-We claim that the fixed point $T$ is the intersection of the circumcircle of the triangle $(ABC) $ and the circumcircle of the triangle $(AXY) $. $\textcolor{red}{\textbf{\underline{Claim.1:}}} BD \parallel CX$ and $CE \parallel BY$. $\textcolor{blue}{\textbf{The proof:}} $ We have: $$\angle{BDX} =\angle{BDP} =\angle{BMP} =\angle{CXP} = \angle{CXD} $$So $BD \parallel CX$. In the same way we can prove that $CE \parallel BY$ $\textcolor{red}{\textbf{\underline{Claim.2:}}} BCXY$ is a cyclic quadrilateral. $\textcolor{blue}{\textbf{The proof:}} $ Let $BD \cap CE=Z$ by radical axis theorem on the quadrilaterals $(BDMP), (ECMP)$ and $(BDCE)$ we can prove that $P, M $ and $Z$ are collinear. Let $BY \cap CX =V $. From Claim.1 we know that $BVCZ$ is a parallelogram, so $Z, M$ and $V$ are collinear, so $P, M$ and $V$ are collinear. We also know that : $$VY \cdot VB= VP \cdot VM= VX \cdot VC$$So $BCXY$ is a cyclic quadrilateral. We will prove that the fixed point $T$ is the intersection of the circumcircle of the triangle $(ABC) $ and the circumcircle of the triangle $(AXY) $. By using radical axis theorem in the quadrilaterals $(BCXY)$, $(ACBT)$ and $(AXYT)$ we conclude that $BC, XY$ and $AT$ are concurrent. So it is enough to prove that as $P$ varies, the line $XY$ passes through a fixed point $K$ on $BC$. $\textcolor{red}{\textbf{\underline{Claim.3:}}} XY$ passes through a fixed point $K$ on $BC$. $\textcolor{blue}{\textbf{The proof:}} $ We will use barycentric coordinates with respect to the triangle $ABC$. We know that $M=(0:1:1)$. Let $P=(m:1:1)$. Let us find the equation of $(PMB)$ it is easy to prove that $v=0$ and that $w=\frac{a^2}{2} $ (since $B, M$ lies on that circle). Now since $P$ lies on that circle so we have : $$(um+\frac{a^2}{2} )(m+2)=a^2+b^2m+c^2 \Longrightarrow u=\frac{2b^2+2c^2-a^2}{2(m+2)}=L$$So the equation of $(PMB) $ is $$ -\sum a^2yz + (\sum x)(Lx+\frac{a^2}{2}z)=0$$Now we will find the coordinates of $D=(x:y:z)$. Since $D \in (ABC) $ so : $$\sum a^2yz=0$$Since $D \in (PMB) $ so : $$- \sum a^2yz + (\sum x)(Lx+\frac{a^2}{2}z)=0 \Longrightarrow (\sum x)(Lx+\frac{a^2}{2}z)=0$$$$\Longrightarrow - 2Lx = a^2z \Longrightarrow D=(a^2:t:-2L)$$Now since $D \in (ABC) $ so : $$-2Lta^2-2La^2b^2+a^2c^2t=0 \Longrightarrow t=\frac{2Lb^2}{c^2-2L}$$$$\Longrightarrow D=(a^2(c^2-2L):2Lb^2:-2L(c^2-2L))$$ Let $P_\infty$ be a point at infinity along the line $BD$, it is easy to prove that $P_\infty=(a^2(c^2-2L):(c^2-2L)(2L-a^2):-2L(c^2-2L))$. Let $X=(x:y:z) $, Since $BD \parallel CX$ so $C, X$ and $P_\infty$ are collinear, we have : \[\begin{vmatrix} x & y & z \\ a^2(c^2-2L) & (c^2-2L)(2L-a^2) & -2L(c^2-2L) \\ 0 & 0 & 1 \end{vmatrix}=0.\]$$\Longrightarrow (2L-a^2)x=a^2y$$So $X=(a^2:2L-a^2:t_{1})$ for some real number $t_{1}$. Similarly $Y=(a^2:t_{2}:2L-a^2)$. Let $XY \cap BC =K=(0:y:z) $, we need to prove that $\frac{z}{y}$ does not depend on $m$. Since $X, Y$ and $K$ are collinear, we have : \[\begin{vmatrix} a^2 & 2L-a^2 & t_{1} \\ a^2 & t_{2} & 2L-a^2 \\ 0 & y & z \end{vmatrix}=0.\] \[\Longleftrightarrow \begin{vmatrix} 1 & 2L-a^2 & t_{1} \\ 1 & t_{2} & 2L-a^2 \\ 0 & y & z \end{vmatrix}=0.\]$$\Longleftrightarrow (2L-a^2-t_{1})y=-(2L-a^2-t_{2})z \Longleftrightarrow \frac{z}{y} =\frac{2L-a^2-t_{1}}{-(2L-a^2-t_{2})}$$So we need to prove that $\frac{2L-a^2-t_{1}}{2L-a^2-t_{2}}$ does not depend on $m$. Since $X, P$ and $D$ are collinear, we have: \[\begin{vmatrix} -a^2 & -(2L-a^2) & -t_{1} \\ a^2(c^2-2L) & 2Lb^2 & -2L(c^2-2L) \\ m & 1 & 1 \end{vmatrix}=0.\] Now we will add $(2L-a^2)$ times the third row to the first row. This gives: \[\begin{vmatrix} -a^2+2Lm-a^2m & 0 & 2L-a^2-t_{1} \\ a^2(c^2-2L) & 2Lb^2 & -2L(c^2-2L) \\ m & 1 & 1 \end{vmatrix}=0.\] $$\Longrightarrow (2L-a^2-t_{1})(a^2(c^2-2L) - 2Lb^2m) +(2Lm-a^2-a^2m)(2Lb^2-2L(2L-c^2))=0$$$$-(2L-a^2-t_{1})=\frac{(a^2+a^2m-2Lm)(2Lb^2-2L(2L-c^2))}{a^2(2L-c^2)+ 2Lb^2m} $$ Let $d=a^2+a^2m-2Lm$, it is easy to see that $d$ is symmetric with respect to $b$ and $c$. Now we have (remember that $L=\frac{2b^2+2c^2-a^2}{2(m+2)}$) : $$-(2L-a^2-t_{1})=\frac{d(2Lb^2-2L(2L-c^2))}{a^2(2L-c^2)+ 2Lb^2m} $$$$=\frac{d(\frac{2b^4+2b^2c^2-a^2b^2}{m+2} - (\frac{2b^2-a^2-c^2m}{m+2})(\frac{2b^2+2c^2-a^2}{m+2}))} {\frac{2a^2b^2-a^4-a^2c^2m+2b^4m+2b^2c^2m-a^2b^2m}{m+2}}$$ $$=\frac{d(m(2b^4+2c^4+4b^2c^2-a^2b^2-a^2c^2)-(2b^2+2c^2-a^2)(2b^2-a^2)+2b^2(2b^2+2c^2-a^2))}{(m+2)(m(2b^2-a^2)(b^2+c^2)+a^2(2b^2-a^2))}$$$$=\frac{d(m(b^2+c^2)(2b^2+2c^2-a^2)+a^2(2b^2+2c^2-a^2)) }{(m+2)(2b^2-a^2)(m(b^2+c^2)+a^2)}$$$$=\frac{d(2b^2+2c^2-a^2)}{(m+2)(2b^2-a^2)}$$$$\Longrightarrow -(2L-a^2-t_{1})=\frac{d(2b^2+2c^2-a^2)}{(m+2)(2b^2-a^2)}$$Similarly we can prove that : $$-(2L-a^2-t_{2})=\frac{d(2b^2+2c^2-a^2)}{(m+2)(2c^2-a^2)}$$So we have : $$\frac{-(2L-a^2-t_{1})}{-(2L-a^2-t_{2})} =\frac{\frac{d(2b^2+2c^2-a^2)}{(m+2)(2b^2-a^2)}}{\frac{d(2b^2+2c^2-a^2)}{(m+2)(2c^2-a^2)}}=\frac{2c^2-a^2}{2b^2-a^2}$$So $\frac{z}{y} =\frac{2c^2-a^2}{a^2-2b^2}$ and it does not depend on $m$. As desired. $\blacksquare$

[asy][asy] size(10cm,0); defaultpen(fontsize(10pt)); pair A = (1,4); pair B = (0,0); pair C = (4,0); pair M = (B+C)/2; pair O = circumcenter(A,B,C); pair P = 0.85*M + 0.15*A; pair Ob = circumcenter(B,P,M); pair Oc = circumcenter(C,P,M); pair E = 2*foot(B,O,Ob) - B; pair F = 2*foot(C,O,Oc) - C; pair X = 2*foot(Oc,P,E) - P; pair Y = 2*foot(Ob,P,F) - P; pair D = 2*foot(O,A,M) - A; pair C1 = 2*foot(O,P,E) - E; pair B1 = 2*foot(O,P,F) - F; pair K = extension(B,E,C,F); pair L = 2*M-K; pair T = extension(X,Y,B,C); pair Q = 2*foot(O,A,T)-A; draw(A--B--C--cycle,linewidth(1.5)); draw(C1--E--F--B1); draw(B--B1,blue); draw(C--C1,blue); draw(A--K,blue); draw(Y--B--K--C--X--L--cycle); draw(X--Y); draw(circumcircle(A,B,C)); draw(circumcircle(B,P,M),red); draw(circumcircle(C,P,M),red); draw(B--T--Y); draw(T--C1,dashed); draw(A--T,dashed); draw(circumcircle(A,X,Y),dashed); dot("$A$",A,N); dot("$B$",B,SW); dot("$C$",C,SE); dot("$D$",E,S); dot("$E$",F,S); dot("$M$",M,1.5*dir(55)); dot("$P$",P,dir(200)); dot("$B_1$",B1,NW); dot("$C_1$",C1,N); dot("$X$",X,2*dir(100)); dot("$Y$",Y,1.5*N); dot("$K$",K,S); dot("$L$",L,dir(150)); dot("$Q$",T,W); dot("$T$",Q,NW); [/asy][/asy] The proof proceed in five steps. Step 1 : Introducing $K,L$. First, by Radical Axis theorem on $\omega_B, \omega_C$ and $\Omega$, lines $BD, CE, AM$ are concurrent at $K$. Moreover, by Reim's theorem on $\overline{XPD}$ and $\overline{CMB}$ gives $CX\parallel BD$. Similarly $BY\parallel CE$. Thus $BY, CX$ meet at point $L$ such that $BLCK$ is parallelogram. Step 2 : $B,C,X,Y$ are concyclic. Clearly $L$ is reflection of $K$ across $M$ so $L\in AM$. Thus $LB\cdot LY = LP\cdot LM = LC\cdot LX$ or $B,C,X,Y$ are concyclic. Step 3 : $XY\parallel DE$ Notice that reflections $X_1, Y_1$ of $X,Y$ across $M$ lies on $BD, CE$ respectively. By reflection, $B,C,X_1, Y_1$ are concyclic hence Reim's theorem on $\odot(BCX_1Y_1)$ and $\odot(BCDE)$ gives $X_1Y_1\parallel DE$. Hence $XY\parallel DE$ as claimed. Step 4 : Key construction. Now let $B_1, C_1$ be points on $\Omega$ such that $BB_1\parallel CC_1\parallel AM$. By Reim's theorem on $PDB_1$ and $BMC$ gives $P, E, B_1$ are colinear. Similarly $P, D, C_1$ are colinear. As $XY\parallel DE$, applying Reim's theorem again gives $B_1, C_1, X, Y$ are concyclic. Step 5 : Conclusion. Now let $Q=B_1C_1\cap BC$, which is a fixed point. By Radical Axis on $\odot(BCXY)$, $\odot(B_1C_1XY)$, $\Omega$, we get $Q\in XY$. By Radical Axis on $\odot(AXY)$, $\odot(ABC)$, $\odot(BCXY)$, we get $\odot(AXY)$ pass through $AQ\cap\Omega$, which is fixed point so we are done.

I neglected to mention this on the results thread out of fear that the AoPS community would hate me, but I'm the author of this problem

juckter wrote: I neglected to mention this on the results thread out of fear that the AoPS community would hate me, but I'm the author of this problem Unfortunately, I did not solve this in the exam but I found the solution quickly afterward with the aid of Geogebra. In my opinion, this problem is a reasonable APMO P3 only if it's given in the statement that the fixed point lies on $\odot(ABC)$ as it's nearly impossible to try degenerate cases or claim the point using compass (Diagram has too many steps to construct). Apart from the difficulty to guess the point, the configuration in this problem is rich and nice. Congrats on finding it.

this problem gives me crippling depression

Quite nice but unfortunately didn't realize this at the exam due to difficulty of constructing the diagram Claim 01. $BY, PM, CX$ concur. Proof. By radical axis on $BDPM$, $PMEC$ and $BDCE$ gives us that $BD, PM, \text{and} EC$ concur. Reim Theorem gives us that $BD \parallel XC$, and $BY \parallel EC$. So, we must have $BY, PM, XC$ concur. Let this three lines concur at point $G$, now notice that this gives us that $BYXC$ is cyclic. Radical axis again on $AZBC$, $BYXC$, $AZYX$ gives us that $AZ, XY, BC$ concur. Therefore, to prove that $Z$ is fixed as $P$ varies, we just need to prove that $XY \cap BC$ is fixed as $P$ varies. By Menelaos, it suffices to prove that $\frac{CX}{XG} \cdot \frac{GY}{BY}$ is constant. Claim 02. $\frac{CX}{XG} \cdot \frac{GY}{BY} $ is constant. Proof. \[ \frac{CX}{XG} \cdot \frac{GY}{BY} = \frac{CX}{BY} \cdot \frac{GY}{GX} = \frac{CX}{BY} \cdot \frac{GC}{GB} = \frac{CX}{BY} \cdot \frac{BE}{CD} \]Let $R_1$ and $R_2$ be radii of circumcircle $CPM$ and $BPM$ respectively. But, we know that \[ \frac{CX}{BY} = \frac{R_1 \cdot \sin \angle XPC}{R_2 \cdot \sin \angle YPB} = \frac{PC \sin \angle DPC}{PB \sin \angle BPE} \]So, \[ \frac{CX}{BY} \cdot \frac{BE}{CD} = \frac{PC \cdot BE \cdot \sin DPC}{PB \cdot CD \cdot \sin BPE} = \frac{\sin (\angle AMC - \angle BAC)}{\sin (180^{\circ} - \angle BAC - \angle AMC)}\]

Rip tried inversion on this for way to long but any inversion doesn't give anything nice. Does anyone have a inversion solution somehow?

Solution. Basically, the official solution, but anyway... Let $T$ be the second intersection point of $(AXY)$ and $\Gamma$. We show that $T$ is the fixed point we need. By the radical axis theorem, $CE$, $MP$ and $DB$ concur at a point, say $Q$. By Reim's theorem, $BD\parallel CY$ and $CE\parallel BX$. Let $R=\overline{BX}\cap \overline{CY}$. The previous parallelisms imply that $QCRB$ is a paralelogram, therefore, $R$ lies on $MP$. Since $BD\parallel YR$, we obtain that $PXRY$ is cyclic; then, the radical axes of $(PXRY)$, $(AXTY)$ and $(ABC)$ are concurrent, i.e. $AT,\ XY$ and $BC$ meet at a common point $S$; hence, it suffices to show that $S$ is a fixed point. Notice that $$SB\cdot SC=SA\cdot ST=XS\cdot SY$$i.e. $BXCY$ is cyclic. Let $L=\overline{EX}\cap \Gamma,\ L\neq E$ and $K=\overline{PY}\cap \Gamma,\ K\neq D$. Using directed angles, we get $$\measuredangle LPM=\measuredangle ECM=\measuredangle ECB=\measuredangle PLB$$in other words, $BL\parallel PM$. Similarly, we prove that $PM\parallel KC$. Moreover, note that \begin{eqnarray*} \measuredangle KLP&=&\measuredangle KLB-\measuredangle PLB=\measuredangle KCB-\measuredangle LPM=\measuredangle AMB-\measuredangle XBC\\ &=&\measuredangle PYC-\measuredangle XYC= \measuredangle PYX \end{eqnarray*}which gives that $KYLX$ is cyclic; therefore, the radical axis theorem applied to $(KYLX),\ (AXTY)$ and $(AKTL)$ leads to infer that $XY$, $AT$ y $KL$ are concurrent, i.e. $S$ coincides with the intersection point of $KL$ and $BC$. Since $K$ and $L$ are fixed, we conclude that $S$ is fixed, as required. $\blacksquare$

Define $U = DP \cap \Gamma$ and $V = EP \cap \Gamma$. Observe that \[\measuredangle BMP = \measuredangle BDP = \measuredangle BCU\]so $AM \parallel UC$. Similarly, $AM \parallel VB$ and it follows that $U, V$ are fixed points. By Reim's, $BD \parallel CX$ and $BY \parallel CE$, and by radical axis, we have a point $R = BD \cap PM \cap CE$. Define $R' = BY \cap CX$. Since $BRCR'$ is a parallelogram, we have $R' \in RM$. Hence $R'$ is on the radical axis of $\odot(BMP), \odot(CMP)$ and thus $BCXY$ is cyclic. Now, as $BC, XY$ antiparallel in $\angle BR'C$ and $BC, DE$ antiparallel in $\angle BRC$, we have $DE \parallel XY$. Then since $UV, DE$ antiparallel in $\angle DPE$ we have $UVYX$ cyclic. Hence by radical axis we have a point $Z = UV \cap XY \cap BC$, which is fixed. Finally, if we define the fixed point $T = AZ \cap \Gamma$, then $ZT \cdot ZA = ZX \cdot ZY$ so $T \in \odot(AXY)$ is the desired fixed point. $\Box$

By proof is almost the same as MarkBcc168's so I won't write down the whole proof.

The fixed point part was tough! APMO 2019 P3 wrote: Let $ABC$ be a scalene triangle with circumcircle $\Gamma$. Let $M$ be the midpoint of $BC$. A variable point $P$ is selected in the line segment $AM$. The circumcircles of triangles $BPM$ and $CPM$ intersect $\Gamma$ again at points $D$ and $E$, respectively. The lines $DP$ and $EP$ intersect (a second time) the circumcircles to triangles $CPM$ and $BPM$ at $X$ and $Y$, respectively. Prove that as $P$ varies, the circumcircle of $\triangle AXY$ passes through a fixed point $T$ distinct from $A$. Solution: Let $G$ be the radical center WRT $\odot (ABC),$ $\odot (BMP)$ and $\odot (CMP)$. Apply Reim's Theorem on $\odot (BMP)$ and $\odot (CMP)$ $\implies$ $BY||CE$ and $XC||BD$. If $G'$ is the reflection of $G$ over $M$, then $BG'CG$ forms a parallelogram $\implies$ $BY$ $\cap$ $CX$ $\cap$ $AM$ $=$ $G'$ $$G'Y \cdot G'B = G'P \cdot G'M = G'X \cdot G'C \implies BYCX \text{ is cyclic}$$$\angle YPD=180^{\circ}-\angle YBD=\angle DGE$ $\implies$ $PDGE$ is cyclic and, $\angle BYP$ $=$ $\angle PMC$ $=$ $\angle G'XP$ $\implies$ $G'XYP$ is cyclic $\implies$ $\angle PED$ $=$ $\angle BGG'$ $=$ $\angle PG'X$ $=$ $\angle PYX$ $\implies$ $XY||DE$. Let $DP, EP$ $\cap$ $\odot (ABC)$ $=$ $K, N$. Apply Reim's Theorem on $\odot (BMP)$, $\odot (CMP)$ and $\odot (ABC)$ $\implies$ $CK|| AM || BN$ $\implies$ $N, K$ are fixed points. Since, $XY$ $,$ $KN$ are antiparallel WRT $\angle NPK$ $\implies$ $NYXK$ is cyclic. Let $\odot (AXY)$ $\cap$ $\odot (ABC)$ $=$ $T$. Apply Radical Axes Theorem on $\odot (ABC)$, $\odot (BYXC)$, $\odot (AXYT)$ and $\odot (NYXK)$, $\odot (AXYT)$, $\odot (ABC)$ $\implies$ $AT$ $\cap$ $NK$ $\cap$ $XY$ $\cap$ $BC$ $=$ $F$ is a fixed point $\implies$ $T$ is the desired fixed point

MarkBcc168 wrote: In my opinion, this problem is a reasonable APMO P3 only if it's given in the statement that the fixed point lies on $\odot(ABC)$ as it's nearly impossible to try degenerate cases It somewhat natural to consider the case when $D$, $P$, $E$ are collinear. In this case $Y=D$, $X=E$, ($AXY$) coincide with circumcircle.

This solution is divided into three main parts. Part I: the parallelogram. First, we let $K$ denote the concurrence point of lines $BD$, $PM$, $CE$ (by radical axis). Note also that $\measuredangle BYP = \measuredangle BMP = \measuredangle PMC = \measuredangle PEC$, which implies $\overline{BY} \parallel \overline{CE}$. Similarly $\overline{CX} \parallel \overline{BD}$. Thus we may construct parallelogram $BLCK$ with $L$ the concurrence point of lines $BY$, $AM$, $CX$. Note now that: $BCDE$ is cyclic. $BYXC$ is cyclic, by power of a point from $L$ ($LY \cdot LB = LP \cdot LM = LX \cdot LC$). Thus $\overline{XY} \parallel \overline{DE}$; say bay $\measuredangle PYX = \measuredangle PLX = \measuredangle PKD = \measuredangle PED$. [asy][asy] size(11cm); pair A = dir(119); pair B = dir(210); pair C = dir(330); pair M = (B+C)/2; draw(A--B--C--cycle, deepcyan); filldraw(unitcircle, invisible, deepcyan); pair O = origin; pair S = extension(O, foot(O, A, M), B, C); pair T = -A+2*foot(O, A, S); pair K = 1.75*M-0.75*A; pair L = B+C-K; pair D = -B+2*foot(O, B, K); pair E = -C+2*foot(O, C, K); filldraw(B--L--C--K--cycle, invisible, orange); draw(circumcircle(B, D, M), dashed+deepgreen); draw(circumcircle(C, E, M), dashed+deepgreen); pair P = -M+2*foot(circumcenter(B, M, D), K, M); pair X = extension(D, P, C, L); pair Y = extension(E, P, B, L); draw(A--K, red); draw(A--S, lightblue); draw(B--S, deepcyan); pair Bp = extension(E, Y, B, B+A-M); pair Cp = extension(D, X, C, C+A-M); draw(D--Cp, orange); draw(E--Bp, orange); draw(C--E, orange); draw(B--Bp, red); draw(C--Cp, red); draw(S--Cp, blue); draw(S--X, blue); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(M)); dot("$O$", O, dir(135)); dot("$S$", S, dir(S)); dot("$T$", T, dir(T)); dot("$K$", K, dir(K)); dot("$L$", L, dir(45)); dot("$D$", D, dir(D)); dot("$E$", E, dir(300)); dot("$P$", P, dir(P)); dot("$X$", X, dir(80)); dot("$Y$", Y, dir(100)); dot("$B'$", Bp, dir(Bp)); dot("$C'$", Cp, dir(Cp)); /* TSQ Source: !size(11cm); A = dir 119 B = dir 210 C = dir 330 M = (B+C)/2 A--B--C--cycle deepcyan unitcircle 0.1 lightcyan / deepcyan O = origin R135 S = extension O foot O A M B C T = -A+2*foot O A S K = 1.75*M-0.75*A L = B+C-K R45 D = -B+2*foot O B K E = -C+2*foot O C K R300 B--L--C--K--cycle 0.1 yellow / orange circumcircle B D M dashed deepgreen circumcircle C E M dashed deepgreen P = -M+2*foot circumcenter B M D K M X = extension D P C L R80 Y = extension E P B L R100 A--K red A--S lightblue B--S deepcyan B' = extension E Y B B+A-M C' = extension D X C C+A-M D--Cp orange E--Bp orange C--E orange B--Bp red C--Cp red S--Cp blue S--X blue */ [/asy][/asy] Part II: the isosceles trapezoid. Let line $EP$ meet $\Gamma$ again at $B'$. Define $C'$ similarly. We apply Reim's theorem three times: Since $\measuredangle BB'E = \measuredangle ECB = \measuredangle MPE$, we have $\overline{BB'} \parallel \overline{PM} \equiv \overline{AM}$. Similarly, $\overline{CC'} \parallel \overline{AM}$. Since $\measuredangle XYP = \measuredangle PED = \measuredangle BCX$ (using $\overline{XY} \parallel \overline{DE}$) we get $B'C'XY$ cyclic as well. Part III: finishing. We now describe the fixed point. By radical axis now, lines $B'C'$, $YX$, $BC$ meet at $S$. Since $BB'C'C$ was an isosceles trapezoid with legs parallel to $\overline{AM}$, we could also have described $S$ as the point on $BC$ such that $\overline{SO} \perp \overline{AM}$, where $O$ is the center of $\Gamma$. This is independent of $P$. Consequently, if $\overline{AS}$ meets $\Gamma$ again at $T$, then $ATXY$ is cyclic and moreover $T$ does not depend on $P$. Remark: In my opinion, Part II is the toughest part, as it requires the introduction of the magical points $B'$ and $C'$. Part I is routine: working nearly only with points already given or which are natural to add. Meanwhile Part III is mostly the identification of $S$, which can eventually be done with a good figure. Parts I and III also motivate each other.

MarkBcc168 wrote: In my opinion, this problem is a reasonable APMO P3 only if it's given in the statement that the fixed point lies on $\odot(ABC)$ as it's nearly impossible to try degenerate cases or claim the point using compass (Diagram has too many steps to construct). Apart from the difficulty to guess the point, the configuration in this problem is rich and nice. Congrats on finding it. My thoughts on finding the point. It's reasonable to guess that the fixed point is the second intersection of $(AXY)$ with the circumcircle by considering the isosceles case, where the diagram remains intact in all regards except that the circle has no fixed point other than $A$. To find the exact point, it's helpful to consider $P=M$, whence $(BMP)$ and $(CMP)$ are tangent to the median, $XYDE$ is a parallelogram, and $X, Y \in (BHC)$. Then, constructing the parallels to $AM$ through $B, C$ (the important step in the general solution) is natural if one attempts the inversion $\odot(M, MB)$.

Here's a coord-bash solution that I submitted at the contest The key claim is to find that $XY \cap BC$ is fixed. Step 1. Let $Z=XY\cap BC$, $U=AZ \cap \odot(ABC)$. Then $U$ is fixed whenever $Z$ is fixed; and $U$ lies on $\odot(AXY)$. Let $T=AM \cap BD \cap CE$; which exists as the radical center; and denote $K$ by the image of reflection of $T$ by $M$. Reim's theorem on circles $\odot(BPM), \odot(CPM)$ gives $BY \parallel CE$, $CX \parallel BD$, hence $K$ lies on both $BY$ and $CX$. Note that $KX \cdot KC=KP \cdot KM =KY \cdot KB$, or $(XYBC)$ is cyclic; then the radical center of $\odot(XYBC), \odot(AXY), \odot(ABC)$ coincides with $Z=XY\cap BC$; and $U \in \odot(AXY)$. Step 2. $Z$ is fixed while $P$ varies on $AM$. Note that $Z$ is fixed iff $\frac{BZ}{ZC}=\frac{BY}{YK}\cdot \frac{KX}{XC}$ is constant; now we calculate this term. Let $X', Y'$ be the reflections of $X,Y$ on $M$; and $U=\odot(MDT) \cap BC$, $V=\odot(MET)\cap BC$. Simple angle chasing gives $X' \in \odot(MET), Y' \in \odot(MDT)$; hence $$\frac{BY}{YK} \cdot \frac{KX}{XC}=\frac{BY\cdot KB}{CK\cdot XC}=\frac{CY' \cdot CT}{BT \cdot BX'}=\frac{CM \cdot CU}{BM \cdot BV}=\frac{CU}{BV}.$$ Hence it remains to prove that $\frac{CU}{BV}$ is constant while $T$ varies on $AM$; since $P$ and $T$ corresponds on $AM$. Now it's time to put coordinates WLOG let $M=(0,0), C=(1,0), B=(-1,0)$ and $\Gamma : x^2+(y-a)^2=a^2+1$ for some $a \in \mathbb{R}$. Let $AM : y=\alpha x$ for some fixed $\alpha \in \mathbb{R}$; and $T=(p, \alpha p)$ varies on $AM$(i.e. $p$ is moving). Since the diagram is symmetric in both $B$ and $C$; our goal is to express $CU$ in the form $p \times F(a, \alpha)$ where $F(a, \alpha)$ is independent on $p$. The equation of line $BP$ is $y=\frac{\alpha p}{p+1}(x+1)$; so if $D=(d, d')$; where $D$ lies on both $BP$ and $\Gamma$; $d^2+(d'-a)^2=a^2+1$ or $d^2-1=d'(2a-d')$. Plugging in $d'=\frac{\alpha p}{p+1}(d+1)$ gives $(d+1)(d-1)=\frac{\alpha p}{p+1}(d+1)\left(2a-\frac{\alpha p}{p+1}(d+1)\right)$; or $d=\frac{1+2a\frac{\alpha p}{p+1}-\left(\frac{\alpha p}{p+1}\right)^2}{1+\left(\frac{\alpha p}{p+1}\right)^2}$. This looks quite complicated; but anyway : $BM \cdot BU=BU=BD \cdot BT$; where $$BD \times BT=(d+1)(p+1) \times \left(1+\left(\frac{\alpha p}{p+1}\right)^2 \right)=\left(2+2a\frac{\alpha p}{p+1}\right) \times (p+1)=2p+2+2a\alpha p;$$hence $BU=2(a\alpha+1)p+2=CU+2$; or $CU=2(a\alpha+1)p$, as desired. $\square$

As several people have mentioned, if $EP$ and $DP$ meet $\Gamma$ again at $B_1$ and $C_1$ then $BB_1 \parallel AM \parallel CC_1$. Now I'll show a different finish. Note that $\angle BB_1P = \angle B_1PA = \angle YBC$ so $\odot(B_1YB)$, say $\omega_1$, is tangent to $BC$, and similarly with $\odot(C_1XC)$, say $\omega_2$. Since $BCC_1B_1$ is an isosceles trapezoid, the line joining the centers of $\omega_1, \omega_2$ is the common perpendicular bisector of $BB_1, CC_1$. Since $M$ clearly lies on the radical axis of $\omega_1$ and $AM$ is perpendicular to the line joining their centers, $AM$ is the radical axis. Hence, $BXYC$ and $B_1YXC_1$ are cyclic, so if $\odot(AXY)$ meets $\Gamma$ again at $T$, then by the radical axis theorem, $AT$ always passes through the intersection of the perpendicular bisector of $BB_1, CC_1$ with $BC$ which is fixed. Hence $T$ is fixed also, so we are done.

IndoMathXdZ wrote: Therefore, to prove that $Z$ is fixed as $P$ varies, we just need to prove that $XY \cap BC$ is fixed as $P$ varies. Alternatively, let $F = XY \cap BC$ and calculate $$\frac{BF}{CF} = \frac{BY\sin \angle FYB}{CX \sin \angle FXC} = \frac{BY \sin XCB}{CX \sin CBY} = \frac{BY\sin DMB}{CX\sin ECM}.$$Since $$\begin{cases}\frac{BY}{CX} = \frac{BP\sin BPE}{CP \sin CPD}\\ \frac{\sin ECM}{\sin DMB} = \frac{BE}{DC} \end{cases}$$it follows that $$\frac{BF}{CF} = \frac{\frac{\sin BPE}{BE}\cdot BP}{\frac{\sin CPD}{CD}\cdot CP} = \frac{\sin PEB}{\sin PDC},$$so it remains to check this ratio is fixed. Note that $\measuredangle BAE = \measuredangle BCE = \measuredangle MPE$, hence $\measuredangle(AB, PM) = \measuredangle(AE, PE)$, from which we compute $\measuredangle PEB = \measuredangle AEB - \measuredangle AEP = \measuredangle ACB - \measuredangle BAM$ is fixed. Likewise $\measuredangle PDC$ is fixed so we are done.

Let the line passing through $B$ and parallel to $AM$ intersect $\Gamma$ again at $V$, and line passing through $C$ and parallel to $AM$ intersect $\Gamma$ again at $U$. Let $UV$ intersect $BC$ again at $W$ and let $AW$ intersect $\Gamma$ again at $T$. $U$ and $V$ do not depend on $P$ (given that $A, B, C$ are fixed), and neither do the points $W$ and $T$. We show that the circle $AXY$ passes through $T$, thus solving the problem. First, notice that $D, P, U$ are collinear. Since $AP\parallel CU$, we have $\angle(BM, AP)=\angle (BC, AP) = \angle (BC, CU)$ and since $B, C, D, U$ is concyclic and so are points $B, D, P, M$, $\angle (BC, CU)=\angle (BD, DU)$ and $\angle (BD, DP)=\angle (BM, MP)=\angle (BM, AP)$. Thus $\angle (BD, DU)=\angle (BM, AP)=\angle (BD, DP)$, showing that $\angle(DU, DP)=0$ and so $D, U, P$ must be on the same line, and similarly for points $C, P, V$. Next, since $\angle (BD, DP)=\angle (BM, MP)=\angle (CM, MP)=\angle (CX, XP)=\angle (CX, DP)$ (since $D, P, X$ collinear), we have $BD\parallel CX$ and similarly $CE\parallel BY$. Let $CX$ and $BY$ intersect at $R$, and $BD$ and $CE$ intersect at $Q$. Since $BD$ is the radical axis of $\Gamma$ and circle $BPM$, and $CE$ the radical axis of $\Gamma$ and circle $CPM$, $Q$ is the radical center of these circles, hence on $PM$ the radical axis of $BMP$ and $CPM$. Since $BQCR$ is a parallelogram, $R$ is also on $PM$. Now consider the following: \[\angle (YP, PX)=\angle (EP, PD)=\angle (EP, PM)+\angle (PM, PD)=\angle(EC, CM)+\angle (BM, BD) \]\[ =\angle (EC, CD)=\angle (EQ, QB)=\angle (BR, RC)=\angle (RY, RX) \]which shows that $R, Y, P, X$ are concyclic. Furthermore, we also have \[\angle (BY, YX)=\angle (RY, YX)=\angle (RP, PX)=\angle (MP, PX)=\angle (MC, CX)=\angle(BC, CX) \]showing that $B, Y, X, C$ are also concyclic. Finally, with $BD\parallel CX$ and $CE\parallel BY$ we also have \[ \frac{DP}{PE}=\frac{DP}{PQ}\div \frac{PE}{PQ}=\frac{PX}{PR}\div \frac{PY}{PR}=\frac{PX}{PY} \]and therefore there exists a constant $k$ such that $PX=kPD$ and $PY=kPE$. Since $D, E, U, V$ are concyclic on $\Gamma$ with the intersection $P$, we have $PX\cdot PU=kPD\cdot PU=kPE\cdot PV=PY\cdot PE$, showing that $Y, X, U, V$ are also concyclic. We can now complete our solution. $UV$ is the radical axis of circle $YXUV$ and $\Gamma$, $XY$ is the radical axis of circle $YXUV$ and $YXBC$, and $BC$ is the radical axis of circle $YXBC$ and $\Gamma$. Thus, $UV, BC, XY$ concur at pre-defined point $W$. Moreover, $XY$ is the radical axis of the circles $AXY$ and $YXBC$ and again $BC$ is the radical axis of $YXBC$ and $\Gamma$. Again, the radical axis of $AXY$ and $\Gamma$, $XY$ and $BC$ will concur at the same point $W$. Since $T$ is the second intersection of $\Gamma$ and $AW$, we have $T$ as the second intersection of $\Gamma$ and $AXY$, so $T$ is on triangle $AXY$. Q.E.D.

APMO 2019 P3 wrote: Let $ABC$ be a scalene triangle with circumcircle $\Gamma$. Let $M$ be the midpoint of $BC$. A variable point $P$ is selected in the line segment $AM$. The circumcircles of triangles $BPM$ and $CPM$ intersect $\Gamma$ again at points $D$ and $E$, respectively. The lines $DP$ and $EP$ intersect (a second time) the circumcircles to triangles $CPM$ and $BPM$ at $X$ and $Y$, respectively. Prove that as $P$ varies, the circumcircle of $\triangle AXY$ passes through a fixed point $T$ distinct from $A$. Solution. Let $\odot(AXY)$ intersect $\odot(ABC)$ at $T,$ $R$ be the radical center of the three circles $\odot(ABC),~\odot(BPM)$ and $\odot(CPM).$ Extend $DY$ and $EX$ to meet $\odot(ABC)$ again at $Y'$ and $X'$ respectively. By Reim's theorem we have $BX\parallel CE$ and $BD\parallel CY,$ therefore if $BX$ and $CY$ intersect at $K,$ then $KBRC$ is a parallelogram and hence $K$ lies on $AM.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(11cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -14.477288666043442, xmax = 13.417871244440672, ymin = -7.205730489221282, ymax = 12.19540149004279; /* image dimensions */ pen qqffff = rgb(0,1,1); pen ffqqff = rgb(1,0,1); /* draw figures */ draw((-3.735729616724736,6.900195121951225)--(-5.62,-1.12), linewidth(0.4)); draw((-5.62,-1.12)--(2.88,-1.02), linewidth(0.4)); draw((2.88,-1.02)--(-3.735729616724736,6.900195121951225), linewidth(0.4)); draw(circle((-1.407550205967138,2.121767507206701), 5.315429412383223), linewidth(0.4) + blue); draw(circle((-3.505369602912248,-0.21358375245892405), 2.3007068761782645), linewidth(0.5) + qqffff); draw(circle((0.7304271282783986,1.0436940963361134), 2.979848461599009), linewidth(0.5) + qqffff); draw((-5.62,-1.12)--(0.19541637010536891,-6.343922188161932), linewidth(0.4)); draw((0.19541637010536891,-6.343922188161932)--(-3.735729616724736,6.900195121951225), linewidth(0.4)); draw((3.667790780668295,0.5423043998796808)--(0.19541637010536891,-6.343922188161932), linewidth(0.4)); draw((-5.62,-1.12)--(-2.9354163701053695,4.203922188161933), linewidth(0.4)); draw((-2.9354163701053695,4.203922188161933)--(2.88,-1.02), linewidth(0.4)); draw(circle((-3.525965077566698,4.440958994514692), 2.468166017993365), linewidth(0.4) + red); draw((-12.606031472926848,-1.2021886055638455)--(-3.735729616724736,6.900195121951225), linewidth(0.4)); draw((-12.606031472926848,-1.2021886055638455)--(-1.551142971366696,2.960445184636171), linewidth(0.4)); draw((-12.606031472926848,-1.2021886055638455)--(-5.62,-1.12), linewidth(0.4)); 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label("$K$", (-2.846261782952315,4.304642588028671), NE * labelscalefactor); dot((-4.0338292953035255,2.0256086206103623),linewidth(2pt) + dotstyle); label("$X$", (-3.9321460355230635,2.1328740828871706), NE * labelscalefactor); dot((-1.551142971366696,2.960445184636171),linewidth(2pt) + dotstyle); label("$Y$", (-1.4466776351944615,3.049843007280249), NE * labelscalefactor); dot((-5.956210864608256,4.8719449652190345),linewidth(2pt) + dotstyle); label("$T$", (-5.862606928982172,4.980303900739361), NE * labelscalefactor); dot((-12.606031472926848,-1.2021886055638455),linewidth(2pt) + dotstyle); label("$Q$", (-12.498566250247857,-1.1006479136568412), NE * labelscalefactor); dot((0.4716013473283409,7.0939480427238735),linewidth(2pt) + dotstyle); label("$Y'$", (0.5561755417693633,7.200333928217339), NE * labelscalefactor); dot((-6.706472594113806,2.5403500706594455),linewidth(2pt) + dotstyle); label("$X'$", (-6.610660525197576,2.6396200674201875), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] By power of a point, $KX\cdot KB=KP\cdot KM=KY\cdot KC$ hence $BXYC$ is cyclic. It is easy to see that $KXPY$ and $DPER$ are cyclic by using Miquel points in $\triangle KBC$ and $\triangle RBC.$ Therefore $\angle YXP=\angle YKP=\angle DRP=\angle PED$ which implies $XY\parallel DE.$ Since $X'Y'ED$ is cyclic so $X'Y'YX$ is also cyclic. Considering the radical center of $\odot(X'Y'YX),~\odot(ABC),~\odot(BXYC)$ we get $X'Y',XY,BC$ are concurrent again considering that of $\odot(AXY),~\odot(ABC),~\odot(BXYC)$ we get $AT,XY,BC$ are concurrent. Thus, $AT,X'Y',BC$ are concurrent. Now note that $\angle Y'CB=\angle BDY'=\angle PMB=\angle AMB$ so $Y'$ is independent of $P$, similarly $X'$ is also independent of $P,$ thus the line $AT$ is independent of $P.$ Therefore as $P$ varies, the circumcircles of $\triangle AXY$ are coaxal. And we are done. $\blacksquare$

Let AXY meet $\Gamma$ at F. we will prove F is our point. By radical axis theorem we have AF,XY and BC meet at single point S so instead we can prove S is fixed. we will prove SB/SC is fixed. SB/SC = BY . Sin ∠SYB . Sin ∠XSC/ CX . Sin ∠SXC . Sin ∠YSB = BY . Sin ∠SYB/CX . Sin ∠SXC. BY/CX = BP . Sin ∠YPB . Sin ∠CXP/ CP . Sin ∠CPX . Sin ∠BYP = BP . Sin ∠YPB/CP . Sin ∠CPX. SB/SC = BP . BE . Sin ∠YPB/ CP . CD . Sin ∠CPD = Sin ∠PEB/ Sin ∠PDC. ∠PEB = ∠PEC - ∠BEC = ∠AMB - ∠BAC and ∠PDC = ∠PDB - ∠CDB = ∠AMC - ∠BAC so both ∠PEB and ∠PDC are fixed and SB/SC is fixed as wanted. we're Done.

This solution contains way too many instances of "radical axis" and "Reim" [asy][asy] size(11cm); pair A,B,C,D,E,P,Bp,Cp,X,Y,M,T,K,L; defaultpen(fontsize(7pt)); A=dir(124); B=dir(220); C=dir(-40); M=0.5B+0.5C; P=0.5A+0.5M; Bp=intersectionpoints(B--(B+A-M),unitcircle)[0]; Cp=intersectionpoints(C--(C+2A-2M),unitcircle)[0]; path wb, wc; wb=circumcircle(B,P,M); wc=circumcircle(C,P,M); D=intersectionpoints(unitcircle,wb)[1]; E=intersectionpoints(unitcircle,wc)[0]; K=intersectionpoint(M--(2M-A),C--(3C-2E)); draw(A--K--E); X=intersectionpoints(P--(2P-D),wc)[1]; Y=intersectionpoints(P--(2P-E),wb)[1]; T=intersectionpoint((3Bp-2Cp)--Bp,B--(3B-2C)); L=intersectionpoint(Y--(2Y-B),A--M); draw(B--L--C); draw(unitcircle); draw(A--B--C--cycle); draw(wb); draw(wc); draw(D--Cp); draw(E--Bp); draw(Cp--T--B); draw(B--K); draw(T--X); draw(D--E); dot("$A$",A,dir(A)); dot("$B$",B,dir(225)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$B'$",Bp,dir(Bp)); dot("$C'$",Cp,dir(Cp)); dot("$X$",X,2*dir(10)); dot("$Y$",Y,1.5*dir(150)); dot("$M$",M,dir(M)); dot("$T$",T,dir(T)); dot("$P$",P,dir(140)); dot("$K$",K,1.5*dir(K)); dot("$L$",L,dir(80)); [/asy][/asy] By radical axis theorem, we have $BD, CE$ concur at some point $K$ on $AM$. By Reim we have $BY \parallel CE$ and $CX \parallel BY$, so we have $BY,CX$ meet at a point $L$ on $AM$. By radical axis this implies $BCXY$ cyclic. Let $B',C'$ be the points on $(ABC)$ such that $BB' \parallel CC' \parallel AM$, and let $BC$ and $B'C'$ intersect at $T$. It suffices to show that $XY$ always passes through $T$: this will imply by radical axis that $(AXY)$ always passes through the point $(AXY) \cap ABC$. To show $XY$ passes through $T$, by radical axis it suffices to show $B'C'XY$ is cyclic, which by Reim is equivalent to $XY \parallel DE$. Consider a reflection across $M$, which sends $D,E$ to points $D',E'$ on lines $CL,BL$ respectively. Then since $BCD'E'$ is cyclic by Reim we have $XY \parallel D'E'$. But clearly $D'E' \parallel DE$, and we win.

Difficult but nonetheless really beautiful problem! Let $O$ and $H$ be point on $\Gamma $ such that $BO\parallel AM\parallel CH$. From Reim's theorem we get that $BD\parallel CY$; $BX\parallel CE$; $E-P-O-X$ and $D-P-H-Y$. Now let $BD$ intersect $CE$ at $Q$ and $BX$ intersect $CY$ at $R$.By radical axis in circles $(BMPD);(CMPE);(BCED)$ we get that $Q$ lies on $AM$, and since $BRCQ$ is a paralellogram we get that $R$ lies on $AM$ as well. Now from angle chasing we have that the following quadrilaterals are cyclic:$XRYP; XYHO; XBCY$. From radical axis on circles $(BCOH);(XYHO); (BCXY)$ we get that $BC; XY$ and $OH$ are concurrent, say at a point $I$, which doesn't depend on $P$ since neither of $BC$ and $OH$ do. Thus we get that $IX*IY=pow_I((AXY))=IB*IC$ is fixed, so let $G$ be the point on ray $OA$ such that $IG*IA=pow_I((AXY))=IX*IY$, thus from power of a point $(AXY) $ always passes trough $G$, which doesn't depend on $P$.

[asy][asy] unitsize(1cm); pair A, B, C, M, P, D, E, X, Y, Z, B1, C1, Q; A=(2,3sqrt(5)); B=(0,0); C=(8,0); M=(B+C)/2; P=(3*M+A)/4; dot(P); draw(A--B--C--A--M); draw(circumcircle(C,P,M)); draw(circumcircle(B,P,M)); draw(circumcircle(A,B,C)); D=intersectionpoints(circumcircle(B,P,M),circumcircle(A,B,C))[1]; E=intersectionpoints(circumcircle(C,P,M),circumcircle(A,B,C))[1]; B1=extension(B,B+A-M,E,P); C1=extension(C,C+A-M,D,P); draw(B--B1--E); draw(C--C1--D); X=intersectionpoints(P--C1,circumcircle(C,M,P))[1]; Y=intersectionpoints(P--B1,circumcircle(B,M,P))[1]; draw(circumcircle(A,X,Y)); Z=extension(B1,C1,B,C); draw(C--Z--C1--Z--A); draw(circumcircle(B,Y,B1)); draw(circumcircle(C,X,C1)); Q=intersectionpoints(A--Z,circumcircle(A,X,Y))[1]; label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE); label("$M$", M, SW); label("$P$", P, SW); label("$D$", D, SW); label("$E$", E, dir(0)); label("$B_1$", B1, NW); label("$C_1$", C1, N); label("$X$", X, SE); label("$Y$", Y, S); label("$Z$", Z, SW); label("$T$", Q, NW); [/asy][/asy] Let the lines through $B$ and $C$ parallel to $AM$ intersect $\Gamma$ again at $B_1$ and $C_1$, respectively. Then, $$\angle BDP=\angle BMA=\angle BCC_1=\angle BDC_1,$$so $DP$ passes through $C_1$. Similarly, $EP$ passes through $B_1$. Now, $$\angle PXC=180^{\circ}-\angle AMC=\angle BMA=\angle BCC_1,$$so the circumcircle of $CXC_1$ is tangent to $BC$ at $C$. Similarly, the circumcircle of $BXB_1$ is tangent to $BC$ at $B$, so $M$ lies on the radical axis of the two circles. Let $B_1C_1$ and $BC$ intersect at $Z$. Then, $BB_1C_1C$ is a cyclic isosceles trapezoid, so $ZC_1=ZC$ implies the circumcircle of $CXC_1$ is also tangent to $ZC_1$. Therefore, the midpoint of $B_1C_1$ also lies on the radical axis, so the radical axis is parallel to $AM$. This implies the radical axis is $AM$, so $P$ lies on the radical axis. Then, we get $PY\cdot PB_1=PX\cdot PC_1$, so $XYB_1C_1$ is cyclic. Now, consider the homothety centered at $Z$ mapping $B$ to $C$. Then, $B_1$ is mapped to $C_1$ and $Y$ is mapped to a point on the circumcircle of $CXC_1$. Therefore, $\angle ZB_1Y=\angle YXC_1$ implies $Y$ is mapped to a point on line $YX$, so $XY$ passes through $Z$. If the circumcircle of $AXY$ intersects $\Gamma$ again at $T$, then since $AXYT$, $XYB_1C_1$, and $ATB_1C_1$ are cyclic, we get $AT$, $B_1C_1$, and $XY$ are concurrent, so $T$ is the intersection of $AZ$ and $\Gamma$. Therefore, the circumcircle of $AXY$ passes through the fixed point $T$.

nice problem but wtf were the test makers smoking? also I think it's fairly easy to get that the fixed point lies on $(ABC)$ by sending $P \to \infty$, in which case $X$ and $Y$ lie on $(ABC)$. I got this very quickly and then went nowhere. Guessing the actual point is still very difficult and did not happen (despite ggb) for a long time Let $K$ and $L$ be the points on $(ABC)$ such that $BK \parallel CL \parallel AM$. Observe that $\measuredangle BDP=\measuredangle BMP=\measuredangle BCL=\measuredangle BDL$, so $D,P,L$ are collinear, and likewise so are $E,P,K$. Note that $\overline{AM}$ also bisects $\overline{KL}$. We focus on the isosceles trapezoid $BCKL$. Draw circles $\omega_1,\omega_2$ tangent to $\overline{BC}$ and $\overline{KL}$ passing through $B,K$ and $C,L$ respectively, and note that their radical axis is $\overline{AM}$. Since $\measuredangle KYB=\measuredangle PMB=\measuredangle KBM$, $Y \in \omega_1$, and likewise $X \in \omega_2$. By radical center this implies that $KLXY$ is cyclic. Lemma: Let $ABCD$ be an isosceles trapezoid with $\overline{AD} \parallel \overline{BC}$. with $\overline{AB} \cap \overline{CD}=T$. Let $\omega_1$ and $\omega_2$ be the circles tangent to $\overline{AB}$ and $\overline{CD}$ through $A,D$ and $B,C$ respectively. Let $\ell$ be a line through $T$ that intersects $\omega_1$ at $P_1$ and $P_2$ (with $P_1$ closer) and $\omega_2$ at $Q_1$ and $Q_2$ (with $Q_1$ closer). Then $ABP_2Q_1$ is cyclic (and symmetric results follow) Proof: Clearly we have $TP_1P_2A \sim TQ_1Q_2B$, so $TP_1\cdot TQ_2=TP_2\cdot TQ_1$. Furthermore this product equals $TA^2\cdot TB^2$, hence $TP_2\cdot TQ_1=TA\cdot TB$ and the result follows. $\blacksquare$ This lemma (with a suitable phantom point argument) thus implies that $\overline{XY},\overline{KL},\overline{BC}$ concur, hence by radical center we have $BCXY$ cyclic, and the radical center of $(KLXY),(BCXY),(ABC)$ is $\overline{KL} \cap \overline{BC}$. Then radical center on $(BCXY),(ABC),(AXY)$ implies that the fixed point $T$ is the second intersection of the line joining $A$ with $\overline{KL} \cap \overline{BC}$, and we're done. $\blacksquare$

Let $DP$ and $EP$ intersect $(BCDE)$ again at $S$ and $T$. As $\measuredangle BDS = \measuredangle BDP = \measuredangle BMP$, $S$ (and similarly $T$) is fixed. Claim: $BCXY$ and $XYST$ are cyclic. Proof: Invert at $M$ with arbitrary radius, using $*$ to denote the image of a point. We have $X^* = (P^*D^*M)\cap P^*C^*$, $Y^* = (P^*E^*M)\cap P^*B^*$, $S =(P^*D^*M)\cap (B^*C^*D^*E^*), T = (P^*E^*M)\cap (B^*C^*D^*E^*)$ . Now invert at $P^*$ swapping $B^*,D^*$ and $C^*,E^*$, using $'$ to denote the image of a point. Then $M'$ still lies on the $P^*-$ median of $P^*B^*C^*$, and $X' = B^*M'\cap P^*C^*$, $Y' = C^*M'\cap P^*B^*$, $S' = B^*M'\cap (B^*C^*D^*E^*)$, $T' = C^*M'\cap (B^*C^*D^*E^*)$. Since $X'Y'\parallel B^*C^*$, $B'C'X'Y'$ and $X'Y'S'T'$ are both cyclic by converse Reim as desired. Going back to the main problem and the original diagram, $XY\cap ST\cap BC$ is fixed, so $(AXY)\cap (ABC)$ is fixed by radical axis and we're done.

Let $R,S$ be the points such that $AM\parallel BS\parallel CR$ and $A,T$ be the intersections of $(AXY)$ and $(ABC)$. We show that $T$ is the required fixed point. Note that $S,R$ are fixed points. Let $R'\in\Gamma$ be the point such that $D,P,R'$ are collinear. We have, \[\angle RCB=\angle AMB=\angle PMB=\angle PDB=\angle R'DB=\angle R'CB\]which means that $R,R',C$ are collinear, but they also lie on $\Gamma$, so this means $R=R'$. Hence, $R,P,D$ are collinear. Similarly, $S,P,E$ are collinear. Applying Pascal's Theorem on $(RDBSEC)$, we get $\underbrace{RD\cap SE}_{=P},DB\cap EC,\underbrace{BS\cap CR}_{=\infty_{AM}}$ are collinear. The line $P\infty_{AM}$ is simply line $AM$. So, lines $AM,BD,CE$ concur at a point, say $U$. Claim 1. $BCXY$ is cyclic. Proof. [Angle Chasing] Let $U'$ be the reflection of $U$ in $M$. Then, since $M$ is the mid-point of $\overline{BC}$ and $\overline{UU'}$, we have $BUCU'$ is a parallelogram. So, $CU'\parallel BD$ and $BU'\parallel CE$. Note that \[\angle BDR=180^\circ-\angle RDB=180^\circ-\angle RCB=\angle CBS=\angle CMA=\angle CMP=\angle CEP=\angle CXP=\angle CXR\]which means $BD\parallel XC$ and hence $C,X,U'$ are collinear. Similarly, $CE\parallel YB$ and $B,Y,U'$ are collinear. Since $BYPM$ is cyclic, we have $U'Y\cdot U'B=U'P\cdot U'M$, while since $CXPM$ is cyclic, $U'P\cdot U'M=U'X\cdot U'C$. Hence, we have $U'Y\cdot U'B=U'X\cdot U'C$, which means that $BCXY$ is cyclic, as claimed. $\blacksquare$ Claim 2. $SRXY$ is cyclic. Proof. [Length Ratios] Let $D',E'$ be the reflections of $D,E$ in $M$. Then, $D'\in U'C$ and $E'\in U'B$. Note that \[\frac{U'X}{U'Y}=\frac{U'B}{U'C}=\frac{UC}{UB}=\frac{UD}{UE}=\frac{U'D'}{U'E'}\]which means $XY\parallel D'E'$. However, $DE\parallel D'E'$ which means $XY\parallel DE$. So, \[\frac{PS}{PR}=\frac{PD}{PE}=\frac{PX}{PY}\Longrightarrow PX\cdot PR=PY\cdot PS,\]which means $SRXY$ is cyclic, as claimed. $\blacksquare$ Claim 3. $AT,SR,XY,BC$ all concur at a point. Proof. Consider the radical center $Z$ of the circles $(AXY),(BCXY),(ABC)$. The radical axis of $(AXY),(BCXY)$ is line $XY$, the radical axis of $(AXY),(ABC)$ is line $AT$, and the radical axis of $(ABC),(BCXY)$ is line $BC$. So, $Z$ is the intersection of lines $AT,XY,BC$. Let the radical center of circles $(AXY),(SRXY),(ABC)$ be $Z'$. The radical axis of $(AXY),(SRXY)$ is line $XY$, the radical center of $(AXY),(ABC)$ is line $AT$. So, $Z'=AT\cap XY$, which means $Z'=Z$. So, $Z$ also lies on the radical axis of $(SRXY),(ABC)$ which is line $SR$. Hence, $Z$ is the intersection of lines $AT,SR,XY,BC$, as desired. $\blacksquare$ Since $Z=SR\cap BC$ and $S,R,B,C$ are fixed points, it follows that $Z$ is a fixed point. Hence, $T=AZ\cap(ABC)$ is a fixed point as well. $\blacksquare$

APMO 2019 p3 HAY trio for the win. Solved with SuperHmm7 and Ammh4 [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13.50522652721632cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -20.069989453768798, xmax = 32.435237073447524, ymin = -11.986443969477834, ymax = 11.503650757456876; /* image dimensions */ pen qqwwzz = rgb(0.,0.4,0.6); pen ttttqq = rgb(0.2,0.2,0.); pen qqqqcc = rgb(0.,0.,0.8); pen qqffff = rgb(0.,1.,1.); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen ffcqcb = rgb(1.,0.7529411764705882,0.796078431372549); pen ffdxqq = rgb(1.,0.8431372549019608,0.); pen bfffqq = rgb(0.7490196078431373,1.,0.); /* draw figures */ draw(circle((6.64,-0.16), 6.07542591099587), linewidth(1.) + qqwwzz); draw((1.66,-3.64)--(11.788571628526674,-3.385369775069239), linewidth(1.) + ttttqq); 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dot((0.577960852826069,0.24308978916707588),linewidth(4.pt) + dotstyle); label("$L$", (0.7097097277503039,0.5230739324400591), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Great problem! Here is a Sketch: Step 1: Show that $BCXY$ cyclic. Step 2: Show that $XY$ is parallel to $DE$. Step 3: Construct $K$ and $L$ Step 4: Prove that $PM$, $BL$ and $CK$ are parallel. The problem follows as this means that $K$ and $L$ are fixed which mean also implies that $Z$ is fixed by radicle axis.

Cute problem! We first prove a claim which reduces the problem to showing $XY\cap BC$ is fixed. Claim: $BXCY$ is cyclic Proof. Let $Q=BD\cap CE \cap AM$ and $Q'=2M-Q$. Then, $Q'$ is on $AM$ and $Q'B \parallel CE$. But by Reim, $CE \parallel BX$ so $Q'$ lies on $BX$. Similarly, $Q'$ lies on $CY$ so $BX, CY, AM$ concur and the claim follows by radical axis. $\blacksquare$ Now, Pascal converse gives that $DYCEXB$ are coconic. Then, Pascal on $EDBXYC$ gives $DE \cap XY$ is at infinity, so $DE\parallel XY$. Thus by Reim $XKLY$ is cyclic where $K,L$ are the second intersections of $EP, DP$ with $(ABC)$, and so by radical axis again it suffices to show that $KL\cap BC$ is fixed. But by Reim, $BK \parallel AM \parallel CL$ so they are fixed, done. Remark: The hardest part about this is convincing yourself the fixed point lis on $(ABC)$. After that it becomes a standard ~20 mohs problem.

MAIN CLAIM: $\odot(ABC)\cap (AXY)$ is fixed. (Kudos to the people who were able to figure this out during the exam or without geogebra.) Let $(AXY)\cap (ABC)=F$, $BD, AM, CE$ intersect at $R$(radical-center). Claim I: $BY \parallel CE$ and $CX \parallel BD$ Proof: Reim's Theorem Let $BY, AM, CX$ intersect at $S$ Claim II: $BD \parallel CX$ Proof: $\measuredangle BDX=\measuredangle BDP=\measuredangle CMP=\measuredangle CXD$ Let $DX \cap (ABC)=L$ and $EY \cap (ABC)=K$. (This is the trickiest part of the solution, rest of the part is just trivial, this it is motivated because later we prove concurrence of lines using radical-axis which requires concylicity of some quadrilaterals.)We have the following claim: Claim III: $\odot (BCXY)$ Proof: $KX\cdot KB=KP\cdot KM=KY\cdot KC$ Claim IV: $RCBS$ is a parallelogram Proof: Direct consequence of Claim I and Claim II. Claim V: $ XY \parallel DE$ Proof: $XY$ and $DE$ are anti-parallel to $BC$ Claim VI: $\odot (KLXY)$ Proof: This equivalent to showing $XY \parallel DE$. \begin{align*}&\angle PDB=\angle PMB=\angle PEC\implies \odot(PDLE)\\ \implies &\angle PED=\angle PRD=\angle PSC=\angle PYX\end{align*}and we are done. Now by applying Radical-axis on $\odot(ABC), \odot (BCXY)$and $\odot(AXY)$, we get that $AF, BC, XY$ concur. Again applying Radical-axis on $\odot(KLXY), \odot(ABC), \odot (BCXY)$ we get that $BC, XY, KL$ concur. Since $KL, BC$ are fixed $\implies T$ is fixed. If $AT \cap (ABC)=F$, we have $AT \cdot TF=BT \cdot TC = XF \cdot FY\implies F$ is fixed.

@above no i dont, will try the problem and post my soln later(If i get one).

yes good yes

this problem is very :blobheart: The main claim is the following. Claim: $BCXY$ is cyclic, and moreover, $XY\parallel DE$. Invert at $P$. Our claim now becomes: Quote: In triangle $\triangle PBC$, $M$ lies on $(PBC)$ such that $PM$ is a symmedian. Let $D$ be on $BM$ and $E$ on $CM$ such that $BCDE$ is cyclic. Let $PD$ meet $CM$ at $X$ and $PE$ meet $BM$ at $Y$. Show that $YX\parallel DE$. Let $PM$ meet $ED$ at $S$. By Ceva, it suffices to show that $S$ is the midpoint of $ED$. However, because $MP$ is a symmedian in $\triangle MBC$, and $BCDE$ is cyclic, it flips over to a median in $\triangle EMD$. Thus, $XY\parallel ED$, so $XY\parallel ED$ in the uninverted diagram as well as $PYE$ and $PXD$ are collinear. Furthermore, in the inverted diagram, $BYXC$ is cyclic by Reim as $XY\parallel ED$, so this is true in uninverted diagram as well. Now, let $XY\cap BC=K$. If $(AXY)\cap(ABC)=F$, then by radical axis theorem, $AFK$ are collinear. Thus, the rest of this solution will be showing that $K$ does not depend on the choice of $P$. This will finish because then $AK\cap (ABC)$ will always lie on $(AXY)$. Let $DP$ and $EP$ meet $(ABC)$ at $D'$ and $E'$ respectively. Claim: $D'$ and $E'$ do not depend on the choice of $P$. Since $PECM$ is cyclic, We have $$\angle E'EC=\angle PMB=\angle AMB$$which does not depend on $P$. Hence, the arc measure of $E'C$ is constant, so $E'$ is also constant. Finally, $EE'DD'$ is cyclic and $YX\parallel DE$, so by Reim, $D'E'XY$ is cyclic. By Radical Axis on $(ABC)$, $(BCXY)$, and $(D'E'XY)$, we have that $BC,XY$, and $D'E'$ concur. Thus, $K=BC\cap D'E'$, which does not depend on $P$ as $D'$ and $E'$ does not depend on $P$. We are done.

Let $F= (AXY) \cap \Gamma$. Observe $$\measuredangle BDX = \measuredangle BMP = \measuredangle CMP = \measuredangle CXP = \measuredangle CXD$$Which implies $CX \parallel BD$, by symmetry $BY \parallel EC$. Let $BD \cap CE = G$ which lies on $AM$ by a radical axis argument. Let $BY \cap CX = H$. Since $M$ is the midpoint of $BC$ and $GBHC$ is a parallelogram we have $H \in MG = AM$. Therefore by PoP $$HY \cdot HB = HP \cdot HM = HX \cdot HC$$Which means $BYXC$ is cyclic. Now define $J= EP \cap \Gamma \neq E$, $K = DP \cap \Gamma \neq D$. We have $$\measuredangle CKP = \measuredangle CKD = \measuredangle CBD = \measuredangle CBD = \measuredangle DPM = \measuredangle KPA.$$Which implies $CK \parallel AM$, by symmetry $BJ \parallel CK \parallel AM$. Let $KC \cap EJ = L$. Now observe $$\measuredangle LED = \measuredangle JED = \measuredangle JBD = \measuredangle(JB, BD) = \measuredangle(LC, XC) = \measuredangle LCX = \measuredangle LDX = \measuredangle LDP.$$Which means $PD$ is tangent to $(LED)$. So $PD^2=PL \cdot PE$. Now by PoP $$PX \cdot PD = PY \cdot PL , PK \cdot PD = PJ \cdot PE \implies PY \cdot PJ \cdot PE \cdot PL = PD^2 PK \cdot PX \implies PY \cdot PJ = PX \cdot PK.$$Therefore $(JYXK)$ is cyclic. To fnish note that by radical axis $AF$, $XY$, $AB$, $JK$ Concur. However $AB \cap JK$ is fixed depending on $P$. Therefore $F$ is fixed and we are finished.

Let $DP$ and $EP$ intersect $(ABC)$ again $C'$ and $B'$, respectively. Applying Reim's theorem, we have $BD\parallel CX$, $BY\parallel CE$, and $BB'\parallel AM\parallel CC'$. In particular, for the first two, the lines are reflections across $M$ because $M$ is the midpoint of $BC$. Claim 1: $BYXC$ is cyclic. By Radical Center, $BD$, $CE$, and $AM$ concur. Reflecting across $M$, we get that $CX$, $BY$, and $AM$ concur. Therefore, we are done by the converse of Radical Center. Claim 2: $B'C'$, $XY$, and $BC$ concur. Let $BY$, $CX$, and $AM$ intersect at $J$. Let $BD$, $CE$, and $AM$ intersect at $K$. Note that $J$ and $K$ are reflections across $M$, so $JBKC$ is a parallelogram. By Miquel Point on a triangle, $YPXJ$ and $DPEK$ are cyclic. Thus, \[\measuredangle PYX=\measuredangle PJX=\measuredangle PJC=\measuredangle PKB=\measuredangle PKD=\measuredangle PED\]so $DE\parallel XY$. By Converse of Reim's, $B'C'XY$ is cyclic, so our claim is true by Radical Center. In particular, $B'C'$ intersect $BC$ is independent of $P$, so $XY$ passes through a fixed point $T'$. Let $AT'$ intersect $(ABC)$ at $T$ then we have by converse of Radical Center that $ATXY$ is cyclic. We are done.