Simple but nice! Let $M$ be the midpoint of $\overline{AB}$ and $N$ be the midpoint of $\overline{AC}$ and $CX \cap \odot(ABC) = K$ It is clear that $Y$ is the $C$-Dumpty point of $\triangle ABC$ and $CX$ is the $C$-Symmedian of $\triangle ABC$. It is obvious that $Y$ is the midpoint of $CK$ and by the midpoint theorem it remains to prove that $YZ \parallel AK$.This amounts to proving $\angle CYZ = \angle BAK$ (as $YC \parallel AB$).Now because $\odot(ACZ)$ is tangent to $BC$ (Well known) we have $\angle AZC = 180^{\circ} - \angle BCA$ which gives us $$\angle YCZ + \angle YAZ = \angle YAC + \angle YCA + \angle ACZ + \angle CAZ = 360^{\circ} - \angle ACB + \angle AZC = 180^{\circ} \implies Z \in \odot(YAC)$$This implies $$\angle ZYC = \angle ZAC = \angle ZCB = \angle KCB = \angle KAB \implies YZ \parallel AK \implies N \in YZ \ \blacksquare$$

Solution. Let $W=\overline{CX}\cap \omega,\ W\neq C$. Observe that $Y$ is the midpoint of the $C$-symmedian chord of $\bigtriangleup ABC$ and the center of spiral similarity carrying $\overline{AC}$ to $\overline{CB}$, so $\angle AYX = \angle ACB = \angle XAB = \angle AZC$, thus $ZCYA$ is cyclic. Therefore: $$\angle CYZ = \angle CAZ = \angle CWA$$which implies that $ZY\parallel AW$ and we infer that $ZY$ bisects segment $AC$. $\Box$

Let $CZ\cap OX=P.$ Now $O,P,Z,A$ and $O,P,C,Y$ is cyclic since $\angle OPZ=\angle OAZ=\angle OYC=90.$ $XY\cdot XC=XO\cdot XP=XA\cdot XZ$ $\implies A,Y,C,Z$ cyclic. So $\angle YZA=\angle YCA$ which means $YZ$ is $Z-median$ of $\triangle CAZ$ (because $CX$ is symmedian of $\triangle ABC$ and $\triangle ABC\sim \triangle CAZ$) $\square$

Let $E$ be the point of intersection of $CX$ with $AB$, and let $F$ be the point of intersection of $OX$ and $AB$. Note that $CE$ is the $C$-symmedian of $ABC$. Moreover, $F$ is the midpoint of $AB$ and $OF\perp AB$, while $Y$ is the midpoint of $CD$ and $OY\perp CX$. The points $A,X,B,O$ lie on the circle $\omega'$ with diameter $OX$, which passes through $Y$, since $O\widehat{Y}X=90^\circ=O\widehat{A}X=O\widehat{B}X$. Since $BF$ is the altitude to the hypotenuse od the right triangle $XBO$ and the quadrilateral $EFOY$ is cyclic, it follows that $XB^2=XF\cdot XO=XE\cdot XY$ (from the power of the point $X$ with respect to the circumcircle of $EFOY$). The power of the point $X$ with respect to $\omega$ is equal to $XB^2=XD\cdot XC$, and so $XE\cdot XY=XD\cdot XC$. Hence $\dfrac{XD}{XY}=\dfrac{XE}{XC}$. Since $AE$ is parallel to $ZC$, it follows that $\dfrac{XE}{XC}=\dfrac{XA}{XZ}$, and so $\dfrac{XD}{XY}=\dfrac{XA}{XZ}$. Hence $AD$ is parallel to $ZY$, and so $YM$ is parallel to $DA$, where $M$ is the point of intersection of $ZY$ with $AC$. Since $Y$ is the midpoint of $CD$, it follows that $M$ is the midpoint of $CA$, as desired. For more properties of point $Y$, see the article Midpoint of Symmedian Chord. Compare this BMO problem to Problem 12, on page 18 of this article (Peru TST 2006/4, Dutch IMO TST 2019 2.3). [asy][asy] import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.3) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.958213035260442, xmax = 5.299702835204805, ymin = -2.3402348108108706, ymax = 5.5571091580229615; /* image dimensions */ pen ududff = rgb(0.30196078431372547,0.30196078431372547,1.); pen ffqqtt = rgb(1.,0.,0.2); pen ffzztt = rgb(1.,0.6,0.2); /* draw figures */ draw((0.,0.)--(4.,0.), linewidth(1.)); draw((1.,5.)--(0.,0.), linewidth(1.)); draw((1.,5.)--(4.,0.), linewidth(1.)); draw(circle((2.,2.2), 2.9732137494637008), linewidth(1.) + ududff); draw((1.,5.)--(2.,-1.8181818181818168), linewidth(1.)); draw((-5.5,5.)--(1.4230769230769231,2.1153846153846154), linewidth(1.)); draw((-5.5,5.)--(2.,-1.8181818181818168), linewidth(1.)); draw((2.,-1.8181818181818168)--(4.,0.), linewidth(1.)); draw((1.4230769230769231,2.1153846153846154)--(2.,2.2), linewidth(1.)); draw(circle((2.,0.1909090909090926), 2.0090909090909093), linewidth(1.) + ffqqtt); draw((1.4230769230769231,2.1153846153846154)--(4.,0.), linewidth(1.)); draw((2.,2.2)--(2.,-1.8181818181818168), linewidth(1.)); draw((2.,2.2)--(4.,0.), linewidth(1.)); draw((0.,0.)--(1.8461538461538465,-0.7692307692307696), linewidth(1.)); draw((-5.5,5.)--(1.,5.), linewidth(1.)); draw(circle((1.8666666666666676,1.1), 1.1080513425729774), linewidth(1.) + ffzztt); /* dots and labels */ dot((0.,0.),dotstyle); label("$A$", (-0.5140865546175465,-0.15586307475044903), NE * labelscalefactor); dot((4.,0.),dotstyle); label("$B$", (4.341939843239851,-0.12225735573413485), NE * labelscalefactor); dot((1.,5.),dotstyle); label("$C$", (1.0653822391492196,5.170643389335348), NE * labelscalefactor); dot((2.,2.2),linewidth(4.pt) + dotstyle); label("$O$", (2.0735538096386446,2.3309601324568003), NE * labelscalefactor); dot((2.,-1.8181818181818168),linewidth(4.pt) + dotstyle); label("$X$", (1.8887223550489167,-2.1890090752374567), NE * labelscalefactor); dot((1.4230769230769231,2.1153846153846154),linewidth(4.pt) + dotstyle); label("$Y$", (1.4854537268531467,2.2469458349160147), NE * labelscalefactor); dot((-5.5,5.),linewidth(4.pt) + dotstyle); label("$Z$", (-5.437324390507572,5.137037670319034), NE * labelscalefactor); dot((0.5,2.5),linewidth(4.pt) + dotstyle); label("$M$", (0.2756478422658365,2.7174259011444133), NE * labelscalefactor); dot((1.8461538461538465,-0.7692307692307696),linewidth(4.pt) + dotstyle); label("$D$", (1.6030737434102462,-1.0968232072072461), NE * labelscalefactor); dot((1.7333333333333336,0.),linewidth(4.pt) + dotstyle); label("$E$", (1.4014394293123613,0.09617981787190731), NE * labelscalefactor); dot((2.,0.),linewidth(4.pt) + dotstyle); label("$F$", (2.0735538096386446,0.12978553688822148), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Let $CX\cap\omega =\{D\}$ and $AD\cap CZ=\{W\}$. Since $CBDA$ is harmonic, projecting through $A$ gives that $(C,W;Z,P_{\infty})=-1$, implying that $Z$ is the midpoint of $CW$. $\triangle COD$ is isosceles, thus $Y$ is the midpoint of $CD$. Since $D,A,W$ are collinear, the homothety of center $C$ and ratio $1/2$ implies the desired collinearity.

Let $M$ be the midpoint of $AC$, let $CY$ intersect the circumcircle again at $T$ and let's use the strandard notations for the angles in $ABC$. We have $\angle CAZ = \beta$, $\angle ACZ = \alpha$, let also $\angle ACT = \varphi$. Then $\triangle ACZ \sim \triangle BAC$, thus $\angle AZM = \angle BCN = \angle ACT = \varphi$ where $N$ is the midpoint of $AB$ (since $CT$ is symmedian). Hence $\angle AMZ = 180^{\circ} - \beta - \varphi$. On the other hand, $MY$ is a midsegment in $CAT$, so $\angle AMY = 180^{\circ} - \angle MAT = \angle ACT + \angle ATC = \angle ACT + \angle ABC = \beta + \varphi$. Therefore $\angle AMZ + \angle AMY = 180^{\circ}$ and 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); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -52.774282072140565, xmax = 57.25377842070913, ymin = -32.584517397502594, ymax = 33.20249527315659; /* image dimensions */ /* draw figures */ draw(circle((-0.39549701806583876,3.893657426391348), 13.400235402563032), linewidth(0.4)); draw((-22.3746890988997,13.868033283960123)--(-6.618834991044153,2.312912690212103), linewidth(0.4)); draw((-9.51439205538458,13.712621839325498)--(-11.775924169509278,-3.1810995796684782), linewidth(0.4)); draw((-22.3746890988997,13.868033283960123)--(-0.696370549482294,-21.003627298320453), linewidth(0.4)); draw((-22.3746890988997,13.868033283960123)--(-9.51439205538458,13.712621839325498), linewidth(0.4)); draw((-11.775924169509278,-3.1810995796684782)--(10.810640545043437,-3.4540490022008368), linewidth(0.4)); draw((10.810640545043437,-3.4540490022008368)--(-0.696370549482294,-21.003627298320453), linewidth(0.4)); draw((-0.696370549482294,-21.003627298320453)--(-9.51439205538458,13.712621839325498), linewidth(0.4)); draw((-9.51439205538458,13.712621839325498)--(10.810640545043437,-3.4540490022008368), linewidth(0.4)); draw((-0.39549701806583876,3.893657426391348)--(-6.618834991044153,2.312912690212103), linewidth(0.4)); draw(circle((-4.954944536725209,8.803139632858423), 6.700117701281517), linewidth(0.4) + linetype("4 4") + red); draw((-11.775924169509278,-3.1810995796684782)--(-6.618834991044153,2.312912690212103), linewidth(0.4)); draw((-6.618834991044153,2.312912690212103)--(10.810640545043437,-3.4540490022008368), linewidth(0.4)); draw(circle((-16.03883091129564,5.987802410440771), 10.111435980779236), linewidth(0.4) + linetype("4 4") + red); draw((-11.775924169509278,-3.1810995796684782)--(-3.7232779267037266,-9.086796458901294), linewidth(0.4)); /* dots and labels */ dot((-11.775924169509278,-3.1810995796684782),dotstyle); label("$A$", (-11.477849445907806,-2.4919864924739894), NE * labelscalefactor); dot((10.810640545043437,-3.4540490022008368),dotstyle); label("$B$", (11.073593762156689,-2.707446140958681), NE * labelscalefactor); dot((-9.51439205538458,13.712621839325498),dotstyle); label("$C$", (-9.251433078232647,14.45750585498842), NE * labelscalefactor); dot((-0.696370549482294,-21.003627298320453),linewidth(4pt) + dotstyle); label("$X$", (-0.41758749036025067,-20.446957199531628), NE * labelscalefactor); dot((-22.3746890988997,13.868033283960123),linewidth(4pt) + dotstyle); label("$Z$", (-22.107192104485975,14.45750585498842), NE * labelscalefactor); dot((-0.39549701806583876,3.893657426391348),linewidth(4pt) + dotstyle); label("$O$", (-0.13030795904732717,4.474542141864374), NE * labelscalefactor); dot((-6.618834991044153,2.312912690212103),linewidth(4pt) + dotstyle); label("$Y$", (-6.306817882275182,2.8945047196433022), NE * labelscalefactor); dot((-10.645158112446929,5.26576112982851),linewidth(4pt) + dotstyle); label("$M$", (-10.32873132065611,5.8391199156007545), NE * labelscalefactor); dot((-5.20316575476915,-3.2605286843783285),linewidth(4pt) + dotstyle); label("$D$", (-4.942240108538796,-2.707446140958681), NE * labelscalefactor); dot((-3.7232779267037266,-9.086796458901294),linewidth(4pt) + dotstyle); label("$E$", (-3.4340225691459474,-8.524856650045356), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $E$ be the second intersection of $CY$ with $(ABC)$ and let $M$ be the midpoint of $AC$. Notice that $Y$ is the $C$-Dumpty point of $ABC$, then we know that $\angle YAC = \angle YCB$, then notice that we have that: $$\angle YAZ + \angle YCZ = \angle ACB - \angle YAC + \angle ABC + \angle YCB + \angle BAC = 180$$giving us a cyclic $YAZC$. Then we have the following relation: $$\angle CYZ = \angle CAZ = \angle ABC = \angle AEC = \angle MYC$$ Thus giving us $Z,M,Y$ colinear.

Only chasing angles! Let $M$ be a midpoint of $\overline{AC}$. We're going to prove that $Y,M,Z$ are collinear. Claim. $O,A,X,B,Y$ are concyclic. Proof. Notice that $\angle OYX=\angle OAX=90^\circ$,$\angle OYX=\angle OBX=90^\circ$ $\Rightarrow O,A,X,Y$ and $O,X,B,Y$ are concyclic. $\Rightarrow O,A,X,B,Y$ are concyclic. $\blacksquare$ Claim. $A,Y,C,Z$ are concyclic. Proof. Notice that $\angle AYX=\angle ABX$ ($A,X,B,Y$ are concyclic) $=\angle ACB=180^\circ-\angle ABC-\angle BAC=180^\circ-\angle ZAC-\angle ACZ=\angle AZC$ $\Rightarrow A,Y,C,Z$ are concyclic. $\blacksquare$ Claim. $Y,M,Z$ are collinear. Proof. Notice that $\angle OYC=\angle OMC=90^\circ\Rightarrow O,Y,M,C$ are concyclic Notice that $\measuredangle AYM=\measuredangle AYO+\measuredangle OYM=\measuredangle ABO+\measuredangle OCM=\measuredangle ABO+\measuredangle OCA$ $=\measuredangle OAB+\measuredangle CAO=\measuredangle CAB=\measuredangle ACZ=\measuredangle AYZ $ $\Rightarrow Y,M,Z$ are collinear. So, the line $YZ$ passes through the midpoint of the line segment $AC$.

If you don't even want to draw the figure.. Set $(ABC)$ as the unit circle. Then it's readily evident that $$y = \frac{ab-c^2}{a+b-2c}\hspace{0.2 cm} \texttt{and} \hspace{0.2 cm} z = \frac{a(ab+c^2 - 2bc)}{c(a-b)}$$Hence $$\frac{y - \frac{a+c}{2}}{z - \frac{a+c}{2}} = \frac{c(a-b)^2}{(a+b-2c)(ca+bc-2ab)} \in \mathbb{R}$$

My solution during contest: Let $CX\cap (ABC)=R$. Since $Y$ is midpoint of $CR$, we get $Y$ is center of spiral similarity that sends $AC$ to $CB$. So $\angle CAY=\angle YCB$. Since $\angle CAZ=\angle CBA$ and $CZ||AB$ we get $\angle ZAY =\angle ZAC+\angle CAY=180-\angle ZCY$, which means $ZCYA$ is cyclic. So $\angle CYZ=\angle CAZ=\angle CRA \implies ZY||AR$. Since $Y$ is midpoint of $CR$ we get $ZY$ is midline in $\triangle CAR$ which means $ZY$ bisects $CA$.

Let $M$ be midpoint of $AC$. Note that $\angle OYC = \angle 90 = \angle OMC$ so $CYOM$ is cyclic so $\angle CYM = \angle COM = \angle ABC$. Note that $\angle CAZ = \angle ABC$ so we need to prove $ZAYC$ is cyclic. Claim $: ZAYC$ is cyclic. Proof $:$ Note that $\angle AYX = \angle AOX = \angle ACB = \angle XAB = \angle AZC$. Now we have $\angle CYM = \angle ABC = \angle ZAC = \angle ZYC$ so $Y,M,Z$ are collinear.

LET XB U CZ = D THEN ,BY SIMPLE ANGLE CHASING, WE CAN PROVE THAT DCYB IS CYCLIC USING THE FACT AYBX AND CYBD IS CYCLIC, WE KNOW THAT Y IS THE MIQUEL POINT OF TRIANGLE DZX HENCE,YAZC MUST BE CYCLIC.

Probably shortest solution? Let $(ABC) \cap CX \equiv T , CX \cap AB \equiv U$. $CX$ is the C-Symmedian in triangle $\triangle ABC$, $Y$ is the midpoint of $CT$ and $(X,U \backslash T,C)=-1$. From MacLaurin relation we get that $XT \cdot XC = XU \cdot XY \iff \dfrac{XT}{XY}=\dfrac{XU}{XC}$ and $AB \parallel CZ \iff \dfrac{XU}{XC}=\dfrac{XA}{XZ}$, and therefore we have that $TA \parallel YZ$, which, combined with $Y$ being the midpoint of $CT$ yields the result.

Since $\measuredangle OYX = 90^\circ$, $Y$ lies on $(OAXB)$. It follows that $$\measuredangle AYC = \measuredangle AYX = \measuredangle AOX = \measuredangle ACB = \measuredangle AZC$$ so $AZCY$ is cyclic. But since $\triangle ABC\sim \triangle CAZ$, $CY$ is the $C$-symmedian of $\triangle ABC$, and $\angle AZY = \angle ACY$, it follows that $ZY$ is the $Z$-median of $\triangle CAZ$, as desired.

Pretty much the same solution as everyone [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 = -6.673729944740892, xmax = 14, ymin = -4.375614084782157, ymax = 10.055298557160818; /* image dimensions */ pen qqffff = rgb(0.,1.,1.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); draw((2.652473130203878,6.5818204259672015)--(-0.3520656975208203,-0.9563528828777813)--(7.096163621151857,-1.6994925341808973)--cycle, linewidth(0.9) + qqffff); /* draw figures */ draw((2.652473130203878,6.5818204259672015)--(-0.3520656975208203,-0.9563528828777813), linewidth(0.9) + qqffff); draw((-0.3520656975208203,-0.9563528828777813)--(7.096163621151857,-1.6994925341808973), linewidth(0.9) + qqffff); draw((7.096163621151857,-1.6994925341808973)--(2.652473130203878,6.5818204259672015), linewidth(0.9) + qqffff); draw(circle((3.6843999941957772,1.8026626572398112), 4.889296680609137), linewidth(0.9) + blue); draw((7.096163621151857,-1.6994925341808973)--(-4.455523958617505,5.047045158005256), linewidth(0.9)); draw((2.652473130203878,6.5818204259672015)--(-4.455523958617505,5.047045158005256), linewidth(0.9) + ffxfqq); draw((-4.455523958617505,5.047045158005256)--(-0.3520656975208203,-0.9563528828777813), linewidth(0.9) + ffxfqq); draw((2.652473130203878,6.5818204259672015)--(11.12615916494295,8.411478428366372), linewidth(0.9)); draw((7.096163621151857,-1.6994925341808973)--(11.12615916494295,8.411478428366372), linewidth(0.9)); draw(circle((7.658344640333894,3.935051581626238), 5.662520189041746), linewidth(0.9) + linetype("4 4") + red); draw((2.652473130203878,6.5818204259672015)--(3.0269941530712456,0.6770264811340093), linewidth(0.9)); draw((3.3720489618155183,-1.3279227085293392)--(4.874318375677867,2.441163945893152), linewidth(0.9)); draw(circle((-0.3855619822108622,3.4248539076225333), 4.381334835665279), linewidth(0.9) + green); draw((3.0269941530712456,0.6770264811340093)--(11.12615916494295,8.411478428366372), linewidth(0.9)); /* dots and labels */ dot((2.652473130203878,6.5818204259672015),dotstyle); label("$A$", (2.4501622872122115,7.1272872964767355), NE * labelscalefactor); dot((-0.3520656975208203,-0.9563528828777813),dotstyle); label("$B$", (-0.739278550318673,-1.4737457817827542), NE * labelscalefactor); dot((7.096163621151857,-1.6994925341808973),dotstyle); label("$C$", (7.1558946704545,-2.2057485969537747), NE * labelscalefactor); dot((3.6843999941957777,1.8026626572398112),linewidth(4.pt) + dotstyle); label("$O$", (3.2344510177525927,1.9771246325949134), NE * labelscalefactor); dot((-4.455523958617505,5.047045158005256),linewidth(4.pt) + dotstyle); label("$X$", (-5.0267236106060915,5.009707724017712), NE * labelscalefactor); dot((3.0269941530712456,0.6770264811340093),linewidth(4.pt) + dotstyle); label("$Y$", (2.894592567851761,0.016402806243965926), NE * labelscalefactor); dot((11.12615916494295,8.411478428366372),linewidth(4.pt) + dotstyle); label("$Z$", (11.234196069264483,8.617435884503456), NE * labelscalefactor); dot((3.3720489618155183,-1.3279227085293392),linewidth(4.pt) + dotstyle); label("$M_A$", (3.103736229329196,-1.9181760624223023), NE * labelscalefactor); dot((4.874318375677867,2.441163945893152),linewidth(4.pt) + dotstyle); label("$M_B$", (5.169029886418867,2.2646971671263856), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Introduce the midpoint of $\overline{AC}$ and $BC$ as $M_b$ and $M_a$ respectively. Observe that $OAXBY$ is cyclic with diameter $\overline{OX}$. Also, $\triangle ABC \sim \triangle CAZ$. Now we claim that $AYCZ$ is cyclic which is true because: \[ \measuredangle ACB = \measuredangle ABX = \measuredangle AYX = \measuredangle AZC \]To complete the proof just notice that \[ \measuredangle CYZ = \measuredangle CAZ = \measuredangle CBA = \measuredangle CM_aM_b= \measuredangle CYM_b\]which is due to $OYM_aM_bC$ being cyclic with diameter $\overline{OC}$. It shows $C-M_b-Z$ are collinear as desired. $\blacksquare$

Clearly, $X,A,O,Y,B$ are concyclic and $Y,O,M,C$ are also concyclic, where $M$ is the midpoint of $AC$. Then, by angles chasing, $$\angle AYC=180^{\circ}-\angle XYA= 180^{\circ}-\angle XBA= 180^{\circ}-\angle XYA= 180^{\circ}-\angle AZC,$$i.e. $A, Y, C, Z$ are concyclic. This implies $$\angle ZYC=\angle ZAC=\angle ABC=\angle MOC=\angle MYC,$$i.e. $Z,M,Y$ are collinear as desired.

Only a beautiful solution can befit a beautiful problem such as this:

"this is a nice bashing exercise " beepbeep 2022 Let $ABC$ be the unit circle we have that $x= \frac{2ab}{a+b}$ , using the complex foot formula on $O$ onto line $XC$, we have $$y= \frac{1}{2}(0+\frac{(\frac{2ab}{a+b}-c)(0)+(\bar{\frac{2ab}{a+b}})(c)+(\frac{2ab}{a+b})(\bar{c})}{ \bar{\frac{2ab}{a+b}}+\bar{c}} = \frac{c^2-ab}{2c-a-b}$$ To compute $z$ (this is so inefficient but whatever), we construct a point $C'$ such that $ABCC'$ is a parallelogram Observe that $c'=a+c-b$ Note that $Z$ lies on $AA$ so $$z+a^2\bar{z}=2a$$ Also, $Z$ lies on $CC'$, so $$\frac{c-p}{b-a}=\bar{(\frac{c-p}{b-a})}=\frac{\frac{1}{c}-\bar{z}}{\frac{1}{b}-\frac{1}{a}} =\frac{\frac{1}{c}- (\frac{2a-z}{a^2})}{\frac{1}{b}-\frac{1}{a}}$$$$c-z=ab(\frac{2a-z}{a^2}-\frac{1}{c})$$$$z=\frac{a(2bc-c^2-ab)}{c(b-a)}$$ note that the midpoint of $AC$ is just $\frac{a+c}{2}$, let it be $K$ $$\bar{\frac{K-Z}{K-Y}}=\bar{\frac{\frac{a+c}{2}-\frac{a(2bc-c^2-ab}{c(b-a)}}{\frac{a+c}{2}-\frac{c^2-ab}{2c-a-b}}}=\frac{\frac{a+c}{2}-\frac{a(2bc-c^2-ab}{c(b-a)}}{\frac{a+c}{2}-\frac{c^2-ab}{2c-a-b}}=\frac{K-Z}{K-Y}$$ So they're collinear Can someone tell me why the conjugate bars are so short on AOPS

Suppose that $YZ$ intersects $CA$ at $N,$ $P$ be midpoint of $AB$. Note that $Y$ is $C$ - Dumpty point of $\triangle ABC,$ we have $\angle{AYC} = 180^{\circ} - \angle{ACB} = \angle{BAC} + \angle{ABC} = \angle{ACZ} + \angle{CAZ} = 180^{\circ} - \angle{AZC}$. Then $Y \in (AZC)$. We also have $\angle{CZY} = \angle{CAY} = \angle{BCX} = \angle{ACP}$. So $\triangle CNZ$ $\sim$ $\triangle APC$. Hence $\dfrac{CN}{CA} = \dfrac{CN}{CZ} \cdot \dfrac{CZ}{CA} = \dfrac{AN}{CA} \cdot \dfrac{CA}{AB} = \dfrac{AN}{AB} = \dfrac{1}{2}$ or $N$ is midpoint of $CA$

Knty2006 wrote: Can someone tell me why the conjugate bars are so short on AOPS Maybe it's better to use \overline{abcde} which will give $\overline{abcde}$.

Switch to $A$-indexing. Quote: In $\triangle ABC$, let $M$ denote the midpoint of $\overline{BC}$, let $O$ denote the circumcenter, and let $\Omega$ denote the circumcircle. Tangents to $\Omega$ at $B$ and $C$ meet at $X$. Let $Y$ be the projection of $O$ onto $AX$ and let the line through $A$ parallel to $BC$ and the line $BX$ meet at $Z$. Prove that $Z,M,Y$ are colinear. Note that by construction, $AX$ is the $A$-symmedian of $\triangle ABC$. Let $AX$ meet $\Omega$ at $F\ne A$. It is well-known that $Y$ is the midpoint of $\overline{AD}$. We now employ Barycentric Coordinates. Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Since $F$ is on the $A$-symmedian, it can be parameterized by $F=(t:b^2:c^2)$ but it is also on $\Omega$ so $\text{Pow}_\Omega (F)=0$. Of course, $\Omega$ has equation $a^2yz+b^2zx+c^2xy=0$, so plugging in $F$ gives $a^2b^2c^2+2tb^2c^2=0 \Rightarrow t=-\frac{a^2}2$. So$$F=\left(-\frac{a^2}2,b^2,c^2\right)=(-a^2:2b^2:2c^2).$$Using the midpoint formula on normalized $A$ and $F$ and then scaling, we find $$Y=(-a^2+b^2+c^2:b^2:c^2).$$Clearly, $$M=(1:1:0).$$ Since $X$ is on the $A$-symmedian, it can be parameterized by $(t:b^2:c^2)$. However, it is also on the tangents to $\Omega$ at $B$ and $C$ so it must simultaneously satisfy $a^2z+c^2z=0$ and $a^2y+b^2x=0$. Hence, $t=-a^2$ and $X=(-a^2:b^2:c^2)$. Since $Z$ is on cevian $BX$, it can be parameterized by $(-a^2:t:c^2)$. However, we also know that $A,Z,P_\infty$ are colinear where $P_\infty=(-2:1:1)$ is the point at infinity along $BC$. Hence, $$\begin{vmatrix}1&0&0\\ -a^2&t&c^2\\ -2&1&1\end{vmatrix}=0\Rightarrow t=c^2$$so $Z=(-a^2:c^2:c^2)$. Finally, it suffices to check $$\begin{vmatrix}-a^2&c^2&c^2\\ 1&1&0\\ -a^2+b^2+c^2&b^2&c^2\end{vmatrix}=-a^2c^2+c^2c^2+c^2(b^2+a^2-b^2-c^2)=0$$which is true.

Here's my solution (might add a bashable one afterwards)

Let $YZ\cap AC=M, CX\cap \odot (ABC)=T$. We need to prove $AM=MC$. Claim: $AYCZ$ is cyclic. Proof: Since $Y$ is the foot of the perpendicular from $O$ onto $CX$, we get that $Y$ is the midpoint of $TC$. Also, since $CX$ is the $C$ symmedian of $\bigtriangleup ABC$ $AB$ is the $A$ symmedian of $\bigtriangleup ACT$. So we get $\angle TCB=\angle TAB=\angle YAC\Rightarrow \angle TAY=\angle CAB=\angle ACZ$(since $ATBC$ is cyclic). So we get $\angle YAZ+\angle YCZ=\angle YAC+\angle CAZ+\angle YCA+\angle ACZ=\angle TCB+\angle ABC+\angle TCA+\angle CAB=\angle ABC+\angle BCA+\angle CAB=180^{\circ}$. Thus, $AYCZ$ is cyclic. So we get $\angle CBA=\angle CTA=\angle CAZ=\angle CYZ\Rightarrow AT\parallel YZ\Rightarrow AT\parallel YM\Rightarrow \bigtriangleup ATC\sim \bigtriangleup MYC$. Thus $\frac{AM}{AC}=\frac{TY}{TC}=\frac{1}{2}\Rightarrow AM=MC$ and thus $YZ$ passes through the midpoint of the line segment $AC$.

Extend $XB$ and $CZ$ to meet at $P$. Furthermore, let $PZ$ meet $\omega$ at $Q$ other than $C$. Let $R$ be the intersection of the C-symmedian with $\omega$. Note that $CX$ is the C-symmedian, so $Y$ is the C-dumpty point of $\triangle ABC$. Therefore, $XAYB$ is cyclic. Claim: $BYCP$ is cyclic. Note that since $CQ\parallel AB$, so $\overarc{AQ}=\overarc{BC}$. Therefore, $$\angle BPC=\frac{1}{2}(\overarc{BQ}-\overarc{BC})=\frac{1}{2}\overarc{AB}=\angle XAB.$$Additionally, from cyclic $XAYB$, we have $$\angle XAB=\angle XYB,$$and since $\angle XYB=\angle BPC,$ $BYCP$ is cyclic. Therefore, $Y$ is the Miquel center of $\triangle XZP$. Therefore, $AZCY$ is also cyclic. Let $YZ$ intersect $AC$ at $M.$ From the circumcircle tangency, $$\angle ARC=\angle ZAC.$$From cyclic $AZCY$, we also have $\angle ZAC=\angle MYC.$ Since $\angle ARC=\angle MYC$, we have $\triangle ARC\sim \triangle MYC$, so $$AC/MC=RC/YC=2,$$so we are done.

$\angle{OYX} = 90^{\circ} = \angle{OAX} \implies OYAX$ is cyclic. $\angle{AZC} = \angle{XAB} = \angle{ACB} = \frac{\angle{AOB}}{2} = \angle{AOX} = \angle{AYX} \implies AZCY$ is cyclic. Let $ M' = ZY \cap AC$. Then, $\angle{ZYO} = \angle{ZYC} + 90^{\circ} = \angle{ZAC} + 90^{\circ} = \angle{ABC} + 90^{\circ} = 180^{\circ} - (90^{\circ} - \angle{ABC}) = 180^{\circ} - \angle{ACO}$. Hence, $YM'CO$ is cyclic and $\angle{OM'C} = \angle{OYC} = 90^{\circ},$ which finishes.

Using directed angles: $$\measuredangle AYC=\measuredangle AYX=\measuredangle ABX=\measuredangle ACB=\measuredangle AZC$$so $AZCY$ is cyclic. Let $M$ be the midpoint of $AC$ and $N$ be the midpoint of $AB$. Since $\triangle CZA\sim \triangle ACB$ and by quoting the properties of $Y$, which is the $C$-dumpty point of $\triangle ABC$: $$\measuredangle YZC=\measuredangle YAC=\measuredangle YCB=\measuredangle ACN=\measuredangle MZC$$so $M,Z,Y$ are collinear.

Maybe, it is a bit late to post this solution, but i will post it. Let $M$ be the midpoint of $AC$. $O$ is the circumcecter of $\omega$ $\Longrightarrow$ $OM$ $\perp$ $AC$. Combining with $OY$ $\perp$ $CX$ $\implies$ $CMOY$ is cyclic. Let $\angle ABC$ = $\beta$ $\Longrightarrow$ $\angle ABC = \angle AOC = 2\angle MOC$ $\implies$ if we can prove that $AYCZ$ is cyclic, the $M, Y, Z$ lie on the same line. We have $OA \perp ZX$ and $OY \perp CX$ $\implies AYOX$ is cyclic. Let $\angle ACB = \gamma$ $\Longrightarrow \angle AOB = 2\gamma = 2 \angle AOX$. Hence $\angle AYX = \gamma$ $ZC \parallel AB$ $\Longrightarrow \angle ZCA = 180^{o} - \beta - \gamma$. Thus $\angle CZA = \gamma = \angle AYX \Longrightarrow ZACY$ is cyclic and we are done.

let $\angle{ACX}$=a , $\angle{BCX}$ =b , $\angle{CAY}$ =t , $\angle{YAB}$ =z , $\angle{CBA}$=s so we can do some angle chasing : $\angle{CBA}$ = $\angle{ZAC}$ = s , ZC || AB so we can say $\angle{CAB}$ = $\angle{ZCA}$= t+z , in $\triangle ABC$ t+z+s+a+b=180 and in $\triangle ZAC$ $\angle{AZC}$ +t+z+s=180 so we get $\angle{AZC}$ = a+b than we know $\angle{AYC}$ =180-a-t; than $\angle{AZC}$ + $\angle{AYC}$ = 180 $\implies OAZC$ is cyclic.So we get $\angle{AZY}$ = $\angle{ACY}$ =a and $\angle{YZC}$ = $\angle{CAY}$ = t. And we can say CX is simmedian of $\triangle ABC$ .Let D point is intersection of CY and circle ABC . AY is medin of $\triangle ACD$ . We know CX is simmedian so we can say AB is simmedian of $\triangle ACD$ and we get $\angle{DCB}$ = $\angle{DAB}$ = $\angle{YAC}$ = b so t=b: let write simmedian sine : sin(a)/sin(s)=sin(b)/sin(b+z) => it is median sine in $\triangle AZC$ .So ZY is median of $\triangle AZC$.

Let $M,N$ be the midpoints of $AC,AB$ respectively. First note the following observations. Claim : Points $A$,$B$,$O$,$X$ and $Y$ are concyclic. Proof : Note that since $XA$ and $XB$ are tangents to $\omega$ \[\measuredangle OAX = \measuredangle XBO = \measuredangle XYO = 90^\circ\]Thus, $XBYOA$ is cyclic as claimed. Claim : $\triangle ACZ \sim \triangle BAC$ and thus, $\triangle AZM \sim \triangle BCN$. Proof : First, note that \[\measuredangle ACZ = \measuredangle CAB \text{ and } \measuredangle ZAC = \measuredangle ABC\]Thus, $\triangle ACZ \sim \triangle BAC$ as claimed. Now, since \[\frac{MA}{AZ}=\frac{AC}{2AZ}=\frac{AB}{2BC}=\frac{BN}{BC}\]we can also conclude that $\triangle BCN \sim \triangle AZM$. Now, armed with these observations, we prove the following key claim. Claim : Quadrilateral $AYCZ$ is cyclic. Proof: Note that, \[2\measuredangle AXO = \measuredangle AXB = 2\measuredangle XAB = 2\measuredangle ACB \]Thus, $\measuredangle AXO = 90+ \measuredangle BCA$. Further, \[\measuredangle XYA = \measuredangle XOA = \measuredangle BCA = \measuredangle CZA\]Thus, $AYCZ$ must indeed be cyclic as claimed. Now, note that, \[\measuredangle YZA = \measuredangle YCA = \measuredangle XCA = \measuredangle BCN = \measuredangle MZA \]Thus, we must have $Y-M-Z$ as was required.

Say $P = CX \cap \omega$, $Q = AB \cap CX$. Since $XA$, $XB$ are tangents, we have $(C, P; X, Q) = -1$ (EGMO Lemma 9.9). Project about $A$, where $C \rightarrow C, P \rightarrow AP \cap CZ = G, X \rightarrow Z, Q \rightarrow P_{\infty}$, since $AQ \parallel CZ$. $$\implies (C, G; Z, P_{\infty}) = -1$$Therefore we have $\frac{ZC}{ZG} = 1$, which means $Z$ is the midpoint of $CG$, and since $Y$ is the midpoint of $CP$ (as $OP, OC$ are radius), we reach our desired conclusion.