my friend attempted a 2.5 hour coordbash on this!

I am very sory for what you are about to read. This is a complete abomination to geometry, and I truly do not know where this solution came from. I fear that it is some almighty power that allowed me to unleash this horrific solution upon everyone. I apologize once again and hope that you will never suffer as much as you will from reading this. [asy][asy] import geometry; unitsize(5cm); pair A = (-sqrt(7)/4,3/4),B=dir(210),C=dir(330),O=circumcenter(A,B,C),H=orthocenter(A,B,C),E=intersectionpoint(line(B,H),line(A,C)),EE=2*E-B,F=intersectionpoint(line(C,H),line(A,B)),M=midpoint(B--C),O1=circumcenter(B,M,E),O11=2*O1-B,O2=circumcenter(C,M,F),K=reflect(line(B,C))*H,L=intersectionpoint(line(A,incenter(A,B,C)),line(O,M)),LL=2*O-L,N=midpoint(A--H); draw(A--B--C--cycle); draw(circle(A,B,C)); draw(circle(B,M,E),purple); clipdraw(circle(C,M,F),purple); draw(line(O2,K),brown+dashed); draw(tangents(circle(B,M,E),K),red); draw(circle(A,E,F),deepgreen); draw(B--O11); draw(B--EE,blue); draw(C--F,blue); draw(A--N,fuchsia); draw(A--L,darkgreen); draw(O--LL--N--K--cycle,Cyan); dot((-2,0),white); dot((1.5,0),white); dot(A); dot(B); dot(C); dot(E); dot(EE); dot(F); dot(H); dot(K); dot(L); dot(LL); dot(M); dot(N); dot(O); dot(O1); dot(O11); dot(O2); label("$A$",A,A); label("$B$",B,B); label("$C$",C,C); label("$E$",E,1.5*S); label("$E'$",EE,dir(B--EE)); label("$F$",F,2*dir(110)); label("$H$",H,dir(125)); label("$K$",K,3*dir(A--K)); label("$L$",L,L); label("$L'$",LL,LL); label("$M$",M,1.5*dir(280)); label("$N$",N,SE); label("$O$",O,S); label("$O_1$",O1,SE); label("$P$",O11,NE); label("$O_2$",O2,dir(135)); [/asy][/asy] We define a bunch of points. Let $O$ and $H$ be the circumcenter and orthocenter of $\triangle ABC$, and let $P$ be the reflection of $B$ in $O_1$, and let $E'$ be the reflection of $B$ in $E$. Define $L$ as the midpoint of arc $BC$ that does not contain $A$, and $L'$ the midpoint of the arc $BAC$. Let $N$ be the midpoint of $AH$. First, note \[n=\frac{a+h}2=\frac{2a+b+c}2\]Now we calculate $O_1$. Note that $O_1$ lies on the perpendicular bisector of $BC$ (which is $LL'$) and the perpendicular bisector of $BE'$ (which is $AC$). Thus, \[p=\frac{\ell(-\ell)(a+c)-ac(\ell-\ell)}{\ell(-\ell)-ac}=\frac{-bc(a+c)}{-bc-ac}=\frac{b(a+c)}{a+b}\]so \[o_1=\frac{b+p}2=\frac{b(2a+b+c)}{2(a+b)}=\frac{bn}{a+b}\]Similarly, $o_2=\frac{cn}{a+c}$. Thus, we know that $K$ is the exsimilicenter of $(BME)$ and $(CMF)$. In addition, \[BO_1^2=\frac 14BP^2=\frac 14\left|b-\frac{b(a+c)}{(a+b)}\right|^2=\frac 14\frac{|b-c|}{|a+b|}=\frac{a(b-c)^2}{c(a+b)^2}\]Thus, we know \[\left(\frac{k-o_1}{k-o_2}\right)^2=\frac{KO_1^2}{KO_2^2}=\frac{BO_1^2}{CO_2^2}=\frac{b(a+c)^2}{c(a+b)^2}\]Note \[(a+b)(k-o_1)=(a+b)k-bn\]so \[c((a+b)k-bn)^2=b((a+c)k-cn)^2\implies (b-c)(bc(n-k)^2-a^2k^2)=0\]As $b\neq c$, this means \[(\sqrt{bc}n-(\sqrt{bc}\pm a)k)(\sqrt{bc}n-(\sqrt{bc}\mp a)k)=0\]so \[k=\frac{\ell n}{\ell\pm a}\]We know $|k|=1$, so $|n|=|\sqrt{bc}\pm a|=|\pm\ell-a|$. Note the reflection of $O$ over $XY$ where $|x|=|y|=1$ is $x+y-0xy=x+y$, so the distance from $O$ to $N$ is the distance from $O$ to its reflection over $AL$ or $AL'$. Note that if $O'$ is the reflection of $O$ over $AL'$, as $OO'=ON$, we must have $O'$ lies inside $(ABC)$ (as $ABC$ is acute), so $\angle AOL'>120^\circ$. This would imply that $|B-C|>150^\circ$, absurd. Thus, the distance from $O$ to $AL$ is half of $ON$. Now, note that if $H'$ is the reflection of $H$ over $BC$, $H'\in(ABC)$. Furthermore $H'L'=AL$ in terms of arcs, as they are diagonals of cyclic trapezoid $AH'LL'$ ($AH',LL'\perp BC$). We claim it suffices to have $OL'NH'$ is a rhombus. Indeed, if this holds, then $h'-\ell=o+n=n$, but if $X=K,Y=O_1O_2\cap (ABC)$, we have \[x+y=o_1+xy\overline o_1=o_2+xy\overline o_2\]Solving for $xy$ we get \[xy=\frac{o_1-o_2}{\overline o_1-\overline o_2}=\frac{n}{\overline n}\cdot\frac{b/(a+b)-c/(a+c)}{a(1/(a+c)-1/(a+b))}=\frac{an}{\overline n}\]Thus $x+y=\frac{bn}{a+b}+\frac{an}{\overline n}\cdot\frac{\overline n}{a+b}=n$, so $x+y=h'-\ell$. However, we also note that \[y=n-k=n-\frac{n\ell}{\ell+a}=\frac{an}{a+\ell}\]so $y/x=-\ell/h'$. Thus, we get that as long as $h'\neq\ell$ (as $ABC$ is not isosceles), we have $k=h'$. To prove this, we note that if $OL'NH'$ is a rhombus, we have $ON$ is the perpendicular bisector of $L'H'$ and vice versa, so we know that $ON$ is twice the distance from $O$ to $L'H'$, or $O$ to $AL$. However, there are only two such points $N$ such that $OL'NH'$ is a rhombus for fixed $L',O,B,C$ ($N$ lies on the intersection of two circles; the one centered at $L'$ with radius $1$ and $M$ with radius $1$), so there are two points $A$ such that $OL'NH'$ is a rhombus. Furthermore, note that if $|a+m|=|n|=|\ell-a|$, \[1+m\overline m-a\overline m-\frac ma=1+1-\frac{\ell}a-\frac a{\ell}\]so this is a quadratic, meaning there are two points $a$ for which $|n|=|\ell-a|$. As $OL'NH'$ implies $|n|=|\ell-a|$, the other must be true, finishing the proof.

[asy][asy] size(10cm); defaultpen(fontsize(10pt)); pen pri=mediumblue; pen sec=purple; pen tri=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=paleblue; pair O,A,B,C,M,D,E,F,MB,MC,S1,Q,P,M1,M2,N,T,A1; O=(0,0); A=dir(132); B=dir(209); C=dir(331); M=(B+C)/2; D=foot(A,B,C); E=foot(B,A,C); F=foot(C,A,B); M1=(B+E)/2; M2=(C+F)/2; MB=intersectionpoint(B--(B+A-D),M--(M+(M1-M)*100)); MC=intersectionpoint(C--(C+(A-D)*100),M--(M+(M2-M)*100)); P=extension(E,F,B,C); Q=intersectionpoint((A+(P-A)*0.1)--P,circumcircle(A,B,C)); N=intersectionpoint(M--(M+(A-D)*100),circumcircle(A,B,C)); S1=intersectionpoint(A--(P-A)*100,M--(M+(A-N)*100)); T=intersectionpoint(M--S1,circumcircle(A,B,C)); A1=-A; filldraw(A--N--M--T--cycle,tfil,tri); draw(A--B--C--cycle,pri); draw(circumcircle(A,B,C),dashdotted+pri); draw(B--E,pri); draw(C--F,pri); draw(B--MB,sec); draw(C--MC,sec); draw(D--MB,dashed+sec); draw(D--MC,dashed+sec); draw(M--MB,tri); draw(M--MC,tri); draw(M--S1,tri); draw(MB--S1,tri); draw(MB--A,tri); draw(A--MC,tri); draw(B--P,pri); draw(P--E,dashed+tri); draw(Q--A1,sec); draw(circumcircle(M,B,E),dotted+pri); draw(circumcircle(M,C,F),dotted+pri); label("$A$",A,dir(120)); label("$B$",B,dir(240)); label("$C$",C,dir(330)); label("$M$",M,dir(270)); label("$D$",D,dir(270)); label("$E$",E,dir(60)); label("$F$",F,dir(120)); label("$M_B$",MB,dir(120)); label("$M_C$",MC,dir(90)); label("$S_1$",S1,dir(210)); label("$Q$",Q,dir(150)); label("$P$",P,dir(180)); label("$N$",N,dir(90)); label("$T$",T,dir(270)); label("$A_1$",A1,dir(330)); label("$M_C$",(MC+(M-MC)*0.6),dir(0)); path clip[]; clip=circle((0,0),(distance(O,S1))*1.1); clip(currentpicture,clip); [/asy][/asy] Claim 1: If $M_B, M_C$ are the antipodes of $M$ in $(BME), (CME)$ and $P = EF \cap BC$, then $P, M_B, A, M_C$ are collinear. Proof: $MM_B || AC, MM_C || AB \implies M_B$ is halfway between $B$ and $B’ = AC \cap$ the perpendicular from $B$ to $BC$. Defining $C’$ similarly, $\triangle{ABB’} \sim \triangle{AC’C} \implies M_B, A, M_C$ collinear. Also, $\triangle{MBM_B} \sim \triangle{CDA}, \triangle{MCM_C} \sim \triangle{BDA}$. Since $P, B, D, C$ forms a harmonic bundle, $\frac{PB}{PC} = \frac{DB}{DC} = \frac{BM_B}{CM_C} \implies P \in M_BM_C$. Thus, by Orthocenter Miquel config, $M_AM_B$ is the line $AQ$ where $Q = A_1M \cap (ABC)$, where $A_1$ is the antipode of $A$. Claim 2: Reflection of $M$ over intersection of common tangents is $M_BM_C \cap$ line through $M$ parallel to $AN$, where $N$ is the antipode of $M$ in $(ABC)$. Proof: The centers $O_B, O_C$ of the two circles are the midpoints of $MM_B, MM_C$. If the intersection of the common external tangents is $K$, then $\frac{KO_B}{KO_C} = \frac{MO_B}{MO_C}$ and $K \in O_BO_C \implies K$ lies on the circle of Apollonius containing the set of all points with distances to $O_B, O_C$ in that ratio. $O_BO_C \cap$ bisector of $\angle{M_BMM_C}$ lies on this circle, so $K$ is the intersection of $O_BO_C$ with the external bisector of $\angle{M_BMM_C}$ which is parallel to $AN$, and a 2x homothety at $M$ finishes. Thus, it suffices to show that if $S_1 = AQ \cap$ line through $M$ parallel to $AN$ and $MS_1 \cap (ABC)$ is the midpoint of $MS_1$, then this point forms a parallelogram with $A, N, M$. By angle chasing $QM \perp TN \implies TN || AQ \implies AN = QT = TM$, so $ANMT$ is a parallelogram and $AT \perp BC$. $\square$

sleepypuppy wrote: Geez, how are you guys so smart. Motivation for solution: I am really really really bad at geometry, so I had to bash. I somehow forgot how to use barycentric so I resorted to complex.

Let $Q=(BME)\cap(CMF)$ be the Miquel point of $BCEF$ and let $T$ lie on $(ABC)$ so that $\overline{AT}\perp\overline{BC}$. We have $\measuredangle QKM=\measuredangle QBM+\measuredangle QCM$ by this lemma to 2017 G7. It is known that $-1=(QT;BC)$, from which $\measuredangle QTM=\measuredangle QBM+\measuredangle QCM$ readily follows. Hence $K=T$.

sad placeholder gone. Here's a solution without diagram because no time: Let $Q = (AEF) \cap (ABC)$ be the $A$-queue point of $\triangle ABC$. It is well known that if $H$ is the orthocenter of $\triangle ABC$ and if $A'$ is the $A$-antipode on $(ABC)$, then $Q-H-M-A'$. Furthermore, if $D$ is the foot of the altitude from $A$ to $BC$, we have \[ \angle MBE = \angle DBE = \angle DAE = \angle HAE = \angle HQE = \angle MQE,\]so $Q \in (BME)$. Similarly, $Q \in (CMF)$, so $(BME) \cap (CMF) = \{Q, M\}$. Now invert at $K$ with radius $KQ = KM$. Clearly this fixes both $Q, M$, so it must swap $(BME)$ and $(CMF)$. Suppose it swaps $B \leftrightarrow B'$ and $C \leftrightarrow C'$. Then $B'-Q-C'$ since $Q \in (ABCK)$, so by Reim's Theorem (since we also have $B-M-C$), $BB' \parallel CC'$. But this means $KB$ and $KC$ are isogonal wrt $\angle QKM$, so $KQ$ is a symmedian of $\triangle BKC$. Thus, $QK$ is a symmedian of $\triangle BQC$, so \[ \angle BAK = \angle BQK = \angle MQC = \angle A'AC,\]i.e. $AK$ and $AA"$ are isogonal wrt $\angle BAC$. We are done. $\blacksquare$

Solution Sketch: Let $Q$ be the $A$-Queue point. One can find the angles of $\triangle QKM$ in terms of $\triangle ABC$'s angles. Then you can angle chase to find $OMKQ$ is cyclic, at which point it's easy to angle chase to finish.

Here is a unique solution using the "forgotten coaxiality lemma". Let $H$ be the orthocenter of $\triangle ABC$. Let $Q$ be the second intersection of $\odot(BME)$ and $\odot(CMF)$. We first prove the following well-known properties of $Q$. Claim: $Q$ is the Miquel point of $BCEF$. In particular, $Q$ lies on both $\odot(AEF)$ and $\odot(ABC)$. Proof. Follows since $BCEF$ is cyclic with $M$ being the circumcenter. $\blacksquare$ Claim: $A(Q,H;B, C) = -1$ Proof. By radical center theorem on $\odot(AEF)$, $\odot(ABC)$, and $\odot(BCEF)$, we get that $AQ$, $EF$, and $BC$ are concurrent. Now, the result follows from a well-known harmonic property. $\blacksquare$ Now, we get to the meat of the solution. Let the circumcircle of $\odot(QMK)$ meet $BC$ again at $T\neq M$. The key claim is the following. Claim: $QT$ is tangent to $\odot(BQC)$. Proof. We use the ``forgotten coaxiality lemma". \begin{align*} \frac{BT}{TC} &= \frac{TB\cdot TM}{TC\cdot TM} \\ &= \frac{\operatorname{pow}(T, \odot(BME))} {\operatorname{pow}(T, \odot(CMF))} \\ &= \frac{\operatorname{pow}(K, \odot(BME))} {\operatorname{pow}(K, \odot(CMF))} \\ &= \left(\frac{r_{\odot(BME)}}{r_{\odot(CMF)}}\right)^2 \\ &= \left(\frac{BQ/\sin\angle QMB}{CQ/\sin\angle QMC}\right)^2 \\ &= \frac{BQ^2}{CQ^2}, \end{align*}implying the result. $\blacksquare$ To finish, let $O$ be the center of $\odot(ABC)$. Then, from the claim, $\angle OQT = 90^\circ = \angle OMT$, so $O$ also lies on $\odot(QMTK)$. Thus, $\angle OKT=90^\circ$, so $KT$ is also tangent to $\odot(ABC)$ as well. This implies that $QBKC$ is harmonic quadrilateral, and the result follows from the second claim.

strange problem. Invert about (BC). Then the two circles go to BE and CF, so K goes to a point K' which passes through both circles tangent to BE, CF tangent to M. So in fact K' is the reflection of M across the angle bisector of BHC. Now, the condition that AKBC cyclic becomes that HK'BC is cyclic. In particular, this implies that HK' is the H-symmedian of HBC, so by HM point stuff we know that the inverse of K' wrt (BC) now goes to the reflection of H across BC; ergo, $AK \perp BC$ as desired.

This was also #3 on USA TST for EGMO 2023.

Pure angle chasing solution (first?) Let $Q$ be the Miquel point of $BCEF.$ Standard configuration knowledge: $\angle HQA = \angle MQA = 90^\circ,$ then $\angle MQE = \angle CFE = \angle MBE$ so $(BMEQ)$ and similarly $(CMFQ)$ cyclic. If $O_1,O_2$ are the centers of $(BMEQ),(CMFQ),$ external angle bisector theorem yields that $K$ is the intersection of $O_1O_2$ and the external angle bisector of $\angle O_1MO_2.$ Let the perpendicular bisector of $BC$ intersect minor arc $BC$ at $L$ and major arc $BC$ at $N.$ Let $J$ be the foot of $AH$ to $(ABC).$ Note that if we reflect $Q$ over $LN$ to $Q'$ then $J,M,Q'$ collinear. If we reflect $J$ over $LN$ to $J'$ then $J',M,Q$ collinear. $MO_1 \parallel AC, MO_2 \parallel AB,$ so the angle bisector of $\angle O_1MO_2$ is parallel to that of $\angle ABC.$ So $\angle NMK = 90 + \angle NLA = 90 + \angle LQM$ so the circumcenter of $LMQ$ is $K.$ Note $\angle MJQ = \angle Q'JQ = 2\angle NJQ = 2\angle NLQ = \angle MKQ$ so $(MKJQ)$ is cyclic, done. $\blacksquare$

oops i got sniped to the angle chase sol First, let ray $MH$ intersect $(ABC)$ at $N$. By PoP through $H$, $(HN)(2HM) = (HB)(2HE)$, so $NMBE$ is cyclic. Similarly, $NMCF$ is cyclic. Therefore, $N$ is the second intersection of $(CMF)$ and $(BME)$, and $KM=KN$. Now, let $O_B$ and $O_C$ be the circumcenters of $(BME)$ and $(CMF)$. By external angle bisector theorem, $KM$ is the external angle bisector of $O_BMO_C$. Since $M$ is the circumcenter of $(BCEF)$, $MB=ME$ so $MO_B$ is perpendicular to $BE$ and thus parallel to $AC$. Therefore, $O_BMB = C$. Similarly, $O_CMC = B$. This means that $KMO_B = \frac{B+C}{2}$, so $KMB = \frac{B-C}{2}$. Now, let $K'$ be the second intersection of $AH$ with $(ABC)$ (e.g. where $K$ is supposed to be). $K'NM$ subtends arc $BC$ minus twice arc $BK$ and thus has angle measure $A-2(90-B) = 2B+A-180 = B-C$. $KM=KN$ gives $$KNM = NMK$$$$KNK' + K'NM = NMB + BMK$$$$KNK' + (B-C) = BMK' + BMK = BMK + BMK + KMK' = (B-C) + KMK'$$$$KNK' = KMK'$$ Therefore, $KK'MN$ is cyclic. This clearly finishes.

Huh. I'll take this as a consolation prize, I guess. Let $H$ be the orthocenter. By properties of Miquel points, we have that $(BME)$ intersects $(CMF)$ again at the $A$-Queue point, call it $Q$. Consider the inversion $\psi$ at $M$ with power $MB\cdot MC = ME^2 = MF^2$. Since $B$, $H$, $E$ are collinear, the inverse of $H$ must be the intersection of line $HM$ with the circumcircle of $(BME)$, which is exactly $Q$. Now, we use the following fact: the intersection of the common tangents of two circles $\omega_1$ and $\omega_2$ intersecting once at $M$ can be redefined as follows: Let $\ell_1$, $\ell_2$ be the inverses of $\omega_1$, $\omega_2$ under $\psi$. Suppose that $\ell_1$ and $\ell_2$ intersect at a point $X$. The common external tangents are then mapped to the two unique circles passing through $M$ and tangent to both $\ell_1$ and $\ell_2$. These two circles intersect at a point $M'$, which is the reflection of $M$ over the angle bisector of the angle formed by $\ell_1$ and $\ell_2$ that "contains" $M$. In particular, the intersection of the common external tangents of $\omega_1$ and $\omega_2$ is the inverse of this point under $\psi$. Now, we apply the fact to the problem using the two circles $(BME)$ and $(CMF)$. The point $X$ in the above logic corresponds to $H$. Thus, the inverse of the intersection of the common external tangents is the reflection of $M$ over the angle bisector of $\angle BHC$, call it $M'$. We are told that the inverse of this point lies on $(ABC)$, so $M'$ lies on circle $(BHC)$. However, this also means that $HM'$ is the $H$-symmedian chord in $\triangle BHC$, so $(H,M'; B,C)= -1$. Since $\psi$ preserves cross-ratio, we have that $(Q, K; B, C) = -1$. Let $A'$ be the $A$-antipode in $(ABC)$. Since $QM$ is the median in $\triangle QBC$ and $Q$, $M$, $A'$ are collinear, the point $K$ on $(QBC)$ such that $K$ is a symmedian in $\triangle QBC$ is the reflection of $A'$ over the perpendicular bisector of $BC$, which is well-known to be the point on $(ABC)$ such that $AK \perp BC$. This concludes the proof.

That's nice problem. I introduce my generalization and proof using complex numbers. General problem. Let $ABC$ be a triangle inscribed in circle $\omega$. Altitude from $A$ meets $\omega$ again at $K$. $(W)$ is a circle passing through $B$ and $C$. Circle $(W)$ meets $CA$ and $AB$ again at $E$ and $F$, respectively. Let $X$ be exsimilicenter of circles $(WBE)$ and $(WCF)$. Prove that line $XK$ bisects arc $\widehat{BC}$. Proof. Let $\omega$ be the unit circle on the complex plane with coordinates $A(1)$, $B(b^2)$, $C(c^2)$. Let $N$ be the midpoint of arc $\widehat{BC}$ then $n=-bc$. Let $M$ be the midpoints of $BC$ then $m=\frac{b^2+c^2}{2}$. Since $W$ lies on the perpendicular bisector of $BC$, $w=\frac{m-sn}{1-s}=\frac{b^2 + c^2+2sbc}{2-2s}$ for some real number $s$. Since $E$ and $F$ are the intersections again of lines $CA$ and $AB$ with $(W)$, we can see $$ e =\frac{2sb(b+c-cb^2)-b^2(b^2+c^2-1)-c^2}{2b^{2}(s-1)} $$and $$ f=\frac{2sc(b+c-bc^2)-c^2(b^2+c^2-1)-b^2}{2c^{2}(s-1)}. $$From these, circumcenters $J$ and $L$ of triangles $WBE$ and $WCF$, respectively, are given by $$j=\frac{\begin{vmatrix} b^2 & 1 & 1\\ w & w\bar{w} & 1\\ e & e\bar{e} & 1\\ \end{vmatrix}}{\begin{vmatrix} b^2 & \frac{1}{b^2} & 1\\ w & \bar{w} & 1\\ e & \bar{e} & 1\\ \end{vmatrix}}=\frac{-b^{2}(2sbc - 2s + b^{2} + c^{2} + 2)}{2\left(b^{2} + 1 \right)\left(s - 1 \right)} $$and $$\label{eq4}l=\frac{\begin{vmatrix} c^2 & 1 & 1\\ w & w\bar{w} & 1\\ f & f\bar{f} & 1\\ \end{vmatrix}}{\begin{vmatrix} c^2 & \frac{1}{c^2} & 1\\ w & \bar{w} & 1\\ f & \bar{f} & 1\\ \end{vmatrix}}=\frac{-c^2(2scb - 2s + c^{2} + b^{2} + 2)}{2\left(c^{2} + 1 \right)\left(s - 1 \right)}. $$Now we get radius $R_J$ and $R_L$ of $(J)$ and $(L)$ as follows $$ R_J^2=\frac{\left(b + c \right)^{2}\left(2sc + b - c \right)(2sb - b + c)}{4\left(b^{2} + 1 \right)^{2}c^{2}\left(s - 1 \right)^{2}} $$and $$ R_L^2=\frac{\left(b + c \right)^{2}\left(2sc + b - c \right)(2sb - b + c)}{4\left(c^{2} + 1 \right)^{2}b^{2}\left(s - 1 \right)^{2}}. $$Thus, $$ t=\frac{R_J}{R_L}=\frac{b(c^{2} + 1)}{c(b^{2} + 1)}. $$We have $$ x=\frac{j-tl}{1-t}=-bc\frac{2scb - 2s + c^{2} + b^{2} + 2}{2\left(s - 1 \right)\left(bc - 1 \right)}. $$It is easily seen intersection $K$ of altitude from $A$ with $\omega$ has coordinates $$ k=-b^2c^2. $$Now with $n=-bc$ and from coordinates of $x$, we can check $$ \frac{x-n}{k-n}=\frac{b^{2} + c^{2} + 2bc}{2b^{2}c^{2}s - 2b^{2}c^{2} - 4bcs + 4bc + 2s - 2}=\frac{\bar{x}-\bar{n}}{\bar{k}-\bar{n}}. $$This completes our proof.

Do you have other problems of 2022 USA TST? Thanks!

$O$ is the circumcenter of $(ABC)$, $T$ is the exsimilicenter of $(BME), (CMF)$. $MH \cap (ABC)=Q, S$ is the midpoint of arc $BAC$ and $R$ is antipode of $S$. $AH \cap (ABC)=X, QS \cap BC=P, QA \cap BC=Z$. $Q =(BME) \cap (CMF)$. $O_1 O_2$ are circumcenters of $(BME),(CMF)$. I will prove that $T \in XR$. It is well known that $(Q,X;B,C)=-1 \implies QR,XS,BC$ are concurrent $\implies P-X-R$ are collinear. Also $QT$ is the external angle bisector of $O_1QO_2$. Easy to see that $BQO_1 \sim CQO_2$. Thus $O_1QO_2 \sim BQC$. With the same reasoning $QO_1M \sim QOC \implies QOO_1 \sim QCM$ and $QO_2M \sim QOB \implies QOO_2 \sim QBM$. Therefore $\frac{O_1O}{O_2O}=\frac{QB}{QC}=\frac{XB}{XC}$. Also $OO_1 \perp QB$ and $OO_2 \perp QC$ gives $\angle O_1OO_2=\angle BXC$. Hence we get $QBXC \cup P \sim QO_1OO_2 \cup T$. Thus $\angle QPT=\angle QXO= 90-\angle QSX=\angle QPX \implies T \in XR$. Rest is obvious.

Proof of the generalization of buratinogigle. General problem. Let $ABC$ be a triangle inscribed in circle $\omega$. Altitude from $A$ meets $\omega$ again at $K$. $(W)$ is a circle passing through $B$ and $C$. Circle $(W)$ meets $CA$ and $AB$ again at $E$ and $F$, respectively. Let $X$ be exsimilicenter of circles $(WBE)$ and $(WCF)$. Prove that line $XK$ bisects arc $\widehat{BC}$. Proof. $O, O_1, O_2$ are the circumcenters of $(ABC), (BWE), (CWF)$. $D= (BWE)\cap (CWF)$, $A'$ is the antipode of $A$. $L$ and $R$ are the midpoints of arcs $BAC$ and $BC$. $LK \cap O_1O_2=J$. $KR$ cuts $BC$ and $O_1O_2$ at $G,X$. I will prove that $XW$ is the ex-angle bisector of $O_1WO_2$. $\angle BDW=\angle BEW=90-\angle C, \angle CDW= 90-\angle B \implies D \in (ABC)$ and $DW$ passes through $A'$. Claim. $D,J,O,W,K,X$ are cyclic. Proof. $\angle KOW= \angle KLR= \angle KDA'=\angle KDW \implies DOWK$ is cyclic. Let $J'=O_1O_2 \cap (DOWK)$.$\angle DKJ'=90- \frac{1}{2} \angle DOW=90- \angle DLR=\angle DKL \implies K, J', L$ are collinear. Thus $J'=J$ and $J \in (DOWK)$. Circumcenter of $DOWKJ$ is on $JX$ and $\angle JKX=90$. Thus $X \in (DOWKJ)$.$\square$ Claim. $\triangle O_1OO_2 \sim \triangle BKC$ Proof. $\angle O_2O_1O=\angle BDW=\angle KBC$ and $\angle O_1O_2O=\angle KCB$.$\square$ $\angle JXO=\angle JKO =\angle KLR=\angle RGB \implies \triangle O_1OO_2 \cup X \sim \triangle BKC \cup G$. Thus $(DJOWKX)$ is the apollonian circle of $O_1OO_2$. Therefore $WX$ is the external angle bisector of $O_1WO_2$. Hence $X$ is the exsimilicenter of $(O_1), (O_2)$.

Another inversion solution, but hopefully the details after the inversion are explained better. We do the most natural thing: invert to remove the circles through $M$ and see what happens with the tangents. Invert with respect to $(BC)$; the circle $(MBE)$ goes to $BE$ and the circle $(CMF)$ goes to $CF$. The image of a common tangent of these two circles will be a circle through $M$, tangent to $BE$ and $CF$. Thus, $K'$ is the intersection of two circles through $M$, tangent to $HB$ and $HC$. Hence $HK'$ and $HM$ are isogonal, so $HK'$ is a symmedian in $\triangle HBC$. In addition, the image of $A$ is the $A$-HM point $H_A$ (follows from $MH_A.MA=MB^2$ and the definition of the $A$-HM point). Hence, $(ABC)$ goes to $(HBC)$, so $K'$ is on $(HBC)$, such that $(B, C, H, K')=-1$. Note that $D$ goes to a point $D'$ on $BC$, such that $(B, C, D, D')=-1$, i.e. $D'$ lies on $EF$ and $HH_A$ as well (it is well-known that $H$ is orthocenter of triangle $AMD'$). Projecting $(B, C, D, D')=-1$ from $H_A$ onto $(HBC)$ yields that $K'=H_AD \cap (HBC)$. We want $K \in AD \iff K' \in (H_AD'M) \iff DK'.DH_A=DB.DC=DH.DA=DD'.DM$, done (the last equation follows by using that $H$ is orthocenter in $\triangle AD'M$, as mentioned above).

Invert around $(BCEF)$. Let $K$ be sent to $K'$ on $(BHC)$. Note the following Circle $(BME)$ is swapped with $BE$ and $(CMF)$ is swapped with $CF$. This means that there are two circle through both $K'$ and $M$ tangent to $BH$ and $CH$ both. Thus the perpendicular bisector of $\angle BHC$ coincides with the perpendicular bisector of $K'M$ so $\measuredangle BHK'=\measuredangle MHC$ To finish we angle chase to obtain \[\measuredangle BCQ=\measuredangle FAQ=\measuredangle FHQ=\measuredangle CHM=\measuredangle K'HB=\measuredangle K'CB\]We can similarly deduce that $\measuredangle K'BC=\measuredangle CBQ$ implying that $\triangle BCK'$ is congruent to $\triangle BCQ$. Then, $K'$ lying on $(BHC)$ combined with above is enough to deduce that $K'$ is the reflection of $Q$ across $BC$, so $MQ=MK'$. Inverting back yields $MH=MK$ as desired.

The main idea of the problem is to invert around $(BC)$. Let $D,E,F$ be the feet of the altitudes. The two circles simply get set to $BE$ and $CF$. Thus, the two tangent lines become two circles that pass through $M$ are are tangent to both lines. There are two such circles, and by symmetry, their second intersection, $K'$, is the reflection of $M$ across the internal angle bisector of $\angle BHC$. Note that $A$ gets sent to the A-Humpty point, which we call $A'$. It is well known that $(BHA'C)$ is cyclic (sketch: $H$ inverts to A-queue point, which we from now on denote as $H'$). Thus, we have that $(ABC)$ gets sent to $(BHC)$. Furthermore, this means that $K'$ is on $(BHC)$. Claim: $$\angle H'CB=\angle K'CB.$$From $AH'BC$ cyclic, we have $$\angle H'CB=\angle H'AB.$$Now, from $(AH)$, we have $$\angle H'AB=180-\angle H'HC=\angle MHC=\angle BHK'=\angle BCK'.$$ Thus, since $(ABC)$ and $(BHC)$ are the same size, this means that $H'$ and $K'$ are symmetric about line $BC$. Hence, $H'M=MK'$, so in the original uninverted diagram, $HM=MK$. It is easy to check that $K$ is not the antipode of $A$ on the circumcircle, as $K'$ is in the same direction of the perpendicular bisector of $BC$ as $A$ is, so we are done.

wait I actually forgot that inversion existed that's probably not good. spent like 30 minutes then like 2 minutes after inverting ;-; Let $Q$ be the $A$-Queue point. Step 1: Notice by angle chase from Miquel configuration, we have $Q\in (BME),(CMF)$. Step 2: Invert around $(BFEC)$, since $K\in (ABC)$ we get $K'\in (BHC)$. Step 3: The external tangents are mapped to two circles passing through $M$ and $K'$ tangent to $BE$ and $CF$. Put another way, $K'$ should be the reflection of $M$ about the angle bisector of $\angle BHC$. Step 4: $BHCK'$ is harmonic. As a result, let $H'$ be the reflection of $H$ across the perpendicular bisector of $BC$; then $H'\in MK'$. We also define $X\in (ABC)$ with $AX\perp BC$. We get $X\in H'M$ which means that $X\in MK'$. Step 5: Since $X$ is both on the circle and on $MK'$ it coincides with $K$. Done!

Here is a computational solution for anyone who doesn't want to do synthetic geometry on Day 1 of the TST. Part 1: Talking about the "wannabe" point Let $H'$ be the reflection of the orthocenter $H$ of $\triangle ABC$ with respect to line $BC$. Let $D$ be the foot of the altitude of $A$ onto $BC$. Define \[ f(X) := \text{Pow}(X, (BME)) - \text{Pow}(X, (ABC)). \]Compute $f(H) = BH \cdot HE = BD \cdot BC - BH^2$ and $f(D) = \tfrac{BD \cdot BC}{2}$, so \[ f(H') = f(2D-H) = 2f(D) - f(H) = BH^2 = 4R^2 \cdot (\cos B)^2 \]by linearity. Thus, \[ \text{Pow}(H', (BME)) = 4R^2 \cdot (\cos B)^2. \]Using a similar function with $B$ swapped with $C$, we can compute \[ \text{Pow}(H', (CMF)) = 4R^2 \cdot (\cos C)^2, \]so we conclude that \[ \frac{\text{Pow}(H', (BME))}{\text{Pow}(H', (CMF))} = \left(\frac{\cos B}{\cos C}\right)^2. \] Part 2: Seeing reality A quick LoS computation shows that $\text{rad}(BME)/\text{rad}(CMF)=\tfrac{\cos B}{\cos C}$. Thus \[ \frac{\text{Pow}(K, (BME))}{\text{Pow}(K, (CMF))}=\left(\frac{\cos B}{\cos C}\right)^2. \] Part 3: The magical and forgotten step to stardom Suppose that $K$ lies on $(ABC)$. Then \[ \frac{\text{Pow}(K, (BME))}{\text{Pow}(K, (CMF))} = \frac{\text{Pow}(H', (BME))}{\text{Pow}(H', (CMF))}. \]Denote $Q$ by the intersection of $(BME)$ and $(CMF)$ distinct from $M$. Leveraging the Unforgotten Coaxiality Lemma, we have that $K$, $H'$, $Q$, and $M$ are concyclic. Since $K$ is external to $(BME)$ and $(CMF)$, it must be $H'$, and we are done.

Note that points $B, C, E, F$ are inscribed inside the circle of diameter $\overline{BC}$ with center $M$. Let $Q$ be the Miquel point of quadrilateral $BCEF$. By the standard facts, $Q$ lies on $\odot(BME)$ and $\odot(CMF)$ and $\angle MQA = 90^{\circ}$. Therefore, line $QM$ meets $\odot(ABC)$ for a second time at the antipode $A'$ of $A$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 5, xmax = 17, ymin = -1.0963716592286714, ymax = 11.483140673868764; /* image dimensions */ draw((8.546847373658695,10.341098226284037)--(7.679547878066041,4.51346100817611)--(15.07954787806604,4.54546100817611)--cycle, linewidth(1.2)); /* draw figures */ draw((8.546847373658695,10.341098226284037)--(7.679547878066041,4.51346100817611), linewidth(1.2)); draw((7.679547878066041,4.51346100817611)--(15.07954787806604,4.54546100817611), linewidth(1.2)); draw((15.07954787806604,4.54546100817611)--(8.546847373658695,10.341098226284037), linewidth(1.2)); draw(circle((9.522388430690755,6.177083213710761), 2.482680032776302), linewidth(1.2) + linetype("2 2") + blue); draw(circle((13.177344976972462,16.609381886066497), 12.212966995162494), linewidth(1.2) + linetype("2 2") + blue); draw(circle((11.369112174924034,6.9427173597649805), 4.417484705393342), linewidth(1.2)); draw((14.191376976189373,3.5443364932459245)--(7.043026886813809,6.048761431108335), linewidth(1.2) + red); draw((8.589827276719056,3.5152850868994365)--(15.714082264524034,6.105239027870612), linewidth(1.2) + red); /* dots and labels */ dot((8.546847373658695,10.341098226284037),linewidth(3.pt) + dotstyle); label("$A$", (8.142302457673382,10.577461503607588), E * labelscalefactor); dot((7.679547878066041,4.51346100817611),linewidth(3.pt) + dotstyle); label("$B$", (7.3707141067430895,4.272932385386087), N * labelscalefactor); dot((15.07954787806604,4.54546100817611),linewidth(3.pt) + dotstyle); label("$C$", (15.126375504519988,4.153614517727982), NE * labelscalefactor); dot((11.37954787806604,4.52946100817611),linewidth(3.pt) + dotstyle); label("$M$", (11.39343078778781,4.153614517727982), E * labelscalefactor); dot((10.95459246676036,8.205011157256111),linewidth(3.pt) + dotstyle); label("$E$", (11.018431775148049,8.31269447609624), NE * labelscalefactor); dot((7.844557731162311,5.622210138257463),linewidth(3.pt) + dotstyle); label("$F$", (7.677531480721077,5.689973845610575), SW * labelscalefactor); dot((8.589827276719056,3.5152850868994365),linewidth(3.pt) + dotstyle); label("$K$", (8.410484096335157,3.0967991184704737), N * labelscalefactor); dot((14.191376976189373,3.5443364932459245),linewidth(3.pt) + dotstyle); label("$A'$", (14.2741050212478,3.2161169861285797), NE * labelscalefactor); dot((7.043026886813809,6.048761431108335),linewidth(3.pt) + dotstyle); label("$Q$", (6.586625262132678,5.8752008939377935), E * labelscalefactor); dot((15.714082264524034,6.105239027870612),linewidth(3.pt) + dotstyle); label("$L$", (15.77410107180685,6.199063677581224), E * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Consider the inversion with pole $M$ fixing $\odot(ABC)$. This inversion swaps $\{B, C\}, \{Q, A'\}$, and say $\{K, L\}$. Since $K$ is the intersection of the two tangent lines to $\odot(BMQ), \odot(CMQ)$, it follows under inversion that $L$ is the second intersection of the two circles through $M$ tangent to lines $CA', BA'$. By symmetry, $L$ must be the reflection of $M$ in the angle bisector of $\angle BA'C$. Thus, arcs $\widehat{QB}$ and $\widehat{LC}$ on $\odot(ABC)$ are equal. It follows that points $Q, L$ are symmetric in the perpendicular bisector of $\overline{BC}$. Therefore, points $A', K$ are also symmetric in the perpendicular bisector of $\overline{BC}$. We conclude that $A'K \parallel BC$, hence $AK \perp BC$.

Note that proving $AK$ is perpendicular to $BC$ is equivalent to proving that $K$ is the reflection of the orthocenter $H$ of triangle $ABC$ in line $BC$. We establish furthermore the following well-known properties: $(AEHF)$, $MH$, and $(ABC)$ concur at some point $K'$; $K'BKC$ is a harmonic quadrilateral; $MK' \cdot MH = ME^2 = MF^2$. Invert about the circle with diameter $BC$. We rephrase the problem as follows: Inverted wrote: Let $BHC$ be an obtuse triangle with $\angle H > 90^{\circ}$, and let $M$ be the midpoint of $BC$. Two circles $\omega_1$ and $\omega_2$ pass through $M$ and are tangent to lines $BH$ and $CH$. Suppose they meet at $T \neq M$. Show that if $T$ lies on $(BHC)$, then quadrilateral $TBHC$ is harmonic. This turns out to be quite simple. Note that $T$ and $M$ are symmetric with respect to $\angle BHC$. Thus $\angle BHT = \angle CHM$ and we are done.

We invert about $(BFEC)$. This sends $(BME)$ to line $BE$, sends $(CMF)$ to line $CF$, fixes $M$, sends $A$ to the $A$-humpty point, sends $H$ to the $A$-queue point and sends $K$ to $K^*$, the reflection of $M$ across the angle bisector of $\angle BHC$. So, $(ABC)$ is sent to $(BHC)$. Since $(ABCK)$ is cyclic, $BHCK^*$ is cyclic. Now, letting $X$ be the midpoint of arc $BK^*C$, we have \[\angle K^*MB = \angle HXM = \angle HBX - 90^{\circ} = \angle HMX - 90^{\circ} = \angle HMB,\]so line $BC$ bisects $\angle HMK^{*}$. So, line $BC$ also bisects $\angle HMK$ which finishes.

We first prove a bunch of claims for any general acute triangle. Let $AD,BE,CF$ be the altitudes and $H$ be the orthocenter of $ABC$. Let $Q$ be the $A-$Queue Point. Note that $\angle EMB=2\angle ECB=2C$ and since $\angle BEM=\angle MBE$, we have \[\measuredangle MBE=90^\circ-C=\measuredangle HAE=\measuredangle HQE=\measuredangle MQE\](it is well-known that $M,Q,H$ are collinear - it follows by $\sqrt{-HA\cdot HD}$ inversion). Since $\angle MBE=\angle MQE$, we have that $Q\in(BME)$. Similarly, $Q\in(CMF)$. So, $MQ$ is the radical axis of the two circles. Let $A,K_1$ be the intersections of $AD$ with $ABC$. Let $O_1,O_2$ be the centres and $r_1,r_2$ be the radii of circles $(BME),(CMF)$, respectively. Claim: $\frac{K_1B}{K_1C}=\frac{r_1}{r_2}$ and $\overline{K_1B},\overline{K_1C}$ are tangents to $(BME),(CMF)$. Proof. Note that \[\frac{r_1}{r_2}=\frac{\frac{MB}{2\sin(\measuredangle BEM)}}{\frac{MC}{2\sin(\measuredangle MFC)}}=\frac{\sin(90^\circ-B)}{\sin(90^\circ-C)}=\frac{\sin(\measuredangle HCB)}{\sin(\measuredangle CBH)}=\frac{HB}{HC}=\frac{K_1B}{K_1C},\]as desired. Moreover, $\measuredangle BEM=90^\circ-C=\measuredangle CBH=\measuredangle K_1BC$, which means $\overline{K_1B}$ is tangent to $(BME)$ by alternate segment theorem. The claim is therefore true. $\blacksquare$ Claim: Call $(BME)$ as $\tau_1$ and $(CMF)=\tau_2$. Then, the locus of all points $X$ such that \[\frac{\text{length of tangent from }X\text{ to }\tau_1}{\text{length of tangent from }X\text{ to }\tau_2}=\sqrt{\frac{Pow_{\tau_1}(X)}{Pow_{\tau_2}(X)}}=\frac{r_1}{r_2} \ \ \ (\star)\]is $(QMK_1)$. Proof. Given equation is simply: \[\frac{O_1X^2-r_1^2}{O_2X^2-r_2^2}=\frac{r_1^2}{r_2^2}\Longrightarrow \frac{O_1X}{O_2X}=\frac{r_1}{r_2}\]which means the equation is that of an Apollonius circle for some harmonic bundle. In particular, points $M,Q,K_1$ lie on this circle - $M,Q$ lie because they are intersection points of the two circles, and $K_1$ lies on the circle by the previous claim. $\blacksquare$ We now return to the original problem. If $K$ is the meeting point of the common external tangents, then it satisfies the equation $(\star)$ in the above claim. Also, from the above claim, we have that $K\in(MQK_1)$ and $K\in(ABC)$, which forces $K\in\{Q,K_1\}$. However, note that $K=Q$ is impossible as $Q$ itself is an intersection point of $\tau_1,\tau_2$. So, $K=K_1$ must be true. Hence, $AK\perp BC$, as claimed. $\blacksquare$

Define $Q$ as the $A$-queue point. First notice that $(ABC)$, $(BME)$, $(CMF)$ all pass through $Q$, which can be proved by inverting about $(BC)$. Define $P$ as the intersection of $BC$ with the tangent to $(ABC)$ at $Q$. We claim $PKMQ$ is cyclic, which follows from Coaxiality Lemma and LOS: \[\frac{\operatorname{pow}(K,(BME))}{\operatorname{pow}(K,(CMF))} = \left(\frac{R_{(BME)}}{R_{(CMF)}}\right)^2 = \left(\frac{QB}{QC}\right)^2 = \frac{PB}{PC} = \frac{\operatorname{pow}(P,(BME))}{\operatorname{pow}(P,(CMF))}.\] Since $O \in (MPQ)$, it follows that $PK$ is also tangent to $(ABC)$, giving the desired harmonic quadrilateral $QBKC$. $\blacksquare$

USA TST 2023 p2 Let $Q$ be the queue point of $\triangle ABC$. And let $Q'$ be the intersection of $(BEM)$ and $(CFM)$. Let $K'$ be the intersection of the $A$-altitude with $(ABC)$ And let line $KM$ intersect $(ABC)$ again at $R$. We first start off with the following claim. Claim 1: $Q=Q'$ Proof: Since $M$ is the center $(BCEF)$, and $BCEF$ is cyclic, this gives us $Q=Q'$.$\blacksquare$ A well know property is that $(QK'BC)$ is harmonic. Hence we want to show that $(QKBC)$ is harmonic. Claim 2: $QR \| BC$ Proof: long angle chase. $\blacksquare$ Now this basically implies that $(QK'BC)$ is harmonic so $K=K'$, hence we done.

BlizzardWizard wrote: To show that $k=\frac1s$, it now suffices to show that $s^2\overline s^2-s^2-\overline s^2=0$. Could you please explain why we have $|s^2-1|=1$?

Solved with stillwater_25. I personally found this problem very challenging and it was pretty hard to avoid accidentally circular reasoning. It turns out the natural way to eliminate the headaching common tangents is via inversion (even though it seems counterintuitive since they will now become circles tangent to two lines). Let $H$ denote the orthocenter, $Q_A$ denote the $A-$Queue Point, $H_A$ the $A-$Humpty Point and $X_A$ the $A-$Ex Point. We first make a minor observation. Claim : $Q_A$ lies on circles $(BME)$ and $(CMF)$. Proof : This is a direct angle chase. It is well known that $M$ is the center of cyclic quadrilateral $BCEF$, $M$ is the intersection of the tangents to $(AEF)$ at $E$ and $F$ and points $Q_A$ , $H$ and $M$ are collinear. Thus, \[\measuredangle MQ_AF =\measuredangle HQ_AF = \measuredangle HFM = \measuredangle CFM = \measuredangle MCF\]which implies that $CMFQ_A$ is cyclic. Similarly we can show that $BMEQ_A$ is also cyclic, proving the claim. Now comes the crux of this solution. We perform an inversion about circle $(BC)$ (with center $M$). Clearly $Q_A$ goes to $H$ and $A$ goes to $H_A$ under this inversion. Further, circles $(BME)$ and $(CMF)$ become lines $\overline{BE}$ and $\overline{CF}$ respectively. Thus, the common external tangents to these circles become the two circles tangent to $\overline{BE}$ and $\overline{CF}$ passing through $M$. Also, $K$ goes to the second intersection of these two circles. Since it is well known that the radical axis of two intersecting circles is bisected by the line passing through its centers, $K$ must be the reflection of the internal $\angle BHC$-bisector. Now, it suffices to prove the following problem. Inverted and Rephrased Version wrote: Let $\triangle ABC$ be a triangle with orthocenter $H$ , $A-$Queue point $Q_A$, $A-$Humpty Point $H_A$ and $M$ the midpoint of side $BC$. If the reflection of $K$ across the internal $\angle BHC$-bisector lies on $(BHC)$, show that it also lies on circle $(MH_AQ_A)$. Note that $MH$ is the $H-$ median of $\triangle BHC$, and its reflection across the $\angle BHC$-bisector must be the $H-$symmedian of $\triangle BHC$. Since $K$ lies on $(BHC)$, this means it is the intersection of the $H-$symmedian of $\triangle BHC$ with $(BHC)$. Now, $X_A$ is also the $H-$Ex Point of $\triangle BHC$ quite clearly, and thus it is a well known lemma that $K$ lies on the circle $(MX_A)$ which is precisely what we needed to show.

Recall that $(ABC) \cap (EMB) \cap (CMF) \cap (AEF) = Q_A$ the A-Queue point. Now when we invert about $(BEFC)$ the following things happen: $(MFC)$ is mapped to $FC$ $(BEM)$ is mapped to $EB$ $Q_A$ is mapped to $H$ the orthocenter of $\triangle ABC$ The external tangents are mapped to the circle through $B$, and $M$ tangent to $CF$, and $EB$, and the circle through $C$, and $M$ tangent to $CF$, and $EB$. Now $$90^\circ - C = \measuredangle CBH = \measuredangle MBH = \measuredangle MK'B =\measuredangle MBK$$Finishing.

me when I forget the forgotten coaxiality lemma Because $HQ \cdot HM = HD \cdot HA = HB \cdot HE$, the queue point $Q$ lies on both $(BEM)$ and $(CFM)$. Let $P$ be the tangent intersection now, and $K$ the point on the circumcircle such that $\overline{AK} \perp \overline{BC}$. The key claim is the following: Claim: $MKPQ$ is cyclic. Proof: Note that $\overline{KB}$ and $\overline{KC}$ are tangent to their respective circles. Now use the forgotten coaxiality lemma: let $k$ be the homothety ratio between the two circles. Then \[\frac{\operatorname{pow}(K, (BEM))}{\operatorname{pow}(K, (CFM))} = \frac{BK^2}{CK^2} = \frac{\cos^2 B}{\cos^2 C} = k^2 = \frac{\operatorname{pow}(P, (BEM))}{\operatorname{pow}(P, (CFM))} \]as $EM=FM$ by tangents lemma. $\blacksquare$ Hence when $P$ lies on $(ABC)$, we must have $P = K$ as $P \neq Q$. The result follows.