Solution. Firstly, by condition (a) we get that if a triangle $\Delta$ is reflected by a point or a line then so is $f(\Delta)$ by the same point or line, now if (b) works for any quadrilateral it works for kites as well. Suppose $f(ABC)$ does not intersect the line $AC$ [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 = -10.920395554384156, xmax = 9.372537609982519, ymin = -8.693153218983554, ymax = 11.599779945383105; /* image dimensions */ /* draw figures */ draw((-4.94,-0.01)--(-3.08,1.95), linewidth(0.4)); draw((-4.94,-0.01)--(-2.3426529872349366,-0.7549754997853285), linewidth(0.4)); draw((-2.3426529872349366,-0.7549754997853285)--(-0.18311501461472002,1.2866753060496035), linewidth(0.4)); draw((-0.18311501461472002,1.2866753060496035)--(-3.08,1.95), linewidth(0.4)); draw(circle((2.24,4.15), 1.4747203124660622), linewidth(0.4)); draw(circle((3.3578616221437723,0.04909737518304785), 1.4747203124660626), linewidth(0.4)); draw((-4.94,-0.01)--(5.17239801437669,2.7465301306365855), linewidth(0.4)); /* dots and labels */ dot((-4.94,-0.01),dotstyle); label("$A$", (-4.843367440990927,0.25961141235467916), NE * labelscalefactor); dot((-3.08,1.95),dotstyle); label("$B$", (-2.971425745347477,2.2129418773739298), NE * labelscalefactor); dot((-2.3426529872349366,-0.7549754997853285),dotstyle); label("$C$", (-2.2389268209652577,-0.4728875120275398), NE * labelscalefactor); dot((-0.18311501461472002,1.2866753060496035),dotstyle); label("$D$", (-0.06855963761053309,1.5618317223675129), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] But this will result in $f(ABC)\cap f(ADC)=\emptyset$ but if this happens there cannot be a point of intersection of all triangles thus $f(ABC)$ intersects the line $AC$ and we can similarly find out it intersects $AB$ and $BC$. [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 = -13.145021917322746, xmax = 17.999747163817553, ymin = -7.770747166057797, ymax = 12.522185998308862; /* image dimensions */ /* draw figures */ draw((-4.94,-0.01)--(-3.08,1.95), linewidth(0.4)); draw((-4.94,-0.01)--(-2.3426529872349366,-0.7549754997853284), linewidth(0.4)); draw((-2.3426529872349366,-0.7549754997853284)--(-0.18311501461472002,1.2866753060496037), linewidth(0.4)); draw((-0.18311501461472002,1.2866753060496037)--(-3.08,1.95), linewidth(0.4)); draw(circle((2.8639805652144834,2.874909279779101), 1.4747203124660626), linewidth(0.4)); draw(circle((3.2484550156811465,1.4644553811381382), 1.4747203124660664), linewidth(0.4)); draw((-4.94,-0.01)--(4.291949872627554,2.506529506873158), linewidth(0.4)); draw((5.698074014204402,2.889823712477165)--(4.291949872627554,2.506529506873158), linewidth(0.4)); /* dots and labels */ dot((-4.94,-0.01),dotstyle); label("$A$", (-4.8433674409909235,0.2596114123546776), NE * labelscalefactor); dot((-3.08,1.95),dotstyle); label("$B$", (-2.971425745347474,2.212941877373928), NE * labelscalefactor); dot((-2.3426529872349366,-0.7549754997853284),dotstyle); label("$D$", (-2.2389268209652546,-0.4728875120275413), NE * labelscalefactor); dot((-0.18311501461472002,1.2866753060496037),dotstyle); label("$C$", (-0.06855963761053001,1.5618317223675113), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Now if we have $ABCD$ as a parallelogram we have that $f(ABC)$ and $f(ADC)$ intersect $\iff$ they contain the midpoint $M$ of $AC$ since $M$ is the point of reflection of $\Delta ABC$ to $\Delta CDA$. [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 = -43.05168336896453, xmax = 2.3915092391940918, ymin = -9.95703251292155, ymax = 35.486160095237096; /* image dimensions */ /* draw figures */ draw((-25.638495616904997,8.991486906239453)--(-23.85523643062513,12.3522446034592), linewidth(0.4)); draw((-23.85523643062513,12.3522446034592)--(-18.471165425895535,8.92290001445946), linewidth(0.4)); draw((-25.638495616904997,8.991486906239453)--(-20.254424612175402,5.562142317239713), linewidth(0.4)); draw((-18.471165425895535,8.92290001445946)--(-20.254424612175402,5.562142317239713), linewidth(0.4)); draw(circle((-11.269541788996074,12.146483928119215), 3.990436998109084), linewidth(0.4)); draw(circle((-32.84011925380446,5.767902992579698), 3.990436998109084), linewidth(0.4)); draw((-40.37842854529166,9.132538895984778)--(-4.292166412511804,8.78721581337445), linewidth(0.4)); /* dots and labels */ dot((-25.638495616904997,8.991486906239453),dotstyle); label("$A$", (-26.8306507133785,9.483905501263967), NE * labelscalefactor); dot((-23.85523643062513,12.3522446034592),dotstyle); label("$B$", (-23.610745354779024,12.946822585040762), NE * labelscalefactor); dot((-18.471165425895535,8.92290001445946),dotstyle); label("$C$", (-18.20373446958368,9.544658432558297), NE * labelscalefactor); dot((-22.054830521400266,8.957193460349457),linewidth(4pt) + dotstyle); label("$M$", (-21.788157415949133,9.423152569969638), NE * labelscalefactor); dot((-20.254424612175402,5.562142317239713),dotstyle); label("$D$", (-20.75535758394553,4.1984004786572795), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Now since $M$ is the point of reflection of both circles we have that the radical axis of these circles passes through $M$ now this also hols true for $\Delta ABD$ and $\Delta ACD$ so their radical axis passes through $M$ so by the Radical Center Theorem and by condition (b) the point they all intersect at $M$. [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 = -35.45882139398872, xmax = -13.05690729981616, ymin = -1.3782373480215024, ymax = 21.02367674615109; /* image dimensions */ /* draw figures */ draw((-27.043129067167197,9.1937889557792)--(-23.85523643062513,12.3522446034592), linewidth(0.4)); draw((-23.85523643062513,12.3522446034592)--(-18.5076404082913,8.864349042629602), linewidth(0.4)); draw((-18.5076404082913,8.864349042629602)--(-27.043129067167197,9.1937889557792), linewidth(0.4)); draw((-27.043129067167197,9.1937889557792)--(-21.695533044833365,5.705893394949603), linewidth(0.4)); draw((-18.5076404082913,8.864349042629602)--(-21.695533044833365,5.705893394949603), linewidth(0.4)); draw(circle((-20.72925628599614,12.47463778448041), 3.990436998109084), linewidth(0.4)); draw(circle((-24.821513189462355,5.583500213928392), 3.990436998109084), linewidth(0.4)); draw(circle((-21.05176550768003,7.656485198775329), 2.204183780394658), linewidth(0.4)); draw(circle((-24.499003967778464,10.401652799633474), 2.204183780394658), linewidth(0.4)); /* dots and labels */ dot((-27.043129067167197,9.1937889557792),dotstyle); label("$A$", (-26.92333273511281,9.493279785915195), NE * labelscalefactor); dot((-23.85523643062513,12.3522446034592),dotstyle); label("$B$", (-23.748729935671246,12.637933502343166), NE * labelscalefactor); dot((-18.5076404082913,8.864349042629602),dotstyle); label("$C$", (-18.387844076236902,9.1638398727656), NE * labelscalefactor); dot((-22.775384737729247,9.029068999204402),linewidth(4pt) + dotstyle); label("$M$", (-23.119799192385653,9.433381619887996), NE * labelscalefactor); dot((-21.695533044833365,5.705893394949603),dotstyle); label("$D$", (-21.56244687567847,6.0191861563376285), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Thus $f(ABC)$ intersect $ABC$ at $M$ and we can similarly find out it intersects all the other midpoints of $ABC$ which is the Euler circle. [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 = -2.5822742133572367, xmax = 10.254940052751257, ymin = -5.713199230398693, ymax = 7.124015035709746; /* image dimensions */ /* draw figures */ draw((2.06,-1.04)--(2.66,3.64), linewidth(0.4)); draw((2.66,3.64)--(7.38,1.12), linewidth(0.4)); draw((7.38,1.12)--(4.46,-2.54), linewidth(0.4)); draw((4.46,-2.54)--(2.06,-1.04), linewidth(0.4)); draw((2.66,3.64)--(4.46,-2.54), linewidth(0.4)); draw((2.06,-1.04)--(7.38,1.12), linewidth(0.4)); draw(circle((3.895039658328249,1.3349949155989425), 1.5354385030857745), linewidth(0.4)); draw(circle((4.652269826800365,-1.403368277119417), 1.444956525213242), linewidth(0.4)); draw(circle((2.099391691394658,-0.4519732937685457), 1.771250155283648), linewidth(0.4)); draw(circle((5.18405056508184,0.7517137568199533), 1.636529461278672), linewidth(0.4)); /* dots and labels */ dot((2.06,-1.04),dotstyle); label("$A$", (2.1201278039071045,-0.8735008038711799), NE * labelscalefactor); dot((2.66,3.64),dotstyle); label("$B$", (2.7207995944335712,3.8117391622352423), NE * labelscalefactor); dot((7.38,1.12),dotstyle); label("$C$", (7.440363662855812,1.288917642024092), NE * labelscalefactor); dot((4.46,-2.54),dotstyle); label("$D$", (4.522814966012972,-2.3665992546083916), NE * labelscalefactor); dot((3.56,0.55),linewidth(4pt) + dotstyle); label("$E$", (3.6303883058022213,0.6882458514976275), NE * labelscalefactor); dot((4.72,0.04),linewidth(4pt) + dotstyle); label("$F$", (4.7802457333814585,0.17338431676065805), NE * labelscalefactor); dot((5.02,2.38),linewidth(4pt) + dotstyle); label("$G$", (5.0891626542236414,2.5245853253928185), NE * labelscalefactor); dot((2.36,1.3),linewidth(4pt) + dotstyle); label("$H$", (2.4290447247492875,1.4433761024451828), NE * labelscalefactor); dot((3.26,-1.79),linewidth(4pt) + dotstyle); label("$I$", (3.3214713849600384,-1.6457931059766342), NE * labelscalefactor); dot((5.92,-0.71),linewidth(4pt) + dotstyle); label("$J$", (5.981589314434392,-0.5645838830289982), NE * labelscalefactor); dot((3.852607184626693,-0.19985715607892462),linewidth(4pt) + dotstyle); label("$K$", (3.9221431754865055,-0.06688439944992772), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Now we must prove for any quadrilateral $ABCD$ the euler circles of $\Delta ABC$, $\Delta ABD$, $\Delta ACD$, $\Delta BCD$ intersect at a point. Let $H$, $G$, $J$, $I$, $E$, $F$, be the midpoints of $AB$, $BC$, $CA$, $DA$, $BD$, $AC$, respectively. Let $K=(HFG)\cap (FIJ)$, we want to prove that $HEKI$ is cyclic, because than similarly it would follow for $GEKJ$ to be a cyclic quadrilateral. We have that $\angle BAC=\angle HGF=180^{\circ}-\angle HKF$, and $\angle CAD=\angle FJI=180^{\circ}-\angle FKI$. Now we have that: $$\angle HKI=\angle HAI=\angle HEI\implies HEKI\text{ is cyclic.}$$Thus we proved our claim. Now we're done with the problem.$\blacksquare$

I was wondering when they said that all $4$ circles meet at one point, do they mean exactly one point? Could the circles be coaxial and still work? Edit: Wait if $A,B,C,D$ concyclic then I think their $9$-point circles are coaxial anyway. Double Edit: I was probably thinking of something else here, the above is obviously untrue

for any of the 3 circles to be coaxial 3 of the points A B C D must be colinear

A B C D being concyclic will only result in the circles being coaxial if 2 of the points A B C D are equal

Isnt it possible that $M$ contain inside $f(ABC)$ and therefore $f(ABC), f(ABD), f(BCD), f(ACD)$ are all coaxial. How do you know they all passes through $M$. I dont think that if the circles are coaxial, then three of the points are collinear as it satisfies the condition of the problem perfectly. Yeah the solution is false i guess.

My proof is indeed incorrect. Here is the correct one at least I think it is: Now we make $ABCD$ such that it is a kite and a parallelogram since we proved in a kite $ABCD$, $\Delta ABC$ and $\Delta ADB$ intersect on the line $AC$ (in the previous solution I posted) so $AC$ is the radical axis of $2$ of these circles and we can similarly find out that $BD$ is the radical axis of the to other circles and and $BD$ and $AC$ intersect at the midpoint of $BD$ and $AC$ thus $f(ABC)$ passes through the midpoint of $BC$ if $AC=AB$ now if we have a kite of two different lengths for sides by the above proof we have that the two radical axis of the two pairs of circles respectively pass through $H$ where is the intersection of $AC$ and $BD$ it also is a foot of the altitude of all the triangles. similarly we can prove that $f(ABC)$ passes through all the other altitudes thus being a $9$-point circle. if I have done something else incorrect pls tell me since I'm not very good at geometry

Still wrong. How do you apply your argument for $BD$. This is a kite.

RevolveWithMe101 wrote: Still wrong. How do you apply your argument for $BD$. This is a kite. for which BD are you talking about the one as a kite and not a parallelogram or the one as a kite and a parallelogram

We will proceed to solve the problem with the following claims. Claim 1 : $f(ABC)$ intersects the lines $\overline{AB},\overline{BC},$ and $\overline{CA}$. Proof : Let the reflection of $A$ with respect to $\overline{BC}$ be $A'$, then since $f(ABC)$, and $f(A'BC)$ intesects and $ABC$ and $A'BC$ are similar by reflection with respect to $\overline{BC}$, by (i) their intersection must lie on $\overline{BC}$, and thus $f(ABC)$ intesects $\overline{BC}$, similarly with reflections of $B,C$ with respect to their opposite sides, we have the result stated above. $\blacksquare$ Claim 2 : For all isosceles triangles $KLM$ with $KL=LM$, then $f(KLM)$ intesects the projection of $L$ on $KM$. Proof : Reflect $L$ with respect to $\overline{KM}$, and let the reflection be $L'$, from Claim 1, we have $f(KLM) \cap f(KL'M) \in \overline{KM}$ and $f(KLL') \cap f(MLL') \in \overline{LL'}$, also from (ii), we know the point $f(KLM) \cap f(KL'M) \cap f(KLL') \cap f(ML'L)$ exists, and thus it must be $ \overline{KM} \cap \overline{LL'} $. Since $ \overline{KM} \cap \overline{LL'} $ is just the projection of $L$ on $ \overline{KM}$, we get the result. $\blacksquare $. Claim 3 : The perpediculars from the points of a triangle $ABC$ to their opposite sides lie on $f(ABC)$. Therefore, $f(ABC)$ is the nine-point circle as desired. Proof : Let the reflection of $A$ with respect to $\overline{BC}$ be $A'$, we have $ABA'$ and $ACA'$ are isosceles triangles. From Claim 2, and (ii) , we have $f(ABA') \cap f(ACA') =\{\overline{BC} \cap \overline{AA'},Q\}$ for some point $Q$ which can be equal to $\overline{BC} \cap \overline{AA'}$, and thus $f(ABC) \cap f(A'BC) \cap f(ABA') \cap f(ACA') = \overline{BC} \cap \overline{AA'} $, which is just the perpendicular from $A$ to $\overline{BC}$, similarly for points $B$ and $C$. Therefore, we get the desired result. $\blacksquare$

SomeUser221104 wrote: We will proceed to solve the problem with the following claims. Claim 1 : $f(ABC)$ intersects the lines $\overline{AB},\overline{BC},$ and $\overline{CA}$. Proof : Let the reflection of $A$ with respect to $\overline{BC}$ be $A'$, then since $f(ABC)$, and $f(A'BC)$ intesects and $ABC$ and $A'BC$ are similar by reflection with respect to $\overline{BC}$, by (i) their intersection must lie on $\overline{BC}$, and thus $f(ABC)$ intesects $\overline{BC}$, similarly with reflections of $B,C$ with respect to their opposite sides, we have the result stated above. $\blacksquare$ Claim 2 : For all isosceles triangles $KLM$ with $KL=LM$, then $f(KLM)$ intesects the projection of $L$ on $KM$. Proof : Reflect $L$ with respect to $\overline{KM}$, and let the reflection be $L'$, from Claim 1, we have $f(KLM) \cap f(KL'M) \in \overline{KM}$ and $f(KLL') \cap f(MLL') \in \overline{LL'}$, also from (ii), we know the point $f(KLM) \cap f(KL'M) \cap f(KLL') \cap f(ML'L)$ exists, and thus it must be $ \overline{KM} \cap \overline{LL'} $. Since $ \overline{KM} \cap \overline{LL'} $ is just the projection of $L$ on $ \overline{KM}$, we get the result. $\blacksquare $. Claim 3 : The perpediculars from the points of a triangle $ABC$ to their opposite sides lie on $f(ABC)$. Therefore, $f(ABC)$ is the nine-point circle as desired. Proof : Let the reflection of $A$ with respect to $\overline{BC}$ be $A'$, we have $ABA'$ and $ACA'$ are isosceles triangles. From Claim 2, and (ii) , we have $f(ABA') \cap f(ACA') =\{\overline{BC} \cap \overline{AA'},Q\}$ for some point $Q$ which can be equal to $\overline{BC} \cap \overline{AA'}$, and thus $f(ABC) \cap f(A'BC) \cap f(ABA') \cap f(ACA') = \overline{BC} \cap \overline{AA'} $, which is just the perpendicular from $A$ to $\overline{BC}$, similarly for points $B$ and $C$. Therefore, we get the desired result. $\blacksquare$ There is a flaw for the point $Q$ you mention. I fix this by considering the reflection $\delta$ with respect to $LL'$ in claim 2. Since $\delta$ maps $\triangle KLM$ to itself, it maps $f(KLM)$ to itself. Thus $f(KLM)$ is symmetric w.r.t. $LL'$, so it tangents $KM$ at the mid point of $KM$. With this observation, in claim 3 now we have $f(ABA')$ and $f(ACA')$ tangent to $BC$ at the same point, which is $BC\cap AA'$.$\blacksquare$

Lemma- Consider $\triangle ABC $ such that $AB = AC.$ Let $M$ be the midpoint of $BC.$ $f(ABC)$ is tangent to $BC$ at $M.$ Let $A'$ be the reflection of $A$ through $BC,$ then four points $A, B, C, A'$ are in general position. Now according to (b), conclude that $f(ABC), f(A'BC), f(ABA'), f(ACA')$ have a common point. Let that common point be $X.$ Notice that if we call the reflection through $BC$ by $g,$ the reflection through $AA'$ by $h,$ then using condition (a) we have $g(f(ABC)) = f(A'BC),$ and $h(f(ABA')) =f(ACA').$ So, with the point $X$ lies on these circles $,\,\, g(X)=X$ and $h(X) = X,$ so $X$ lies on $BC$ and $AA', $ or $X$ is the midpoint of $BC.$ We have $f(ABC)$ passes through the midpont of $BC.$ Let $M$ be the midpoint of $BC.$ Let $t$ be the reflection through $AM.$ We have $t(ABC) = ABC,$ then $t(f(ABC)) = f(ABC). $ Notice that the center of $f(ABC)$ is unchanged after the reflection, so we have the center of $f(ABC)$ lies on $AM.$ Combining with the fact that $M\in f(ABC),$ $ f(ABC)$ is tangent to $BC$ at $M.$ Back to the problem Consider an arbitrary triangle $ABC$ and let $A'$ be the reflection of $A$ through $BC.$ Then by conidion (b), $f(ABC),f(ABA'),f(ACA')$ have a common point. $(\star)$ According to Lemma, we have $f(ABA')$ touches $f(ACA')$ at the midpoint of $AA'$ Both of them touch $AA'$ at the midpoint of $AA', $ as $ABA'$ and $ACA'$ are the isosceles triangles. So the common point in $(\star)$ is the midpoint of $AA'$. The only intersection of $f(ABA') $ and $f(ACA').$ So $f(ABC)$ passes through the midpoint of $AA'.$ Similarly, we have that $f(ABC)$ is the Euler circle of $\triangle ABC. $

amar_04 wrote: The mapping $f$ assigns a circle to every triangle in the plane so that the following conditions hold. (We consider all nondegenerate triangles and circles of nonzero radius.) (a) Let $\sigma$ be any similarity in the plane and let $\sigma$ map triangle $\Delta_1$ onto triangle $\Delta_2$. Then $\sigma$ also maps circle $f(\Delta_1)$ onto circle $f(\Delta_2)$. (b) Let $A,B,C$ and $D$ be any four points in general position. Then circles $f(ABC),f(BCD),f(CDA)$ and $f(DAB)$ have a common point. Prove that for any triangle $\Delta$, the circle $f(\Delta)$ is the Euler circle of $\Delta$. One of the most beautiful functional geometry problem I've seen. Claim 01. Let $\triangle ABC$ be an isosceles triangle with apex $A$. Let $M$ be the midpoint of $BC$. Then $f(ABC)$ is tangent to $BC$ at $M$. Proof. We first claim that $f(ABC)$ passes through $M$. Reflect $A$ wrt $BC$ to get $A'$. Take $\sigma$ to be the reflection wrt line $BC$. Since $\sigma : \triangle ABC \mapsto \triangle A'BC$, then $\sigma : f(ABC) \mapsto f(A'BC)$. Take $\sigma'$ to be the reflection wrt line $AA'$. Since $\sigma' : \triangle ABA' \mapsto \triangle ACA'$, then $\sigma' : f(ABA') \mapsto f(ACA')$. By the second condition of the problem, $f(ABC), f(A'BC), f(ABA'), f(ACA')$ concur at a point: let it be $M'$. Therefore, we know that $\sigma : M' \mapsto M'$ and $\sigma' : M' \mapsto M'$. Therefore, we conclude that $M' = AA' \cap BC \equiv M$, which is what we wanted. Now, take $\sigma$ to be the reflection wrt line $AM$. Then, $\sigma$ preserves $\triangle ABC$, which means $\sigma$ preserves $f(ABC)$, and this means that the center lies on $AM$. Since $M \in f(ABC)$, we are done with this claim. Claim 02. For any arbitrary triangle $\triangle ABC$, $f(ABC)$ passes through the midpoint of $BC$. Proof. Reflect $A$ wrt $BC$ to get $A'$. Let the midpoint of $BC$ be $M$. Since $\triangle ABA'$ is isosceles, by the previous claim, $f(ABA')$ is tangent to $AA'$ at $M$. Since $\triangle ACA'$ is isosceles, by the previous claim, $f(ACA')$ is tangent to $AA'$ at $M$. Take $\sigma$ to be the reflection wrt line $AA'$. Since $\sigma : ABA' \mapsto ACA'$, then $\sigma : f(ABA') \mapsto f(ACA')$. The last result gives us $f(ABA') \cap f(ACA') \in AA' $. However, from the first two results, it follows that $f(ABA') \cap f(ACA') = \{ M \}$. However, $f(ABA'), f(ABC), f(A'BC), f(ACA')$ concur at a point. It then follows that $M \in f(ABC)$, from which the result follows. Since $f(ABC)$ passes through midpoint of $AB$, $BC, CA$, then it must be the circumcircle of these three points, which means that $f(ABC)$ is the Euler circle of $\triangle ABC$ for any $\triangle ABC$.