Let the circle with diameter $AS$ intersect $AB, AC$ again at $E, F$ respectively. It’s clear that $(KEF)$ is the A-mixtilinear incircle. Let the tangent to $\omega$ at $K$ intersect $BC$ at $X$. Since $\angle XKO=\angle XMO=90^{\circ}$, $XKMO$ is cyclic with diameter $XO$. It suffices to prove that $\angle XTO=90^{\circ}$, which is equivalent to proving that $XT$ is tangent to $\omega$, or $(T, X; B, C)=-1$. By radical axis, $AT$, $EF$ and the tangent to $\omega$ at $K$ are concurrent at a point $Y$. So if $U$ and $V$ are the second intersections of lines $KE$ and $KF$ with $\omega$, we have \[(T, X; B, C)\stackrel{A}{=} (Y, AK\cap EF; E, F)\stackrel{K}{=} (K, A; U, V) =-1,\]as desired.

Inversion about $\omega$ affirms that all we want is $(\overline{BC}, \overline{KT})=-1$. Let $DE$ be the touch-chord of the $A$-excircle; with $D,E$ on $\overline{AB}, \overline{AC}$ respectively. Apply $\sqrt{bc}$-inversion. Note that $K'$ is the $A$-extouch point on $\overline{BC}$ and $S'=\overline{DE} \cap \overline{BC}$. Clearly, $(\overline{BC}, \overline{K'S'})=-1$ and we're done. Note: The last line is a degenerate case of the purely projective fact: $ABCD$ is cyclic in that order, then $\overline{AA} \cap \overline{BB}, \overline{CC} \cap \overline{DD}, \overline{AC} \cap \overline{BD}, \overline{BC} \cap \overline{AD}$ form a harmonic bundle, on the "quadrilateral" $DK'K'E$. In order to prove it, we consider a homography fixing the circle and mapping $ABCD$ to the vertices of some rectangle. Then it becomes obvious.

By homothety, $K,S,O$ are collinear. Let $I$ be the incircle of $ABC$. Let $E,F$ be the tangency points of the A-mixtilinear incircle and $AB$, $AC$ respectively. It's known that $E,I,F$ are collinear and $EF\perp AI$. Also $A,T,E,S,F$ lie on a circle with diameter $AS$. Let $L,N$ be the midpoints of arcs $\overarc{BC}$ and $\overarc{BAC}$ in $(ABC)$, respectively. It's well known that $K,I,N$ are collinear. Since $T=(AEF)\cap (ABC)$, $T$ is the center of spiral similarity which sends $A,E,I,F,S$ to $N,B,M,C,L$ respectively. $\angle TKI=\angle TKN=\angle TLN=\angle TSI$, so $T,I,S,K$ are concyclic. $\angle TMO=\angle TIA=\angle TKO$, so $K,T,O,M$ are concyclic, as desired.

This problem is equivalent to this https://artofproblemsolving.com/community/c6h1480699

I've been promoting，Look at my post, I have a very beautiful proof. My English is not good.

PARISsaintGERMAIN wrote: This problem is equivalent to this https://artofproblemsolving.com/community/c6h1480699 How?

My first post!! Let the tangent line in $K$ $l$ and let the intersection of $l$ and $BC$ $P$ Let the intersection of $\omega$ and $AB$ $D$, $\omega$ and $AC$ $E$. Then, radical axis of $\omega$ and circle $S$ is $l$, radical axis of $\omega$ and circle with diameter $AS$ is $AT$, radical axis of the circle with diameter $AS$ and circle $S$ is $DE$. So, $DE, AT$ and $l$ intersect at one point. Let the point $Q$. Point $A$'s polar is $DE$. By La Hire's theorem, point $Q$'s polar is $AK$. Let the intersection of $AK$ and $DE$ $R$. Then $(D, R, E, Q)$is harmonic division. By the pencil of $A$, $BKCT$ is harmonic quadrilateral. Since $PK$ is tangent line, $PT$ is tangent line. $\angle OMP=\angle OKP=\angle OTP$ $(O, M, K, P, T)$is concyclic. I solved this problem in Korea Winter Program! Sorry for my poor English and poor Latex...

Inversion with respect to the circle centered at A with radius $\sqrt{bc}$ implies the fact that $BKCT$ is a harmonic quadrilateral thus $TT,KK,BC$ are concurrent and let's call the point of concurrency X. then $\angle XMO=\angle XTO=\angle XKO=\frac{\pi}{2}$ so $XKTMO$ are concyclic.

Here's another solution: From the Lemma given here, we get that $BKCT$ is harmonic. Let $P$ be the midpoint of arc $\widehat{BAC}$, and $Q$ be the antipode of $P$ in $\omega$. Suppose $KQ$ meets $BC$ at $X$. Then we have $$\frac{XB}{XC}=\frac{KB}{KC}=\frac{TB}{TC} \Rightarrow X \text{ lies on } TP \text{ also.}$$Now, as $\angle PTQ=\angle PKQ=\angle XMP=90^{\circ}$, so $\triangle KMT$ is infact the orthic triangle of $\triangle XPQ$. As $O$ is the midpoint of $PQ$, so we get that $K,M,O,T$ lie on the nine-point circle of $\triangle XPQ$. Hence, done. $\blacksquare$ REMARK: This is true in general for any two points $T$ and $K$ such that $BKCT$ is harmonic as shown in this solution.

Here is my solution for this problem Solution Let $E$, $F$ be tangency points of $A$ - Mixtilinear ($S$) with $AC$, $AB$; $N$, $P$ be second intersections of $KE$, $KF$ with ($O$) It's easy to see that: $P$ is midpoint of $\stackrel\frown{AB}$, $N$ is midpoint of $\stackrel\frown{AC}$ Hence: $\dfrac{KB}{KC}$ = $\dfrac{KB}{KA}$ . $\dfrac{KA}{KC}$ = $\dfrac{FB}{FA}$ . $\dfrac{EA}{EC}$ = $\dfrac{FB}{EC}$ On the other side, the spiral similar: $S_T^{k ; \alpha}$: $E$ $\mapsto$ $F$, $B$ $\mapsto$ $C$, $FB$ $\rightarrow$ $EC$ (With $k$ = $\dfrac{TC}{TB}$, $\alpha$ = ($\overrightarrow{TB}$ ; $\overrightarrow{TC}$)) So: $\dfrac{TE}{TF}$ = $\dfrac{TB}{TC}$ = $\dfrac{FB}{EC}$ = $\dfrac{KB}{KC}$ It means that: $TBKC$ is harmonic quarilateral or tangents at $T$, $K$ intersect at a point on $BC$. Let $G$ be concurrent point of these 3 lines Then: $\widehat{OTG}$ = $\widehat{OKG}$ = $90^o$ But: $M$ is midpoint of $BC$ so: $\widehat{BMO}$ = $90^o$ Hence: $T$, $O$, $M$, $K$, $G$ lie on a circle with diameter $OG$ or $T$, $O$, $M$, $K$ lie on a circle

Attachments:

Solution We invert twice in this solution: first at $A$, then at $B$ with arbitrary radius Let the mixtilinear incircle touch $AB$ and $AC$ at $E$ and $F$ respectively. Before inverting, notice that $T$,$E$,$F$,$A$ lie on a circle. Now we are ready to invert at $A$ with arbitrary radius. The inverted problem reads as follows: Problem after 1st inversion wrote: Let $ABC$ be an acute-angled triangle. The $A$-excircle touches $BC$, $AB$ and $AC$ at $D$, $E$ and $F$ respectively.$EF \cap BC=T$. Let the $A$-symmedian intersect the circumcircle of $ABC$ at $M$.Prove that the reflection of $A$ across $BC$(point $O$) lies on the circle passing through $T$,$M$ and $D$. We make a claim (relevant to the above problem) before inverting again. Claim 1: $\frac{CT}{BT}=\frac{CD}{BD}$ Proof: $\frac{BD}{BT}=\frac{BE}{BT}=\frac{sin \angle BTE}{sin \angle BET}=\frac{sin \angle CTF}{sin \angle CFT}=\frac{CF}{CT}=\frac{CD}{CT}$ We invert at $B$ with arbitrary radius

Claim 2

Claim 3

Claim 4

Claim 5

Main proof(assuming you had the patience to read all claims) From claim 4, $T'D'$ is parallel to $O'M'$ Also, $T'O'=T'A'=D'M'$ by reflections and claim 5 Therefore,$T'D'M'O'$ is an isosceles trapezium and hence cyclic. Inverting back at $B$ and $A$, we get the required conclusion.

Remarks

Attachments:

Let $I$ be the incenter WRT $\Delta ABC$. Let $F,E$ $\in$ $AB, AC$ such $FE \perp AI$ at $I$. Let $M$ be the midpoint of $\overline{BC}$. Let $K$ be the $A-$mixtilinear incircle touch point. Let $D,G$ be the midpoint of arc $BC$ and arc $BAC$. Also, $FE$ is tangent to $\odot (BIC)$. Let $EF$ $\cap$ $BC$ $=$ $L$. By La Hire's Theorem, $KI$ is polar of $L$ WRT $\odot (BIC)$ $\implies$ $L$ $\in$ $KD$. Let $LG$ $\cap$ $\odot (ABC)$ $=$ $T'$ and Let $T'D$ $\cap$ $BC$ $=$ $N$. Since, $N$ is orthocenter WRT $\Delta LDG$ $\implies$ $N$ $\in$ $KG$. Apply Pascal on $DAT'GKK$ $\implies$ $KK$ $\cap$ $AT$ $\in$ $EF$ $\implies$ By Radical Axes Theorem, $AT'FE$ is cyclic $\implies$ $T' \equiv T$. Now Pascal on $TTGKKD$ $\implies$ $TBKC$ is harmonic $\implies$ Reflection of $K$ over $GD$ lies on $TM$ $\implies$ $KTOM$ is cyclic

Excellent Problem. 2018 Korea Winter Program Practise Test #5 wrote: Let $\Delta ABC$ be a triangle with circumcenter $O$ and circumcircle $w$. Let $S$ be the center of the circle which is tangent with $AB$, $AC$, and $w$ (in the inside), and let the circle meet $w$ at point $K$. Let the circle with diameter $AS$ meet $w$ at $T$. If $M$ is the midpoint of $BC$, show that $K,T,M,O$ are concyclic. Solution:- If $X,Y$ are the points of tangency points made by the $A-\text{Mixtillinear Incircle}$ with $AB,AC$ be $X,Y$, then $\odot(AS)=\odot(AXY)$. Let $\gamma$ be the $A-\text{Mixtillinear Incircle}$ Lemma :- $TBKC$ is a harmonic quadrilateral. By Radical axis Theorem on $\odot(ABC),\odot(AXY),\gamma$ we get $AT,XY,KK$ are concurrent at $N$. Let $AK\cap\gamma=J$. So, $(J,K;X,Y)=-1$ and as $NK$ is tangent $\gamma$. Hence, $NJ$ is also tangent to $\gamma$. So, $$-1=(K,J;X,Y)\overset{K}{=}(N,NY\cap JK;X,Y)\overset{A}{=}(T,K;B,C)\implies TBKC\text{ is a harmonic quadrilateral}$$ Main Proof:- As $TBKC$ is a harmonic quadrilateral, hence, $BB,CC,TK$ concurs at a point $Z$. So, $ZO\cap BC=M$. Hence, $$ZK.ZT=ZC^2=ZM.ZO\implies T,O,M,K\text{ are concyclic.}\blacksquare$$. imn8128 wrote: Inversion with respect to the circle centered at A with radius $\sqrt{bc}$ implies the fact that $BKCT$ is a harmonic quadrilateral. May I know how to get that $TBKC$ is a harmonic quadrilateral using $\sqrt{bc}$ Inversion? Thanks.

I found it just a nine-point circle

let $E,F$ be points on $AC,AB$ such that $IE \perp AI $ and $IF \perp AI $ and let $L,G$ be the midpoints of minor ,major arc $BC$ claim: $TG,BC,LK,EF$ are concurrent proof: clearly $EF$ is tangent to $(BIC),(IKL)$ so by radical axis on $(IBC),(IKL),(ABC)$ we have that $BC \cap EF \cap LK =J$ let $EF \cap GC= Q$ easily get $QFBC$ is cyclic combining with the fact $(TFBJ)$ is cyclic by inversion with center $J$ with radius $IJ$ we have that $T^*=G$ $\blacksquare$ now $LT \perp AJ$ $GK \perp JL $ $JM \perp GL $and $O$ is the midpoint of $GL$ so $(KTOM)$ is the nine point circle of $\triangle GJL$

To solve this problem, we have a powerful lemma for mixtilinear circles: $\rule{18cm}{0.3mm}$ $\text{Lemma 1.}$ Given a triangle $ABC$ inscribed in $\omega$ with $H$ be the $A-$mixtilinear center. Let $S$ be the touch point of the $A-$mixtilinear center and $\omega$. The circle $(AH)$ intersects $\omega$ at $G\ne A.$

Prove that: $GBSC$ is a harmonic quadrilateral.Proof

$\rule{18cm}{0.3mm}$ Let the tangent of $(O)$ drawn from $K$ intersects $BC$ at $Q.$ By $\text{Lemma 1},$ we have $TBKC$ is a harmonic quadrilateral $\Rightarrow QT$ is the tangent of $(O)$ $\Rightarrow Q,T,O,M,K$ all lie on $(OQ) \Rightarrow K,T,M,O$ are concyclic.

lol hsolved at school Let $M_a$ and $M_{BC}$ be the midpoints of arc $BC$. Firstly, it is well known that $TBKC$ is harmonic, so $BB,M_AM_{BC},CC,KT$ concur so we are done by PoP.

Denote by $\Omega$ the $A$-mixtilinear incircle. Inverting about $\omega$, it suffices to show that $K,T,BB \cap CC$ are collinear, which is equivalent to showing that $(K,T;B,C)=-1$. Let $D$ and $E$ be the tangency points of $\Omega$ with $AB$ and $AC$ respectively, so $D$ and $E$ lie on $(AS)$. Next we perform $\sqrt{bc}$ inversion. It is well-known that $\Omega'$ is the $A$-excircle. $D'$ and $E'$ are the tangency points of the $A$-excircle with $AC$ and $AB$ respectively, and $K'$ is the $A$-extouch point. $(AS)'$ is the line $D'E'$, so $T'$ is the intersection between $D'E'$ and $BC$. It thus remains to show that $(K',D'E' \cap BC;B,C)=-1$, or that cevians $AK', BD', CE'$ concur. Let $a,b,c$ denote the lengths of segments $BC, AC, AB$ respectively, and let $s=\frac{a+b+c}{2}$. Using directed lengths, it is well-known that $$\frac{AE'}{E'B} \cdot \frac{BK'}{K'C} \cdot \frac{CD'}{D'A} = \frac{s}{c-s} \cdot \frac{s-c}{s-b} \cdot \frac{s-b}{-s} = 1.$$By Ceva's theorem, $AK', BD', CE'$ concur, so we are done.

oops Perform a $\sqrt{bc}$ inversion followed by a reflection about $\overline{BC}$ - then it suffices to show that if $H_a$ is the $A$-HM point, $K'$, $E$, and $F$ are the extouch points on $\overline{BC}$, $\overline{CA}$, $\overline{AB}$, and $\overline{EF} \cap \overline{BC} = T'$, then $AH_aK'T'$ is cyclic. Inverting back, it suffices to show that if $X = \overline{BB} \cap \overline{CC}$, then $\overline{TKX}$ collinear - but this is equivalent to $(T, K; B, C) = -1$, and after $\sqrt{bc}$ inverting again, this is equivalent to $(T', K'; B, C) = -1$ which is clear by extraversion.

It suffices to prove that $BKCT$ is harmonic. Observe that a $\sqrt{bc}$ inversion maps the $A$-mixtilinear incircle to the $A$-excircle and therefore maps $K$ to the point where $BC$ is tangent to the $A$-excircle and $(AS)$ to the polar of $A$ wrt the $A$-excircle, so it maps $T$ to the point where said polar intersects $BC$. Hence $\{B,C;K,T\}=\{B',C';K',T'\}=-1$, as wanted.

Invert with respect to $(O)$. $K$ and $T$ are obviously preserved and $M$ is sent to the intersection of tangents at $B$ and $C$ to $(O)$ which we call $X$. Now it suffices to prove that $\overline{X-T-K}$ are collinear but it is well known that $(KT;BC)=-1$ so $KT$ passes through pole of $BC$ which is obviously $X$.