Let $D'$ be the the touch point of $A$-excircle and let $Q$ be the isogonal conjugate of harmonic conjugate of $D'$ wrt $BC$.Note that $E,F$ are intouch points of $A$-mixtlinear incircle(well-known) so by $\sqrt{bc}$-inversion($E,F$ got carried to the touch points of the A-excircle) $A,X,Q$ are collinear $\implies$ $X(B,C;Q,D)=(B,C;A,XD\cap \odot ABC,A)$ but $D,D'$ are are isotomic conjugates by Steiner's(and $BD=CD'$) $(B,C;Q,D)=\tfrac{AB^2}{AC^2}=(B,C;A,M)$ and hence $XD\cap \odot ABC\equiv M$.

This problem was proposed by me. First Solution (Joshua Lee): Letting $D_1$ be the extouch point, we observe that \[ \measuredangle\left( \overline{XM}, \overline{AD_1} \right) = \measuredangle XMI = \measuredangle XBF = \measuredangle XBA \]since $\overline{IM} \parallel \overline{AD_1}$, and $\triangle XBF \sim \triangle XMI$ by spiral similarity. Thus $\overline{XM}$ and $\overline{AD_1}$ meet on $\Gamma$. Now by the Butterfly Theorem, $\overline{XD}$ and $\overline{AM}$ meet on $\Gamma$ as well. Second Solution (mine): Let $\omega_A$ denote the $A$-mixtilinear incircle, tangent to $\overline{AB}$ and $\overline{AC}$ at $F$ and $E$. Suppose $\omega_A$ is tangent to $\Gamma$ at $T$. Again let line $AM$ meet $\Gamma$ again at $K$. Let $D_1$ and $T_1$ be the reflections of $D$ and $T$ with respect to the perpendicular bisector of $\overline{BC}$. [asy][asy] size(10cm); pair A = dir(115); pair B = dir(210); pair C = dir(330); draw(A--B--C--cycle, red); pair I = incenter(A, B, C); pair O = circumcenter(A, B, C); pair M_A = -A+2*foot(O, A, I); pair M_B = -B+2*foot(O, B, I); pair M_C = -C+2*foot(O, C, I); draw(unitcircle, blue); pair L = dir(90); pair T = -L+2*foot(O, I, L); pair E = extension(T, M_B, A, C); pair F = extension(T, M_C, A, B); pair D = foot(I, B, C); draw(incircle(A, B, C), red); pair M = midpoint(B--C); pair K = -A+2*foot(O, A, M); draw(circumcircle(A, E, F), blue); pair X = -K+2*foot(O, K, D); draw(circumcircle(E, F, T), heavygreen); pair D_1 = 2*M-D; pair T_1 = -A+2*foot(O, A, D_1); draw(A--T, red); draw(A--T_1, red); pair S = -T+2*foot(circumcenter(T, E, F), A, T); pair R = extension(A, X, F, E); draw(R--A, purple); draw(R--S, purple); draw(R--E, purple); draw(R--T, purple); draw(X--T_1, grey+dashed); draw(A--K--X, orange+dotted); draw(B--X--C, lightcyan); draw(I--M, grey+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(-100)); dot("$T$", T, dir(T)); dot("$E$", E, dir(E)); dot("$F$", F, dir(230)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M)); dot("$K$", K, dir(K)); dot("$X$", X, dir(X)); dot("$D_1$", D_1, dir(45)); dot("$T_1$", T_1, dir(T_1)); dot("$S$", S, dir(130)); dot("$R$", R, dir(R)); /* TSQ Source: !size(10cm); A = dir 115 B = dir 210 C = dir 330 A--B--C--cycle 0.1 lightred / red I = incenter A B C R-100 O := circumcenter A B C M_A := -A+2*foot O A I M_B := -B+2*foot O B I M_C := -C+2*foot O C I unitcircle 0.1 lightcyan / blue L := dir 90 T = -L+2*foot O I L E = extension T M_B A C F = extension T M_C A B R230 D = foot I B C incircle A B C 0.1 lightred / red M = midpoint B--C K = -A+2*foot O A M circumcircle A E F 0.1 lightcyan / blue X = -K+2*foot O K D circumcircle E F T 0.2 lightgreen / heavygreen D_1 = 2*M-D R45 T_1 = -A+2*foot O A D_1 A--T red A--T_1 red S = -T+2*foot circumcenter T E F A T R130 R = extension A X F E R--A purple R--S purple R--E purple R--T purple X--T_1 grey dashed A--K--X orange dotted B--X--C lightcyan I--M grey dashed */ [/asy][/asy] It is well known that $\angle BAT = \angle CAD_1$. (To prove this, notice that an inversion at $A$ with $\sqrt{AB \cdot AC}$ plus a reflection around the angle bisector swaps the $A$-excircle and $A$-mixtilinear incircle, so it also swaps $T$ and $D_1$.) Thus $A$, $D_1$, $T_1$ are collinear. Now we will prove that $X$, $M$, and $T_1$ are collinear. Let $R$ be the radical center of $\omega_A$, $\Gamma$, and the circumcircle of triangle $AEF$. Moreover, let $\overline{AT}$ meet the $A$-mixtilinear circle again at $S$. We see that $SFTE$ is harmonic quadrilateral, and therefore that $(R, \overline{AT} \cap \overline{EF}; F, E) = -1$ is harmonic. Upon projecting onto $\Gamma$ from $A$ we see that $XBTC$ is a harmonic quadrilateral. Since $\overline{TT_1} \parallel \overline{BC}$, using perspectivity at $T_1$ we conclude that $X$, $M$, $T_1$ are collinear. Now, applying the Butterfly Theorem, we see that $X$, $D$, $K$ are collinear as well, as required.

Thank you @v_Enhance for your document A Guessing Game: Mixtilinear Incircles. When I tried this problem at home, I managed to do it in a very short time (:

Let $T$ be the $A$-mixtilinear touchpoint on $\odot(ABC)$; note by $\sqrt{bc}$ inversion that $XBTC$ is harmonic, so projecting through $D$ we want to show that $TD$ passes through the point $A'$ on $\odot(ABC)$ with $AA'\parallel BC$, which is well-known.

As per my personal hobby it's time to trigo-bash the whole thing again WLOG let $AB<AC$. First, well-known spiral similarity property should dictate the similarity of triangles $BXF$ an $CXE$, so $\frac{CX}{CE}=\frac{BX}{BF}$. Also, let's also invoke an identity for triangles (feel free to verify it; I'm not gonna do this): $$\frac{BX}{XC}\cdot \frac{\sin\angle BXD}{\sin\angle CXD}=\frac{BD}{DC}.$$Denoting $N_1$ as the other intersection of $XD$ and $\Gamma$ gives $\frac{\sin\angle BXD}{\sin\angle CXD}=\frac{BN_1}{CN_1}.$ Similarly we have $\frac{AB}{AC}\cdot \frac{\sin\angle ABM}{\sin\angle ACM}=\frac{BM}{CM}=1$. ALso let $N_2$ as the other intersection of $AM$ and $\Gamma$ and we have $\frac{\sin\angle ABM}{\sin\angle ACM}=\frac{BN_2}{CN_2}.$ Therefore all we need is $\frac{\sin\angle ABM}{\sin\angle ACM}=\frac{\sin\angle BXD}{\sin\angle CXD}$, and it's not hard to see that $\frac{\sin\angle ABM}{\sin\angle ACM}=\frac{AC}{AB}$, so we are left with proving the fact $\frac{BF}{EC}\cdot\frac{AC}{AB}=\frac{BD}{DC}$. Now, $\frac{BD}{DC}=\frac{\tan\frac12\angle C}{\tan\frac12\angle B}$, $\frac{AC}{AB}=\frac{\sin\angle B}{\sin\angle C}=\frac{2\sin\frac 12\angle B\cos\frac 12\angle B}{2\sin\frac 12\angle C\cos\frac 12\angle C}$. Also $IE=IF$, and by angle chasing we have $\angle FIB=\angle ICE=\frac12\angle C$, $\angle EIC=\angle IBF=\frac12\angle B$. Therefore $BIF$ and $ICE$ similar, yielding $\frac{BF}{EC}=(\frac{BF}{FI})^2=(\frac{\frac12\angle C}{\frac12\angle B})^2$, now it's no longer difficult to prove that $(\frac{\frac12\angle C}{\frac12\angle B})^2\cdot \frac{2\sin\frac 12\angle B\cos\frac 12\angle B}{2\sin\frac 12\angle C\cos\frac 12\angle C}=\frac{\tan\frac12\angle C}{\tan\frac12\angle B}$.

MathStudent2002 wrote: Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$. The points $D$, $E$, $F$ are selected on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$, $\overline{IE}\perp \overline{AI}$, and $\overline{IF}\perp \overline{AI}$. Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $XD$ and $AM$ meet on $\Gamma$. Proposed by Evan Chen, Taiwan Consider the following Claim: $\frac{BX}{CX}=\left(\frac{BA}{CA}\right) \cdot \left(\frac{BD}{CD}\right)$ (Proof) Note that $X$ is the spiral center for $BE \mapsto CF$ so $$\frac{BX}{CX}=\frac{BE}{CF}=\frac{\left(\tfrac{BE}{IE}\right)}{\left(\tfrac{CF}{IF}\right)}=\left(\frac{\sin \tfrac{C}{2}}{\sin \tfrac{B}{2}}\right)^2,$$by applying sine law to $\triangle BEI$ and $\triangle CFI$. Next, we see that $$\left(\frac{BA}{CA}\right) \cdot \left(\frac{BD}{CD}\right)=\left(\frac{\sin C}{\sin B}\right) \cdot \left(\frac{BI\cos \tfrac{B}{2}}{CI\cos \tfrac{C}{2}}\right).$$Apply sine law to $\triangle IBC$ and use the double-angle formula for sines to conclude that both sides are equal. $\blacksquare$ Let $AM$ meet $\Gamma$ again at point $T$ and let $TD$ meet $\Gamma$ again at $Y$. Project line $BC$ onto circle $\Gamma$ from $T$ to get $$(BC, XA)=(BC, DM) \overset{T}{=} (BC, YA),$$so $X=Y$ just as we desired. $\blacksquare$

Here is a fun solution I found: We make use of the following Pingpong Lemma: Let $P_1,P_2,P_3,P_4$ be points on secant $XY$ of circle $\omega$. Say a point $P$ on $\omega$ is good if $P_1(P_2(P_3(P_4P\cap\omega)\cap\omega)\cap\omega)\cap\omega=P$. If a point $P$ different from $X$ and $Y$ on $\omega$ is good, then all points on $\omega$ are good. Let $Y=EF\cap BC$, $N=AI\cap\Gamma$, $Z=AM\cap\Gamma$, $A'\in\Gamma$ satsify $AA'\parallel BC$, $K$ be the foot of the $A$-external-angle-bisector, $L=AK\cap\Gamma$, $T=NY\cap\Gamma$, and $P=A'M\cap\Gamma$. First, note that $X$ is the center of spiral similarity taking $LM$ to $AI$, $X\in(IMN)=(NY)$, so $Y\in ML$. We apply the Pingpong Lemma with $M,K,Y,D$ on secant $BC$ of circle $\omega$. Note that $ABPC$ is harmonic, so $P\in NK$, and that $T$ is the $A$-mixtilinear-touchpoint, so $D\in A'T$. Thus, $A'P,PN,NT,TA'\cap\Gamma=M,K,Y,D$, so $A'$ is good. Hence, $Z$ is also good. As $ZM\cap\Gamma=A,AK\cap\Gamma=L,LY\cap\Gamma=X$, we have that $XD\cap\Gamma=Z$, as desired.

Cute problem. My solution. The parallel through A cuts the circumcircle again at A'. AM meets the circumcircle again at Z. Then By reflection in the perpendicular bisector of BC, noting that the A median and the A symmedian are isogonal, A'B'C' is harmonic. By a $\sqrt{bc}$ inversion at $A$, noting that AT is isogonal to the A Nagel cevian and that X goes to the harmonic conjugate of the excircle touchpoint with BC, we have that XTBC is harmonic. Inversion at the incircle touchpoint swapping B and C preserves cross ratios of point s on the circumcircle, and thus we are done, by noting that the line TA' meets BC at the incircle touchpoint, from a well known property that if D is the incircle touchpoint, then $\angle DTB = \angle ABC$ and angle chasing.

Can't believe that no one has posted this solution yet.

Another solution for this problem.

Let $N$ be the midpoint of arc $BAC$, $N'$ be the midpoint of another arc $BC$, $D_1$, $A_1$ be the reflections of points $D$, $A$ across the line $MN$. By easy angle-chasing problems condition is reduced to $\angle DXM=\angle IMA$. It's well-known that $MI \parallel D_1A$, so $\angle IMA=\angle MAD_1 = \angle DA_1M$, so it's enough to proof that $XDMA_1$ is cyclic. Let $L= BC \cap EF$. Consider spiral similarity which takes $BC$ to $EF$. It's well-known that $X$ is it's center. It also maps $N$ into $A$ and $M$ into $I$ so $X,I,M,L$ are concylic. Let it map $A_1$ into $A'$. By spiral similarity $\angle AXN=\angle NXA_1=\angle AXA'$, so $X$,$A'$,$N$ are collinear. Let $R=BE \cap CF$, $R'=AR \cap BC$. By Ceva's theorem $\frac{AF\cdot BR'\cdot CE}{FB\cdot R'C\cdot EA}=1$, so $\frac{XB}{XC}=\frac{BF}{CE}=\frac{BR'}{R'C}$, so $XR'$ is a bisector of $\angle BXC$,so it's passes through $N'$. Projecting harmonic quadrilateral $BNCN'$ from $X$ we get that $L\in XN$. From the above $\angle A_1DM=\angle AD_1M=\angle IML= \angle IXA'=\angle MXA_1$, so $A_1,D,M,X$ are concylic.

This is a great example of circular reasoning: XD

ABCDE wrote: Here is a fun solution I found: We make use of the following Pingpong Lemma: Let $P_1,P_2,P_3,P_4$ be points on secant $XY$ of circle $\omega$. Say a point $P$ on $\omega$ is good if $P_1(P_2(P_3(P_4P\cap\omega)\cap\omega)\cap\omega)\cap\omega=P$. If a point $P$ different from $X$ and $Y$ on $\omega$ is good, then all points on $\omega$ are good. Let $Y=EF\cap BC$, $N=AI\cap\Gamma$, $Z=AM\cap\Gamma$, $A'\in\Gamma$ satsify $AA'\parallel BC$, $K$ be the foot of the $A$-external-angle-bisector, $L=AK\cap\Gamma$, $T=NY\cap\Gamma$, and $P=A'M\cap\Gamma$. First, note that $X$ is the center of spiral similarity taking $LM$ to $AI$, $X\in(IMN)=(NY)$, so $Y\in ML$. We apply the Pingpong Lemma with $M,K,Y,D$ on secant $BC$ of circle $\omega$. Note that $ABPC$ is harmonic, so $P\in NK$, and that $T$ is the $A$-mixtilinear-touchpoint, so $D\in A'T$. Thus, $A'P,PN,NT,TA'\cap\Gamma=M,K,Y,D$, so $A'$ is good. Hence, $Z$ is also good. As $ZM\cap\Gamma=A,AK\cap\Gamma=L,LY\cap\Gamma=X$, we have that $XD\cap\Gamma=Z$, as desired. Where can I learn more about this Ping Pong Lemma?

is slightly cleaner than a pure bary-bash, albeit a bit wordy. I just don't know the mixtilinear incircle configuration

Just realized that this problem is just equivalent to 2018 Korea Winter Mop practice test #5. Uhh seriously, nobody makes the problem on their own in Korea?

here is a length bash: Let $XD\cap \Gamma=P$ and $PM\cap \Gamma=A'.$ We show that $A'=A.$ Note that $\dfrac{s-b}{s-c}=\dfrac{DB}{DC}\cdot\dfrac{MC}{MB}=(B,C;D,M)\stackrel{P}{=}(B,C;X,A')=\dfrac{XB}{XC}\cdot\dfrac{A'C}{A'B}=\dfrac{FB}{EC}\cdot\dfrac{A'C}{A'B},$ where the last equality follows from spiral similarity at $X$ sending $FB$ to $EC.$ Let $Y,Z$ be the intouch points on $AB,AC.$ Note that $\triangle AIF\sim \triangle AYI\ (\spadesuit)$ and similar on the other side. Furthermore, by trivial pythagoras we know that $AI^2=r^2+(s-a)^2.$ But $\dfrac{FB}{EC}=\dfrac{AB-AF}{AC-AE}\stackrel{\spadesuit}{=}\dfrac{c-\frac{AI^2}{s-a}}{b-\frac{AI^2}{s-a}}=\dfrac{c(s-a)-r^2-(s-a)^2}{b(s-a)-r^2-(s-a)^2}=\dfrac{(s-a)(s-b)-r^2}{(s-a)(s-c)-r^2}$ $=\dfrac{s-b}{s-c}\cdot\dfrac{s(s-a)(s-b)(s-c)-r^2s(s-c)}{s(s-a)(s-b)(s-c)-r^2s(s-b)}=\dfrac{s-b}{s-c}\cdot \dfrac{K^2-Kr(s-c)}{K^2-Kr(s-b)}=\dfrac{s-b}{s-c}\cdot \dfrac{c}{b},$ whence it follows that $\dfrac{b}{c}=\dfrac{A'C}{A'B}$ upon substitution. But the function $f(S)=\dfrac{SC}{SB}$ is monotonic as $S$ varies along the arc $BC$ not containing $P,$ so $f(S)=\dfrac{b}{c}$ has the unique solution $S=A.$ It follows that $A'=A,$ and we are done.

The following lemma trivializes this problem. LEMMA In a $\Delta ABC$ let $P$ be a point on $\odot (ABC)$ and $X\in \odot (ABC)$ such that $XBCP$ is harmonic quadrilateral and $L$ is a fixed point on $\odot (ABC)$ and $T=PL\cap BC$ then as $P$ varies $XT$ passes through fixed point. PROOF Consider negative inversion at $T$ swapping $B,C$ so we get $XT$ meets $\odot (ABC)$ at a point $G$ such that $LGBC$ is harmonic since $L,B,C$ are fixes so $G$ is also fixed. _______________________________________________ Now let line through $A$ parallel to $BC$ meet $\odot (ABC)$ at $A_1$ ,and $T$ be mixtilinear touch point since $XBCT$ is harmonic by lemma we have $XD$ meets circumcircle at a point $G$ such that $A_1BCG$ is harmonic now by symmetry in perpendicular bisector of $BC$ we have $AG$ is the $A-$ median which is what we want $\blacksquare$

good solution

i dont think this solution has been posted before? tough for a g2, could maybe fit as a g3 or easy g4 [asy][asy] size(10cm); pair A = dir(125); pair B = dir(215); pair C = dir(325); pair I = incenter(A,B,C); pair D = foot(I,B,C); pair E = extension(I, I + (A-I)*dir(90), A, C); pair F = extension(I, I + (A-I)*dir(90), A, B); pair X = intersectionpoints(circumcircle(A,B,C), circumcircle(A,E,F))[1]; pair M = (B+C)/2; pair K = extension(A,M,X,D); pair H = 2M-D; pair Z = -A + 2*foot(circumcenter(A,B,C), A, H); pair G = dir(90); pair T = -G + 2*foot(circumcenter(A,B,C),G,I); pair Y = extension(B,C,A,T); draw(A--B--C--cycle, red+1.3); draw(E--F, red); draw(circumcircle(A,E,F), blue); draw(circumcircle(A,B,C), blue); draw(X--K, dotted+lightblue+1.3); draw(A--K, dotted+lightblue+1.3); draw(A--T, orange); draw(A--Z, orange); draw(T--Z, orange); draw(circumcircle(T,E,F), dotted+orange); draw(incircle(A,B,C), dashed+lightred); draw(circumcircle(A,X,Y), heavygreen+0.7); draw(X--Z, dashed+cyan); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(-100)); dot("$T$", T, dir(T)); dot("$E$", E, dir(E)); dot("$F$", F, dir(230)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M)); dot("$X$", X, dir(X)); dot("$K$", K, dir(K)); dot("$T'$", H, dir(45)); dot("$Z$", Z, dir(Z)); dot("$Y$", Y, dir(Y)); [/asy][/asy] Let $T$ be the $A$-mixtillinear incircle touchpoint and let $Y = \overline{AT} \cap \overline{BC}$. The main part of the problem is the following: Claim: $AXYM$ is cyclic. Proof. Perform a $\sqrt{bc}$ inversion about $A$ followed by a reflection about the $A$-angle bisector. Now our inverted claim is the following: Inverted Claim wrote: Let $ABC$ be a triangle, and suppose the touchpoints of the $A$-excircle with $\overline{BC}$ and the extensions of $\overline{AB}$ and $\overline{AC}$ be $T$, $E$ and $F$ respectively. Let $\overline{EF} \cap \overline{BC} = X$, and let $\overline{AT} \cap (ABC) = Y$. If the intersection of the $A$-symmedian with $(ABC)$ is $M$, show that $M$, $X$, $Y$ are collinear. To solve this problem, let $\overline{XY} \cap (ABC) = M'$ - we wish to show that $M \equiv M'$. Let $\omega_A$ denote the $A$-excircle. Note that \[-1 = (E, F; T, \overline{AT} \; \cap \; \omega_A) \overset{T}{=} (E, F; \overline{AT} \; \cap \; \overline{EF}, X) \overset{A}{=} (B, C; T, X) \overset{Y}{=} (B, C; M', A) = (A, M'; B, C); \]since $M$ is the unique point on circle $(ABC)$ with $(A, M; B, C) = -1$, we have $M \equiv M'$, as desired - inverting back gives the claim. $\blacksquare$ Now let $T'$ be the $A$-excircle touchpoint on $\overline{BC}$, and $Z = \overline{AT'} \cap (ABC)$. Claim: $X, M, Z$ are collinear. Proof. Note that $\overline{TZ} \parallel \overline{BC}$, so if we define $Z' = \overline{XM} \cap (ABC)$ and show that $\overline{TZ'} \parallel \overline{BC}$, we will be done. However, this is just angle chasing: \[\measuredangle XMB = \measuredangle XMY = \measuredangle XAY= \measuredangle XAT =\measuredangle XZ'T\]so $\overline{TZ'} \parallel \overline{BC}$ as desired. (Alternatively, Reim's theorem should work if I'm not messing up.) $\blacksquare$ Finally, the Butterfly Theorem tells us $X$, $D$, $\overline{AM} \cap (ABC)$ are collinear, so we are done. $\square$

Let $K$ be the excircle touchpoint on $BC$. $E,F$ are the touchpoints of the $A$-mixtillinear incircle. Hence under $\sqrt{bc}$-inversion, we see that $X'$ is the harmonic conjugate of $K$. Using the inversion distance formula, we have: \[ \frac{XB}{XC} = \frac{\frac{r^2}{AX' \cdot AC}}{\frac{r^2}{AX' \cdot AB}} \cdot \frac{X'C}{X'B} = \frac{AB}{AC} \cdot \frac{CK}{BK} = \frac{AB}{AC} \cdot \frac{BD}{CD}\]But by ratio lemma, \[\frac{XB}{XC} \cdot \frac{\sin \angle BXD}{\sin \angle CXD} = \frac{BD}{CD} \text{ and } \frac{AB}{AC} \cdot \frac{\sin \angle BAM}{\sin \angle CAM} = 1\]therefore $\frac{\sin\angle BXD}{\sin \angle CXD} = \frac{\sin \angle BAM}{\sin \angle CAM} $, combined with $\angle BXD + \angle CXD = \angle BAM + \angle CAM$ this implies that $\angle BXD = \angle BAM$ and $\angle CXD = \angle CAM$, which means $XD$ and $AM$ concur on $(ABC)$.

Worse ping pong app. Let $S$ be the Sharkey-Devil point, and let $T = AS \cap BC \cap EF$. Let $N$ be the midpoint of arc $\widehat{BC}$, let $L$ be the midpoint of $\widehat{ABC}$. Claim: $T$ lies on $AX'$. Proof. Well known, but note that $X$ is the center of the spiral sim from $FI$ to $BM$. As such, $TXIM$ is cyclic which gives that $\angle AXM_A$ is a right angle. $\blacksquare$ We now ping pong on $D, M, T, SX \cap BC$ with $S \to M_A \to N \to X \to S$, so it follows that $X \to XD \cap \Gamma \to A \to S \to X$ is a valid cycle.

Let $T$ be the $A$-mixtilinear intouch point, $U$ be the $A$-extouch point. First we claim that $(B,C;X,T)=-1$. Proof: Taking $\sqrt{bc}$ inversion, the $A$-mixtilinear incircle is taken to the $A$-excircle. Since $E$ and $F$ lie on the $A$-mixtilinear incircle, $E', F',T'$ are the tangency points between the $A$-excircle and lines $AB$, $AC$, $BC$ respectively, so $T'=U$. Thus, $X'$ is the intersection between $BC$ and $E'F'$. It suffices to show that $(B,C;X',U)=-1$, or that lines $AU, BF', CE'$ are concurrent, which is well-known. Let $\perp_{BC}$ denote the perpendicular bisector of $BC$, $A'$ be the reflection of $A$ across $\perp_{BC}$. We claim that $T,D,A'$ are collinear. Proof: Let $V$ be the second intersection between line $AU$ and $\Gamma$. Since $\angle BAT = \angle CAU$, $V$ is the reflection of $T$ across $\perp_{BC}$. It is well-known that $DM=UM$, so reflecting the line $AUV$ gives the line $TDA'$. Now we can prove the desired result. Let $P$ be the second intersection between line $XD$ and $\Gamma$, $\infty_{BC}$ denote the point at infinity along $BC$. Note that $$-1=(C,B;X,T) \stackrel{D}{=} (B,C;P,A') \stackrel{A}{=} (B,C;AP \cap BC,\infty_{BC})$$from which is follows that $A,M,P$ are collinear, so we are done.

i am a geo(gebra) main :skull: Reflect $D$ across $M$ to get $D',$ and let $X'$ be such that $BXX'C$ is an isosceles trapezoid. Let $AM$ intersect $\Gamma$ at point $G,$ and let $G'$ be such that $BG'GC$ is an isosceles trapezoid. We desire to show that $X', D', G'$ are collinear. It is well-known that $(B,C;A,G') = -1,$ so if we let $Z = G'D' \cap \Gamma \ne G',$ then $$(B,C;A,G') \stackrel{D'}{=} (C,B;AD' \cap \Gamma, Z) = -1.$$Hence we would like to show that $(C,B; AD' \cap \Gamma, X') = -1.$ Now, let $T$ be the $A$-mixtilinear touchpoint (that is, $(EFT)$ is tangent to $\Gamma$). It is well-known (say by force-overlaid inversion) that $AT$ and $AD$ are isogonal in $\angle BAC,$ so reflecting our cross ratio across the perpendicular bisector of $BC,$ our problem has reduced to showing that $(B,C;X,T) = -1.$ By the Radical Center Theorem on $\Gamma, (AEF),$ and $(TEF) = \omega,$ we see that $AX$ and $EF$ meet on the tangent to $\Gamma$ at $T,$ which is thereby the tangent to $\omega$ at $T.$ Thus projecting from $A$ onto $EF$ gives \[ (B,C;X,T) \stackrel{A}{=} (F,E;AX \cap EF, AT \cap EF) = (F,E; TT \cap EF, AT \cap EF) \stackrel{T}{=} (F,E;T,AT \cap \omega). \]However, since $AF$ and $AE$ are both tangent to $\omega,$ this cross ratio equals $-1,$ and we conclude.

MathStudent2002 wrote: Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$. The points $D$, $E$, $F$ are selected on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$, $\overline{IE}\perp \overline{AI}$, and $\overline{IF}\perp \overline{AI}$. Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $XD$ and $AM$ meet on $\Gamma$. Proposed by Evan Chen, Taiwan Let $S$ be the Sharky-Devil point then we have that $AS,GN,BC,TL$ are concur. Also $A',D,T$ are collinear Now it is enought to prove that $(G,T/B,C)=-1$ $(G,T/B,C)=(J,Q/B,C)=(S,R/B,C)=-1$ The last one is true from here: USA TSTST 2020 P2

OTIS Walkthrough Writeup: Note that by EGMO Lemma 4.40 in $A$-mixtilinear incircle touches $\overline{AC}$, and $\overline{AB}$, at $E$, and $F$ respectively. Let $T$ denote the mixtilinear intouch point, and let $D_1$ denote the reflection of $D$ about $M$ then $\overline{AT}$, and $\overline{AD_1}$ are isogonal. by EGMO 2013/5. Let ray $AD_{1}$ meet $\Gamma$ again at $T_1$. Note that $\overline{TT}$, $\overline{AX}$, $\overline{EF}$ concur at the radical center of $(ABC)$, $(AEF)$, and $(TEF)$, call it $R$. Let $\overline{AT}$ meet the $A$-Mixtilinear incircle again at $S$. Note that $-1=(ST; FE) \stackrel{T}= (R, \overline{AT} \cap \overline{EF}; EF) \stackrel{A}= (XT ; BC)$. So quadrilateral $XTBC$ is harmonic. Notice that since $-1 \stackrel{T_1}= (BC; \overline{TT_1} \cap \overline{BC}, \overline{T_1X} \cap \overline{BC})= (BC; \infty \overline{T_1X} \cap \overline{BC})$ Then $\overline{T_1X} \cap \overline{BC} = M$. Then by butterfly theorem we can conclude.

Suppose I don't want to draw the mixtilinear incircle, how can I solve this? Construct $D$ such that $ADBC$ is an isosceles trapezoid. We know that $E$ and $F$ are the tangency of the mixtilinear incircle, so if $X'$ is the image of $X$ under $\sqrt{bc}$-inversion, then $X'$ is $BC$ intersect the polar of $A$-excircle wrt to $A$. However, if we consider the polar of incircle wrt to $A$ and intersect with $BC$, call this intersection $T$, then $T$ and $X'$ are symmetric wrt to $M$, because their harmonic conjugates wrt to $B$ and $C$ are the intouch point and extouch point to $BC$ respectively, which is known to be symmetric wrt to $M$. So if we consider $X''$ where $XX''BC$ is an isosceles trapezoid, then $AX$ and $AX''$ are isogonal, so $A$, $X''$, $X'$ are colinear. Reflecting across the perpendicular bisector of $BC$, then $D$, $X$, $T$ are colinear. But $(B,C;D,T)=-1$ and if $AM$ intersect $(ABC)$ at $Y$, then $DBYC$ is harmonic. So $X$, $D$, $Y$ are colinear. QED

Another idea: Denote $f(P)=BP/CP$ for any point $P$. You want $f(X)f(Y)=f(D)$, where $Y$ is $AM$ intersect $(ABC)$. But $f(Y)f(A)=f(M)=1 \Rightarrow f(Y)=\frac{1}{f(A)}$. So you want $f(X)=f(A)f(D)$ But $f(X)=\frac{BF}{CE}=\frac{b-\frac{bc}{s}}{c-\frac{bc}{s}}=\frac{bs-bc}{cs-bc}=\frac{b}{c}\cdot \frac{s-c}{s-b}=f(A)f(D)$ QED

MathStudent2002 wrote: Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$. The points $D$, $E$, $F$ are selected on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$, $\overline{IE}\perp \overline{AI}$, and $\overline{IF}\perp \overline{AI}$. Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $XD$ and $AM$ meet on $\Gamma$. Proposed by Evan Chen, Taiwan

Let $T$ be the $A-$Mixtillinear touch-point so $\odot(TEF)$ is the $A-$Mixtillinear Incircle. Also let $D'$ be $D-$Antipode and finally consider the reflections of $D,T$ over the perpendicular bisector of $BC$ to be $X_A, T'$. $\textbf{Claim I:}$ It is enough to show that $AX_A$ and $XM$ concur. (hint given to me by $\textbf{kotmhn}$.) $\textbf{Proof:}$ Butterfly Theorem. $\textbf{Claim II:}$ $AX, EF$ and tangent at $T$ to $\odot(TEF)$ concur at $J$. $\textbf{Proof:}$ Radax on $\odot(ABC), \odot(TEF)$ and $\odot(AXE)$. $\textbf{Claim III:}$ $\overline{A-D'-X_A-T'}\iff \overline{A'-D-T}$ where $A'$ is the $A-$Antipode. $\textbf{Proof:}$ $\sqrt{bc}$ inversion takes $T\to X_A$ and also because $\angle BAT=\angle CAX_A$. $\textbf{Claim IV:}$ $X$ is the miquel point of $BEFC$. $\textbf{Claim V:}$ $XTBC$ is harmonic $\textbf{Proof:}$ $-1=(TK; EF) \stackrel{T}= (\overline{AT} \cap \overline{EF},J; EF) \stackrel{A}= (XT ; BC)$. $\textbf{Claim VI:}$ $\overline{X-M-T'}$ $\textbf{Proof:}$ $\angle AT'X=\angle ABX=\angle IMX=\angle AT'M$.

The key is to draw in the $A$-mixtilinear touch point $T$, and let $\gamma = (TEF)$ be the $A$-mixtilinear incirle. Also define $K = XD \cap \Gamma$, $L = TD \cap \Gamma$ as the reflection of $A$ over the perpendicular bisector of $BC$ from mixtilinear properties, and $A' = KM \cap \Gamma$ which we would like to show is simply $A$. First notice radical axis theorem on $(AEF)$, $(ABC)$, and $\gamma$ tells us $AX$, $EF$, and the tangent at $T$ concur, say at point $R$. Since $TA$ is a symmedian in $\triangle TEF$, and $TR$ is the tangent to $(TEF)$, we get \begin{align*} -1 &= (R, AT \cap ER; FE) \\ &\overset{A}{=} (XT;BC) \\ &\overset{D}{=} (KL;CB) \\ &\overset{M}{=} (A',LM \cap \Gamma;BC) \\ &\overset{L}{=} (LA' \cap BC, M; BC). \end{align*} Thus $LA'$ is parallel to $BC$, giving the desired $A = A'$. $\blacksquare$

Let $AM \cap \Gamma = M'$ and $T_A$ be the A-mixti touch point. Claim: $T_A,D,M,M'$ are concyclic. Proof: \[ \measuredangle TDM = -\measuredangle TDB = \measuredangle DBT + \measuredangle BTD = \measuredangle DBT + \measuredangle ABC = \measuredangle AM'T \]$\blacksquare$ Now spi sim concludes.

The following solution is atrociously long and unnecessarily complicated. This problem reminded me heavily of 21CHNTST42 which I suppose is because it's the exact same configuration, just with a bunch of unnecessary points deleted. Let $T$ denote the $A-$mixtillinear intouch point , $X_a$ the $A-$extouch point and $L$ and $N$ the major and minor $BC$ arc midpoints in $\Gamma$ respectively. First, note that $X$ is the Miquel point of quadrilateral $BCEF$ by definition. Note that since $AL \perp AI \perp EF$, lines $\overline{AL}$ and $\overline{EF}$ are parallel. Also it is well known that points $T$ , $I$ and $L$ are collinear. We start off with the following lemma. Claim : Lines $\overline{LX}$ , $\overline{EF}$ , $\overline{BC}$ and $\overline{TN}$ concur, at say $Q$. Proof : Let $Q = EF \cap BC$, and let $L' = QX \cap \Gamma$. Then note, \[\measuredangle L'QF = \measuredangle XQF = \measuredangle XBF = \measuredangle XBA = \measuredangle XL'A\]which implies that $AL' \parallel EF$ so $L' \equiv L$ which implies that line $\overline{LX}$ also passes through $Q$. Now note that $XQIMN$ is cyclic with diameter $QN$ due to all the right angles. Let $T'$ be the reflection of $T$ across the perpendicular bisector of segment $BC$. Then since arcs $TN$ and $T'N$ are equal , lines $\overline{AT}$ and $\overline{AT'}$ are isogonal. Thus points $A$ , $X_a$ and $T'$ are collinear. Now letting $Q' = TN \cap BC$ note, \[\measuredangle MQ'N = \measuredangle NLT = \measuredangle NAT = \measuredangle T'AN = \measuredangle MIN\]since it is well known that $AX_a \parallel IM$ by the Midpoint of the Altitude Lemma. But this implies that $Q'IMN$ is cyclic, which indicates that $Q' \equiv Q$ so $\overline{TN}$ also passes through $Q$ proving the claim. Now note that by Pascal's Theorem on hexagon $NNXLLT$, points $Y=NX \cap LT$ , $Q = XL \cap TN$ and $\infty = LL \cap NN$ are collinear. Thus, $Y$ must also lie on line $\overline{BC}$. Further, due to the right angles $MYTN$ is also cyclic. Now let $R = XD \cap \Gamma$. Then, we can observe the following. Claim : Points $T$ , $D$ , $M$ and $R$ are concyclic. Proof : This is a direct angle chase. Note, \[\measuredangle TMD = \measuredangle TMY = \measuredangle TNY = \measuredangle TNX = \measuredangle TRX = \measuredangle TRD\]which implies the claim. Further note, \[\measuredangle XT'A = \measuredangle XNA = \measuredangle XNI = \measuredangle XMI\]which implies that points $X$ , $M$ and $T'$ are collinear. Thus, lines $\overline{AX_a}$ and $\overline{XM}$ concur on $\Gamma$. Now, since $D$ and $X_a$ are well known to be reflections of each other across $M$ , an application of the Butterfly Theorem suffices. But the following finish is also possible (and is the one I found). Let $A'$ and $X'$ denote the reflections of $A$ and $X$ across the perpendicular bisector of segment $BC$. Reflecting the previous observation across the perpendicular bisector of segment $BC$ it follows that $\overline{A'D}$ and $\overline{X'M}$ concur on $\Gamma$ (at $T$). But then, since arcs $AX$ and $AX'$ are equal, \[\measuredangle DRM = \measuredangle DTM = \measuredangle A' TX' = \measuredangle XT'A = \measuredangle XRA = \measuredangle DRA\]which implies that points $A$ , $R$ and $M$ are collinear. Thus, lines $\overline{AM}$ and $\overline{XD}$ intersect at $R$, which lies on $\Gamma$ as desired.

We define some new points $\bullet$ $T$ is the $A$-Mixtilinear intouch point of $\triangle ABC$. $\bullet$ $\triangle D'E'F'$ is the $A$-extouch triangle of $\triangle ABC$. $\bullet$ $Y=\overline{AM} \cap (ABC)$. $\bullet$ $\ell$ is the perpendicular bisector of $\overline{BC}$. $\bullet$ $A'$ is reflection of $A$ across $\ell$. Claim: $(X,T;B,C)=-1$. Proof: By $\sqrt{bc}$ inverting at $A$ in $\triangle ABC$, then $X^*=\overline{BC} \cap \overline{E'F'}$. So we need to prove that $(B,C;D,X^*)=-1$ which is just because $\overline{AD'}$, $\overline{BE'}$, $\overline{CF'}$ concur by Ceva. $\square$ Well known that $T$, $D$, $A'$ are collinear (just reflect across $\ell$). And so we have \[-1=(X,T;B,C) \overset{D}= (\overline{DX} \cap (ABC),A';C,B)\]See that $Y$ is the reflection of the $A$-harmonic conjugate of $\overline{BC}$ across $\ell$ and so we win.