Ah yes, this problem. Didn't solve :'(

Typo fixed

Let $l$ be the parallel line to $BC$ that passes through A, and define $J=BD\cap l$, $Q=PM\cap l$, and $F'=AE\cap BC$. Let $\omega$ the circumcircle of $\triangle CDE$. $\angle ABE = \angle EDC$ and also $\angle ABE = \angle EAJ = \angle AF'B$, since $l$ is tangent to $(ABE)$ and $l\parallel BC$. From the previous angle chasing we can conclude that $CF'ED$ is cyclic, therefore $F'\in \omega$. Now notice that $Q\in \omega$. $MD\cdot MA = ME\cdot MP = ME \cdot MQ = MD \cdot MC$, done. $\angle AJD = \angle DBC = \angle DEQ$, so $J$ also lies on $\omega$. To finish, we apply Pascal on $EAAPBD$. $K-F'-J$ collinear. Then, $KP\cdot KA = KE\cdot KD = KF' \cdot KJ$, therefore $APF'J$ is an isosceles trapezoid since $AJ\parallel PF'$. Then it is easy to see that $KP=KF'$, which means $F=F'$. We are done

Justpassingby wrote: There’s a typo in “the intersection of $AC$ with the circumcircle of $\bigtriangleup ABC$“. Thanks. Fixed.

Redefine $F$ as the intersection of $AE$ with $BC$. Note that by angle chase we get $\angle ABD = \angle ACB$. Hence, $\angle FCD = \angle ACB = \angle ABD = \angle AED \implies DEFC$ is cyclic. Let $L$ be the reflection of $P$ across $M$. Note that $\angle ALC = \angle APC = 180 - \angle APB = 180 - \angle ABC$, hence $L \in (ABC)$. Also, $\angle ELC = \angle APE = \angle EDC$, hence $L \in (CDEF)$. Let $J = AL \cap (CDEF)$. We have $\angle JDA = 180 - \angle JDC = 180 - \angle JLC = \angle ALC = 180 - \angle ABC = 180 - \angle ADB = \angle BDC$, hence $J,D,B$ are collinear. Also note that $\angle JFC = \angle JDC = \angle BDA = \angle ABC= \angle AEL = \angle FCL = \angle PCL = \angle PAJ$, hence $PFJA$ is cyclic. Applying radical axis theorem on $(ABPED) , (CFEDLJ) , (PFJA)$ we get that $J,F,K$ are collinear. Now since $BC \parallel AL$, we have $\frac{KP}{KA} = \frac{KF}{KF}$ and by PoP through $K$ we have $KP \cdot KA = KF \cdot KJ$. Multiplying them together gives $KP = KF$, as desired.

Solution. Let $Q$ and $R$ be points such that $ABFQ$ and $APCR$ are parallelograms. Clearly, $R= \overline{AQ}\cap \overline{PM}$ and $Q\in KF$. According to Reim's theorem, $EDCR$ is cyclic. By symmetry, so do $RCFQ,\ ABCR$ and $APFQ$, thus $$KF\cdot KQ = KD \cdot KE = KP\cdot KA$$implying that $EDFQ$ is cyclic as well. By the radical axis theorem applied to $(ABCR)$, $(ABDE)$ and $(EDCR)$, we conclude that $AB,\ ED$ and $CR$ concur at a point, say $S$. Suppose that $(EDCR),\ (EDFQ)$ and $(RCFQ)$ are distinct circles. The aforementioned theorem also implies that $ED,\ CR$ and $FQ$ concur at $K$. We infer that $AB,\ ED,\ CR$ and $FQ$ are concurrent at $S\equiv K$, which is a contradiction since $AB\parallel FQ$. Hence, $E,\ D,\ C,\ F,\ Q$ and $R$ must be concyclic, as desired. $\blacksquare$

My problem, hope you all enjoy(ed) it. The original statement asked you to prove that $A$, $E$, and $F$ are collinear, which is pretty easy once you solve this version. Not the most insane problem ever but I imagine it was hard enough for centro

Solved with MrOreoJuice.

Redefine $F$: let $F$ be the intersection of $BC$ and $(CED)$. We'll show that $KF=KP$. Also, let $N$ be the midpoint of $PF$ and $R$ be the on $(ABC)$ such that $ARCB$ is isosceles trapezoid. Hence, $ARPC$ is a parallelogram, thus $M$ is the midpoint of $RP$. By Reim's theorem, $REDCF$ is cyclic. Also, by PoP, $MECN$ is cyclic. We have $$\measuredangle EFP=\measuredangle EFC=\measuredangle ERC=\measuredangle PRC=\measuredangle EPA,$$which implies that $KP$ is tangent to $(EPF)$. Also, $$\measuredangle PEN=\measuredangle MCP=\measuredangle DCP=\measuredangle DCF=\measuredangle DEF,$$which implies that $ED$ is the symmedian of $\triangle EPF$. We conclude that $KF$ is tangent to $(EPF)$. We are done. [asy][asy] import olympiad; size(12cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta; pair O,A,B,C,M,P,D,E,R,K,F,N; O=(0,0);A=dir(109);B=dir(200);C=dir(340);path w=circumcircle(A,B,C);M=midpoint(A--C);P=2foot(A,B,C)-B;D=intersectionpoints(circumcircle(A,B,P),A--C)[1];R=intersectionpoints(w,P--100M-99P)[0];E=intersectionpoints(circumcircle(A,B,P),P--100M-99P)[0]; K=extension(E,D,A,P);F=intersectionpoints(circumcircle(D,E,C),B--100C-99B)[1];N=midpoint(F--P); draw(A--B--C--cycle,deep);draw(w,deep);draw(circumcircle(A,B,P),deep);draw(C--F,deep);draw(circumcircle(D,E,C),med+dashed); draw(A--R--C,deep);draw(A--P--R,deep);draw(E--K,lightred);draw(A--K,lightred);draw(K--F,lightred); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$M$",M,dir(M)); dot("$D$",D,dir(D)); dot("$P$",P,dir(P)); dot("$R$",R,dir(R)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$K$",K,dir(K)); dot("$N$",N,dir(N)); [/asy][/asy]

SatisfiedMagma wrote: Solved with MrOreoJuice.

Btw how did u make that claims and things?(Formatting)

Looks like they used some style template on overleaf @above

Let $\Gamma$ be circumcircle of $\triangle ABP$. The parallel to $BC$ through A intersects $KF$ at $S$. Now, $ \angle SAP = \angle APB = \angle ABP$, so $AS$ is tangent to $\Gamma$. As $KP = KF$, we have $KA = KS $ and $ASFP$ is an isosceles trapezoid. By Power of Point $K$ wrt $\Gamma$, $KD \cdot KE = KP \cdot KA = KF \cdot KS$ so $EDFS$ is cyclic. As $AB = AP = FS$ we have $ABFS$ is a parallelogram and let $X$ be a point on the line $AS$ such that $AXCP$ is a parallelogram. So $P, M, X$ are collinear. By Power of Point $M$ wrt $\Gamma$, $MC \cdot MD = MD \cdot MA = MP \cdot ME = ME \cdot MX$ so $EXCD$ is cyclic. Notice that $SF = AB = AP = XC$ and $XS \parallel FC$ so $XSFC$ is an isosceles trapezoid and cyclic. Finally, $DE \cap SF = K$, $EX \cap FC = P$ and $XS \cap CD = A$ are collinear, then applying Pascal's Theorem in the hexagon $DEXSFC$ we have that $C, D, E, F, S, X$ lie on the same circle.