Truly marvelous problem. Congratulations to the proposer! Main claim: $X, Y, Z, T$ form an orthocenter pair and their NPC is $\odot ABC$. Pick any two of $X, Y, Z, T$, wlog $X, Y$. Consider the midpoint of $XY$: By the equal chords condition, its three pedals on $AB, BC, CA$ are on the radical axis of $\omega_1, \omega_2$, which means that they are collinear. Simson Theorum yields the claim. Recall the well-known property that $P-$ Simson line bisects $PH$, perform a homothety with center ${X}$ and ratio $2:1$, which sends the radical axis to the perpendicular bisectors of $YZ, ZT, TY$, and sends the radical center to the orthocenter ${H}$ of $\triangle YZT$. By Menelaus we only need the circumcenter of $\triangle {ABC}$ is the midpoint of $OX$, which is evident, once seen from $\triangle YZT$.

LoloChen wrote: Truly marvelous problem. Congratulations to the proposer! Main claim: $X, Y, Z, T$ form an orthocenter pair and their NPC is $\odot ABC$. Pick any two of $X, Y, Z, T$, wlog $X, Y$. Consider the midpoint of $XY$: By the equal chords condition, its three pedals on $AB, BC, CA$ are on the radical axis of $\omega_1, \omega_2$, which means that they are collinear. Simson Theorum yields the claim. Recall the well-known property that $P-$ Simson line bisects $PH$, perform a homothety with center ${X}$ and ratio $2:1$, which sends the radical axis to the perpendicular bisectors of $YZ, ZT, TY$, and sends the radical center to the circumcenter ${O}$ of $\triangle YZT$. By Menelaus we only need the circumcenter of $\triangle {ABC}$ is the midpoint of $OX$, which is evident, once seen from $\triangle YZT$. Excuse me, how can we construct the picture with this problem? how to draw it on paper?

LoloChen wrote: Recall the well-known property that $P-$ Simson line bisects $PH$, perform a homothety with center ${X}$ and ratio $2:1$, which sends the radical axis to the perpendicular bisectors of $YZ, ZT, TY$, and sends the radical center to the circumcenter ${O}$ of $\triangle YZT$. I believe you meant a homothety with center $H$ (orthocenter of $ABC$)

Very Amazing Problem, Solved with rjp08 Claim: $\odot(ABC)$ is Nine point circle of $\triangle ZTY$ SubClaim : If $M$ is midpoint of $ZT$. Then $M$ lie on $\cdot(ABC)$. Let $M_a,,Z',T'$ be feet of perpendicular from $M,Z,T$ to $BC$. We denote circle center at $Z$ and $T$ by $\omega_2$ and $\omega_3$. $Z_1,Z_2,T_1,T_2$ be intersection of $\omega_2$ and $\omega_3$ with $BC$. such that $Z_1M_a<Z_2M_A$ and similarly for $T$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15.79725855019218cm); 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 = -11.869818776940178, xmax = 40.927439773252004, ymin = -13.944530582610264, ymax = 13.879376798887742; /* image dimensions */ pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen qqwuqq = rgb(0.,0.39215686274509803,0.); /* draw figures */ draw(circle((10.273771473203443,-1.4035772082811433), 5.97362794451791), linewidth(0.4)); draw((xmin, 0.25834920340981005*xmin + 1.7204773111081308)--(xmax, 0.25834920340981005*xmax + 1.7204773111081308), linewidth(0.4)); /* line */ draw((xmin, -0.022177419354838683*xmin-4.503790322580646)--(xmax, -0.022177419354838683*xmax-4.503790322580646), linewidth(0.4)); /* line */ draw((xmin, 45.09090909090914*xmin-43.019899527136374)--(xmax, 45.09090909090914*xmax-43.019899527136374), linewidth(0.4)); /* line */ draw((xmin, 45.09090909090914*xmin-487.0347450841058)--(xmax, 45.09090909090914*xmax-487.0347450841058), linewidth(0.4)); /* line */ draw((xmin, 45.09090909090914*xmin-931.0495906410753)--(xmax, 45.09090909090914*xmax-931.0495906410753), linewidth(0.4)); /* line */ draw(circle((0.9979438370352675,1.9782953064539217), 9.112889309044746), linewidth(0.4) + yqqqyq); draw(circle((20.805639256219848,7.095597639384402), 13.643759460781377), linewidth(0.4) + yqqqyq); draw((xmin, -3.8707299531082167*xmin + 36.66044542837814)--(xmax, -3.8707299531082167*xmax + 36.66044542837814), linewidth(0.4) + qqwuqq); /* line */ draw((7.904612220187914,6.063826139991542)--(7.34,3.8), linewidth(0.4)); draw((8.948639082210121,2.0226800933126)--(10.901791546627557,4.536946472919162), linewidth(0.4)); draw((10.901791546627557,4.536946472919162)--(7.904612220187914,6.063826139991542), linewidth(0.4)); draw((7.34,3.8)--(5.24,-4.62), linewidth(0.4)); draw((7.34,3.8)--(15.16,-4.84), linewidth(0.4)); /* dots and labels */ dot((7.34,3.8),dotstyle); label("$A$", (6.658796606284543,4.119320200700802), NE * labelscalefactor); dot((5.24,-4.62),dotstyle); label("$B$", (4.954659739934444,-3.626756464526928), NE * labelscalefactor); dot((15.16,-4.84),dotstyle); label("$C$", (14.962590791408665,-4.02955245111877), NE * labelscalefactor); dot((10.901791546627557,4.536946472919162),dotstyle); label("$M$", (11.02758384547298,4.831959253901753), NE * labelscalefactor); dot((0.9979438370352675,1.9782953064539217),dotstyle); label("$Z$", (1.1126057139814913,2.2912461077070576), NE * labelscalefactor); dot((20.805639256219848,7.095597639384402),linewidth(4.pt) + dotstyle); label("$T$", (20.94256197696447,7.341688093435538), NE * labelscalefactor); dot((0.8537679902658104,-4.522724693332509),linewidth(4.pt) + dotstyle); label("$Z'$", (-0.033813632472212266,-4.060536757779681), NE * labelscalefactor); dot((-5.5290735432257865,-4.381169739968784),dotstyle); label("$z_2$", (-5.332130071487977,-3.5338035445441953), NE * labelscalefactor); dot((20.538293253503944,-4.959276664896257),linewidth(4.pt) + dotstyle); label("$T'$", (20.663703217016273,-4.71120719765881), NE * labelscalefactor); dot((26.921134786995534,-5.100831618259982),linewidth(4.pt) + dotstyle); label("$T_2$", (26.79859593587663,-4.246442597745146), NE * labelscalefactor); dot((7.236609523757408,-4.664279646696234),linewidth(4.pt) + dotstyle); label("$Z_1$", (7.092576899537296,-3.905615224475126), NE * labelscalefactor); dot((10.696030621884876,-4.741000679114383),linewidth(4.pt) + dotstyle); label("$M_a$", (10.810693698846604,-4.494317051032434), NE * labelscalefactor); dot((14.155451720012348,-4.817721711532527),linewidth(4.pt) + dotstyle); label("$T_1$", (13.816171444954962,-4.122505371101503), NE * labelscalefactor); dot((7.246374866525965,8.611685181065532),linewidth(4.pt) + dotstyle); dot((9.677479965423643,-0.7984661443918248),linewidth(4.pt) + dotstyle); dot((8.948639082210121,2.0226800933126),linewidth(4.pt) + dotstyle); label("$M_b$", (9.509352819088347,1.795497201132483), NE * labelscalefactor); dot((7.904612220187914,6.063826139991542),linewidth(4.pt) + dotstyle); label("$M_c$", (8.022106099364624,6.319205973625477), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note $M_a$ is midpoint of $Z_2T_2$ and $Z_1T_1$. Hence by $M_aZ_1.M_aZ_2 = M_aT_1.M_aT_2$ we have $M_a$ lie on radical axis. If $M_b,M_c$ be feet of perpendicular from $M$ to $AC$ and $AB$. Then $M_a,M_b,M_c$ lie on radical axis of $\omega_2$ and $\omega_3$. hence $M_a,M_b,M_c$ are collinear $\implies$ It's simson line of $M$, hence $M$ lie on $\odot(ABC)$. If $M,N,K$ are midpoints of $ZT,ZY,YT$, then they lie on $\odot(ABC)$.Therefore $\odot(ABC)$ is nine point circle of $\triangle ZTY$. Notice, as midpoint of all segment made by $X,T,Y,Z$ lie on $\odot(ABC)$ we have that $X,T,Y,Z$ is orthic system. Hence $X$ is orthocenter of $\triangle ZTY$. Now we define some points. Let $H,G$ and $O$ be orthocenter, centroid and circumcenter of $\triangle ABC$. Also note simson line of $M,N,K$ are radical axis of $\omega_2,\omega_3,\omega_4$. Hence they will concurrent at point $L$. Define $P$ as circumcenter of $\triangle ZTY$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20.42738160390522cm); 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 = -7.104959819977746, xmax = 43.322421783927474, ymin = -16.13837058445351, ymax = 10.436622772534099; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((9.36,3.74)--(6.12,-4.4)--(19.52,-4.58)--cycle, linewidth(0.) + ffxfqq); draw((9.302867064505303,-0.5132296423829497)--(15.675312627943786,-1.0592764411471924)--(10.021820307550918,-3.667493916469863)--cycle, linewidth(0.) + qqwuqq); /* draw figures */ draw(circle((12.84856646774735,-2.3633851788085267), 7.030036034103298), linewidth(0.4)); draw((1.7175980965542124,0.6341908964178558)--(25.92638568303088,8.563671536600104), linewidth(0.4)); draw((1.7175980965542124,0.6341908964178558)--(13.728461783853401,-14.983909231782206), linewidth(0.4)); draw((13.728461783853401,-14.983909231782206)--(25.92638568303088,8.563671536600104), linewidth(0.4)); draw((xmin, 0.7676380021378651*xmin-10.378907185682543)--(xmax, 0.7676380021378651*xmax-10.378907185682543), linewidth(0.4) + red); /* line */ draw((xmin, -3.0621001914181716*xmin + 37.44816830229984)--(xmax, -3.0621001914181716*xmax + 37.44816830229984), linewidth(0.4) + red); /* line */ draw((xmin, -0.5191263145084511*xmin + 5.699476006261947)--(xmax, -0.5191263145084511*xmax + 5.699476006261947), linewidth(0.4) + red); /* line */ draw((9.302867064505303,-0.5132296423829497)--(15.675312627943786,-1.0592764411471924), linewidth(0.4)); draw((10.021820307550918,-3.667493916469863)--(12.488340735264044,-0.7923822536475345), linewidth(0.4)); draw((10.021820307550918,-3.667493916469863)--(15.675312627943786,-1.0592764411471924), linewidth(0.4)); draw((9.302867064505303,-0.5132296423829497)--(12.848566467747348,-2.3633851788085254), linewidth(0.4)); draw((9.302867064505303,-0.5132296423829497)--(10.021820307550918,-3.667493916469863), linewidth(0.4)); draw((13.821991889792546,4.59893121650898)--(15.675312627943786,-1.0592764411471924), linewidth(0.4)); draw((15.675312627943786,-1.0592764411471924)--(19.82742373344214,-3.210118847591051), linewidth(0.4)); /* dots and labels */ dot((9.36,3.74),dotstyle); label("$A$", (9.467419110883123,4.04441947063062), NE * labelscalefactor); dot((6.12,-4.4),dotstyle); label("$B$", (5.294730844362798,-3.3539639806465544), NE * labelscalefactor); dot((19.52,-4.58),dotstyle); label("$C$", (19.854749476476275,-5.632666083639924), NE * labelscalefactor); dot((13.821991889792546,4.59893121650898),dotstyle); label("$M$", (13.936042715454537,5.0505996200043155), NE * labelscalefactor); dot((7.723029940203807,-7.174859167682175),dotstyle); label("$N$", (8.017335954432797,-6.816407435844272), NE * labelscalefactor); dot((19.82742373344214,-3.210118847591051),dotstyle); label("$K$", (20.47621368638356,-3.2355898454261194), NE * labelscalefactor); dot((13.728461783853401,-14.983909231782206),dotstyle); label("$Y$", (12.367585423783776,-14.777068029418512), NE * labelscalefactor); dot((25.92638568303088,8.563671536600104),dotstyle); label("$T$", (26.039798041743992,8.868165480863338), NE * labelscalefactor); dot((1.7175980965542124,0.6341908964178558),dotstyle); label("$Z$", (1.8322873891650804,0.9370984210942067), NE * labelscalefactor); dot((11.609868386818087,1.8975880926844015),linewidth(4.pt) + dotstyle); dot((7.760008242973783,-4.4220299614727825),linewidth(4.pt) + dotstyle); dot((13.730412684813263,0.16107937621591192),linewidth(4.pt) + dotstyle); dot((19.808970202732745,-4.58388168929044),linewidth(4.pt) + dotstyle); dot((13.699743027193414,-4.501817443648867),linewidth(4.pt) + dotstyle); dot((19.03252726426687,-4.180809728218539),linewidth(4.pt) + dotstyle); dot((12.488340735264044,-0.7923822536475345),linewidth(4.pt) + dotstyle); label("$L$", (12.663520761834862,-0.18745586349992377), NE * labelscalefactor); dot((9.302867064505303,-0.5132296423829497),linewidth(4.pt) + dotstyle); label("$H$", (9.408232043272905,-0.2762364649152499), NE * labelscalefactor); dot((12.848566467747348,-2.3633851788085254),linewidth(4.pt) + dotstyle); label("$O$", (12.781894897055297,-3.501931649672098), NE * labelscalefactor); dot((11.666666666666666,-1.7466666666666668),linewidth(4.pt) + dotstyle); label("$G$", (11.59815354485095,-2.495751500298402), NE * labelscalefactor); dot((10.021820307550918,-3.667493916469863),linewidth(4.pt) + dotstyle); label("$X$", (10.118476854595514,-4.212176460994707), NE * labelscalefactor); dot((15.675312627943786,-1.0592764411471924),linewidth(4.pt) + dotstyle); label("$P$", (15.918809480396819,-0.7793265396020977), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Now take homothety at $H$ send simson line of $M$ to line $\lambda_M$ which pass through $M$. Observe that as simson line of $M$ is radical axis of $\omega_2$ and $\omega_3$, therefore they are perpendicular to $ZT$, Hence we have $\lambda_M \perp ZT$. But as $M$ is midpoint of $ZT \implies \lambda_M$ pass through $P$.Similarly we get homothety at $H$ send all simson lines to lines pass through $P$. Hence it also send $L$ to $P$ $\implies HL=LP$. Take $\triangle HPX$ as reference triangle. $L$ is midpoint of $HP$. As $O$ is nine point center of $\triangle ZTY$, we have $O$ is midpoint of $PX$. Let $G'=HO \cap XL$. Then $G'$ is centroid of $\triangle HPX$. Therefore we have $HG':G'O = 2:1$, But $HG:OG=2:1$. Hence we have $G' \equiv G$. hence $G$ divide $XL$ in ratio $2:1$. $\blacksquare$.