This was also P3 from the first Romanian TST from 2022.Spoiler alert!

Sorry for no latex. If we proof this two claims we finish the problem: Claim 1. Let Y be the Humpty point. Then DA=DY(i.e. Y is on the circle centered at D with radius DA) Proof: It is well known that Y is on the apollonius circle so BY/YC=AB/AC. Since DA is tangent to (ABC) we have DB/DC=(AB/AC)^2=(YB/YC)^2 so DY is tangent to (BYC). If we apply Pop we have DY^2=DB*DC=DA^2 so DY=DA and claim 1 is proved. Claim 2. X,Y,C are collinear. Proof:It is well known that A,Y,M are collinear. To finish the claim,we want to prove that <AYC+<AYX=180. We have that A,Y,E,X are concyclic so <AYX=<AEX=<AEB=<ACB. So we want <AYC+<ACB=180 or <ACB=<MYC or <ACB=<YAC+<YCA or <YCM=<YAC which is true by the definition of the Humpty point,so the claim is proved. Now from claim 1 and 2 we see that the point Y in the problem is unique and we also proved that the humpty point satisfies the same conditions,so the point Y in the problem is the humpty point and is obviously on the median AM.

Here is a quick solution which uses nothing about Humpty. By uniqueness, it must be the case that $DE$ is the second tangent from $D$ to $\omega$ and hence that $ABEC$ is harmonic -- so the assertion becomes equivalent to $\angle BAE = \angle CAY$. Since $\angle ECD = \angle BED = \angle XED = \angle EXD$, we deduce that $CEDX$ is cyclic. Hence (due to $DE = DX$) $\angle YCD = \angle XCD = \angle ECD$. Therefore $\angle CAY = \angle AYX - \angle ACY = \angle AEX - (\angle ACB - \angle BCE) = \angle AEB - \angle ACB + \angle BAE = \angle BAE$, as desired. Remark. It is also true that $CD$ is the perpendicular bisector of $YE$. Indeed, $\angle YDE = 2\angle YXE = 2\angle CXE = 2\angle CDE$, i.e. $\angle YDC = \angle CDE$, which together with $\angle YCD = \angle ECD$ concludes.

Note that $DE^2 = DA^2 = DB.DC$ so $DE$ is tangent to $\omega$ so $CB$ is symmedian in $ACE$ so $BE$ is symmedian in $ABC$. Now Note that $\angle BAM = \angle EAC = \angle EBC$. Claim $: XDEC$ is cyclic. Proof $:$ Note that $\angle DCE = \angle BCE = \angle BED = \angle XED = \angle EXD$ Now we have $\angle EBC = \angle BED + \angle BDE = \angle BAE + \angle EAY = \angle BAY$ so $\angle BAM = \angle BAY$ so $A,Y,M$ are collinear.

First, lets $M = AY \cap BC$, instead of midpoint. Lets prove that $M$ is the midpoint of $BC$. We have $\angle XYA = \angle XEA = \angle BEA = \angle BCA = \angle C = \angle DAB$, then, $\triangle MYC \sim \triangle MCA \Rightarrow \angle MAC = \angle MCY$, but $\angle XDC = \angle XDA + \angle ADC = 2\angle C + \angle B-\angle C =\angle B+ \angle C = 180 - \angle A$ and $\angle XEC = \angle BEC = 180 - \angle BAC \Rightarrow XDEC$ is cyclic, then $\angle XCD = \angle XED$ and $DA = DE \Rightarrow DE$ and $(ABC)$ are tangent to each other, so $\angle DEB = \angle BAE$, then $AE$ and $AM$ are isogonal, but, since $D = AA \cap EE$, so, the polar of $D$ is $AE$, then, let $T = BB \cup AE$, so, $T$ is in the polar of $D$, so $D$ is in the polar of $T$, then, $BD = BC$ is the polar, so $T$ is the symmedian of $ABC$, so, $M$ is the midpoint of $BC$.

Solved with Cookierookie, sevket12 and hakN It will well-known that $\Omega$ is the Apollonius center which satisfies $\frac{|WB|}{|WC|}=\frac{|AB|}{|AC|}$. Then, $\frac{|XB|}{|XC|}=\frac{|EB|}{|EC|}=\frac{|AB|}{|AC|}$. Hence, $AE$ is symmedian and $\angle XCB=\angle BCE$. Let $Y^*$ be the reflection of $E$ over $BC$. We know that $Y^*$ is the $A$-Humpty point and hence $AY^*$ bisects $BC$. Also, since $\frac{|Y^*B|}{|Y^*C|}=\frac{|AB|}{|AC|}$ we have $Y^*\in\Omega$ and since $\angle BCX=\angle BCE=\angle BCY^*$ we get $Y^*\in XC$, yields $Y^*=Y$. Done.

Applying inversion via $\Omega$, we have $C$ is mapped to be $B$. Hence, $XY$ is mapped to be the circumcircle of $\triangle DXY$ that also passes through $B$. Moreover, since $D,A,M,E$ are concyclic by angles chasing, we have $AE$ meets $DC$ at $M'$ the image of $M$ under the inversion. Hence, $$\angle M'DY=\angle BDY=\angle BXY=\angle EXY=\angle EAY=\angle M'AY,$$i.e. $A,D,M',Y$ are concyclic. Note that this circle is mapped to be $AY$ , so we are done.

Claim1 $AE\text{ is A-symmedian in } \triangle ABC$ Since $DA=DE$ by uniqueness $DE$ is tangent to $\omega$. So $ABEC \text { is harmonic quarilateral }$ Hence $CB$ is symmedian in $\triangle ACE \implies AE$ is symmedian in $\triangle ABC \ \square$ Claim2 $AE \text{ and } AY \text { are isogonal lines }$ It suffices to show that $\angle BAE=\angle CAY.$ Easily note that $DECX$ is cyclic since $$\angle DXE=\angle DEX=\angle DEB=\angle DCE$$Then $$\angle BAE= \angle BCE = \angle BED$$$$\angle CAY=\angle CAE-\angle YAE=\angle CBE-\angle YXE=(\angle BDE+\angle BED)-\angle BDE=\angle BED \,\,\, \square$$ Claim $1$ and $2$ yields $A,Y,M$ are collinear.... so we are done

Angle chasing and similar triangles

Here's a solution using $\sqrt{bc}$ inversion : First we claim that $\Omega$, or in other words the A-Apollonius circle, is mapped to the perpendicular bisector of $BC$ by the inversion. Indeed, note that both the foot of the internal bisector of $\angle BAC$ and the external bisector of $\angle BAC$ lie on $\Omega$. Thus, since these two points are mapped to the midpoints of minor and major arcs $BC$ of $\omega$ respectively, $\Omega$ is mapped to the line passing through these midpoints, i.e. the perpendicular bisector of $BC$. Back to the problem, which we start by rephrasing in the following way : let $Y$ be the intersection of line $AM$ and the A-Apollonius circle, prove that points $C,Y$ and $X$ are collinear. Then, since $AM$ is mapped to $AE$ by the inversion, we have that $Y$ is mapped to the intersection of the perpendicular bisector of $BC$ and line $AE$, i.e. the intersection of the tangents to $\omega$ passing by points $B$ and $C$ because $A,E,B,C$ are harmonic by construction. Let $P$ be this point. Now let $X'$ be the image of $X$ by the inversion, which thus lies on the perp bisector of $BC$ since $X\in \Omega$. Because points $B,E$ and $X$ are collinear and because $E$ is mapped to $M$, we have that points $A,C,M$ and $X$ are concyclic. Finally, we have $\angle AXP=\angle AXM=\angle ACM=\angle ACB=180-\angle ABP$, thus points $A,B,P$ and $X'$ are concyclic, and thus before inversion we had that points $C,Y$ and $X$ were collinear, as needed. $\square$

Attachments:
BMO 2021 G6.pdf (293kb)

Solved with starchan, mueller.25, AdhityaMV Observe that $\angle AYC = 180 - \angle AYX = 180 - \angle AEX = 180 - \angle ACB$ so $(AYC)$ is tangent to $BC$. Since $\Omega$ is the $A$-apollonian circle of $\triangle ABC$, this forces that $Y$ is actually the $A$-humpty point of $\triangle ABC$, and so $Y$ lies on the median $AM$, so we're done. $\blacksquare$

By $DA=DE$ we know that $DE^2=DA^2=DB\times DC$ so $DE$ is tangent to $(ABC)$. Now we have $\measuredangle DCE=\measuredangle BCE=\measuredangle BED=\measuredangle XED=\measuredangle DXE$ so $DXCE$ cyclic. Now we have $\measuredangle BCY=\measuredangle DCX=\measuredangle DEX=\measuredangle EXD=\measuredangle ECD=\measuredangle ECB$, so $Y$ is the reflection of $E$ through $BC$, which is the $A$-humpty point, and we're done.

It is easy to show $DE$ is tangent line to $\omega$ From that observe $AEBC$ is harmonic Thus $AE$ is symmedian of $ABC$ $DXCE$ is cyclic then it is easy to get $Y$ is $A-humpty$ and we are done.

Note $\Omega$ and $w$ are orthogonal to each other. hence inversion at $\Omega$ persevere $w$ and swaps $B$ and $C$ which give us $$XE \leftrightarrow (DXCE)$$$$ XY \leftrightarrow (DXYB)$$ Now angle chasing by considering $DX=DE=DY$ $$\angle BAE = \angle BCE = \angle YCB$$$$\angle EAC = \angle EBC = \angle EBD = \angle DYX = \angle DXY = \angle YBC$$ As $AE$ is symmedian in $\triangle ABC$ we get $Y$ is Humpty point of median $AM$ hence $Y$ lie on $AM$.

We use complex numbers, and take $(AXYE)$ to be the unit circle. So $d=0$. Let $b,c\in\mathbb R$. The tangent condition gives $DB\times DC=DA^2=1$, so $bc=1$. From $A,B,C,E$ cyclic, we have $\frac{(e-b)(a-c)}{(a-e)(b-c)}\in\mathbb R$. Since $b-c\in\mathbb R$, $$\frac{(e-b)(a-c)}{a-e}=\overline{\frac{(e-b)(a-c)}{a-e}}=\frac{\left(\frac1e-b\right)\left(\frac1a-c\right)}{\frac1a-\frac1e}=-\frac{(1-be)(1-ac)}{a-e}$$Cancelling denominators and simplifying, $e=\frac{ab+ac-bc-1}{a-b-c+abc}=\frac{ab+ac-2}{2a-b-c}$. Now, $X$ lies on $BC$, so $b=x+e-xeb\implies x=\frac{b-e}{1-be}$ $C$ lies on $XY$, so $c=x+y-xyc$, so $$y=\frac{c-x}{1-xc}=\frac{c-bce-b+e}{1-be-bc+ce}=\frac1e \text{ (since }bc=1\text{)}$$so $Y$ is the reflection of $E$ about $BC$! Of course, this information is useless to us, since we're bashing, but it's nice to know Finally, to show $A,M,Y$ collinear, we show $z=\frac{a-m}{a-\frac1e}\in\mathbb R$. $$2z=\frac{(2a-b-c)(ab+ac-2)}{a^2b+a^2c-4a+b+c}$$which is equal to its conjugate, so we're done.

Solved with erkosfobiladol. Let the parallel to $BC$ passing through $A$ intersect $\Omega$ at $K$ and $\omega$ at $L$. $AB\cap \Omega=F,AC\cap \Omega =G,FC\cap \Omega=N$ We want to show that $-1=(AY,AL;AB,AC)=(Y,K;F,G)$ By inversion centered at $D$ with radius $AD$ gives that $(DBAG),(DBNF)$ are cyclic. $\angle DFB=\angle FAD=\angle C=\angle BGD\implies BG=BF$ $\angle DNB=\angle DFB=\angle C=\angle DCA=\angle KAG\implies N,B,G$ are collinear. $\angle KAD=\angle DAK=\angle ADC=\angle ADB=\angle AGN\implies N,D,K$ are collinear. By Pascal at $AEXYNG$ gives that $AE\cap YN, B, C$ are collinear hence $AE, YN, BC$ are concurrent. Let that point be $P$. Also the tangents to $\Omega$ at $G$ and $F$ intersect on $BC$ since $DB\perp FG$. $(Y,K;F,G)=(NY,NK;NF,NG)=(P,D;C,B)=-1$ as desired.$\blacksquare$

I hope this solution isn’t wrong Let $O$ be the circumcenter of triangle $ABC$. Then $DAOME$ is cyclic quadrilateral. By $DA=DE$ we know that $\angle AMD=\angle EMD$. Let $AM\cap \Omega= Y$, then we will show that $C, Y, X$ are collinear. By easy angle chasing we know that $XDEC$ is cyclic quadrilateral. Since $DX=DE$, $\angle DCX=\angle DCE$. By $\angle YMD=\angle EMD$ and $DY=DE$ we know that triangles $YMD$ and $EMD$ are congruent, so $Y$ and $E$ $BC$ are symmetrical about the line. Then $\angle YCD=\angle DCE=\angle XCD$, so $C, Y, X$ are collinear and we are done.

Since $DA$ is a tangent to $\omega$, then by symmetry $DE$ must also be a tangent. Define $P$ as the intersection between $\Omega$ and $BC$. Since the tangents of $A$ and $E$ intersect on $BC$, then $(A,E;B,C)=-1$. This implies that $AE$ is the symmedian line of $\Delta ABC$. We also know from power of a point that: $DA^2=DC \cdot DB$, which implies that $\omega$ and $\Omega$ are orthogonal. This also implies that $\Omega$ is the Apollonius circle of $\Delta ABC$ from $A$. Thus, $P$ is the angle bisector of $\angle BAC$. Since $\omega$ and $\Omega$ are orthogonal, $B$ and $C$ switch. And since $BXE$ is colinear, $CEXD$ must be cyclic. Thus we get: $1/2\angle YDE = \angle YXE = \angle CXE = \angle CDE = \angle PDE$, and since $\Delta YDE$ is an isosceles triangle, this implies: $\angle YDP = \angle PDE$, this also implies that $\angle YAP = \angle PAE$. Since $P$ is the angle bisector of $\angle BAE$, then $AY$ and $AN$ must be isogonal conjugates. But since $AN$ and $AM$ also are isogonal conjugates, $AY$ must be $AM$.

Let $H_A$ denote the $A$-Humpty point of $\Delta ABC$. It suffices to show that $Y=H_A$. Since $DA$ is tangent to $\omega$, and $D$ lies on $BC$, $\Omega$ is the $A$-Apollonius circle of $\Delta ABC$. It is well-known that $H_A$ lies on $\Omega$. Now take $\sqrt{bc}$-inversion. $\Omega'$ is the perpendicular bisector of $BC$ (denoted by $\perp_{BC}$), and $E, M$ are images of each other. Thus, $X'$ is the second intersection between $(AMC)$ and $\perp_{BC}$. $H_A'$ is the intersection between the tangents to $\omega$ at $B$ and $C$, which also lies on $\perp_{BC}$. Since $\measuredangle H_A'BA = \measuredangle BCA = \measuredangle MCA = \measuredangle MX'A = \measuredangle H_A'XA$, we have that $A, B, X', H_A'$ are concyclic. Inverting back, this means that $C,X,H_A$ are collinear. Hence $Y=H_A$, which lies on the $A$-median.

$\color{blue}\textbf{Claim 1:}$ $DE, DY$ are tangent to $\odot (ABC),\odot (BYC)$ respectively Proof: $DY^2=DE^2=DA^2=DB \cdot DC. \square$ $\color{blue}\textbf{Claim 2:}$ $XCED $ is cyclic and $BC$ is the bisector of $\angle XCE$ Proof: $\angle DCE \overset{\text{Claim 1}}{=} \angle DEB = \angle DEX = \angle DXE = \beta \implies D \in \odot (XCE) \implies \angle XCD = \beta. \square$ $\color{blue}\textbf{Claim 3:}$ $XYBD $ is cyclic Proof: $\angle DYB \overset{\text{Claim 1}}{=} \angle DYC = \angle XCD = \beta =\angle DXE. \square$ $\color{blue}\textbf{Claim 4:}$ $E$ is the reflection of $Y$ over $DC$ Proof: $\angle EDC \overset{\text{Claim 2}}{=} \angle EXC = \angle BXY \overset{\text{Claim 3}}{=} \angle BDY$, since Claim 2 and ASA criterion $\implies \triangle DYC \cong \triangle DEC . \square$ $\color{blue}\textbf{Claim 5:}$ $AB \cap XD \cap \odot (AXEY) = P$ Proof: Let $P' := XD \cap \odot (AXEY), P'' := AB \cap \odot (AXEY) \implies \angle EAP' = \angle P'XB = \beta = \angle BCE = \angle EAB = \angle EAP'' \implies P' \equiv P'' . \square$ $\color{blue}\textbf{Claim 6:}$ $BC$ is tangent to $\odot (AYB), \odot (AYC)$ Proof: $\angle BAY = \angle PAY =\angle PXY = \overset{\text{Claim 3}}{=} \angle CBY \overset{\text{Claim 4}}{=} \angle EBC = \angle EAC \implies \angle BAY = \angle CBY, \angle YAC = \beta = \angle YAC. \square$ By Claim 6, we have that $M = AY \cap BC \implies MB^2 = Pow(M, \odot (AXEY)) = MY \cdot MA = Pow(M, \odot (ABC)) = MC^2 \implies M$ is the midpoint and we are done