalefactor); dot((-5.67321199943157,3.4154163471390664),dotstyle); label("$Q$", (-5.586859786967153,3.6199370477186985), NE * labelscalefactor); dot((-5.681709497259271,4.567770618546749),dotstyle); label("$R$", (-5.607151102721249,4.7765420457021674), NE * labelscalefactor); dot((-8.40695499534037,2.2428407052581405),dotstyle); label("$X$", (-8.326187413770086,2.4430407339811326), NE * labelscalefactor); dot((-4.696221497533215,2.27020377460163),dotstyle); label("$T$", (-4.612876630770555,2.4633320497352287), NE * labelscalefactor); dot((-4.3673857463578285,4.277534600102148),dotstyle); label("$V$", (-4.288215578705022,4.472172309390729), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] 3: $V$ lies on $IQ$ $V$ is actually $X(22)$ (Exeter point) of $\triangle RXT$'s intouch triangle, i.e. $\triangle PLM$. Hence it suffices to be able to show that $V$ lies on the line $IQ$. This is actually well-known to be true as $IQ$ is just the Euler Line and $X(22)$ is literally documented to be on that line. We are thus done