It's a special case of yetti's old problem. There's a short solution using Desargues' involution theorem: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=209898&p=1156398

Interesting. I thought that TS had created this, but it looks like this is in fact from Cosmin. Small world. No one solved this during the TST, but we did find a solution afterwards using inversion at the incircle. One uses the fact that $TI$ (here $I$ is the incenter and $T$ is the mixtilinear tangency point) passes through the midpoint of arc $BAC$ in order to control the image of $T$ under inversion.

Nice, this is one of my favorite problems. Congratulations for the IMO result by the way, Evan!

Thanks for the congratulations! Yes, this is a really beautiful result indeed, one of my favorites as well. I just read through Vladmir's inversive solution and it looks a bit different from the one I mentioned above. So let me post the Taiwan team's solution, in its full glory. [asy][asy] defaultpen(fontsize(8pt)); pointfontsize=8; size(11cm); pointpen = black; pair A = Drawing("A", dir(110), dir(110)); pair B = Drawing("B", dir(210), dir(210)); pair C = Drawing("C", dir(330), dir(330)); draw(A--B--C--cycle, heavyred); pair O = circumcenter(A,B,C); pair I = incenter(A,B,C); pair Ta = foot(I,B,C); pair Tb = foot(I,C,A); pair Tc = foot(I,A,B); draw(unitcircle, heavyred); draw(incircle(A,B,C), blue+1); pair L = Drawing("L", dir(90), dir(90)); pair T = Drawing("T", 2*foot(O,I,L)-L, dir(I-L)); pair M = Drawing("M", dir(-80), dir(-80)); pair D = Drawing("D", Ta, dir(Ta-I)); pair E = Drawing("E", Tb, dir(Tb-I)); pair F = Drawing("F", Tc, dir(Tc-I)); draw(D--E--F--cycle, dashed+blue); pair Ac = Drawing("A^\ast", (E+F)/2, dir(60)); pair Bc = Drawing("B^\ast", (F+D)/2, dir(B-I)); pair Cc = Drawing("C^\ast", (D+E)/2, dir(0)); draw(circumcircle(Ac,Bc,Cc), heavycyan+1); real p = abs(D-I)*abs(D-I); pair Mc = Drawing("M^\ast", I+p/conj(M-I), dir(100)*1.5); pair Lc = Drawing("L^\ast", I+p/conj(L-I), dir( 10)*1.5); pair Tc = Drawing("T^\ast", I+p/conj(T-I), dir(110)*1.5); Drawing("I", I, dir(45)); pair H = Drawing("H", orthocenter(D,E,F), dir(90)); draw(Ac--Bc--Cc--cycle, heavycyan); pair[] K = IPs(circumcircle(D,E,F), L(Mc-(M-I)*dir(90), Mc+(M-I)*dir(90))); pair K1 = Drawing("K_1", K[1], dir(K[1]-K[0])); pair K2 = Drawing("K_2", K[0], dir(K[0]-K[1])); pair X1 = Drawing("X_1", extension(K1,M,B,C), dir(-45)); pair X2 = Drawing("X_2", extension(K2,M,B,C), dir(225)); pair X1c = Drawing("X_1^\ast", (D+K1)/2, dir(100)); pair X2c = Drawing("X_2^\ast", (D+K2)/2, dir(80)); draw(K1--D--K2, darkgreen); draw(K1--M--K2, darkgreen); draw(K1--K2, darkgreen); draw(Ac--Lc); draw(L--T, dashed); draw(I--A--L, dashed); pair Dp = Drawing("D'", 2*Mc-D, dir(115)); draw(D--Dp, dotted); pair Hp = Drawing("H'", 2*Mc-H, dir(-20)); draw(H--Hp, dotted); [/asy][/asy] Let $DEF$ be the intouch triangle. Let the tangents to the incircle be $\overline{MK_1}$ and $\overline{MK_2}$. Moreover, denote by $H$ the orthocenter of triangle $DEF$. After inverting around the incircle, denote the inverse with a star. We know that $(A^\ast B^\ast C^\ast)$ is the nine-point circle of $\triangle DEF$. Using the image of $L$, we see that $T^\ast$ is diametrically opposite $A^\ast$ on this circle, meaning that $T^\ast$ is in fact the midpoint of $\overline{DH}$. Meanwhile, $\overline{K_1K_2}$ is a chord of the incircle whose midpoint $M^\ast$ lies on the nine-point circle, and $X_1^\ast$ and $X_2^\ast$ are the midpoints of $\overline{DK_1}$ and $\overline{DK_2}$. Now it's easy to show that $T^\ast$, $M^\ast$, $X_1^\ast$ and $X_2^\ast$ are concyclic -- take a homothety at $D$ with ratio $2$, then a homothety at $M^\ast$ with ratio $-1$, which sends the four points to $K_1$, $K_2$, $D$, and $H'$. Here $H'$ is the image of $M^\ast$ under a homothety at $H$ with ratio $2$. As $M^\ast$ lies on the nine-point circle of $\triangle DEF$, this implies $H'$ lies on the circumcircle of $\triangle DEF$, as required.

What is the A-mixtilinear incircle? v_Enhance wrote: Let $DEF$ be the intouch triangle. Let the tangents to the incircle be $\overline{MK_1}$ and $\overline{MK_2}$. Moreover, denote by $H$ the orthocenter of triangle $DEF$. After inverting around the incircle, denote the inverse with a star. What is the ratio of this inversion around the incircle?

Dear Mathlinkers, for a proof http://jl.ayme.pagesperso-orange.fr/ vol. 4 A new mixtilinear incircle adventure III, p. 12..... with a complete history and a generalization Sincerely Jean-Louis

MathPanda1 wrote: What is the A-mixtilinear incircle? Circle tangent to $\overline{AB}$, $\overline{AC}$, and internally tangent to the circumcircle of $ABC$. MathPanda1 wrote: What is the ratio of this inversion around the incircle? Whatever the radius of the incircle is.

I have a solution here, and I guess that it's less "magical" than the above inversive solution, as in, just through working backwards, you can actually arrive at a solution. Note that this took about 2.5 hours though, so maybe that's why no one solved this during the TST. Let $T$ be the mixtilinear incircle tangency point. Also, extend the tangents from $M$ to the incircle to hit the circumcircle of $ABC$ at $Y_1, Y_2$, such that $M, X_1, Y_1$ are collinear and $M, X_2, Y_2$ are collinear. By a well-known lemma about the incircle, $Y_1Y_2$ is also tangent to the incircle. Now note that by the spiral-similarity lemma, since $X_1Y_1 \cap X_2Y_2 = M$, if $TX_1X_2M$ are concyclic, then $T$ is the spiral center sending $X_1X_2 \rightarrow Y_1Y_2$. So it would be nice to show that $T$ is indeed the spiral center. As for more motivation (working backwards), I considered the case where $M \equiv T$. In this (degenerate) case, $TX_1X_2$ should be tangent to the circumcircle. If this were true, the the circle $TX_1X_2$ should be homothetic to the circumcircle at $T$. So once again to exploit this homothety, we extend $TX_1, TX_2$ to hit the circumcircle at $Y_1, Y_2$. So if they were homothetic, $Y_1Y_2 || BC$, so $\angle BTY_1 = \angle CTY_2$. By the tangency, $\angle Y_1TI = \angle Y_2TI$. So $\angle BTI = \angle BTY_1 + \angle Y_1TI = \angle CTY_2 + \angle Y_2TI = \angle CTI$. So $T$ should satisfy $\angle BTI = \angle CTI$. Let's prove this to characterize $T$. Let the mixtilinear incircle hit $AB, BC$ at $T_1, T_2$. By a well-known lemma $T_1, T_2, I$ are collinear. Also, $\angle IT_1B = 90 + \angle A/2$. But by the above, $\angle BTI = \angle BTC / 2 = 90 - \angle A/2$. So $IT_1BT$ should be concyclic. You can probably prove this many ways. I just did an inversion at $A$, and this becomes clear: in the new triangle $AB'C'$, $T$ maps to where the excircle hits $B'C'$, $I$ maps to the excenter, $B$ maps to $B'$, and $T_1$ maps to where the excircle hits $AB'$. So we get that $\angle BTI = \angle CTI$ from the cyclic quads $BTIT_1$ and similarly $CTIT_2$. As an added bonus, $\angle BIT = \angle BT_1T = \angle TT_2T_1 = \angle TCI$, from cyclic quads and the fact that circle $TT_1T_2$ is tangent to $AB$. So in fact, $T$ is the spiral center sending $BI$ to $IC$. Now we have almost enough to finish. Note that to this point, we have barely even used $M, Y_1, Y_2$. Let's phantom point $T'$ to be the point that sends $X_1X_2$ to $Y_1Y_2$. Of course we're going to try to show that $T' \equiv T$. I'll do this by proving that $T'$ in fact also will send $BI$ to $IC$, and because spiral centers are unique, we'll be done. Just angle chasing around a bit shows that $T'$ in fact sends $BY_1$ to $X_2C$. If $T$ sends both $BY_1$ to $X_2C$ and $BI$ to $IC$, then it also sends $Y_1I$ to $IX_2$. We'll prove this now. Note that almost all points in the diagram are gone now, and we can reduce it to the following problem: $Y_1X_1X_2Y_2$ is a tangential quadrilateral (remember that $Y_1Y_2$ is tangent to the incircle). Let $T'$ be its Miquel Point. Prove that $T'$ also sends $Y_1I$ to $IX_2$. This can be done with complex numbers with the incircle as the unit circle. Let the tangency points where $X_1Y_1, Y_1Y_2, Y_2X_2, X_2X_1$ hits the incircle be $a, b, c, d$. Then $y_1 = \frac{2}{1/a+1/b}$, $y_2 = \frac{2}{1/b+1/c}$, $x_2 = \frac{2}{1/c+1/d}$, $x_1 = \frac{2}{1/a+1/d}$. If $T'$ sends $Y_1Y_2$ to $X_1X_2$, then $\frac{t'-y_1}{t'-y_2} = \frac{t'-x_1}{t'-x_2}$. Solving for $t'$ gives $t' = \frac{2}{1/a+1/b+1/c+1/d}$. To show that $T'$ sends $Y_1I$ to $IX_2$, it suffices to prove that $\frac{t'-y_1}{t'-0} = \frac{t'-0}{t'-x_2}$, which is an easy computation. Having proven this last thing, we are done. $\blacksquare$ Notes: This looks long because I tried to work backwards and explain motivation. Hopefully nothing here seems "magical". I will try to add diagrams for the parts when I have time. Also, a synthetic solution would be appreciated for the last lemma. I could not find one.

mathocean97 wrote: $Y_1X_1X_2Y_2$ is a tangential quadrilateral (remember that $Y_1Y_2$ is tangent to the incircle). Let $T'$ be its Miquel Point. Prove that $T'$ also sends $Y_1I$ to $IX_2$. Here's a synthetic solution for this lemma (if inversion is counts as synthetic). Inverting about the incircle of the quadrilateral gives the following problem: Let $ABCD$ be a cyclic quadrilateral. Let $M,N$ be the midpoints of $AD,BC$, respectively. Let the nine-point circles of $\triangle ABC$ and $\triangle BCD$ intersect at $P \neq N$. Prove that $MPNO$ is a parallelogram. Solution: Let the orthocentres of $\triangle ABC, \triangle BCD$ be $H_1, H_2$ and the nine-point centres be $N_1, N_2$ (respectively). We can angle chase to show that $\angle AH_1H_2=\angle H_2DA$, and since $AH_1 || DH_2$, it follows that $H_1H_2 || AD$. Since $N_1$ is the midpoint of $OH_1$ and $N_2$ is the midpoint of $OH_2$, $N_1N_2 || H_1H_2$ and thus $N_1N_2 || AD$. $NP$ is the common chord of the nine-point circles of $\triangle ABC, \triangle BCD$, so $NP \perp N_1N_2$. Thus $NP \perp AD$, so $NP || MO$. Similarly, $MP || NO$ so $MPNO$ is a parallelogram and we're done.

Here's another inversive solution, though I believe the most efficient one posted yet. Rename $X_1, X_2$ as $X, Y.$ Call the $\omega(O)$ the common incircle of $\triangle ABC$ and $\triangle MXY$, and let $D, E, F, S, T$ be the points of tangency of lines $BC, CA, AB, MX, MY$ with $\omega.$ We compose an inversion $\mathcal{I} : X \to X'$ about $\omega.$ Since $EF, FD, DE$ are the polars of $A, B, C$ with respect to $\omega$, we find that $A', B', C'$ are the midpoints of $\overline{EF}, \overline{FD}, \overline{DE}.$ Similarly, $M', X', Y'$ are the midpoints of $\overline{TS}, \overline{SD}, \overline{DT}.$ Also, note that $A, B, C, M$ are concyclic, so under inversion, $A', B', C', M'$ are concyclic as well. Now, here is the key to the solution: the tangency of the $A$-mixtilinear incircle with $\odot (ABC)$ is sent to the antipode of $A'$ WRT $\odot (A'B'C')$ (see below for a sketch of the proof of this). Thus, it suffices to prove that $\Gamma_1 \equiv \odot (A'B'C')$ and $\Gamma_2 \equiv \odot (M'X'Y')$ meet for a second time at the antipode of $A'$ w.r.t. $\Gamma_1.$ Let $H_1, H_2, N_1, N_2$ be the orthocenters and nine-point centers of $\triangle DEF, \triangle DST$, and let $D'$ be the antipode of $D$ w.r.t. $\omega.$ Recall that the reflection of the orthocenter of a triangle in the midpoint of a side is the antipode of the vertex opposite that side w.r.t. the circumcircle. Therefore, $A', M'$ are the midpoints of $\overline{D'H_1}, \overline{D'H_2}$, so $A'M' \parallel H_1H_2.$ Now, recall that the nine-point center of a triangle is the midpoint of the segment connecting the orthocenter and the circumcenter. Since $O$ is the center of $\omega$, we find that $N_1, N_2$ are the midpoints of $\overline{OH_1}, \overline{OH_2}.$ Thus, $N_1N_2 \parallel H_1H_2 \parallel A'M'.$ Now, note that the radius of the nine-point circle is half the radius of the circumcircle, so $\Gamma_1$ and $\Gamma_2$ have the same radius. Therefore, $N_1A' = N_2M'$ and $N_1N_2 \parallel A'M'.$ This is enough to imply that $N_1N_2M'A'$ is either an isoceles trapezoid or a parallelogram. If the former were the case, we would have $\angle N_1A'M' = \angle N_2M'A'.$ But since $\angle N_1A'M' = \angle N_1M'A' \ne \angle N_2M'A'$, this is obviously false. Therefore, $N_1N_2M'A'$ is a parallelogram, so $N_1A' \parallel N_2M'.$ Now, if $Z$ is the second intersection of $\Gamma_1$ and $\Gamma_2$, note that $N_1ZN_2M'$ is a rhombus. Therefore, $N_1Z \parallel N_2M'$, and it follows that $N_1, A', Z$ are collinear, as desired. $\square$

Let $D$ be where the incircle is tangent to $BC$ and let $AM$ meet $BC$ at $N$. By the dual of Desargues' Involution Theorem (see Deduction 1 here: http://apollonius.math.nthu.edu.tw/d1/ne01/whw/conic/involution/inv(ded2).htm for details), $M(A,D;B,C;X_1,X_2)\rightarrow(N,D;B,C;X_1,X_2)$ is an involution. Hence, it suffices to show that the circumcircle of $MDN$ passes through the mixtilinear touchpoint $T$. But note that $TD$ meets the circumcircle at $A'$, where $AA'\parallel BC$, so $\angle ANB=\angle A'AM=\angle A'TM=\angle DTM$ as desired.

I have searched everywhere for the dual of Desargues' Involution Theorem and it appears the link does not work for me. Could someone please post the statement of the dual of Desargues' Involution Theorem? Thank you very much for all your help!

Dual of Desargues' Involution: Consider a complete quadrilateral $ABCDEF$. Choose an arbitrary point $P$. If $(A,C), (B,D), (E,F)$ are opposite points in the quadrilateral, then there is an involution swapping the pairs $(PA, PC), (PB, PD), (PE, PF)$. Moreover, if $\omega$ is a conic tangent to the sidelines of the complete quadrilateral, and $PX$ and $PY$ are tangent to $\omega$, then this involution also swaps $(PX, PY)$.

Thank you very much for all your help K6160! Sorry, I am a little new to this, but what does it mean that there is an involution swapping the pairs $(PA, PC), (PB, PD), (PE, PF)$? Thanks!

This is probably not the right way to think about this and it might not even be correct (someone verify), but the way I like to think about involutions (in that case) is that there are two lines $\ell_1$ and $\ell_2$ passing through $P$ such that $(\ell_1,\ell_2;PX,PY)$ is harmonic, where the involution swaps the pair $(PX,PY)$. $\ell_1$ and $\ell_2$ are the fixed points of the involution. Involutions of lines can be projected onto another line or circle like harmonic bundles.

If you project $PA, PB, PC, PD, PE, PF$ onto a line $l$, such that their images are $A', B', C', D', E', F'$, then $(A',C'), (B',D'), (E',F')$ are swapped under an involution of $l$. Essentially, an involution in this case is a mapping $f$ such that $f(f(x))=x$ that preserves cross ratio.

Interesting, thank you very much ABCDE and K6160! Just wondering, how does one go from $M(A,D;B,C;X_1,X_2)\rightarrow(N,D;B,C;X_1,X_2)$ is an involution to it suffices to show that the circumcircle of $MDN$ passes through the mixtilinear touchpoint $T$? Thanks!

But, I guess I don't really understand swapping points. What is the best way I should view an involution and the transformation swapping points/lines? Sorry I am a really slow learner. Thank you very much!

If you have a point $P$ then all circumcircles of $PXY$ where $(X,Y)$ is a pair of an involution on a fixed line different from $P$ are coaxial. This follows from inversion about $P$; the involution on the line becomes an involution on the circle and the circumcircles become lines passing through pairs of the involution. These pass through the intersection of the tangents to the circle at the fixed points of the involution by harmonics.

To be fair, 90% of the difficulty of the problem is figuring out what's going on in the diagram. Invert about the incircle, and define some new points: Let $DEF$ be the contact triangle, so the images of $A, B, C$ are $X, Y, Z$, the midpoints of $\overline{EF}$ and cyclic permutations. Let $T_1$ and $T_2$ be the tangency points from $M$, which remain fixed. Then, $X_1$ and $X_2$ invert to $Y_1$ and $Y_2$, the midpoints of $\overline{DT_1}$ and $\overline{DT_2}$. $M$ inverts to $Q$, the midpoint of $\overline{T_1T_2}$. Let $H$ be the orthocenter of triangle $DEF$ and $M_A$ the midpoint of major arc $\widehat{BAC}$ (note that $M_A$ lies on $\overline{IT}$, where $T$ is the mixtilinear tangency point). By considering the inverse of $M_A$, we note $\angle XM_A^*I = 90^\circ$, hence $T$ inverts to the $X$-antipode in $(XYZ)$, which is the midpoint of $\overline{DH}$. So by taking a homothety at $D$ with ratio $2$, it suffices to show that $T_1, T_2, H, Q'=2Q-D$ are concyclic. On the other hand, by taking a homothety at $Q$ with ratio $-1$, this is equivalent to $T_1, T_2, H'=2Q-H, D$ concyclic. This is clear as $(T_1T_2H)$ has radius equal to that of the incircle, as both radii are equal to twice the radii of $(XYZ) = (ID)$.

We invert about the incircle. Let $H_1$ and $N_1$ be the orthocenter and nine point center of the intouch triangle $DEF$, and let $T$ be the $A-$ mixtilinear incontact point of $\triangle ABC$. First, if $L$ is the midpoint of arc $\overarc{BAC}$, we have $\angle IL^*A^* = \angle IAL = 90^\circ$, and since $T$ lies on $LI$, it follows that $T^*$ is the antipode of $A^*$ in the nine point circle of $\triangle DEF$ (or equivalently the midpoint of $DH_1$). Next, note that $MX_1$ and $MX_2$ touch the incircle at points $P$ and $Q$ such that $M^*$ is the midpoint of $PQ$. Then $X_1^* = (IP)\cap (ID)$ is just the midpoint of $DP$, and similarly $X_2^*$ is the midpoint of $DQ$. For ease of notation we will use $Y_1 = X_1^*$ and $Y_2 = X_2^*$. Let $H_2$ and $N_2$ be the orthocenter and nine point center of $\triangle DPQ$. We are now ready to tackle the actual problem. We want to show that $(M^*Y_1Y_2)$ passes through $T^*$. Phantom $Z = (M^*Y_1Y_2)\cap (A^*B^*C^*)$. Since the midpoint $K$ of $DH'$ is the antipode of $M^*$ in $(M^*Y_1Y_2)$, we have by midlines at $I$ and $M$ respectively that $$KZ\perp MZ\perp N_1N_2\parallel H_1H_2\parallel KT^*$$ as desired.

Denote the circumcircle of $(ABC)$ by $\Omega$ and the incirlce by $\omega$.Let $f(Z) = \frac{ZB}{ZC}$ . Let $MX_1, MX_2 \cap \Omega = N,K$ respectively. By ratio lemma , It's enough to show that $\frac{f(X_1)f(X_2)}{f(M)}$ is a fix value(and then consider a special case which is easy). Which is , again by ratio lemma , proving that $\frac{MB.NB.KB}{MC.NC.KC}$ is a fix value. Now , fix $BC$ and move $M$ on $\Omega$. It's well-known that the map $M \to P = NK \cap \omega$(Which is a single point , by poncelet) is a projective map.Also , the map $P \to PP \cap BC = Q$ is also a projective map since $BC$ is already tangent to $\omega$. Now let $T$ be the midpoint of arc $BC$ not containing $A$.Assume that $M$ is beyond $BC$ for now , the other case is similar. Let $MT \cap BC = S$. We have $f(M) = f(S)$ by ratio lemma. Also , by ratio lemma , $f(N)f(K) = f(Q)$. So proving that $\frac{MB.NB.KB}{MC.NC.KC}$ is a fix value , is just proving that $(MSQ)$ passes through a fix point on $\Omega$(by ratio lemma , again). But clearly the map $M \to (TS\bullet) \cap BC$ (where $\bullet$ is any arbitrary fix point on $\Omega$) is a projective map by inverting through $T$. But the Map $M \to Q$ was also projective .So by the moving point lemma, it's enough to show the statement of our claim for three cases which is ok by considering $M= B,C,T,T',A ,...$(where $T'$ is the midpoint of arc $(BAC)$) . So we're done. Btw this is probably one of the few solutions for this problem which prevents any kind of angle chasing ,or maybe the only one.I hope it's true

This is a truly amazing problem. Let $\omega$ be incircle of $ABC$. Let $\omega$ touches $BC, CA, AB$ at $D, E, F$, respectively. Consider inversion around $\omega$. Then $(ABC)$ becomes nine point circle of $DEF$. Let $\omega_A$ be $A$ mixtilinear incircle. Claim: $\omega_A^*$ is circle centered $A^*$ with radius $r$, where $r$ is radius of $\omega$. Proof: Note that $\omega_A^*$ tangent to $(IA^*B^*)$ and $(IA^*C^*)$, where $I$ is center of $\omega$. Let $\omega_A$ touches $AB, AC$ at $P, Q$, respectively. Then it's well-known that $P, I, Q$ are collinear. Note that $P^*$ lies on $(IA^*B^*)$ and $Q^*$ lies on $(IA^*C^*)$. Then since $PQ \perp AI$ and $EF \perp AI$, so $EFP^*Q^*$ is rectangle. Since $\omega_A^*$ passes through $P^*, Q^*$, thus $\omega_A$ touches $(IA^*B^*)$ and $(IA^*C^*)$ at $P^*, Q^*$, respectively. Hence the center of $\omega_A^*$ is $A^*$, so $\omega_A^*$ is circle centered $A^*$ with radius $r$. $\blacksquare$ Note that $\omega_A^*$ touches $(A^*B^*C^*)$ at $T^*$. Let $H$ be orthocenter of $DEF$. Claim: $T^*$ is midpoint of segment $DH$. Proof: Let $X = DT^* \cap EF$. Since $A^*T^* = r$, so $A^*T^*$ is diameter of $(A^*T^*X)$. Hence $\angle{T^*XA^*} = 90^{\circ}$, so $DX$ is altitude of $DEF$. Note that $(A^*T^*X)$ is nine point circle of $DEF$, so $T^*$ is midpoint of $DH$. $\blacksquare$ Since $M$ lies on $ABC$, so $M^*$ lies on $A^*B^*C^*$. Let the tangents from $M$ to $\omega$ touches $\omega$ at $Y, Z$. Then $M^*$ is midpoint of $YZ$ and $X_1^*, X_2^*$ is midpoint of $YD, ZD$, respectively. We need to show that $M^*X_1^*X_2^*T^*$ is cyclic. Note that $(M^*X_1^*X_2^*)$ is nine point circle of $DYZ$. Thus it's enough to show that $T^*$ lies on nine point circle of $DYZ$. Let $\gamma$ be nine point circle of $DEF$ and let $N_9$ be center of $\gamma$. Let $N = M^*N_9 \cap \gamma$. Let $DS$ be altitude of $DYZ$ and let $K = T^*N \cap DS$. Claim: $M^*K = M^*N = r$. Proof: Note that $IH$ is Euler line of $DEF$. Thus $I, N_9, H$ are collinear and $IN_9 = HN_9$. Thus $M^*INH$ parallelogram and note that $M^*I \perp YZ$. Since $DK \perp YZ$, so $DK \parallel NH$. Thus $\frac{T^*K}{T^*N} = \frac{T^*D}{T^*H} = 1$, so $T^*K = T^*N$. Therefore $M^*K = M^*N = T^*A^* = r$. (Note that $\gamma$ is nine point circle of $DEF$) $\blacksquare$ Now we'll prove that $T$ lies on nine point circle of $DYZ$. Note $M^*K = r$ and $M^*$ is midpoint of $YZ$ and $DK \perp YZ$, thus circle with diameter $M^*K$ is nine point circle of $DYZ$. Since $\angle{KT^*M^*} = 90^{\circ}$, so $T$ lies on nine point circle of $DYZ$. This completes proof. $\blacksquare$

Invert about the incircle. If $DEF$ is the intouch triangle of $ABC$ with $D$ opposite $A,$ then $(ABC)$ inverts to the nine-point circle of $DEF.$ If the tangents from $M$ meet the incircle at $Y_1,Y_2$ then $(MX_1X_2)$ inverts to the nine-point circle of $DY_1Y_2.$ The mixtilinear incircle inverts to a circle internally tangent to the images of $AB,AC$ and the nine-point circle. Since these circles all have radius $\tfrac{r}{2}$ and pass through $A,$ the mixtilinear incircle inverts to a circle centered at $A$ with radius $r.$ In particular, the tangency point to the nine-point circle of $DEF$ is simply the antipode $K$ of $A$ with respect to it. Let $H$ be the orthocenter of $DEF.$ If $N$ is the reflection of $H$ over $M,$ then notice that since $M$ is on the nine-point circle we have that $N$ is on the incircle. Also, notice that since $IM \perp Y_1Y_2$ we get that the reflection $P$ of $H$ over $Y_1Y_2$ also lies on the incircle, so using directed angles we get $\measuredangle Y_1DY_2=\measuredangle Y_1PY_2=-\measuredangle Y_1HY_2.$ Next, take a homothety at $D$ with scale factor $\tfrac{1}{2}.$ By properties of nine-point circles, we get that $K$ is the midpoint of $HD,$ so this takes $\measuredangle Y_1HY_2$ to $\measuredangle X_1KX_2.$ Also, by the medial triangle we get $\measuredangle Y_1DY_2=-\measuredangle X_1MX_2,$ so we get that $K$ lies on the circumcircle of $MX_1X_2.$ Inverting back, we get our desired result.

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Let $Y_1$ and $Y_2$ be the tangency points of $M$ to the incircle, and let $\triangle DEF$ be the contact triangle. Lastly, denote by $H$ the orthocenter of $\triangle DEF$. Now invert about the incircle, where the prime of each point represents its image under inversion. Note, $A'$, $B'$, and $C'$ are the midpoints of $\overline{EF}$, $\overline{DF}$ and $\overline{DE}$, so $(A'B'C')$ is the nine-point circle of $\triangle DEF$. $M'$ is the midpoint of $\overline{Y_1Y_2}$. $X_1'$ is the midpoint of $DY_1$ as the tangents from $X_1$ meet the incircle at $Y_1$ and $D$. Similarly $X_2'$ is the midpoint of $DY_2$. $N'$ is defined on $(A'B'C')$ such that $\angle A'N'I = 90$. This follows as $\angle NAI = 90$, and under inversion $\angle NAI = \angle A'N'I$. $T'$ is the point diametrically opposite $A'$, and is the midpoint of $HD$, due to the fact that $N'$, $I$ and $T'$ are collinear. Now it suffices to show $X_1'$, $T'$, $M'$ and $X_2'$ are concyclic. Taking a homothety at $D$ with ratio $2$, followed by a homothety with ratio $-1$ at $M'$. Then we wish to show the reflection of $H$ over $M'$ lies on the circumcircle of $\triangle DEF$. However noting that from the homothety with ratio $2$, sending the nine point circle of $\triangle DEF$ to the circumcircle, we must have the reflection of $H$ about $M'$ lie on the circumcircle of $\triangle DEF$, so we are done.

My pay grade only allows me to perform one homothety. Let $T$ be the tangency point and $N$ be the midpoint of arc $BAC$. It's well-known that $T,I,N$ are collinear. Let $\overline{MX_1}$ and $\overline{MX_2}$ be tangent to the incircle at $P$ and $Q$ respectively. Now invert about the incircle. $A,B,C$ get sent to the midpoints of the intouch triangle, so $(ABC)$ gets sent to its 9-point circle $\omega$. $M$ gets sent to the midpoint of $\overline{PQ}$. $X_1$ and $X_2$ get sent to the midpoints of $\overline{DP}$ and $\overline{DQ}$ respectively. To figure out where $T$ gets sent, we first note that since $\angle IAN=90^\circ$, $N$ gets sent to the point $N'$ on $\omega$ such that $\angle IN'A'=90^\circ$, where $A'$ is the midpoint of $\overline{EF}$. $T$ then gets sent to the second intersection of $\overline{IN'}$ with $\omega$, which is the point diametrically opposite $A'$, i.e. the midpoint of $\overline{DH}$ where $H$ is the orthocenter of $\triangle DEF$. Now use complex numbers with respect to the incircle (i.e. circumcircle of $\triangle DEF$), letting lowercase letters denote the position of their uppercase points. We wish to show that $d+\tfrac{e+f}{2},\tfrac{d+p}{2},\tfrac{d+q}{2},\tfrac{p+q}{2}$ are concyclic given that $\tfrac{p+q}{2}$ lies on $\omega$. The condition on $\tfrac{p+q}{2}$ is equivalent to having $p+q-(d+e+f)$ lie on the unit circle, by a homothety at $H$. Now, $d+\tfrac{e+f}{2}$, $\tfrac{d+p}{2}$, $\tfrac{d+q}{2}$, and $\tfrac{p+q}{2}$ are concyclic iff $2d+e+f,d+p,d+q,p+q$ are, which is equivalent (upon subtracting $d+p+q$) to $d+e+f-p-q,-q,-p,-d$ being concyclic. This last statement is true by our condition, so we're done. $\blacksquare$

Let $\triangle DEF$ be the contact triangle with orthocenter $H$. We invert about the incircle: $\triangle ABC$ is mapped to the midpoints of $\triangle DEF$. The mixtilinear incircle touch point is mapped to the midpoint of $DH$, which we denote $K$. $M$ is mapped to a point on the nine-point circle of $\triangle DEF$. $X_1$ and $X_2$ are mapped to the midpoints of the respective contact chords in $\triangle DEF$. Hence we need to show that $TX_1'X_2'M'$ is cyclic. Under a homothety with scale factor 2 from $D$ composed with a reflection about $M$, we find that these four points are mapped onto the incircle. $\blacksquare$

Invert about the incenter with arbitrary radius. Then the problem becomes: Inverted problem wrote: Consider two triangles $ABC$ and $A'B'C$ with the same circumcircle, and orthocenter $H$. Suppose $(BHC) \cap (A'HC') = P$ and $(BHC) \cap (A'HB') = Q$. Prove that $A'PQ$ passes through the antipode $A_1$ of $A$. In fact we claim that $P$ is the orthocenter of $A_1A'C'$. Redefine it as such; then we may readily verify that $P \in (A'HC')$; now a vector calculation (using $A + B + C = A' + B' + C'$) gives us that $BPCB'$ is a parallelogram, so it also lies on $(BHC)$. With this lemma it is now clear that $(A'A_1QP)$ is precisely the reflection of $(ABC)$ across $A'A_1$ and hence we are done.

Let $M' \in (ABC)$ satisfy $MM' || BC$. Let $D$ be the intersection of $BC$ and the incircle. Clearly, $M'$ is projective in $M$, and the tangency point $X$ of $PQ$ to the incircle is projective to $M$ by Poncelet's porism. Now let $T = PQ \cap BC$, then $T$ is projective in $X$ by projecting the incircle to a pencil at the antipode of $D$, translating to the incenter, then projecting onto line $BC$. Thus $T$ is projective in $M'$. When $M' = B, C, A'$, we have that $T = B, C, D$ respectively where $A' \in (ABC)$ satisfies $AA' || BC$. Note that $A'D \cap (ABC) = S$, where $S$ is the A-mixtilinear in-touch point. This is because $A'S$ reflects to $AS'$ where $AS'$ is collinear with the $A$-extouch point by $\sqrt{BC}$ inversion. Thus since three points define a projective map, it follows that $\overline{M'TS}$ is always collinear. Now, let $R$ be the intersection of the radical axis of $(MX_1X_2)$ and $(ABC)$ with $BC$ and let $S' = MR \cap (ABC)$. It follows that $R$ is the antiversion center of the involution swapping $X_1 \leftrightarrow X_2, B \leftrightarrow C, R \leftrightarrow \infty_{BC}$. Thus by conjugating with projection through $M$, it follows that there exists an involution on $(ABC)$ swapping $P \leftrightarrow Q, B \leftrightarrow C, S' \leftrightarrow M'$. It follows that $S'TM$ are collinear, so $S = S'$, finishing.

Anantmudgal's first solution is in fact valid but I also don't have a proof. Claim: In a poncelet configuration, the map from $A$ to $BC$ to the incircle is projective as $A$ varies. Proof. Well known'', i3435 finds $A \mapsto AI \cap (ABC) \overset{\tau}\mapsto D \mapsto BC$ where $\tau$ follows by homothety. $\blacksquare$ Define $Y_i \coloneq AX_i \cap \Gamma$. Then $Y_1Y_2$ is tangent to the incenter by Poncelet's Porism. Let $N = Y_1Y_2 \cap BC$. Claim: $U \coloneq AN \cap MD$ lies on $\Gamma$. Proof. Consider the projective maps $M \mapsto MD \cap \Gamma$ and $M \mapsto Y_1Y_2 \mapsto E \mapsto AE \cap \Gamma$. It remains to check three cases. $M = B, C, T$ work. $\blacksquare$ Let $A_1$ be on $\Gamma$ such $AA_1BC$ is an isoceles trapezoid. Then since $\measuredangle NDT = \measuredangle AA_1T = \measuredangle AUT = \measuredangle NUT$ it follows that $(NUDT)$ is cylic. As such, since $\measuredangle TY_1X_1 =\measuredangle TY_1M = \measuredangle TUM = \measuredangle TUD = \measuredangle TND$ it follows that $T \in (NUD)$ and thus $T$ is the Miquel point of $X_1X_2Y_1Y_2$, which finishes.

We invert with respect to the incircle and restate in terms of the intouch triangle as follows: Quote: In $\Delta ABC$ , $K$ and $L$ are two points on $(ABC)$. Let the orthocenter of $\Delta ABC$ be $H$ and the midpoint of $AH$ be $T$. Prove that $T$ lies on the nine point circle of $\Delta AKL$. We take a homothety with center $A$ and factor $2$. Then it suffices to prove that $K$ , $L$, $H$ and the point $P$ such that $AKPL$ is a parallelogram are concyclic. Reflecting about the midpoint of $KL$, it suffices to prove that the image of $H$ lies on $(ABC)$. Let this image be $Q$. Then $Q$ is also the image of the midpoint of $KL$ with center $H$ and scale factor $2$, which famously maps the nine point circle to the circumcircle implying $Q$ lies on $(ABC)$ which finishes.

Not that bad for a sweep but anyways.... it was a beautiful problem though can't lie. We will invert about the incircle and using the fact that the mixtilinear incircle, incenter and major arc midpoint of $\widehat{BC}$ are collinear, we can exploit the incenter-orthocenter duality to get this new problem with some reformed labels. Quote: Let $\triangle ABC$ be a triangle with $O$ circumcenter, $H$ orthocenter, $\Gamma$ circumcircle, $\Omega$ nine point circle, $\triangle DEF$ orthic triangle and $\triangle MNL$ medial triangle. Let $Z \in \Omega$ be an arbitary point and $U$ and $V$ be the unique distinct points such that $(OZU)$ and $(OZV)$ are both tangent to $\Gamma$. Let $X_1=(ALON) \cap (OZU)$ and define $X_2$ similarly. Prove that if $P$ is the midpoint of $\overline{AH}$, then $X_1$, $Z$, $X_2$, $P$ are concyclic. See that because of Pole Polar/La Hire shenanigans, we have that $\triangle X_1ZX_2$ is the medial triangle of $\triangle AUV$. Hence we have that $\measuredangle X_2ZX_1=\measuredangle UAV$ and $\measuredangle X_2PX_1=\measuredangle VHU$ (by taking a double homothety at $A$) and so if we prove that the reflection of $H$ over $\overline{UV}$ lies on $\Gamma$ or equivalently proving if $T$ is the unique point on $\Gamma$ such that $\measuredangle OZT=90^{\circ}$ and then $\overline{HT} \parallel \overline{OZ}$, then we are done. But this is trivial if we take a double homothety at $H$.

Let $MX_1$ and $MX_2$ intersect $(ABC)$ again at $Y_1$ and $Y_2$, respectively. Let $Y_1Y_2$ intersect $X_1X_2$ at $P$. By Poncelet's Porism, $MY_1PX_2$ has an incircle. Let $T$ be the Miquel Point of the complete quadrilateral $MY_1X_1PX_2Y_2$. $T$ is the intersection of the circles $(ABC)$ and $(MX_1X_2)$. It suffices to show that $TI$ passes through the midpoint of arc $BC$ in $(ABC)$ containing $A$. Let $TP$, $TX_2$ intersect the circumcircle again at $M_1$ and $Y_3$, respectively. Claim: $MM_1\parallel BC$. $TPX_2Y_2$, $TY_1Y_3Y_2$ are cyclic, so by Reim's Theorem $Y_1Y_3\parallel BC$. There is a spiral similarity centered at $T$ taking $Y_1P$ to $MX_2$. Therefore, $$\measuredangle Y_1TM_1=\measuredangle Y_1TP=\measuredangle MTX_2=\measuredangle MTY_3$$ implies $Y_1Y_3\parallel MM_1$, verifying the claim. It suffices to show that $TI$ bisects $\angle PTM$. Indeed, $$\begin{align*} \measuredangle PTI - \measuredangle ITM &= \measuredangle TPI + \measuredangle PIT - \measuredangle TMI - \measuredangle MIT \\ &= \measuredangle TPY_2 + \measuredangle Y_2PI + \measuredangle PIM + \measuredangle MIT - \measuredangle TMI - \measuredangle MIT \\ &= \measuredangle TY_1M + \measuredangle IPY_1 + \measuredangle PIM - \measuredangle TMI \\ &= \measuredangle PY_1M + \measuredangle TY_1P + \measuredangle IPY_1 + \measuredangle PIM - \measuredangle TMI \\ &= \measuredangle PY_1M + \measuredangle TMY_2 + \measuredangle IPY_1 + \measuredangle PIM + \measuredangle IMT \\ &= \measuredangle PY_1M + \measuredangle TMY_2 + \measuredangle IPY_1 + \measuredangle PIM + \measuredangle IMY_2 \\ &= \measuredangle PY_1M + \measuredangle IPY_1 + \measuredangle PIM + \measuredangle Y_1MP \\ &= 0 \end{align*}$$