Very beautiful problem. This Lemma kills it. https://tieba.baidu.com/p/7951343410#/

My solution for this one too is again a huge one, I believe there exists a much simpler solution to this ? @above noo sniped . @below lmao ded $\textbf{LEMMA: }$If $ABC$ be a triangle, $M$ be the midpoint of $BC$, $\odot(I)$ and $\odot(I_A)$ denote the incircle and $A-$excircle respectively. Then $$\operatorname{Pow}_{\odot I}(M)=\operatorname{Pow}_{\odot I_A}(M)$$i.e., $M$ lies on the Radical Axis of $\{\odot(I),\odot(I_A)\}$. $\textbf{PROOF: }D,E$ be the points where $\odot(I),\odot(I_A)$ touches $BC$. Using our previous lemma, we know that $MD=ME$. Now as both $\odot(I),\odot(I_A)$ are tangent to $BC$, we get $$\operatorname{Pow}_{\odot I}(M)=MD^2=ME^2=\operatorname{Pow}_{\odot I_A}(M).$$ Apart from this Lemma, we use the fact that the if $DEF$ is the contact triangle, then $AM,EF,ID$ concur which I refer to as Incircle Concurrency Lemma. Let $I,I_A$ be the incenter and $A-$excenter. $\odot(I), \odot(I_A)$ denote the incircle and $A-$excircle. $D,E$ be the point where $\odot(I)$ touches $AB,AC$. $K,L$ be the points where the $\odot(I_A)$ touches $AB,AC$. Let $\ell$ be the line through $M$ such that $\ell\perp II_A$. Also let $R=DP\cap QK,S=PE\cap QL$. Firstly by first Lemma, note that as $M$ lies on the Radical Axis of $\{\odot(I),\odot(I_A)\}$ and $I,I_A$ are the centers respectively, from $\ell\perp II_A$ we get that $\ell$ itself is the Radical Axis of $\{\odot(I),\odot(I_A))$. $\textbf{CLAIM: }EPQL$ and $PDKQ$ is cyclic. $\textbf{PROOF: }$This could be proved aswell using inversion at $A$ with radius $\sqrt{AE\cdot AL}$, but for the sake of simplicity, we provide a simpler proof. Let $P'=AM\cap \odot(I_A)(\not=Q)$. Now note that the homothety $\Psi$ centred at $A$ that maps $\odot(I)\mapsto\odot(I_A)$ also maps $D\mapsto K,E\mapsto L$ and $P\mapsto P'$, which implies $EP\parallel LP'$ and $DP\parallel KP'$. Now $$\measuredangle LEP=\measuredangle AEP=\measuredangle ALP'=\measuredangle LQP'=\measuredangle LQP\implies PELQ\text{ is cyclic.}$$ Similarly, $PDKQ\text{ is cyclic.}$ $\textbf{CLAIM: }R,S\in\ell$. $\textbf{PROOF: }$We have that $$\operatorname{Pow}_{\odot I}(S)=SE\cdot SP=\operatorname{Pow}_{\odot EPQL}(S)=SL\cdot SQ=\operatorname{Pow}_{\odot I_A}(S)$$and so $S\in\ell$ and similarly $R\in \ell$. $\textbf{CLAIM: }PQRS$ is cyclic. $\textbf{PROOF: }$We use the same idea of homothety from the previous claim. We get $D\xmapsto{\Psi}K,E\xmapsto{\Psi}L,P\xmapsto{\Psi}P'$ and so we get $$\measuredangle RPS=\measuredangle DPE=\measuredangle KP'L=\measuredangle KQL=\measuredangle RQS.$$ $\textbf{CLAIM: }\{\odot(PEFD),\odot(PQRS)\}$ are tangent to each other at $P$. $\textbf{PROOF: }$Let $\ell_1,\ell_2$ be the tangents at $P$ to $\odot(PEFD),\odot(PQRS)$. Now firstly notice that as $AD=AE$ and $AI$ is the angle bisector of $\angle DAE$, we get $AI\perp DE$. Moreover as $\overline{A-I-I_A}$ are collinear $\implies DE\perp II_A$ and $RS\perp II_A\implies DE\parallel RS$. So $$\measuredangle(\ell_1,PD)=\measuredangle PED\stackrel{DE\parallel RS}{=}\measuredangle PSR=\measuredangle(\ell_2,PR)=\measuredangle(\ell_2,PD)\implies\ell_1\equiv\ell_2$$ Now since the tangents at $P$ to $\{\odot(PEFD),\odot(PQRS)\}$ are the same line, we conclude the circles are tangent at $P$. Now let $U,V=BC\cap\odot(PQRS)$. $\textbf{CLAIM: }M$ is the midpoint of $UV$. $\textbf{PROOF: }$Let $M'=PM\cap DE$ and $F$ be the point where $\odot(I)$ touches $BC$. Also $O$ be the centre of $\odot(PQRS)$. Now by Incircle Concurrency Lemma we know that $\overline{M'-I-F}$ are collinear $\implies M'I\perp BC$. Now as $\overline{P-D-R}$ and $\overline{P-E-S}$ are collinear and $DE\parallel RS$, we can say there exists a homothety $\mathcal{H}$ centred at $P$ that sends $DE\xmapsto{\mathcal{H}} RS$. Also as $P$ is the centre of $\mathcal{H}$, $P\xmapsto{\mathcal{H}}P\implies\odot(PED)\xmapsto{\mathcal{H}}\odot(PSR)$ and so their centers also get mapped, i.e. $I\xmapsto{\mathcal{H}}O$. Also, observe that the line $PM\xmapsto{\mathcal{H}}PM\implies PM\cap DE\xmapsto{\mathcal{H}}PM\cap SR\implies M'\xmapsto{\mathcal{H}}M$. Therefore from $M'\xmapsto{\mathcal{H}}M$ and $I\xmapsto{\mathcal{H}}O$, we can conclude that $M'I\parallel MO$ which together with the fact $MI'\perp BC\implies MO\perp BC$. So, we get that $OM\perp UV$ from which we can conclude that $M$ is the midpoint of $UV$. Now let $T_1=QR\cap BC,T_2=PR\cap BC,T_3=QS\cap BC,T_4=PS\cap BC$. $\textbf{CLAIM: }MT_1=MT_4$ and $MT_2=MT_3$. $\textbf{PROOF: }$We already know that $M$ is the midpoint of $UV$. So by Butterfly Theorem on $\odot(PQRS)$ with $\{PQ,RS,UV\}$ as the chords, we get $T_4=UV\cap PS,T_1=UV\cap QR$ so $MT_1=MT_4$. Similarly by Butterfly Theorem on $\odot(PQRS)$ with $\{QP,RS,UV\}$ as the chords, we get $T_2=UV\cap PR,T_3=UV\cap SQ$ so $MT_2=MT_3$. Now finally we change the definitions of $P',Q'$ a little (because yeah, I messed them up and changing the definitions here would have needed to rewrite the entire Length Bash below which I am too lazy to do ). Let $P'=AM\cap\odot(I)(\not=P)$ and $Q'=AM\cap\odot(I_A)(\not=Q)$. As $AD,AE$ are tangents to $\odot(I),\odot(I_A)$, we get $(P,P';D,E)=-1$ and $(Q,Q';L,K)=-1$. So, projecting onto $BC$ from $P$ and $Q$, we get $-1=(P,P';D,E)\stackrel{P}{=}(X,M;T_2,T_4)$ and $-1=(Q,Q';L,K)\stackrel{Q}{=}(Y,M;T_3,T_1)$. We use signed lengths for the rest of the solution in order to avoid configuration issues. So using our signed conventions, we have $MT_1=T_4M$ and $MT_3=T_2M$. So as $(X,M;T_2,T_4)=(Y,M;T_3,T_1)=-1$, we get \begin{align*} &(X,M;T_2,T_4)=(Y,M;T_3,T_1)\\ &\implies \dfrac{T_2X}{T_2M}\div \dfrac{T_4X}{T_4M}=\dfrac{T_3Y}{T_3M}\div \dfrac{T_1Y}{T_1M}\\ &\implies \dfrac{T_2X\cdot T_4M}{T_2M\cdot T_4X}=\dfrac{T_3Y\cdot T_1M}{T_3M\cdot T_1Y}\\ &\implies \dfrac{T_2X\cdot \cancel{T_4M}}{\cancel{T_2M}\cdot T_4X}=\dfrac{T_3Y\cdot -\cancel{MT_1}}{-\cancel{MT_3}\cdot T_1Y}\\ &\implies \dfrac{T_2X}{T_4X}=\dfrac{T_3Y}{T_1Y}\\ &\implies \dfrac{T_2X}{T_4X}=\dfrac{-YT_3}{-YT_1}\\ &\implies \dfrac{T_2X}{T_4X}=\dfrac{YT_3}{YT_1}\\ &\implies \dfrac{MX-MT_2}{T_4M+MX}=\dfrac{YM-T_3M}{YM+MT_1}\\ &\implies (MX-MT_2)(YM+MT_1)=(YM-T_3M)(T_4M+MX)\\ &\implies MX\cdot YM + MX\cdot MT_1 -MT_2\cdot YM-MT_2\cdot MT_1=YM\cdot T_4M +YM \cdot MX-T_3M\cdot T_4M-T_3M\cdot MX\\ &\implies \cancel{MX\cdot YM} + MX\cdot MT_1 -MT_2\cdot YM+\cancel{T_2M\cdot MT_1}=YM\cdot T_4M +\cancel{YM \cdot MX}+\cancel{MT_3\cdot T_4M}-T_3M\cdot MX\\ &\implies MX\cdot MT_1 -MT_2\cdot YM=YM\cdot T_4M -T_3M\cdot MX\\ &\implies MX\cdot MT_1 +T_3M\cdot MX=YM\cdot T_4M +MT_2\cdot YM\\ &\implies MX(MT_1 +T_3M)=YM(T_4M +MT_2)\\ &\implies MX\cancel{(MT_1 +T_3M)}=YM\cancel{(MT_1 +T_3M)}\\ &\implies MX=YM\\ \end{align*}And we are done .

@above indeed. Rather i feel the problem is misplaced as far as difficulty is concerned. Let AQ intersect the excircle again at R. We consider the homothety centered at A and mapping the incircle to the excircle as $\mathbb{H}$. So $\mathbb{H}(P)=R$ and hence $PX || \text{Line tangent to excircle at R}$. So $\angle XPM= 180^o - \angle MQY$ We know $BW=CZ$ where $W,Z$ are the intouch and extouch points to BC respectively. By Sine Rule we have $$\frac{XP}{XM} = \frac{\sin \angle XMP}{\sin \angle XPM} = \frac{\sin \angle YMQ}{\sin \angle AQY} = \frac{YQ}{YM}$$$$\implies XW/XM=YZ/YM \implies MX=MY$$ @above is this correct tho?After seein the solution of yours i feel i did something horrible...

Yay cross ratio chasing!

Consider labellings as in the following diagram . $G=FE \cap PX$ so $(F,E;H,G)=-1$. Projecting through $A$ on $BC$ we have $AG \parallel BC$ similarly $AG' \parallel BC$ so $A-G-G' \parallel BC$. The homothety at $A$ sending incircle to excircle maps $Q'$ to $Q$ so $\angle PQI_A=\angle AQ'I=\angle IPQ \implies \angle JPQ=\angle JQP \implies JP=JQ$ so $JM$ is radax of the incircle and excircle .Since the radax of incircle and excircle is also midline of trapezoid $FEE'F'$ which bisects $GG'$ so $JM$ bisects $GG'$ and since $GG' \parallel XY$ we get $JM$ is median in $XJY$ and hence $M$ is midpoint of $XY$.

It suffices to show that <MPX + <MQY = 180 degrees, because by Law of Sines on triangles MPX, MQY this would imply that MX/PX = MY/QY, or MX/DX = MY/EY if D, E are tangency points of the incircle, excircle on BC, and that would imply that MX = MY since MD = ME. However, this is obvious because if P’, Q’ are the intersection points of AM with incircle, excircle other than P, Q then homothety at A taking incircle to excircle takes PP’ to Q’Q, so arcs PP’ and Q’Q of the incircle, excircle respectively have the same measure, thus <MPX + <MQY = 180.

Let the incircle and excircle touchpoints be $D,E,F$ and $D', E', F'$, respectively. It is well-known (easily provable with projective) that if $PP \cap EF=T$, then $AT \parallel BC$. Similarly, if $QQ \cap E'F'=T'$, then $AT' \parallel BC$. Let the second tangent from $T$ to the incircle meet $BC$ at $Z$; define $W$ similarly but for the excircle. Then the homothety taking the incircle to the excircle takes $T$ to $T'$ and thus $TX \parallel T'W, T'Y \parallel TZ$. Notice that $TXT'W, TYT'Z$ are parallelograms, so $XW=YZ$. Thus, need to prove that $M$ is midpoint of $WZ$, or equivalently, $ZD=WD'$. This easily follows by length chasing using the equal segments from the parallelograms since the incircle of $ABC$ is also incircle for $\triangle TXZ$ and the excircle of $ABC$ is also excircle for $\triangle T'YW$.

$\textbf{Generalization:}$ Let $\triangle{ABC}$ be a triangle with $(O_1),(O_2)$ are two different circles with centers $O_1,O_2$,they are tangent to $AB,AC$ and satisfying $\overline{BX}=\overline{YC}$ where $X,Y$ are orthogonal projections of $O_1,O_2$ on $BC$. $A-$median intersects $(O_1),(O_2)$ at $P,Q$ with $P$ closer to $A$ and $Q$ further to $A$. Tangents from $P,Q$ to $(O_1),(O_2)$ intersects $BC$ at $Z,T$. Prove that: $\overline{ZB}=\overline{CT}$.

Noob_at_math_69_level wrote: $\textbf{Generalization:}$ Let $\triangle{ABC}$ be a triangle with $(O_1),(O_2)$ are two different circles with centers $O_1,O_2$,they are tangent to $AB,AC$ and satisfying $\overline{BX}=\overline{YC}$ where $X,Y$ are orthogonal projections of $O_1,O_2$ on $BC$. $A-$median intersects $(O_1),(O_2)$ at $P,Q$ with $P$ closer to $A$ and $Q$ further to $A$. Tangents from $P,Q$ to $(O_1),(O_2)$ intersects $BC$ at $Z,T$. Prove that: $\overline{ZB}=\overline{CT}$.[/quot ...

GeoKing wrote: Consider labellings as in the following diagram . $G=FE \cap PX$ so $(F,E;H,G)=-1$. Projecting through $A$ on $BC$ we have $AG \parallel BC$ similarly $AG' \parallel BC$ so $A-G-G' \parallel BC$. The homothety at $A$ sending incircle to excircle maps $Q'$ to $Q$ so $\angle PQI_A=\angle AQ'I=\angle IPQ \implies \angle JPQ=\angle JQP \implies JP=JQ$ so $JM$ is radax of the incircle and excircle .Since the radax of incircle and excircle is also midline of trapezoid $FEE'F'$ which bisects $GG'$ so $JM$ bisects $GG'$ and since $GG' \parallel XY$ we get $JM$ is median in $XJY$ and hence $M$ is midpoint of $XY$. This proof works for generalisation too. But in the generalisation proving that $M$ lies on radax is a little different. Consider the labellings as below $N$ be midpooint of $O_1O_2$. Note that $N$ is midpoint of arc $BC$ in $(ABC)$. The projections of $N$ on $AB,AC$ are $M_1,M_2$ which are midpoints of $T_1T_1'$ and $T_2T_2'$ and $M_1M_2$ is radax of the $2$ circles.$M \in M_1M_2$ since $M-M_1-M_2$ is simson line of $N$ wrt $(ABC)$. So we are done

A nice problem indeed, here is my solution: Let $I_1$ be the center of the incircle of $ABC$ and let $I_2$ be the center of the $A-$excircle of $ABC$. Let $R$ be the intersection of $PI_1$ and $QI_2$. Lemma 1. $RP=RQ$. Proof. Let $P'$ be the second intersection of $AM$ and the $A-$excircle of $ABC$. By homothety centered at $A$ sending the incircle to the excircle it follows that $I_1P$ and $I_2P'$ are parallel, so $\angle P'QI_2 = \angle QP'I_2 = \angle QPI_1$ so $\angle PQR=\angle QPR$ and our lemma is proved. Lemma 2. Given a circle $\omega$ centered at $R$ and circles $\omega_1$ and $\omega_2$ both internally tangent to $\omega$ at points $P$ and $Q$ respectively. The internal tangent of $\omega_1$ and $\omega_2$ touches $\omega_1$ and $\omega_2$ at points $D$ and $E$ respectively. If $PQ$ passes through the midpoint $DE$, $R$ lies on the perpendicular bisector of $DE$. We will prove this lemma in the end of the solution. Lemma 3. $R$ lies on the perpendicular bisector of $BC$. Proof: Lemma 3 follows directly from lemma 2 applied to the incircle of $ABC$, the $A-$excircle and the circle centered at $R$ passing through $P$ and $Q$. Now to finish the problem, $RX=RY$, $RP=RQ$ and $\angle RQY=\angle RPX=90\circ$ so triangles $RPX$ and $RQY$ are congruent and so $YQ=PX$. Finally, let $D$ and $E$ be the touching points of $BC$ with the incircle and the $A-$excircle of $ABC$ respectively. It is well-known $BE=CD$ and so $M$ is the midpoint of both $BC$ and $DE$. $YQ=YE$ and $XD=XP$, so $MX=XD+DM = YE+EM = YM$. $\Box$ Now we prove lemma 2: Let $QE$ intersect $\omega$ at $W_1$ and $PD$ intersect $\omega$ at $W_2$. It is very well-known that $W_1$ and $W_2$ are the midpoints of the respective arcs of $\omega$. Now assume $PQ$ passes through the midpoint of $DE$, but $W_1W_2$ does not pass through this midpoint. Let $L$ be the midpoint of $DE$ and let $PQ$ and $W_1W_2$ intersect at $L'$. Let the line parallel to $DE$ through $L'$ intersect $W_1Q$ and $W_2P$ at $S$ and $T$ respectively and $\omega$ at $F$ and $G$. Then because $L'F=L'G$ (because $W_1W_2$ is a diameter of $\omega$), by Butterfly theorem $SL'=L'T$. But also $LE=LD$ and $ST\parallel DE$ (because $ST$ and $DE$ are both perpendicular to $W_1W_2$) by Steiner's theorem on trapezoid $ESTD$ $LL'$, $ES$ and $TD$ must have a common point (by assumption $L\neq L'$) which is clear contradiction and with this we have proved our lemma.

Let $PP \cap QQ = K$, incircle and excircle touches $BC$ at $D, D'$ respectively. Claim : $\angle KPQ = \angle KQP$ Proof : If $AM$ intersects the excircle again at $Q'$, then by considering homothety taking incircle to excircle, line $KP$ goes to the tangent of excircle through $Q'$. Let this tangent intersects $QQ$ at $S$, then $\angle KPQ=\angle SQ'Q = \angle SQQ' = \angle KQP$ Now $\frac{PX}{XM}=\frac{\sin \angle PMX}{\sin \angle MPX}=\frac{\sin \angle QMY}{\sin \angle MQY}= \frac{QY}{YM}\implies \frac{XM-MD}{XM}=\frac{YM-MD'}{YM} \implies 1-\frac{MD}{XM}=1-\frac{MD'}{YM}$ and since $MD=MD'$, so $XM=YM$

Interesting noone gave a DDIT solution. Proof: Let incircle and excircle touch $BC$ at $D$ and $D’$. $ABCD$ is degenerate quadriple. Apply DDIT from $P$. $(X,X),(B,C),(D,M)$ is an involution. Similarly $(Y,Y),(B,C),(D’,M)$ is an involution. If $X’$ is refletcion of $X$ across $M$, $(X’,X’),(B,C),(D’,M)$ is also involution which implies $X’=Y$ since any involution on a line has $2$ stable points and the other case is not possible

kamatadu wrote: So as $(X,M;T_2,T_4)=(Y,M;T_3,T_1)=-1$, we get \begin{align*} &(X,M;T_2,T_4)=(Y,M;T_3,T_1)\\ &\implies \dfrac{T_2X}{T_2M}\div \dfrac{T_4X}{T_4M}=\dfrac{T_3Y}{T_3M}\div \dfrac{T_1Y}{T_1M}\\ &\implies \dfrac{T_2X\cdot T_4M}{T_2M\cdot T_4X}=\dfrac{T_3Y\cdot T_1M}{T_3M\cdot T_1Y}\\ &\implies \dfrac{T_2X\cdot \cancel{T_4M}}{\cancel{T_2M}\cdot T_4X}=\dfrac{T_3Y\cdot -\cancel{MT_1}}{-\cancel{MT_3}\cdot T_1Y}\\ &\implies \dfrac{T_2X}{T_4X}=\dfrac{T_3Y}{T_1Y}\\ &\implies \dfrac{T_2X}{T_4X}=\dfrac{-YT_3}{-YT_1}\\ &\implies \dfrac{T_2X}{T_4X}=\dfrac{YT_3}{YT_1}\\ &\implies \dfrac{MX-MT_2}{T_4M+MX}=\dfrac{YM-T_3M}{YM+MT_1}\\ &\implies (MX-MT_2)(YM+MT_1)=(YM-T_3M)(T_4M+MX)\\ &\implies MX\cdot YM + MX\cdot MT_1 -MT_2\cdot YM-MT_2\cdot MT_1=YM\cdot T_4M +YM \cdot MX-T_3M\cdot T_4M-T_3M\cdot MX\\ &\implies \cancel{MX\cdot YM} + MX\cdot MT_1 -MT_2\cdot YM+\cancel{T_2M\cdot MT_1}=YM\cdot T_4M +\cancel{YM \cdot MX}+\cancel{MT_3\cdot T_4M}-T_3M\cdot MX\\ &\implies MX\cdot MT_1 -MT_2\cdot YM=YM\cdot T_4M -T_3M\cdot MX\\ &\implies MX\cdot MT_1 +T_3M\cdot MX=YM\cdot T_4M +MT_2\cdot YM\\ &\implies MX(MT_1 +T_3M)=YM(T_4M +MT_2)\\ &\implies MX\cancel{(MT_1 +T_3M)}=YM\cancel{(MT_1 +T_3M)}\\ &\implies MX=YM\\ \end{align*}And we are done . For the last part you could simply reflect points across $M$ Btw I loved your solution.

This problem was proposed by me and Anant.

Awsome geo!, the conditions over incircle and excircle are really similar, suggesting the use of an homothety sending the incircle to the excircle, congrats to the proposer for this nice problem. Let the incircle and A-excircle touch $BC$ at $D,Z$ respectivily. Let $AP$ hit the incircle again at $Q'$, notice that by homothety we get that the tangent from $Q'$ to the incircle is parallel to $QY$, this gives us that $\angle MQY=180-\angle XPM$. Now by LoS we get: $$\frac{YZ}{MY}=\frac{YQ}{MY}=\frac{\sin(YMQ)}{\sin(MQY)}=\frac{\sin(\angle PMX)}{\sin(XPM)}=\frac{XP}{MX}=\frac{XD}{MX} \implies \frac{MX}{MY}=\frac{MX-MD}{MY-MZ} \implies \frac{MX}{MY}=\frac{MD}{MZ}=1$$Hence $MX=MY$ as desired, thus we are done .