Let the C-median, C-symmedian of $ \triangle ABC $ cuts $ \Omega $ at $ U, $ $ V, $ respectively and let $ Z $ $ \equiv $ $ AV $ $ \cap $ $ BU, $ $ Y $ $ \equiv $ $ AC' $ $ \cap $ $ MZ. $ Obviously, $ MZ $ is the perpendicular bisector of $ AB, $ so $ \measuredangle BYM $ $ = $ $ \measuredangle MYA $ $ = $ $ \measuredangle BAC $ $ = $ $ \measuredangle BUM $ $ \Longrightarrow $ $ B, $ $ M, $ $ U, $ $ Y $ lie on a circle with diameter $ BY. $ On the other hand, from $ \measuredangle ACK $ $ = $ $ \measuredangle MBZ $ $ \Longrightarrow $ $ \triangle ACK $ $ \sim $ $ \triangle MBZ $ $ \Longrightarrow $ $ \measuredangle UKY $ $ = $ $ \measuredangle UZY, $ so we get $ K, $ $ U, $ $ V, $ $ Y, $ $ Z $ lie on a circle with diameter $ YZ $ $ \Longrightarrow $ $ KZ $ $ \perp $ $ AC'. $ Let $ AB $ cuts $ KZ $ at $ E. $ From Reim's theorem (K-Z-E and V-Z-A) we get $ A, $ $ E, $ $ K, $ $ V $ are concyclic, so $ AV $ $ \perp $ $ EV $ $ \Longrightarrow $ $ E, $ $ V, $ $ Y $ are collinear and $ EV $ passes through the antipode $ A' $ of $ A $ (in $ \Omega $). Analogously, if the perpendicular from $ L $ to $ BC' $ cuts $ AB $ at $ F, $ then $ FV $ passes through the antipode $ B' $ of $ B $ (in $ \Omega $). Finally, let $ EK $ cuts $ FL $ at $ D. $ From $ \measuredangle EDF $ $ = $ $ \measuredangle ACB $ $ = $ $ \measuredangle A'VB' $ $ \Longrightarrow $ $ V $ lies on the circumcircle of $ \triangle DEF, $ so notice $ A'B' $ $ \parallel $ $ EF $ we get $ \odot (DEF) $ is tangent to $ \Omega $ at $ V. $

My solutions, Lemma: Let $\triangle ABC$. Let $E\in CA, F\in AB$ such that $B, C, E, F$ are cyclic. $BE\cap CF=H$. Let $P\in CA, Q\in AB$ such that $H, P, Q$ are conlinear. Let $M, N\in BC$ such that $PM\parallel BE, QN\parallel CF$. Let $K=PM\cap QN$. Then circle $(KMN), (ABC)$ are tangent to each other. Proof. $BE, CF$ intersect $(ABC)$ again at $X, Y$; $XP\cap YQ=Z$. Because $H\in PQ$, by Pascal's theorem for $AXYBCZ$ $\Rightarrow Z\in (ABC)$. We have $(ZB, ZQ)\equiv (ZB,ZY)\equiv (CB,CY)\equiv (NB,NQ)(mod \pi)$ $\Rightarrow N\in (BQZ)$. SImilarly, $M\in (CPZ)$. Let $ZM, ZN\cap (ABC)=R, S (\neq Z)$. According to Reim's theorem, $PM\parallel AR, QN\parallel AS$ $\Rightarrow (KM,KN)\equiv (AR, AS)=(ZR,ZS)\equiv (ZM,ZN)(mod \pi)\Rightarrow Z\in (KMN)$ (1). We have $(ZB, ZS)\equiv (ZB, ZN)\equiv (QB, QN)\equiv (FB, FC)\equiv (EB, EC)\equiv (PM, PC)\equiv (ZM, ZC)\equiv(ZR, ZC) (mod \pi) $ $\Rightarrow RS\parallel BC\equiv MN$ $\Rightarrow (ZMN)$ and $(ZBC)$ are tangent (2). From (1), (2)$\Rightarrow$ $(KMN), (ABC)$ are tangent to each other.

______________________________________________________________________________________________________________________________________________________ Back to main problem. Denote by $\mathbb{R}_{M}$ be the symmetry in center $M$. Let $H$ be orthocenter of $\triangle ABC, BH\cap CM= K', AH\cap CM=L'$. Let $\Delta \equiv \triangle DEF$. We have $\mathbb{R}_{M}$: $(HAB) \mapsto (C'BA), E \mapsto E', F \mapsto F', D \mapsto D'$. By Lemma, $(D'E'F')$ is tangent to $(HBC)$. Therefore, $(DEF)$ is tangent to $(C'AB)$. DONE.

Nice tangent problem, here is the general problem Let $ABC$ be a triangle inscribed incircle $(O)$. Circle $(K)$ passes through $B,C$ which intersect $CA,AB$ again at $E,F$. $AD$ is diameter of $(O)$. $AK$ cuts $DB,DC$ at $M,N$. $P,Q$ lie on $BC$ such that $MP\perp CF$ and $NQ\perp BE$. $MP$ cuts $NQ$ at $R$. Prove that circle $(PQR)$ is tangent to $(O)$.

Dear Buratinogigle, nice general problem. My solution: Let $BD, CD\cap (K)=X, Y$; $BY\cap CX=Z$. Acording to Pascal's theorem for $BEFCXY$ $\Rightarrow A=BE\cap CF, K=EY\cap FX$ and $Z=CX\cap BY$ are collinear. Applying lemma (in #3) $\Rightarrow$ $(RPQ)$ and $(DBC)$ are tangent to each other.

buratinogigle wrote: Nice tangent problem, here is the general problem Let $ABC$ be a triangle inscribed incircle $(O)$. Circle $(K)$ passes through $B,C$ which intersect $CA,AB$ again at $E,F$. $AD$ is diameter of $(O)$. $AK$ cuts $DB,DC$ at $M,N$. $P,Q$ lie on $BC$ such that $MP\perp CF$ and $NQ\perp BE$. $MP$ cuts $NQ$ at $R$. Prove that circle $(PQR)$ is tangent to $(O)$. Let the polar of $ A $ WRT $ \odot (K) $ cuts $ AK $ at $ T. $ From $ \measuredangle BTM $ $ = $ $ \measuredangle KBA $ $ = $ $ 90^{\circ} $ $ - $ $ \measuredangle FCB $ $ = $ $ \measuredangle BPM $ we get $ B, $ $ M, $ $ P, $ $ T $ are concyclic. Similarly, we can prove $ T $ $ \in $ $ \odot (CNQ), $ so if $ S $ is the second intersection of $ \odot (BMP) $ and $ \odot (CNQ), $ then we get $ \measuredangle CSB $ $ = $ $ \measuredangle CST $ $ + $ $ \measuredangle STB $ $ = $ $ \measuredangle (CN,AK) $ $ + $ $ \measuredangle (AK,BM) $ $ = $ $ \measuredangle CAB $ $ \Longrightarrow $ $ S $ $ \in $ $ \odot (O), $ hence $ S $ is the Miquel point of the complete quadrilateral formed by any four lines of $ BC, $ $ BD, $ $ CD, $ $ PR, $ $ QR. $ Let $ Y $ $ \equiv $ $ BD $ $ \cap $ $ QR, $ $ Z $ $ \equiv $ $ CD $ $ \cap $ $ PR. $ Since $ S $ lies on $ \odot (BQY) $ and $ \odot (CPZ), $ so $ \measuredangle (SQ,SB) $ $ = $ $ \measuredangle (QY,BY) $ $ = $ $ \measuredangle EBF $ $ = $ $ ECF $ $ = $ $ \measuredangle (CZ,PZ) $ $ = $ $ \measuredangle (SC,SP) $ $ \Longrightarrow $ $ SP, $ $ SQ $ are isogonal conjugate WRT $ \angle BSC, $ hence notice $ S $ $ \in $ $ \odot (PQR) $ we conclude that $ \odot (PQR) $ is tangent to $ \odot (O) $ at $ S. $ ____________________________________________________________ Remark : $ AS $ is the A-symmedian of $ \triangle ABC. $ Proof : From $ \measuredangle MYN $ $ = $ $ \measuredangle FBE $ $ = $ $ \measuredangle FCE $ $ = $ $ \measuredangle MZN $ we get $ M, $ $ N, $ $ Y, $ $ Z $ are concyclic, so notice $ S $ is the Miquel point of $ MZNY $ $ \Longrightarrow $ $ MN, $ $ YZ $ and the perpendicular from $ S $ to $ DS $ are concurrent $ \Longrightarrow $ $ A $ $ \in $ $ YZ, $ hence note that $ S $ lies on $ DR $ we conclude that $ (A,S;B,C) $ $ = $ $ D(A,R;M,N) $ $ = $ $ -1 $ $ \Longrightarrow $ $ AS $ is the A-symmedian of $ \triangle ABC. $

Proof of the original problem: $A$ and $C$ are swapped in my solution($C'$ from the original problem is $A'$ in my solution). Let $Z$ be on $BC$ such that $ZL||AC$ and $Y$ on $BC$ such that $KY||AB$. Let $B',C'$ be points on circumcircle such that $\angle BAB'=\angle CAC'=90$.Now suppose circles touch at point $T$. We see that homotethy with center at $T$ sends $YZ$ to points on circumcircle parallel to $BC$ such that they make angle $180-\angle BAC$ at point $A$ so $T$ has to send $Y,Z$ to $B',C'$ in order for problem to be true so we only need to prove that $C'Z$ and $B'Y$ intersect on the circumcircle. To prove this it is enoguh to prove $\tan \angle BAC= \tan (\angle BC'Z +\angle CB'Y)$. Now we easily have $BC'=a \cdot \tan \alpha$. We also have $BZ=a- \frac{b \tan \phi}{\sin \angle ACB}$. Now we get $\tan BC'Z=\frac{BZ}{BC'}$ and after easy computations we get $\tan BC'Z=\tan \phi$ so $\angle BC'Z=\angle CAM$. By analogy we get $\angle CB'Y=\angle BAM$ so we are finished. Nice result is that $AT$ is actually $A$-symmedian.

Let the perpendiculars from $K$ and $L$ meet at $X$, $\{Y\}=KX\cap AB, \{Z\}=LX\cap AB,\ \{P\}=BL\cap XY,\ \{N\}=AK\cap ZX,\ \{T\}=C^\prime X\cap KL$. We have to prove that $(XYZ)$ is tangent to $(ABC)$. Note that $$\dfrac{CL}{CK}=\dfrac{BC\cos{\widehat{ACM}}}{AC\cos{\widehat{BCM}} }=\dfrac{\sin{\widehat{ACM}} \cos{\widehat{ACM}}}{\sin{\widehat{BCM}} \cos{\widehat{BCM}} }=\dfrac{C^\prime L}{C^\prime K}\cdot \dfrac{\sin{\widehat{LC^\prime T}} }{\sin{\widehat{KC^\prime T}} }=\dfrac{TL}{TK}$$so $(L,K,T,C)=-1$, whence $C-P-N$ are collinear. As $X$ is the orthocenter of $\triangle{C^\prime PN}$, $C^\prime X\perp PN$, so $\{R\}=C^\prime X \cap PN \in (ABC)$ because $CC^\prime$ is diameter. Also note that $$(P,N,R,C)=-1\Leftrightarrow C^\prime (P,N,R,C)=-1\Leftrightarrow RBCA \mathrm{\ is\ harmonic}$$i.e. $CR$ is symmedian. Let $O_1,\ O_2$ be the centers of $(ABC)$ and $(XYZ)$. As $\widehat{O_2XR}=\widehat{CC^\prime R}$, we have $O_2X\parallel O_1C^\prime$. By law of sines within $\triangle{C^\prime NP}$, we infer that $\dfrac{RX}{RC^\prime}=\dfrac{\sin{\widehat{ACM}} \sin{\widehat{BCM}} } {\cos{\widehat{ACM}} \cos{\widehat{BCM}}}$. Note that $\dfrac{O_2X}{O_1C^\prime }=\dfrac{2R_{\triangle{XYZ}}}{2R}=\dfrac{\frac{YZ}{\sin{\hat{C}}}}{2R}=\dfrac{YZ}{2R\sin{\hat{C}}}=\dfrac{BZ+AY-AB}{2R\sin{\hat{C}}}$. By law of sines in $\triangle{LZB}$ and $\triangle{LBC}$ and the analogous, $BZ=\dfrac{BC\sin{\widehat{BCM}} }{\sin{\hat{B}} \cos{\widehat{BCM}} }$ and $AY=\dfrac{AC\sin{\widehat{ACM}} }{\sin{\hat{A}} \cos{\widehat{ACM}} }$, so $$\dfrac{O_2X}{O_1C^\prime }= \dfrac{\sin{\hat{A}} \cdot \sin{\widehat{BCM}} }{\sin{\hat{C} } \sin{\hat{B}} \cos{\widehat{BCM}} }+\dfrac{\sin{\hat{B}}\cdot \sin{\widehat{ACM}} }{\sin{\hat{C} } \sin{\hat{A}} \cos{\widehat{ACM}} }-1$$

we have $\dfrac{RX}{RC^\prime}=\dfrac{O_2 X}{O_1C^\prime}$ and $O_2X\parallel O_1C^\prime$, whence $R-O_2-O_1$ are collinear. As $O_1C^\prime=O_1R$, we infer $O_2X=O_2R$, so $R\in (XYZ)$. Therefore, $(XYZ)\cap (ABC)=\{R\}\in O_1O_2$, so $(XYZ)$ and $(ABC)$ are tangent.

Another generalization of problem, In a triangle $ABC$,$AC<BC$,$M$ is midpoint of $AB$. Let $C'$ be point such that the tangent lines at $C'$ of $(ABC')$ and at $C$ of $(ABC)$ are parallel to each other. $AC'$ and $BC'$ intersect with $CM$ at $K,L$,respectively.The parallel drawn from $K$ to $AC$ and parallel drawn from $L$ to $BC$ intersect with $AB$ and each other and form a triangle $\Delta$. Prove that circumcircles of $\Delta$ and $(ABC')$ are tangent.

Let $Q$ be the intersection of $C$-symmedian with $(O)$. $AC'$ meets $ AQ$ at $T$,$ BC'$ meets $CP$ at $R$. Let $S$ be the projection of $T $ onto $CB.$ Since $T$ lies on $C$-symmedian of triangle $ABC$ and $\triangle CAT\sim\triangle CBL$, we get $\frac{TA}{TS}=\frac{CA}{CB}=\frac{TA}{LB}$. This follows that $TS=LB$ or $TL\perp RB$ or $Z,T,L$ are collinear. Similarly $R,Y,X$ are collinear. We get $X$ is the orthocenter of triangle $TRC'$. But $\angle CQC'=90^\circ$ so $Q,X,C'$ are collinear. On the other side, $ZT\parallel CB$ then applying Reim theorem, $AZTQ$ is cyclic. Similarly, $BYQR$ is cyclic. Hence $ \angle ZQY=\angle ZQT+\angle TQY=\angle C'AB+\angle C'BA=180^\circ-\angle AC'B=\angle ZXY$. We get $Q\in (XYZ).$ Let $Qt$ be the tangent of $(O)$. Then $\angle tQC'=\angle QAC'=\angle QZX$ or $Qt$ is also a tangent of $(J)$. We are done. Remark. The common tangent of two circles is the line joining centers of $(AZT)$ and $(BYR).$

Another proof which is base on properties of paralogic triangles (See here Let $J,T$ be the intersections of $OM$ and $AC',BC'$. Since $XYZ$ and $TJC'$ is paralogic triangles wrt $(M,K,L)$ we get $(XYZ)$ is orthogonal to $(JTC')$. Let $S$ be the second intersection of $(TJC')$ and $(O)$. Since $TJC'$ and $ABC$ are paralogic triangles wrt $(A,M,B)$ we get $S$ is Miquel point of complete quadrilateral $ABCAMB$ or the intersection of circles through $M,A$ and tangent to $CA$, through $M,B$ and tangent to $CB$. Then $\angle MSA=\angle CAM=\angle CSB$ or $CABS$ is harmonic quadrilateral. Let $P,Q$ be the intersections of $MO$ and $ZL,YK$ then $XPQ$ and $AC'B$ are paralogic triangles wrt $(M,K,L).$ Let $S'$ be the intersection of $(XPQ)$ and $(AC'B)$ such that $C'X, BP,AQ$ concur at $S'$. Then $\angle S'CA=\angle S'C'A=\angle XLK=\angle LCB$ or $ CS'$ is symmedian of triangle $ABC$. This means $S'\equiv S$ or $S,X,C'$ are collinear. We know that $XC', ZT, YJ$ concur at the intersection of $(XYZ) $ and $(JTC'). $ Therefore $ (XYZ), (JTC'), (O)$ concur at $S$. But $ (XYZ)$ and $(O) $ are orthogonal to $(JTC')$ then $(XYZ)$ is tangent to $(O)$ at $S.$

MRF2017 wrote: In acute triangle $ABC$,$AC<BC$,$M$ is midpoint of $AB$ and $\Omega$ is it's circumcircle.Let $C^\prime$ be antipode of $C$ in $\Omega$. $AC^\prime$ and $BC^\prime$ intersect with $CM$ at $K,L$,respectively.The perpendicular drawn from $K$ to $AC^\prime$ and perpendicular drawn from $L$ to $BC^\prime$ intersect with $AB$ and each other and form a triangle $\Delta$.Prove that circumcircles of $\Delta$ and $\Omega$ are tangent.(M.Kungozhin) My solution : Lemma: $\triangle ABC$ with $E$, $F$ lies on $CA,AB$ such that $BCEF$ is cyclic, $BE$ cuts $CF$ at $H$ An abitrary line passes through $H$ cuts $CA,AB$ at $M,N$ . Lines through $M,N$ and parallel $BE,CF$ cut $BC$ at $P,Q$ and cut each other at $R$ then $(PQR)$ tangent $(O)$ Prove : Let $R'$ be the intersection of $(BNQ)$ and $MN$ => $\angle NR'B = \angle NQB = \angle HCB$ => $R'$ lies on $(HCB)$. Similary, Let $R_{1}'$ be the intersection of $(CMP)$ with $MN$ so $R_{1}'$ lies on $(HCB)$ => $(HCB)$ , $MN$ , $(BNQ)$ and $(CMP)$ are concurrent $X$ is the second intersection of $(BNQ)$ and $(CMP)$ => $\angle QXP = \angle CXP -\angle CXQ = 180 - \angle CRP - \angle CXR' - \angle R'XQ =180 - \angle CR'P - \angle CPR' -\angle CBR' = \angle CR'B$ And $\angle QRP$ = $\angle BHC$ => $QXPR$ is cyclic=> $\angle CXR = \angle CXQ +\angle QXR = \angle CMR' +\angle CHR' + \angle QPR =\angle HCA + \angle QBH = \angle HBA + \angle HBC - \angle ABC$ => $ABCX$ is cyclic => $X$ lies on $(O)$ and $(PQR)$. Let $T$ be the circumcenter of $(PRQ)$ => $\angle OXT = 90 - \angle XBC + 90 - \angle XPQ + \angle CXQ$ $\angle CXQ = \angle CMR' + \angle CHR' = \angle HCA$ $Ct$ is the ray parallel with $AX$ => $\angle tCA = \angle CAX = \angle CBX $ , $\angle tCH = \angle ARN = \angle XRB$ =>$\angle ACH = \angle XBC + \angle XPQ = \angle CXQ$ => $\angle OXT = 180$ Bach to main problem : Let $AD, BE$ be 2 altitudes of $\triangle ABC$ $AD$ cuts $AM$ at $R$ . By Thales: $ \frac{XM}{MB} = \frac{ML}{MC} = \frac{MR}{MC} = \frac{MX}{MA}$ =>$RX \parallel CA$ Similary , if $BE$ cuts $CM$ at $S$ so $YS \parallel CB$ $YS$ cuts $XR$ at $U$. By applying the lemma => $(XUY)$ tangent $(AHB)$ $U'$ reflects $U$ about $BC$ , $CH$ cuts $(O)$ at $H'$ so $(XYU')$ tangent $(AH'B)$ $XUYZ$ is a parallelogram => $XYZU'$ is an isoceles trapezoid => $(XYZ)$ tangent $(O)$ =>$(XZY)$ tangen $(ACB)$

Point $F$ is chosen on $\Omega$ such that $\angle FCA=\angle BFM$. Let $A',B'$ be the antipodes of $A,B$, respectively, in circle $\Omega$. Let $\ell_1$ and $\ell_2$ denote the lines from $K,L$ respectively, perpendicular to $AC^\prime$ and $BC^\prime$, respectively. Let $\ell_1$ and $\ell_2$ meet at $X$, $\ell_1$ and $AB$ meet at $Y$ and $\ell_2$ and $AB$ meet at $Z$. Note that the corresponding sides of triangle $XYZ$ and triangle $C'A'B'$ are parallel, hence, they are homothetic. The main idea is to show that the homothety center is $F$ from where the result will follow as $F$ lies on $\Omega$. Indeed, we have that $$\angle AC'X=\angle KC'X=\angle KLX=\angle BCM=\angle ACF$$implying that points $F,X,$ and $C'$ are collinear. Let $C_1$ be a point such that $CAC_1B$ is a parallelogram. We observe that $\triangle BB'Z$ and $\triangle BC_1L$ are similar. Indeed, $\angle LBC_1=\angle B'BZ=90^{\circ}-\angle CBA$ and $$\frac{BB'}{BC_1}=\frac{BB'}{CA}=\sin \angle CBA=\frac{BZ}{BL}$$as $\angle LBZ=90^{\circ}-\angle CBA$. Thus, $\angle BB'Z=\angle BC_1L=\angle ACM=\angle BCF$ yielding that $B',Z,$ and $F$ are collinear. It follows that $F$ is the stated homothety center and hence, the tangency point of the circumcircle of triangle $\Delta$ and of circle $\Omega$.

Symmedian AG cuts C'A,C'B at H,I, line through H,I perp BC',AC' cuts AB,BC',BA at E,F,L,I, cuts at D. From AGE=AHE=90-HC'I=FIB=FGB so (EGF) tangent (O). EDF=90-FGB=EGF from EGBL cyclic so is EGDF. Notice C(KLGA)=-1 so A,K,L collinear. HAK=AHC-AKC=90-HCA-(180-HIL)=GCA-GCB=A-2GAC so AKL conjugate AG so pass M

Here is a short spiral similarity solution to the original problem. 2016 All Russian 10.8, A-labelled wrote: In acute triangle $ABC$ ,$M$ is the midpoint of $BC$ and $\Omega$ is it's circumcircle. Let $A'$ be antipode of $A$ in $\Omega$. Lines $A'B$ and $A'C$ intersect with $AM$ at $K,L$, respectively. The perpendicular drawn from $K$ to $A'B$ and perpendicular drawn from $L$ to $A'C$ intersect with $BC$ and each other and form a triangle $\Delta$. Prove that circumcircles of $\Delta$ and $\Omega$ are tangent. Let $B',C'$ denote $B$-antipode and $C$-antipode. Place points $E,F$ on $BC$ such that $KE\parallel AB$ and $LF\parallel AC$. Let $D$ be the remaining vertices on $\Delta$ and let $A$-symmedian meet $\Omega$ at $P$. Claim: $B',E,P$ are colinear. (Similarly $C',F,P$ are colinear) Proof: Let $E'$ be the reflection of $E$ across $M$. Since $\triangle ABC\cup M$ and $\triangle KEE'\cup M$ are homothetic, we get $KE'\parallel AC$. Thus $\angle BE'K = 180^{\circ}-\angle C = \angle BC'A$ which means $\triangle BE'K\sim\triangle BC'A$. This similarity induces $\triangle BE'C'\sim\triangle BKA$. However, by symmetry, $\triangle BE'C'\cong\triangle CEB'$. Thus $\angle EB'C = \angle BAK = \angle PAC$, which implies the claim. $\blacksquare$ Now we are almost done. Note the homothetic triangles $\triangle DEF\sim\triangle A'B'C'$, which has the center $P\in\odot(A'B'C')\equiv\Omega$ so $\odot(DEF)$ and $\Omega$ are indeed tangent.

Let $X$ be the intersection of $C$-symmedian and circumcircle of $ABC$ ,$R$ be the intersection of those perpendicular lines and $P,Q$ intersection of them with $AB$. we have $PR$ is parallel to $BC$ and $KC'LR$ is cyclic.so $C',R.X$ are collinear and now with a little angle chasing we have $PQRX$ is cyclic. It's enough to prove $X$ is the center of $(PQR), (ABC)$ homothety. Let $Y$ be the intersection of $XP$ with $(ABC)$. we had $C',R.X$ are collinear ,and again with a little angle chasing we'll have $PR$ is parallel to $YC'$. but $X$ was on both $(ABC),(PQR)$ and we're done....

Solved with: Alan Lee, Alex Zhao, Ankit Bisain, Derek Liu, Eric Shen, Ethan Zhou, Evan Chen, Kevin Wu, Luke Robitaille, Mason Fang, Raymond Feng, Reagan Choi, Rey Li, Samuel Wang, Tristan Shin, and intense sleep deprivation. This solution was written while I am definitely not low ofn sleep totally okay. Define $T$ on $(ABC)$ such that $(AB;CT) = -1$. Let $Z,Y,X$ be the intersections of the perpendiculars with each other and $AB$, respectively. Then, I claim that $C', T, Z$ are collinear. Observe that if $P = CM\cap (ABC)$, then $C'P\perp CP$, which means $C'P$ is the altitude of $CP$. Since $CZ$ is the line that contains the circumcircle of $(C'KLZ)$, this means $\angle ZCK = \angle PC'L$, so they are isogonal. Now, since $TP$ is parallel to $AB$, by isogonality, $T, Z, C'$ are collinear. Now, I claim $Z$ is the incenter of $\triangle ZKL$. First, note that \[-1 = (AB;CT) \overset{C'}{=} (KL;C C'T\cap AP)\]Now, since $\angle C'TC = 90$, this means by lemma 9.18 in EGMO, we get $C'T$ bisects $\angle KTL$. Now, consider $N$, the midpoint of $C'Z$. Since $N$ is the circumcenter of $(ZKC'L)$, $N$ lies on the angle bisector and the perpendicular bisector, so by fact $5$, we have $Z, C'$ are the incetners and excentteres of $\triangle KTL$. Now, we have \[\angle TKZ = \angle ZKP = \angle ZC'L = \angle PC'K = \angle ACP = \angle TCB = \angle YAT\]Therefore, $(ATKY)$ is cyclic. Similarly, $(BLTX)$ is also cyclic. To conclude, observe that $\angle ATY = \angle AKY = 90$, so $TY\cap (ABC)$ is the $A'=$ A-antipode. Similarly, $TX\cap (ABC)$ is the $B'=$ B-antipode. Since $XY, XZ, YZ || AB, CB, AC ||| A'B', C'B', A'C'$, this means $T$ is the center of homothety sending $(XYZ)$ to $(ABC)$. Therefore, $T$ is the tangency point of $(ABC), (XYZ)$, which proves that these two circles are tangent.

Solved with Alex Zhao, Ankit Bisain, Jeffrey Chen, Kevin Wu, Luke Robitaille, Mason Fang, Reagan Choi, Rey Li, and Tristan Shin. Let the two perpendiculars intersect at \(Z\), and let \(\overline{ZK}\) and \(\overline{ZL}\) meet \(\overline{AB}\) at \(X\) and \(Y\). If \(T\) is the harmonic conjugate of \(C\) with respect to \(\overline{AB}\), then I contend the tangency point is \(T\). [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair C,A,B,Cp,M,Tp,T,K,L,Z,X,Y; C=dir(125); A=dir(210); B=dir(330); Cp=-C; M=(A+B)/2; Tp=2*foot(origin,C,M)-C; T=reflect(M,origin)*Tp; K=extension(A,Cp,C,M); L=extension(B,Cp,C,M); Z=extension(K,K+C-A,L,L+C-B); X=extension(Z,K,A,B); Y=extension(Z,L,A,B); draw(circumcircle(Z,X,Y),linewidth(.3)); draw(T--Cp,gray+dashed); draw(K--T--L--cycle,linewidth(.8)); draw(C--K); draw(circumcircle(A,X,K),dashed+linewidth(.4)); draw(A--Cp--B,gray+linewidth(.4)); draw(Cp--L,gray+linewidth(.4)); draw(Z--L,linewidth(.4)); draw(X--Z--Y,linewidth(.4)); draw(circumcircle(Cp,K,L),linewidth(.4)); draw(unitcircle); draw(C--A--B--cycle,linewidth(.8)); dot("\(C\)",C,C); dot("\(A\)",A,W); dot("\(B\)",B,B); dot("\(C'\)",Cp,SE); dot("\(K\)",K,dir(160)); dot("\(L\)",L,S); dot("\(T\)",T,S); dot("\(X\)",X,NE); dot("\(Y\)",Y,NW); dot("\(Z\)",Z,dir(200)); dot("\(M\)",M,SW); [/asy][/asy] Claim: \(T\), \(Z\), \(C'\) collinear. Proof. Let \(\overline{CM}\) intersect \(\Omega\) at \(T'\). Then \(\overline{C'T}\) and \(\overline{C'T'}\) are isogonal in \(\angle AC'B\), but \(\overline{C'P}\) contains the center of \((C'KL)\), so \(\overline{C'T}\) contains \(Z\). \(\blacksquare\) Claim: \(Z\) and \(C'\) are the incenter and \(T\)-excenter of \(\triangle TKL\). Proof. Note \(-1=(CC';AB)\stackrel{C'}=(C,\overline{C'T}\cap\overline{CM};K,L)\), and since \(\angle CTC'=90^\circ\), we have \(\overline{TZ}\) bisects \(\angle KTL\) by a well-known Apollonius lemma. Then the midpoint of \(\overline{ZC'}\) is equidistant from \(K\), \(L\), \(Z\), \(C'\), and it lies on \(\overline{TZ}\), so it is the arc midpoint of \(KL\) on \((TKL)\), and the conclusion follows. \(\blacksquare\) Claim: \(ATZX\) is cyclic. Proof. Angle chase: \(\measuredangle TKX=\measuredangle TKZ=\measuredangle ZKL=\measuredangle ACM=\measuredangle TCB=\measuredangle TAX\). \(\blacksquare\) Finally we conclude \(\angle ATX=\angle AKX=90^\circ\), so line \(TX\) passes through the antipode \(A'\) of \(\Omega\). Since \(\overline{ZX}\parallel\overline{C'A'}\), etc., the conclusion readily follows by homothety.

[asy][asy] size(8cm); 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 = -6.379658795187723, xmax = 3.888871689988824, ymin = -2.8962293855541033, ymax = 3.1359566842132742; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures */ draw(circle((-1.868380703800035,-0.8461481053606017), 2.6411502538089437), linewidth(0.8) + blue); draw((-4.41676140760007,-0.15229621072120336)--(-2.864548318315209,1.599935447379614), linewidth(0.8)); draw(circle((-3.9110878936382556,0.41853851934488756), 0.762599496309902), linewidth(0.8)); draw(circle((-3.8246319465798817,1.9055808087489228), 1.0075612438896373), linewidth(0.8)); draw((-4.48,-1.24)--(-4.243849513483322,2.8217883680868674), linewidth(0.8) + linetype("4 4") + blue); draw((0.68,-1.54)--(-4.243849513483322,2.8217883680868674), linewidth(0.8) + linetype("4 4") + qqwuqq); draw((-4.48,-1.24)--(-0.727362306149961,1.5358157229581875), linewidth(0.8)); draw((-4.48,-1.24)--(0.68,-1.54), linewidth(0.8)); draw((-4.41676140760007,-0.15229621072120336)--(-0.727362306149961,1.5358157229581875), linewidth(0.8) + linetype("4 4") + qqwuqq); draw((-4.243849513483322,2.8217883680868674)--(-3.1485469041907104,0.42798474528302843), linewidth(0.8) + linetype("4 4") + blue); draw((-4.3472018410248205,1.0441283343730934)--(-1.607420988247321,0.884838749909285), linewidth(0.8) + linetype("4 4") + qqwuqq); draw((-3.1485469041907104,0.42798474528302843)--(-2.9150213320399927,-0.08238960778526279), linewidth(0.8)); draw((-4.3472018410248205,1.0441283343730934)--(0.68,-1.54), linewidth(0.8)); draw(circle((-4.330305460541695,1.334746078682832), 1.4895534234319208), linewidth(0.8)); /* dots and labels */ dot((-0.727362306149961,1.5358157229581875),dotstyle); label("$A$", (-0.6813150842433385,1.6509337777247404), NE * labelscalefactor); dot((-4.48,-1.24),dotstyle); label("$B$", (-4.434163669632973,-1.123411342149187), NE * labelscalefactor); dot((0.68,-1.54),dotstyle); label("$C$", (0.7231251839086106,-1.4227182845422248), NE * labelscalefactor); dot((-2.6036811530749806,0.14790786147909374),linewidth(4pt) + dotstyle); label("$M$", (-2.557739376938156,0.23498170409613844), NE * labelscalefactor); dot((-1.868380703800035,-0.8461481053606017),linewidth(4pt) + dotstyle); label("$O$", (-1.8209838264322153,-0.7550335668962174), NE * labelscalefactor); dot((-4.41676140760007,-0.15229621072120336),dotstyle); label("$C'$", (-4.365092836773042,-0.041301627343588776), NE * labelscalefactor); dot((-3.1485469041907104,0.42798474528302843),linewidth(4pt) + dotstyle); label("$K$", (-3.098794234340956,0.522776841012521), NE * labelscalefactor); dot((-4.3472018410248205,1.0441283343730934),linewidth(4pt) + dotstyle); label("$L$", (-4.29602200391311,1.1329025312752519), NE * labelscalefactor); dot((-1.607420988247321,0.884838749909285),linewidth(4pt) + dotstyle); label("$Y$", (-1.556212300469143,0.9717372546020776), NE * labelscalefactor); dot((-2.9150213320399927,-0.08238960778526279),linewidth(4pt) + dotstyle); label("$Z$", (-2.8685581248078496,0.004745594563032426), NE * labelscalefactor); dot((-3.4054143796764413,0.9893732494109783),linewidth(4pt) + dotstyle); label("$W$", (-3.3635657603040285,1.0868553093686306), NE * labelscalefactor); dot((-2.864548318315209,1.599935447379614),linewidth(4pt) + dotstyle); label("$X$", (-2.8225109029012283,1.6969809996313616), NE * labelscalefactor); dot((-4.243849513483322,2.8217883680868674),linewidth(4pt) + dotstyle); label("$D$", (-4.1924157546232115,2.917232380156823), NE * labelscalefactor); dot((-2.0889045240024915,0.9128319787322602),linewidth(4pt) + dotstyle); label("$E$", (-2.0397081304886666,1.0062726710320435), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let the line through $L$ perpendicular to $BC'$, the line through $K$ perpendicular to $AC'$ and $AB$ be $l_1,l_2,l_3$ respectively. Let $l_1\cap l_2=W$, $l_2\cap l_3=Z$ and $l_3\cap l_1=Y$. Let $C'W$ meet $(ABC)$ for the second time at $X$. Claim 1. $CX$ is the $C$-symmedian. Proof. $$\angle XCA=\angle XC'A=\angle WC'K=\angle WLK=\angle MCB$$$\blacksquare$ Claim 2. $CX$, $LW$ and $AC'$ are concurrent. Proof. Let $CX\cap BC'=D$. Let $AC'\cap CX=E_1$, then $$-1=(X,C;A,B)\overset{C}{=}(X,C;E,D)$$Now let $H'$ be the orthocenter of $\triangle DC'E$. Let $L',K'$ be the projection of $H'$ on $D'$ and $C'E$. Notice that if $L'K'\cap DE=C_1$, then from $\angle C'XC=90^{\circ}$ we have $$(X,C_1;E,D)=-1$$Hence $C_1=C$. Meanwhile, $L'K'$ and $LK$ are both perpendicular to the isogonal of $C'W$ w.r.t. $\angle DC'E$, which implies that they are parallel. Hence $L'K'=LK$, which implies $L,W,E$ are collinear, and $D,W,K$ are collinear as desired. $\blacksquare$ Therefore, $W$ is the orthocenter of $\triangle DC'E$, hence $LXEC'$ is cyclic. This implies $X$ is the miquel point of $C'EYB$, so $XEYA$ is cyclic, meanwhile, $WXEK$ is cyclic as well, hence $X$ is also the miquel point of $KEYZ$. Therefore, $X$ lies on $WYZ$, the circumcircle of $\triangle \Delta$. Let $XZ\cap \Omega= B'$, then $$\angle XBC'=\angle XAC'=\angle XYE=\angle XYW=\angle XZW$$Hence $BC'\|ZW$, and the two circles are tangent by Reim's theorem.

Let $R$ be the intersection of $(ABC)$ and $C$-symmedian of $\triangle ABC$. I claim that $R$ is the desired tangency point. Let $X$ be the intersection of the perpendicular from $K$ to $AC'$ and the perpendicular from $L$ to $BC'$. Let $P=\overline{XL}\cap \overline{CR}$ and $Q=\overline{XK}\cap \overline{CR}$. Let $Y=\overline{XK}\cap \overline{AB}$, $Z=\overline{XL}\cap \overline{AB}$ and let $U$ be the circumcenter of $\triangle XYZ$. [asy][asy] import geometry; size(8cm);defaultpen(fontsize(10pt)); pair A,B,C,M,K,L,C1,X,P,Q,R,O,Y,Z,U; C=dir(120); A=dir(210); B=dir(330); M=midpoint(A--B);C1=-C;K=extension(C,M,A,C1);L=extension(C,M,B,C1); X=intersectionpoint(perpendicular(L,line(B,L)),perpendicular(K,line(K,A)));P=extension(A,K,X,L);Q=extension(B,L,X,K);R=extension(P,Q,X,C1); O=(0,0);Y=extension(X,K,A,B);Z=extension(X,L,A,B);U=circumcenter(X,Y,Z); draw(A--B--C--cycle, red);draw(circumcircle(A,B,C),royalblue);draw(C--C1,red);draw(C--L,orange);draw(A--C1--B,red);draw(C--Q,red+0.3);draw(R--C1,red+0.3);draw(C1--Q,red+0.3);draw(Q--K,red+0.3);draw(P--L,red+0.3);draw(Z--P,red+0.3);draw(K--Y,red+0.3);draw(circumcircle(X,Y,Z),royalblue+dotted);draw(O--R,cyan+dashed);draw(circumcircle(K,L,C1),royalblue+dotted); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$C'$",C1,dir(C1)); dot("$M$",M,dir(M)); dot("$K$",K,dir(K)); dot("$L$",L,dir(L)); dot("$X$",X,dir(X)); dot("$R$",R,dir(R)); dot("$Q$",Q,dir(Q)); dot("$P$",P,dir(P)); dot("$O$",O,dir(90)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$U$",U,dir(U)); [/asy][/asy] Claim: $R$ lies on $C'X$. Proof. As $KC'LX$ is cyclic and $XL\parallel BC$,\begin{align*} \measuredangle AC'R=\measuredangle ACR=\measuredangle MCB=\measuredangle KLX=\measuredangle AC'K.\,\square \end{align*} Claim: $P$ lies on $AC'$. Proof. Note that $CP$ and $CM$ are isogonal, hence \begin{align*} \frac{CP}{CL}=\frac{\sin{\angle CLP}}{\sin{\angle CPL}}=\frac{\sin{\angle LCB}}{\sin{\angle PCB}}=\frac{\sin{\angle MCB}}{\sin{\angle ACM}}=\frac{CA}{CB} \end{align*}and as $\measuredangle ACP=\measuredangle LCB$, we conclude that $\triangle PAC\sim\triangle LBC$, which implies that $\measuredangle PAC=90^\circ$, the claim follows. $\square$ Similarly, we get that $Q$ lies on $BC'$. Claim: $R$ is the Miquel point of quadrilateral $ZPKY$. Proof. We have $AZPR$ cyclic, since \begin{align*} \measuredangle PRA=\measuredangle CBA=\measuredangle XZY=\measuredangle PZA. \end{align*}Additionally, $PRKX$ is cyclic due to $\measuredangle XRP=\measuredangle C'RC=90^\circ=\measuredangle XKP$, we conclude that $R$ is the Miquel point of quadrilateral $ZPKY$. $\square$ Final claim: $R,O,U$ are collinear. Proof. Indeed, \begin{align*} \measuredangle RZX=\measuredangle RZP=\measuredangle RAP=\measuredangle RCC'\implies \measuredangle XRU=\measuredangle C'RO, \end{align*}the claim follows. $\blacksquare$