From Brocard's theorem $X$ and $Y$ belongs to the polar of $I$ wrt $\Gamma$, then $X$ and $Y$ are midpoints of $A_1B_1$ and $A_1C_1$, respectively. Thus $$MY=\frac{AC_1}{2}=\frac{AB_1}{2}=MX$$done, this problem is too easy for P6.

By brokard theorem on $C_1C_2A_1A_2$ we have that $I$ lies on the polar of $Y$ w.r.t. $\Gamma$ and by brokard on $A_2A_1B_2B_1$ we have that $I$ lies on the polar of $X$ w.r.t. $\Gamma$. Also note that $Y \in \mathcal P_B$ and $X \in \mathcal P_C$ thus by La'Hire we have that $C \in \mathcal P_X$ and $B \in \mathcal P_Y$ thus we have that $\mathcal P_X=BI$ and $\mathcal P_C=CI$ meaning that $XI_A \perp CI$ and $YI_A \perp BI$ but since $\angle IBI_A=\angle ICI_A=90$ we have that $C,X,I_A$ and $B,Y,I_A$ are colinear meaning that $X$ is midpoint of $A_1B_1$ and $Y$ is midpoint of $A_1C_1$. Now by midbase we have $2MY=AC_1=AB_1=2MX$ meaning that $MY=MX$ are desired thus we are done . Girls that knew projective geo should've killed this problem...

No need for projective, elementary methods suffice. Lemma: $X$ is the midpoint in $A_1B_1$. Similarly, $Y$ is the midpoint in $A_1C_1$. Proof: Since $I$ is inside $ABC$ and lines $AB,AC$ are tangent to $\Gamma$, $A_1$ is inside segment $IA_2$ and $B_2$ is inside segment $IB_1$. By Menelaus' theorem applied to $A_2,X,B_2$ on the sides of triangle $IA_1B_1$, and using that $IA_1\cdot IA_2=IB_1\cdot IB_2$ is the power of $I$ with respect to $\Gamma$, the Lemma is equivalent to $IB_1\cdot B_1B_2=IA_1\cdot A_1A_2$. Note that the power of $I$ with respect to $\Gamma$ also equals $II_A^2-\rho^2$, where $\rho=I_AA_1$ is the radius of $\Gamma$. Or, denoting by $\alpha$ the angle between $IA_1$ and $BC$, we have $\angle IA_1I_A=90^\circ+\alpha$, and using the Cosine Law we obtain $$IA_1\cdot A_1A_2=IA_1\cdot IA_2-IA_1^2=II_A^2-I_AA_1^2-IA_1^2=-2I_AA_1\cdot IA_1\cos\angle IA_1I_A=2\rho\cdot IA_1\sin\alpha=2\rho\cdot r.$$Similarly, $IB_1\cdot B_1B_2=2\rho\cdot r=IA_1\cdot A_1A_2$. The first part of the Lemma follows, and the second one is proved analogously, exchanging $B$ and $C$. Since by the Lemma $X$ is the midpoint in $A_1B_1$, $Y$ is the midpoint in $A_1C_1$ and by definition $M$ is the midpoint of $A_1A$, triangle $XYM$ is similar to triangle $B_1C_1A$. Or since $AB_1C_1$ is isosceles at $A$, $MXY$ is isosceles at $M$. The conclusion follows.

This problem took me ages to solve as I somehow forgot about proyective geo. Anyways here is my solution. Note that $BI\parallel C_1A_1$ as $\angle ABI=\angle AC_1A_1=\angle \frac{B}{2}$. Then we use the tangents and parallel lines to get that \[\angle BIC_2= \angle C_2C_1A_1=\angle BA_1C_2 \text{ and } \angle ABI=\angle AC_1A_1=\angle C_1A_1A_2 \]thus both $BIA_1C_2$ and $BIA_2C_1$ are cyclic. Then we get that $\angle C_2BA_1=\angle C_2IA_1=\angle C_1IA_2=\angle C_1BA_2$.Thus $\angle C_1BC_2=\angle A_2BA_1 (*)$. Let $I_a$ be the $A$-excenter. $(*)$ together with the fact that $BI_a$ is the angle bisector of $\angle C_1BA_1$ imply that $BI_a$ is the angle bisector of $\angle C_2BA_2$. The perpendicular bisector of $C_2A_2$ also goes through $I_a$ as $C_2$ and $A_2$ both lie on the excircle with center $I_a$. Thus by the incenter excenter lemma, $BC_2A_2I_a$ is cyclic. Clearly $C_1BA_1I_a$ is cyclic. Then we get that $BI_a$, $A_1C_1$ and $C_2A_2$ are the radical axii of $(BC_2A_2I_a),(A_1A_2B_1B_2),(C_1BA_1I_a)$, which implies that $BI_a$ goes through $X$. As $BA_1I_aC_1$ is a kite, $X$ is the midpoint of $A_1C_1$. Analogously $Y$ is the midpoint of $A_1B_1$. Thus $MXY$ is similar to $AC_1B_1$ which is clearly isosceles as desired.

Very nice problem. The main claim is that $X,Y$ are midpoints of segments $A_1B_1$ and $A_1C_1$. In order to prove that, we see that $X$ and $Y$ belong to the polar of $I$ wrt $\Gamma$, so the polar of $I$ is $XY$, hence $II_A \perp XY$, and since $II_A \perp C_1B_1$, we have that $XY \parallel C_1B_1$ (fact I)). Now, notice that $X$ belongs to the polar of $C$, and hence the polar of $X$ is $CI$, so $I_AX \perp CI$ hence $X$ belongs to the segment $CI_A$. So $X,Y$ are the midpoints of segments $A_1B_1$ and $A_1C_1$, as claimed. Now, let $N$ be the midpoint of segment $A_1I_A$, $P$ be the midpoint of $XY$ and $L$ be the midpoint of $C_1B_1$. Take an homothety centered at $A_1$ and scaling factor $\frac12$. Due to fact I, $L$ goes to $P$; $I_A$ clearly goes to $N$ and $A$ goes to $M$. Since $A,L,I_A$ are collinear points, then $M,P,N$ must also be, and we're done.

FabrizioFelen wrote: From Brocard's theorem $X$ and $Y$ belongs to the polar of $I$ wrt $\Gamma$, then $X$ and $Y$ are midpoints of $A_1B_1$ and $A_1C_1$, respectively. Thus $$MY=\frac{AC_1}{2}=\frac{AB_1}{2}=MX$$done, this problem is too easy for P6. Lol this sol is much easier than mine. Nicely done.

Clearly, $\angle{ABI}=\angle{BC_1A_1}$, so $BI\parallel{A_1C_1}$. By property of angles between parallels, $\angle{BIC_1}=\angle{IC_1A_1}$ and by property of semi-inscribed angle, $\angle{BC_1I}=\angle{C_1A_1C_2}$. This implies that $\triangle{IBC_1}\sim{\triangle{A_1C_1C_2}}$. It then follows that $\angle{IC_2A_1}=180^\circ-\angle{C_1C_2A_1}=180^\circ-\angle{C_1BI}=\angle{ABI}=\angle{A_1BI}=\frac{\angle{ ABC}}{2}$. From which it follows that $IBC_2A_1$ is cyclic. Let $N$ denote the second intersection of $BC_2$ with $\tau$. By Reim's Theorem, $BI\parallel{A_2N}$. Note that $A_1NC_1C_2$ is a harmonic quadrilateral, so if we project from $A_2$ onto the line $A_1C_1$, also parallel to $A_2N$ (remember that $BI\parallel{A_1C_1}$), we have $$-1=(N, C_2; C_1, A_1)\stackrel{A_2}{=}(P_\infty, Y; C_1, A_1).$$ This implies that $Y$ is the midpoint of $A_1C_1$. Similarly, $X$ is the midpoint of $A_1B_1$. By the Midsegment Theorem, $MY=\frac{AC_1}{2}$ and $MX=\frac{AB_1}{2}$. But $AC_1=AB_1$ for being common tangents, therefore $MY=MX$.

Claim: $BIA_{2}C_1$ and $BIC_{2}A_{1}$ are cyclic. This is just an angle chasing: $\angle C_{1}BA_{1} = 180 - B \Rightarrow \angle BC_{1}A_{1} = \angle C_{1}A_{2}A_{1} = \frac{B}{2}$. Since $\angle ABI = \frac{B}{2}$, $BIA_{2}C_{1}$ is cyclic. Note that $\angle IC_{2}A_{1} = \angle IA_{2}C_{1} = \frac{B}{2} = \angle IBA_{1}$, then $BIC_{2}A_{1}$ is cyclic. A known fact in a cyclic complete quadrilateral $ABCDPQ$ is that the intersection of the diagonals (let $R$) is the inverse of the Miquel Point M. This is because with an angle chasing we get $\angle AR'B = \angle APB$, so $R'$ lies on $(PAB)$. Similarly, we can achieve that R' lies on $(PCB), (QBC), (QDA)$, so $R' = M$. Note that in this problem, $B$ is the Miquel Point of the cyclic quadrilateral $A_{1}A_{2}C_{1}C_{2}$, so $Y$ is the inverse of $B$ wrt $\Gamma$ $\Rightarrow$ Y is the midpoint of $A_{1}C_{1} \Rightarrow MY = \frac{AC_{1}}{2}$. Similarly, $X$ is the midpoint of $A_{1}B_{1} \Rightarrow MX = \frac{AB_{1}}{2}$. Since $AC_1 = AB_1$, $MX = MY$ and we are done.

The main content of the problem is the following claim. Claim: Point $Y$ coincides with the midpoint of $\overline{A_1C_1}$. [asy][asy]size(6cm); pair A_1 = dir(40); pair C_1 = dir(140); pair B = 2*A_1*C_1/(A_1+C_1); pair Y = midpoint(A_1--C_1); pair C_2 = dir(80); pair I = extension(C_1, C_2, B+dir(0), B); pair A_2 = extension(C_2, Y, A_1, I); filldraw(unitcircle, invisible, blue); draw(C_1--B--A_1, blue); draw(C_1--I--A_2--C_2, lightred); draw(B--I, deepgreen); draw(C_1--A_1, deepgreen); pair J = origin; dot("$A_1$", A_1, dir(0)); dot("$C_1$", C_1, dir(C_1)); dot("$B$", B, dir(B)); dot("$Y$", Y, dir(Y)); dot("$C_2$", C_2, dir(C_2)); dot("$I$", I, dir(I)); dot("$A_2$", A_2, dir(A_2)); dot("$J$", J, dir(90)); /* TSQ Source: A_1 = dir 40 R0 C_1 = dir 140 B = 2*A_1*C_1/(A_1+C_1) Y = midpoint A_1--C_1 C_2 = dir 80 I = extension C_1 C_2 B+dir(0) B A_2 = extension C_2 Y A_1 I unitcircle 0.1 lightcyan / blue C_1--B--A_1 blue C_1--I--A_2--C_2 lightred B--I deepgreen C_1--A_1 deepgreen J = origin R90 */ [/asy][/asy] Proof. By Brokard theorem both $I$ and $B$ lie on the polar of $Y$, hence $\overline{BI}$ is exactly the polar of $Y$. In particular, if $J$ is the excenter, then $\overline{YJ} \perp \overline{BI}$. As $\overline{BI} \parallel \overline{A_1C_1}$, we're done. $\blacksquare$ Similarly $X$ is the midpoint of $\overline{A_1B_1}$. All that remains is the extraction $MX = \frac{1}{2} AC_1 = \frac{1}{2} AB_1 = MY$.

Notice that $\angle BA_2C_1=\tfrac{1}{2}\angle ABC=\angle IBA_1$ so $\overline{IB}\parallel\overline{A_1C_1}.$ Also, by Brokard, $I$ lies on the polar of $Y$ and $B$ lies on the polar of $Y$ by La Hire. Hence, the line perpendicular to $\overline{IB}$ and therefore $\overline{A_1C_1}$ through $Y$ passes through the center of $\Gamma,$ so $Y$ is the midpoint of $\overline{A_1C_1}.$ Similarly, $X$ is the midpoint of $\overline{A_1B_1}$ so \[MX=\tfrac{1}{2}AB_1=\tfrac{1}{2}AC_1=MY,\]as desired. $\square$

Solution. An alternative way to show that $Y$ is the midpoint of $A_1B_1$ without projective geometry. Since $BI \parallel C_1A_1$, we have $\angle ABI = \angle IA_1B = \angle BA_1C_1 = \angle IC_2A_1$ where the last equality holds by virtue of the semi-inscribed angle theorem. Thus $IBC_2A_1$ is cyclic. Therefore $$\angle IC_2B = \angle IA_1B = \angle CA_1A_2 = \angle A_1CA_2 =\angle A_1C_2Y$$which means that $BC_2$ and $YC_2$ are isogonal conjugates of $\bigtriangleup C_1C_2A_1$. In particular, $BC_2$ is the $C_2$-symmedian of $\bigtriangleup C_1C_2A_1$, which forces $C_2Y$ to be a median and hence $Y$ is the midpoint of $A_1C_1$. $\square$

oops By using Brocard on $A_1A_2B_1B_2$, $\overline{IP}$ is the polar of $X$, so $\overline{I_AX} \perp \overline{IP}$. Since $X$ clearly lies on the polar of $C$, $C$ lies on $\overline{IP}$ by La Hire's. Furthermore, since $\angle ICI_A=90^\circ$, it follows that $I_A,X,C$ are collinear, hence $X$ is the midpoint of $\overline{A_1B_1}$. Likewise, $Y$ is the midpoint of $\overline{A_1C_1}$, so a homothety of scale factor $2$ at $A_1$ sends $\triangle MXY$ to $\triangle AB_1C_1$. Since $AB_1=AC_1$, we are done. $\blacksquare$

lol why does the statement say "$A$-excenter $\Gamma$" Let $I_A$ be the $A$-excenter. By Brokard on $A_1B_2B_1A_2$, we find that $I$ lies on the polar of $X$, and by La Hire's we can see that $C$ lies on the polar of $X$, too (as line $A_1B_1$ is the polar of $C$). Therefore since $\overline{IC} \perp \overline{I_AC}$, $X$ must lie on $\overline{I_AC}$, which implies that $X$ is the midpoint of $\overline{A_1B_1}$. Similarly, $Y$ must be the midpoint of $\overline{A_1C_1}$. By taking a homothety at $A_1$, we find that $MX = \tfrac{1}{2}AB_1 = \tfrac{1}{2}AC_1 = MY$, so we are done.

Brocard's on $A_1A_2C_1C_2$ implies that I lies on the polar of Y. Since B also lie on the polar of Y (tangents construction), we get that BI is the polar of Y; in particular, $T_aY$ perp. IB ($T_a$ is the the center of circle T), but since we know $T_aB$ perp. IB we get that $T_aB$ passes through Y, which readily implies that Y is the midpoint of $A_1C_1$. Doing analogous work on $A_1A_2B_1B_2$, ... gives that X is the midpoint of $A_1B_1$ as well. Then $MY=\frac{AC_1}{2}=\frac{AB_1}{2}=MX$, as desired.

By Brokard's on $A_1A_2C_1C_2$ and $A_1A_2B_1B_2$ it follows that $XY$ is the polar of $I$. Claim: $XA_1 = XB_1$. Similarily, $YA_1 = YC_1$. Proof. The polar of $X$ is $IC$. As such, $IC \perp XI_A$. However, $CI \parallel A_1B_1$ follows since $\measuredangle ICA_1 = \measuredangle B_1A_1C$. Thus, $XI_A \perp A_1B_1$ so $XA_1 = XB_1$. The proof of $YA_1 = YC_1$ is similar. $\blacksquare$ Thus, taking a homothety of size $2$ at $A_1$ means that $MX = MY$ is the same as $AC_1 = AB_1$, which holds.

who else is here because of OTIS lol Since $A_1C$ and $B_1C$ are both tangent to $\Gamma$, with $X$ lying on $A_1B_1$, we find that $C$ is on the polar of $X$. However, observe that since $A_1B_2B_1A_2$ is a cyclic quadrilateral inscribed in $\Gamma$, with intersection points $I=A_1A_2\cap B_1B_2$, $K=A_1B_2\cap A_2B_1$, and $X=A_1B_1\cap A_2B_2$, the line $KI$ must be the polar of $X$ due to Brokard's Theorem, forcing $K$, $C$, and $I$ to be collinear. Angle chasing on the angle bisectors of $\angle A_1CA$, we find that \begin{align*} \angle ICA_1&=\frac{\angle C}2\\ &=\frac{180^\circ-\angle A_1CB_1}2\\ &=\angle CA_1B_1, \end{align*}implying that $CI\parallel A_1B_1$. Again using Brokard's Theorem on $A_1B_2B_1A_2$, we find that the orthocenter of $\triangle IKX$ is exactly $I_A$, so that $KI\perp XI_A$. Combined with the results that $CI\parallel A_1B_1$ and that $K$, $C$, and $I$ are collinear, as found previously, we find that $XI_A\perp A_1B_1$. Since $A_1$ and $B_1$ both lie on $\Gamma$, centered at $I_A$, it thus follows that $A_1I_A=B_1I_A$, and we therefore conclude that $X$ is the midpoint of $A_1B_1$. Symmetrical logic yields that $Y$ is the midpoint of $A_1C_1$. Then, a homothety centered at $A_1$ with ratio $2$ thus maps $\triangle MXY$ to $\triangle AB_1C_1$, implying that $\frac{MX}{MY}=\frac{AB_1}{AC_1}$. Since $AB_1$ and $AC_1$ are both tangents to $\Gamma$ from $A$, respectively at $B_1$ and $C_1$, it follows that $AB_1=AC_1$, so that $MX=MY$, as desired. $\blacksquare$ - Jörg

Brockards sobad. We first claim that $XY$ is the polar of $I$ with respect to $\Gamma$. Indeed note that by Brokards on $A_1A_2B_1B_2$ we have that $I$ lies on the polar of $X$. Also we have that $B$ lies on the polar of $X$, so $BI$ is the polar of $X$ with respect to $\Gamma$. Similarly $CI$ is the polar of $Y$. Then by La Hire's we have $X$ and $Y$ lie on the polar of $I$, so $XY$ is the polar of $I$. Now some simple angle chasing gives $IB \parallel A_1C_1$, hence $XI_A perp A_1B_1$ from which we can conclude that $X$ is the midpoint of $A_1B_1$. Now a homothety centered at $A_1$ with scale factor $2$ finishes.

By Brokard's we have $CI$ being the polar of $X$ wrt $\Gamma$. Note that since $\angle ICI_A = 90^{\circ}$, we have $X$ lying on $I_AC$ which implies that $X$ is the midpoint of $A_1B_1$, similarly for $Y$. Then taking a homothety centered at $A_1$ sending $\triangle MYX \to \triangle AC_1B_1$ with factor of $2$ finishes.

By definition, $Y$ lies on $A_1C_1$, which is the polar of $B$ with respect to $\Gamma$. By Brokard's Theorem on $A_1A_2C_1C_2$, $Y$ also lies on the polar of $I$ with respect to $\Gamma$. La Hire's Theorem then tells us that the polar of $Y$ with respect to $\Gamma$ must be the line $IB$. We note that $\angle IBA_1 = \angle BA_1C_1 = \frac12 \angle B$, so $IB \parallel A_1C_1$. Since $IB$ is the polar of $Y$, $IB\perp OY$, so we also have $OY \perp A_1C_1$, so $Y$ must be the midpoint of $A_1C_1$. We can follow the same steps to show that $X$ is the midpoint of $A_1B_1$. Then we note that $\triangle MXY$ is the image of $\triangle AB_1C_1$ under a homothety centered at $A_1$ with scale factor $\frac12$. $AB_1 = AC_1$, so we also have $MX = MY$, as desired.

Let $I_A$ be the center of $\Gamma$. Note that $B$ lies on the polar of $Y$ because of the tangents construction. Also, Brokard on $A_1A_2C_1C_2$ gives that $I$ lies on the polar of $Y$ as well. Hence, the polar of $Y$ is simply $\overline{BI}$. Since $\overline{IB} \parallel \overline{A_1C_1}$, we must have that $Y$ is the midpoint of $\overline{A_1C_1}$ as $\overline{I_AY} \perp \overline{IB}$. Similarly, we can determine that $X$ is the midpoint of $\overline{A_1C_1}$. Thus, a homothety centered at $A_1$ with scale factor $2$ maps $\triangle MXY$ to $\triangle AB_1C_1$; we consequently have \[MX = \frac{1}{2} AB_1 = \frac{1}{2} AC_1 = MY. \ \square\]

ok bluds lets use projective geometry from brocard twice I is the pole of XY. now reverse reconstruct with X'Y' as midpoints to see that IB and IC are polars of X and Y, so I is the pole of X'Y' midpoints = XY. now dilate MXY to get the equal lengths.