EDIT: Please do not discuss the problems untill Aug 30

Cute. Here's sketch, I won't bother writing up as I am retired.

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Let $DM\cap BC=T,TO\cap AD=R$. If $B'$ is the reflection of $B$ with respect to $AD,$ then $P=AD\cap (ACB')$ hence $P\neq A$ is unique. $OE\perp DM,OM\perp BC$ so $TO\perp AM$. We will show that $\angle ABR=\angle RCA$. Note that $0<\angle ABR+\angle RCA<180$ since $P,R$ lies between $A,D$. Let's prove that $\sin ABR=\sin RCA$. \[\frac{DB}{DC}=\frac{\sin RAB}{\sin CAR}\overset{?}{=}\frac{\sin RAB}{\sin ABR}.\frac{\sin RCA}{\sin CAR}=\frac{RB}{RC}\]By $\angle MOR=\angle C+\frac{\angle A}{2}=\angle TDR,$ we can see that $R,O,D,M$ are cyclic. Also $TB.TC=TD.TM=TR.TO$ thus, $R,O,B,C$ are cyclic. \[\frac{\sin DMB}{\sin DMC}=\frac{DB}{DC}\overset{?}{=}\frac{RB}{RC}=\frac{\sin BOR}{\sin COR}\]\[\frac{\sin COR}{\sin DMC}=\frac{\sin OTC}{OC}.\frac{MC}{\sin CTM}=\frac{\sin OTB}{OB}.\frac{MB}{\sin BTM}=\frac{\sin BOR}{\sin DMB}\]As desired.$\blacksquare$

Very Beautiful Geometry Let $Q$ be the reflection of $M$ about $BC$. We wish to prove that $P$ and $Q$ are isogonal conjugates. First we have $AEOQ$ is cyclic as $$\angle EQO=\angle EMO=\angle EAO$$Notice that we also have $$\angle MAQ=\angle EOM=\frac{1}{2} \angle DOM=\angle DAM=\angle MAP$$Notice that the point $P$ is uniquely defined. Let $Q^{*}$ be the isogonal conjugate of $Q$. Then $Q^{*}$ lies on $AD$ and $\angle ABQ^{*}=\angle ACQ^{*}$ so $Q^{*}=P$ as desired. Now notice that $BCOP$ is cyclic as $$\angle BPC=\angle BAC+\angle ABP+\angle ACP=\angle BAC+\angle CBQ+\angle BCQ=$$$$\angle BAC+\angle CBM+\angle BCM=2\angle BAC=\angle BOC$$Also $DMOP$ is cyclic as $$\angle MOP=\angle COP-\angle COM=180^{\circ}-\angle CBP-\angle BAC=$$$$180^{\circ}-\angle BAC-\angle ABC+\angle QBC=\angle ACM=180^{\circ}-\angle MDP$$To finish we have $OA=OM$ so it suffices that $$\angle OAP=\angle ODP=\angle OMP$$

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This problem is either too easy for P3 or I made a mistake. $AOED$ is cyclic because $\angle DAE = \frac{1}{2}\angle DOM = \angle DOE$. $MD \cap BC = K$ is on $(AOED)$ because $ME \cdot MA = MD \cdot MK$. $KO \perp AM$ because $\angle KEA = \angle MCA = 90 -\angle OMA = 90 - \angle OAE = 90 - \angle OKE$. Redefine $P = KO \cap AD$. Now $KP \cdot KO = KD \cdot KA = KD \cdot KM = KB \cdot KC \implies BPOC$ is cyclic. Let $Q = MO \cap (AOED)$. Note that $Q$ is the reflection of $M$ across $BC$ (write $ME \cdot MA$ in different ways. $\angle DAE= \angle EAQ$ from angle chasing and $\angle BQC + \angle BPC = 180 + \angle A$. This means $P$ and $Q$ are isogonal conjugates, which gives $\angle PBA = \angle PCA$.

(Tedious ) Complex bash: Take ABC as the unit circle. there exist $x,y,z$ such that $x^2=a,y^2=b,z^2=c$. Then M would be $m=-yz$ ($m^2=bc$). $$e=\frac{am(b+c)-bc(a+m)}{am-bc} = \frac{am(b+c)-m^2 (a+m)}{am-m^2} = \frac{a(b+c)-m(a+m)}{a-m}$$and: $$\bar{e}=\frac{b+c-a-m}{bc-am}=\frac{b+c-a-m}{m(m-a)}$$$ED=EM \Rightarrow |e-d|=|e-m| \Rightarrow (e-d)(\bar{e} - \frac{1}{{d}})=(e-m)(\bar{e} - \frac{1}{{m}}) \Rightarrow \frac{1}{d}(d-m)(\bar{e}d-\frac{e}{m})=0 $ $$\Rightarrow d=\frac{e}{\bar{e}m}=\frac{ab+ac-am-bc}{a+m-b-c}$$Now: $\measuredangle ACP = \measuredangle PBA \implies \frac{a-b}{p-b} \div \frac{p-c}{a-c} \in \mathbb{R} \implies \frac{a-b}{p-b}.\frac{p-c}{a-c}=\bar{ \left (\frac{a-b}{p-b}.\frac{p-c}{a-c} \right)} \implies \frac{a-b}{p-b}.\frac{p-c}{a-c}=\frac{\frac{(b-a)(c-a)}{(ab.ac)}}{(\bar{p}-\frac{1}{b})(\bar{p} -\frac{1}{c})}$ $$\implies a^2bc(\bar{p}- \frac{1}{b})(\bar{p} - \frac{1}{c}) =(p-c)(p-b)$$we note that since P lies on AD: $p+\bar{p}ad=a+d$. Therefore we can rewrite above as: $a^2bc \frac{(ab+db-pb-ad)(ac+dc-pc-ad)}{(adb)(adc)}= (p-b)(p-c) \implies (ab+db-pb-ad)(ac+dc-pc-ad)=(p-b)(p-c)d^2 \implies p^2 (d^2-bc)+p(-d^2 b-d^2 c+2abc+2bcd-abd-acd)-a^2 bc-2abcd+a^2 bd+d^2 ab+a^2 cd+d^2 ac-a^2 d^2=0 \implies (p-a)(p(d^2-bc)-d^2 b-d^2 c+d^2 a+abc+2bcd-abd-acd)=0$ $$\implies p=\frac{+d^2 b+d^2 c-d^2 a-abc-2bcd+abd+acd}{d^-bc}$$also we get: $$\bar{p}=\frac{ac+ab-bc-d^2-2ad+dc+db}{-a(d^2-bc)}$$To prove the problem it suffices to show $PA=PM$ (since $OA=OM$) or equivalently $|p-a|=|p-m| \iff (p-a)(\bar{p}-\frac{1}{a})=(p-m)(\bar{p}-\frac{1}{m}) \iff am\bar{p}=p$ but: $\frac{p}{\bar{p}}=\frac{\frac{+d^2 b+d^2 c-d^2 a-abc-2bcd+abd+acd}{d^-bc}}{\frac{ac+ab-bc-d^2-2ad+dc+db}{-a(d^2-bc)}}=\frac{a(d^2 b+d^2 c-d^2 a-abc-2bcd+abd+acd)}{-ac-ab+bc+d^2+2ad-dc-db}=am$ $\iff m(-ac-ab+bc+d^2+2ad-dc-db)=(d^2 b+d^2 c-d^2 a-abc-2bcd+abd+acd) \iff d^2 (b+c-a-m)+d(ac+ab-2bc-2am+cm+bm)-abc-bcm+acm+abm=0$ plugging in $d$: $\frac{(ab+ac-am-bc)^2}{b+c-a-m}+\frac{(ab+ac-am-bc)(ac+ab-2bc-2am+cm+bm)}{a+m-b-b}=abc+bcm-acm-abm $ $\iff (ab+ac-am-bc)^2+(bc+am-ac-ab)(ac+ab-2bc-2am+cm+bm)=(b+c-a-m)(abc+bcm-acm-abm)$ $\iff (ab+ac-am-bc)(+am+bc-mc-mb)=(b+c-a-m)(abc+bcm-acm-abm)$ $\iff b^2 ac-abcm-b^2 am+a^2 bm+c^2 ab-c^2 am-abcm+a^2 mc-abcm+m^2 ac+m^2 ab-a^2 m^2-b^2 c^2+c^2 bm+b^2 cm-abcm =b^2 ac+b^2 cm-abcm-b^2 am+c^2 ab+c^2 bm-c^2 am-abcm-a^2 bc-abcm+a^2 cm+a^2 bm-abcm-m^2 bc+m^2 ac+m^2 ab$ which the last one is true, proving the problem(note $m^2=bc$ in last couple lines).$\blacksquare$

Here is the solution that I have written last month but forgot to post it... [asy][asy] unitsize(4cm); import geometry; pair A,B,C,M,E,X,D,M_1,Y,P,O; A=dir(135); B=dir(210); C=dir(330); M=intersectionpoints(perpendicular((B+C)/2,line(B,C)),circle(A,B,C))[0]; E=intersectionpoint(line(B,C),line(A,M)); X=intersectionpoint(line(B,C),perpendicular((A+M)/2,line(A,M))); D=intersectionpoints(line(M,X),circle(A,B,C))[1]; M_1=intersectionpoints(perpendicular((B+C)/2,line(B,C)),circle(A,B,C))[1]; Y=intersectionpoint(line(B,C),line(M_1,D)); O=circumcenter(A,B,C); P=intersectionpoints(line(A,D),circle(X,Y,D))[1]; draw(A--B--C--cycle); draw(circle(A,B,C)); draw(A--M); draw(X--B); draw(M--X); draw(M_1--D); draw(circle(X,Y,D),purple+dashed); draw(A--D); draw(X--O,red+dashed); draw(M_1--M); draw(P--Y); draw(X--A); draw(D--E); draw(B--P--C); draw(B--D--C); dot(A); dot(B); dot(C); dot(M); dot(E); dot(X); dot(D); dot(Y); dot(P); dot(O); dot(M_1); label("$A$",A,dir(110)); label("$B$",B,dir(225)); label("$C$",C,dir(320)); label("$M$",M,dir(270)); label("$E$",E,dir(235)); label("$D$",D,dir(270)); label("$X$",X,dir(180)); label("$M_1$",M_1,dir(90)); label("$O$",O,dir(135)); label("$P$",P,dir(60)); label("$Y$",Y,dir(315)); [/asy][/asy] Let $M_1$ be the midpoint of arc $BAC$ and let $X$ and $Y$ be the intersections of $BC$ with $MD$ and $M_1D$ respectively. Note that $\angle BDY=\angle CDY$ and $\angle XDM_1=90$ so $DY$ and $DX$ are bisectors of $\angle BDC$. We have $$\frac{BP}{CP}=\frac{sin\angle BAD}{sin\angle CAD}\times \frac{sin\angle ACP}{sin\angle ABP}=\frac{sin\angle BAD}{sin\angle CAD}=\frac{BD}{CD}=\frac{BX}{CX}=\frac{BY}{CY}$$Thus $PYDX$ is cyclic. Now Note that $\angle DPY=\angle DXY=\angle DM_1M=\angle DAM$ so $AM\parallel PY$. Since $ED=EM$ and $AXDE$ is cyclic(by shooting lemma) we get that $AX=XM$ so $OX$ is the perpendicularbisector of $AM$ and note that $PY$ and $PX$ are bisectors of $\angle BPC$ so $PX\perp PY\parallel AM\perp OX$ and hence $X-P-O$ are collinear and $PO\perp AM$ and we are done!

i3435 wrote: Let me continue in this theme of elegant solutions. Note that $P$ lies on the isogonal conjugate of the $BC$-perpendicular bisector, which is a hyperbola. Now, we claim that $Q$, the isogonal conjugate of $P$ which lies on the perpendicular bisector of $BC$, also lies on $\odot(E,M)$. To show this, reconstruct $Q'$ as the intersection of the circle with the perpendicular bisector, we wish to show that $AE$ bisects $\angle DAQ'$. Simply note that $AOED$ is cyclic as \[\measuredangle AED=2\measuredangle AMD = \measuredangle AMO\]and $AQOD$ is cyclic as \[\measuredangle DQM =\frac{1}{2}\measuredangle DEM = \frac{1}{2}\measuredangle DEA = \measuredangle DAO\]Fact $5$ finishes as $ADEQ$ is cyclic, with $DE=QE$. Now for the fun part! We'll take the isogonal conjugate of $OP$, call it $\mathcal{C}$. Let $P_\infty$ be the point at infinity on this line, then note that $AP_\infty$ desiredly is the external $A$-bisector. Then its conjugate is $N$, the midpoint of arc $BAC$. Thus we wish to show that $N$ lies on $\mathcal{C}$. We first claim that $X$, the intersection of the parallel through $Q$ to $BC$ with $AM$, lies on $\mathcal{C}$. To see this, we will use converse of Pascal on $XQHBCA$, where $H$ is the orthocentre. It suffices to show thus that if $K=BH\cap AM, L=HQ\cap AC$, that $KL\parallel BC$. Magically we have this orthocentre config instead of a incentre-ish config. Let $D_1$ be the foot from $A$ to $BC$. \[\measuredangle HAK+ \measuredangle LHA = \frac{\angle A}{2}-90^\circ+\angle B + 180^\circ-\angle B- \frac{\angle A}{2}=90^\circ\]Thus $AK\perp HL$, yet $HK\perp AC$. Hence $K$ is orthocentre of $\triangle AHL$, so $KL\perp AD_1$, or $KL\parallel BC$. Now use converse of Pascal on $QXABCN$. Let $CN\cap AM=U, AB\cap MN=V$. We want $UV\parallel BC$. \[\measuredangle AVN=90^\circ-\angle B=90^\circ-\measuredangle CMA=\measuredangle AUN\]thus $AUVN$ cyclic, so $\angle UVN=90^\circ$, thus $UV\parallel BC$

I got baited into believing that this was a P1 level problem, but I guess despite it's position it actually is, it would serve as a really tough Problem 1. We present a slightly different solution to the ones above. Let $T = \overline{MD} \cap \overline{BC}$. We start off with locating $T$. Claim : Point $T$ lies on $(ADE)$ with $TO \perp AM$. Proof : Note that, \[\measuredangle ETD = \measuredangle TEA + \measuredangle EMT = \measuredangle BCA + CAE + \measuredangle EMD = \measuredangle EAD\]which implies that $T$ lies one $(ADE)$. Now, \[\measuredangle MAT = \measuredangle EAT = \measuredangle EDM = \measuredangle DME = \measuredangle TMA\]which implies that $T$ lies on the perpendicular bisector of segment $AM$. Thus, $OT \perp AM$ as claimed. Further note that in particular, $O$ also lies on the circle $(ADE)$ since \[\measuredangle DOA = 2 \measuredangle DMA = 2\measuredangle DME = \measuredangle DEA\] We now define $P' = \overline{TO} \cap \overline{AD}$. The following claim is the punchline. Claim : Points $B$ , $P'$ , $O$ and $C$ are concyclic. Proof : First of all note that $DPOM$ is cyclic since, \[\measuredangle DOP' = \measuredangle DOT = \measuredangle MOT + \measuredangle DOM = \measuredangle MCA + 2\measuredangle DAM = \measuredangle MDA + \measuredangle DAM + \measuredangle PAM = \measuredangle DMA + \measuredangle AMP = \measuredangle DMP'\]But then, \[TB \cdot TC = TD \cdot TM = TP \cdot TO \]which implies that $BPOC$ is cyclic, as claimed. What remains is a simple angle chase. Note that, \[\measuredangle ABP' = \measuredangle ABC + \measuredangle CBP' = \measuredangle ABC + \measuredangle COP' = \measuredangle ABC + \measuredangle COM + \measuredangle MOP = \measuredangle ADC +2\measuredangle CAM + \measuredangle MDP =\measuredangle MDC + 2 \measuredangle CAM = \measuredangle CAM\]A similar angle chase yields that $\measuredangle P'CA = \measuredangle MAB$ which since $\measuredangle MAB = \measuredangle CAM$ implies that $P'\equiv P$ and we are done.