Some of the solutions include complex bash and inversions. Below are hints for a synthetic approach (Credits to Zilin Jiang):

The problem is easy without bash, using projections of orthocenter on bisectors, the circle of diameter $HA$ cut $AM_a,AK_a,AL_a$ at $A_1,A_2,A_3$. also $\angle HAA_2=\angle AL_aB =\angle A_3A_2A \Rightarrow A_3A_2K_aL_a$ is cyclical, then from the link we get $M_aA_2.M_aA_3=M_aA_1,M_aA=M_aL_a.M_aK_a$ i.e. $A_1 \in \odot (AL_aK_a)$ i.e. $A_1 \equiv X_a$ hence $X_a$ belongs to circle of diameter $HG$ Consider $H$ ortocenter and $G$ centroid.

Here is my solution: Let $D,E,F$ feet of $A,B,C$ onto $BC,AC,AB$ let $X$ be the projection of $H$ onto the $AM_a$. Lemma1: $BC.EF,HX$ are collinear. Prove: let $EF\cap BC=T$ then $AH$ is polar of $T$ WRT $\odot (BFEC)$ so $AM_a\perp TH$ so the lemma proved. Lemma 2: the locus of point $X$ such that $\frac{XB}{XC}=\frac{AB}{AC}$ is the $A$-appolonus circle(a circle with diameter $K_al_a$) Prove: it's well known. Lemma 3: in cyclic quadrilateral $ABCD$ let $AB\cap CD=R$ then $\frac{RD}{RC}=\frac{AD}{AC}.\frac{BD}{BC}$. Prove: it's well known. BACK TO THE MAIN PROBLEM: $HXM_aD$ is cyclic$\longrightarrow TD.TM_a=TH.TX$ and $(TDBC)=-1\longrightarrow TD.TM_a=TB.TC$ so $TB.TC=TH.TX$ hence $HXCB$ is cyclic. using Menelaus theorem for $T,E,F$ and $\triangle ABC$ we have $\frac{TB}{TC}=\frac{AE}{CE}.\frac{BF}{AF}$(1). Using lemma 3 for cyclic quadrilateral $HXCB$ we have $\frac{TB}{TC}=\frac{HB}{HC}.\frac{XB}{XC}$(2) From (1) , (2) we deduce that $\frac{XB}{XC}=\frac{AE}{CE}.\frac{BF}{AF}.\frac{HC}{HB}$ From Menelaus for three concurrent lines $AD,BE,CF$ we have $\frac{AE}{CE}.\frac{BF}{AF}=\frac{BD}{CD}$ in triangle $HBC$: $\frac{BD}{CD}=\frac{HB}{HC}.\frac{\sin C}{\sin B}$ so $\frac{XB}{XC}=\frac{AE}{CE}.\frac{BF}{AF}.\frac{HC}{HB}=\frac{BD}{CD}\frac{HC}{HB}=\frac{\sin C}{\sin B}=\frac{AB}{AC}$ Hence by lemma (2) because $\frac{XB}{XC}=\frac{AB}{AC}$ we get that $X\in \odot (K_aL_a)$ so $X\equiv X_a$ thus $X_a\in \odot (GH)$ similarly $X_b,X_c\in \odot (GH)$ DONE

My second solution: (using inversion) Let $P,Q,X$ projections of $H$ onto $AK_a,AL_a,AM_a$ and let $D,E,F$ feet of $A,B,C$ on $BC,CA,AB$. Apply an inversion with center $A$ and radius $\sqrt{AH.AD}$ then $X\longleftrightarrow M_a$ and $A$-appolonus circle is taken to line $PQ$ so in order to prove the problem we have to show that $P,Q,M_a$ are collinear.let $M$ midpoint of $AH$ then because $M_aE=M_aF,ME=MF$ line $MM_a$ is perpendicular bisector of $EF$(1).let $S,T$ projections of $P$ on $AC,AB$ so $PS=PT$.Because $AFPE$ is cyclic we have $\angle PES=\angle PFT$ thus $\triangle PES=\triangle PFT\longrightarrow PE=PF$ similarly $QE=QF$ so $PQ$ is perpendicular bisector of $EF$(2). From (1),(2) we deduce that $P,Q,M_a$ are collinear. DONE

The main difficulty of the problem is claiming that $\angle HX_aA = 90^\circ$ (ie that the circumcenter is the midpoint of $HG$). To prove this we just compute $X_a = (a^2:2S_A:2S_A)$ from the givens and then observe this lies on the circle with diameter $AH$. Unlike the recent TST problem the Euler Line is a red herring.

Oh, I see the synthetic solution now. Just show that $M_a$ has the same power with respect to the circle with diameter $AH$ and the circle with diameter $K_aL_a$. Then $AM_a$ is the radical axis, so $X_a$ lies on both circles.

If you've done this problem, you should be able to recognize $X_a$ with some effort, and then the crucial claim about the point lying on both the circle through $AK_aL_a$ and circle with diameter $\overline{AH}$ is pretty easy to make.

Since $(L_{a}BK_{a}C)=-1$, we find $M_{a}B^{2}=M_{a}X_{a} \cdot MA$ $\Longrightarrow$ $BC$ is tangent to $\odot ABX_{a}$ and $\odot ACX_{a}$. By angle chasing we find $X_{a} \in \odot BHC$. Let the $R$ be the antipode of $H$ in $\odot BHC$. By angle chasing we find $RCAB$ is a parallelogram $\Longrightarrow$ $A,M_{a},R$ are collinear. So $\angle HX_{a}G=90^ {\circ} $ $\Longrightarrow$ $X_{a}\in (HG)$ $\Longrightarrow$ center of $X_{a}X_{b}X_{c}$ is the midpoint of $HG$.

Let $H, G$ be the orthocenter, centroid, respectively, of $\triangle ABC.$ From the Internal and External Angle Bisector Theorems, we have $K_aB / K_aC = L_aB / L_aC = AB / AC.$ Thus, the division $(B, C; K_a, L_a)$ is harmonic. Hence, it is well-known (Lemma 1.5) that $M_aK_a \cdot M_aL_a = M_aB^2.$ Meanwhile, from Power of a Point, we find $M_aK_a \cdot M_aL_a = M_aX_a \cdot M_aA.$ Therefore, $M_aB^2 = M_aX_A \cdot M_aA$, which can be rearranged into $M_aX_a / MB = M_aB / M_aA.$ Combining this relation with $\angle X_aM_aB = \angle BM_aA$, it follows that $\triangle X_aM_aB \sim \triangle BM_aA.$ Similarly, we have $\triangle X_aM_aC \sim \triangle CM_aA.$ By using these similar triangles, we find \[\angle BX_aC = \angle BX_aM_a + \angle CX_aM_a = \angle ABM_a + \angle ACM_a = 180^{\circ} - \angle BAC = \angle BHC.\] It follows that $B, C, H, X_a$ are concyclic. Therefore, $\angle HX_aB = \angle HCB = 90^{\circ} - \angle ABM_a.$ Meanwhile, from $\triangle X_aM_aB \sim \triangle BM_aA$, it follows that $\angle ABM_a = \angle BX_aM_a.$ Therefore, $\angle HX_aB = 90^{\circ} - \angle BX_aM_a \implies \angle HX_aM_a = 90^{\circ}.$ Therefore $HX_a \perp GX_a.$ Similarly, $HX_b \perp GX_b$ and $HX_c \perp GX_c$, so that $\odot (X_aX_bX_c)$ is just the circle of diameter $\overline{HG}.$ Consequently, its center is the midpoint of $\overline{HG}$, which clearly lies on the Euler line. $\square$

For a problem involving the same key point, see 2005 USA TST #6.

Here's a solution which explicitly does the barycentric calculation.

Let $H$ be the orthocenter of $\triangle ABC$ and let $A'$ be the antipode of $A$ WRT $\odot(ABC).$ Let $AM_a$ cut $\odot(ABC)$ for a second time at $X.$ By the Angle-Bisector Theorem, $\tfrac{K_aB}{K_aC} = \tfrac{L_aB}{L_aC} = \tfrac{AB}{AC} \implies (B, C; K_a, L_a) = -1.$ Then by Power of a Point and a well-known result of harmonic divisions (Lemma 1.5), we have \[M_aA \cdot M_aX_a = M_aK_a \cdot M_aL_a = -M_aB \cdot M_aC = -M_aA \cdot M_aX.\]Therefore, $M_aX_a = -M_aX \implies M_a$ is the midpoint of $\overline{XX_a}.$ But since $M_a$ is also the midpoint of $\overline{HA'}$ (well-known), it follows that $HXA'X_a$ is a parallelogram. Hence, $\angle HX_aX = \angle A'XX_a = 90^{\circ}.$ Then if $G$ is the centroid of $\triangle ABC$, we have $\angle HX_aG = 90^{\circ} \implies X_a$ lies on the circle $\omega$ of diameter $\overline{HG}.$ Similarly, $X_b, X_c \in \omega$, so the circumcenter of $\triangle X_aX_bX_c$ is just the midpoint of $\overline{HG}.$ $\square$

It suffices to show that $X_a$ lies on the circle with diameter $\overline{AH}$, where $H$ is the orthocenter of $\triangle ABC$. From this, we derive that $\angle AX_aH=\angle GX_aH$, where $G$ is the centroid of $\triangle ABC$, so that the circumcenter of $\triangle X_aX_bX_c$ is simply the midpoint of $\overline{HG}$. Denote the feet of the altitudes from $B$ and $C$ as $E$ and $F$; note that the circle with diameter $AH$ passes through $E$ and $F$. Invert about $A$ with radius $\sqrt{AB \cdot AC}$, composed with a reflection about the $A$-angle bisector, and denote inverses with a '. We get that $F'$ is the intersection of the line through $B$ perpendicular through $AB$ with $AC$, and define $E'$ similarly. $K_a'$ is the midpoint of minor arc $BC$ in the circumcircle of $\triangle ABC$, and $L_a'$ is the midpoint of major arc $BAC$. The conclusion we desire is that the $A$-symmedian, line $E'F'$, and line $K_a'L_a'$ are concurrent; we claim that the concurrency point is the intersection of the tangents to $(ABC)$ at $B$ and $C$, and moreover that this point is the circumcenter of $(BCEF)$. Let $X_a'$ be the intersection of the tangents from $B$ and $C$; angle chasing, we get \begin{align*} \angle E'BF'&=180^{\circ}-\angle ABF' \\ &=\angle BAF'+\angle BF'A \\ &=\angle BAC+\angle BF'C \\ &= \frac{1}{2}\big(\angle BOC+\angle BX_a'C\big)\\ &=90^{\circ} \end{align*}so that $E'$ and $F'$ are indeed are our desired points, hence we are done$.\:\blacksquare\:$ I believe that we can actually use circle $(BEFC)$ to prove that the intersection of the tangents from $B$ and $C$ in $(ABC)$ lies on the $A$-symmedian.

Throughout the solution, let $Q$ denote $Q_a$ for almost every point $Q$. Let $H$ be the orthocenter and $\triangle DEF$ the orthic triangle of $\triangle ABC$. Let $\Omega = (ABC)$, $\omega = (DEF)$. Let $P$ be the intersection of $(AKL)$ and $\Omega$. $(AKL)$ is the Apollonius circle WRT $BC$, so $(A, P, B, C)$ is harmonic, which means $AP$ is the $A$-symmedian. $AX$ and $AP$ are isogonal conjugates, so we get $KX = KP$ and $LX = LP$, hence $X$ is the reflection of $P$ in $BC$. Now we forget about $K$ and $L$. Let $AM$ intersect $(AEHF)$ again in $X'$. It's trivial to prove that the tangents in $E$ and $F$ to $(AEHF)$ intersect in $M$, hence $(A, X', E, F) = -1$. Let $T$ be the reflection of $H$ in $BC$. By some cross-ratio chasing, $(D, HX' \cap BC, B, C)$ and $(D, TP \cap BC, B, C)$ are both harmonic, so $HX'$, $TP$, $BC$ are concurrent in $R$. Reflecting $TP$ in $BC$ yields $X' \equiv X$. Let $\psi$ be the inversion that swaps $\Omega$ and $\omega$. It preserves cross-ratios, so $X$ maps to $R$. $RA \cdot RB = RE \cdot RB$, so $R_a$, $R_b$, $R_c$ lie on the radical axis of $\Omega$ and $\omega$, which is perpendicular to the Euler line of $\triangle ABC$. Under $\psi$, the Euler line maps to itself, while the line $R_aR_bR_c$ maps to the circumcircle of $\triangle X_aX_bX_c$. $\psi$ preserves angles, so the Euler line passes through the center of $(X_aX_bX_c)$. $\blacksquare$

Hmmm so I found a very interesting finish to the bary-bash, without needing to claim anything synthetic! Compute $X_a=(a^2:2S_A:2S_A)$ as above. Now we'll actually compute the equation of $(X_aX_bX_c)$. Our equations for $u,v,w$ quickly become \[2a^2S_A=a^2u+2S_A(v+w), 2b^2S_B=b^2v+2S_B(u+w), 2c^2S_C=c^2w+2S_C(u+v).\]After staring at this for a bit, we realize it is satisfied simply by \[u=2S_A/3,v=2S_B/3,w=2S_C/3\](!!!!) So the circle's equation is \[a^2yz+b^2zx+c^2xy=\frac{2}{3}(x+y+z)(S_Ax+S_By+S_Cz).\]Now comes the interesting part: Observe that the nine-point circle has equation\[a^2yz+b^2zx+c^2xy=\frac{1}{2}(x+y+z)(S_Ax+S_By+S_Cz)\]and additionally the circumcircle has equation \[a^2yz+b^2zx+c^2xy=0(x+y+z)(S_Ax+S_By+S_Cz).\]Now $N_9$ and $O$ lie on the Euler Line, so proceed by a linearity argument. By working out the euler line ratios we can actually find that the center is the midpoint of $HG$. I find this solution rather mind-blowing (I've never seen this "linearity of circle equation" before, though it's obvious in retrospect.)

Gah finally getting back to MO By harmonics, $M_aK_a \cdot M_aL_a = M_aB^2 = M_aX_a \cdot M_aA \implies \triangle M_aBX_a \sim \triangle M_aBA$, $\triangle M_aCX_a \sim \triangle M_aAC$. Angle chasing, $$\angle BX_aC = 180-\angle X_aBC -\angle X_aCB = 180-\angle BAM_a - \angle CAM_a = 180-\angle BAC = \angle BHC$$so $B, X_a, H, C$ are cyclic. Now $\angle HX_aC + \angle M_aX_aC = \angle HBC + \angle ACB = 90-\angle C + \angle C = 90 \implies HX_a \perp GX_a$ This implies that $X_a, X_b, X_c$ lies on a circle with center the midpoint of $HG$, as required. $\blacksquare$

Worth noting that the angle bisectors in this problem were irrelevant; the problem holds for any points such that $(L_a,B;K_a,C)=-1$ and similar.

We use the following Lemma (#2). Quote: Lemma: In triangle $ABC$ let $M$ be the mid-point of $BC$ and let us consider points $E$ and $F$ on $AC,AB$ respectively such that $BE\cap CF = D$ and $A,E,D,F$ are on the same circle. Then, $\circ{AEF}\cap \circ{BDC}={D,X}$ where $X$ is the point where the $A-$ Apollonian circle intersects the median $AM$. We just allow \(D=H\), where H is the orthocenter. Also note that \((AK_a L_a) \) is just the A-appolonius circle of \(\triangle ABC\). It follows from the lemma that \(X_a=AM \cap (AEHF)\). It immediately follows that \(\angle AX_aH =90 \implies \angle HX_aG=90\) where G is the centroid. Note that this means the \(X_a\) lies on the circle with diameter \(HG\), This holds true for \(X_b, X_c\) as well. So the circumcenter of \(X_aX_bX_c\) is the midpoint of \(HG\) and hence lies on the Euler Line!

Another nice property of the orthocentroidal circle!!!

Claim: $X_a$ is the A-Humpty Point. Of course, we will phantom this. Let $X_a'$ be the A-Humpty point. Then, letting $D,E,F$ be the feet of the altitudes, we have by power of a point on $M_a$ and $(AH)$ that $$M_aX_a'\cdot M_aA=M_aE^2=M_aF^2=M_aB^2.$$However, since $\angle L_aAK_a=90$ and $AK_a$ bisects $\angle BAC$, we have that $$(L_aK_a;BC)=-1.$$This means that $$M_aB^2=M_aK_a\cdot M_aL_a,$$hence $$M_aX_a'\cdot M_aA=M_aK_a\cdot M_aL_a$$and thus $AX_a'K_aL_a$ is cyclic, completing the proof of the claim that $X_a$ is the A-Humpty point. Now, it suffices to show that the circumcenter of the three Humpty points lies on the Euler line. More specifically, we claim that it is the midpoint of $HG$. By definition of Humpty points, $\angle AX_aH=90$. Since it also lies on the median, we have $$\angle GX_aH=90.$$Similarly, $$\angle GX_bH=\angle GX_cH=90.$$Thus, $X_a$, $X_b$, and $X_c$ all lie on the circle with diameter $GH$, so we are done.

One of the problems of all time. We claim that $(X_AX_BX_C)$ is centered at the midpoint of $HG$, where $H$ and $G$ are the orthocenter and centroid. To do this we will show that $HX_A \perp X_AG$ or equivalently that $X_A$ is the $A$ humpty point. To see this let $H_A$ be the $A$ humpty point. It is well known that $H_A$ lies on $AM$. Then by power of a point it suffices to show that, \begin{align*} M_AK_A \cdot M_AL_A = M_AH_A \cdot M_AA. \end{align*}However, letting $E = BH \cap AC$ we know that \begin{align*} M_AH_A \cdot M_AA &= M_AE^2\\ &= M_AB \cdot M_AC\\ &= MK_A \cdot M_AL_A \end{align*}where the first line follows from the Three Tangents lemma. Thus in a similar fashion we can show $X_B$ and $X_C$ lie on the circle with diameter $GH$. Therefore their circumcenter lies on the Euler Line as desired.

The $A$-Humpty point can be characterized as the point of intersection of the $A$-Apollonius circle and the $A$-median distinct from $A$, which is exactly $X_a$. We further note the $A$-Humpty point is also the foot from $H$ to the $A$-median, implying \[\angle HX_aA = \angle HX_aG = 90,\] so $X_a$ lies on the circle with diameter $HG$, or the Euler line, and symmetrically for $X_b$ and $H_c$. $\blacksquare$

Note that $(AK_a L_a)$ is the $A$-Apollonian circle. Let $H$ be the orthocenter of triangle $ABC$, and further let $X_a'$, $X_b'$ and $X_c'$ be the feet from $H$ to $AM_a$, $BM_b$, $CM_c$ respectively. Now it is well-known that $X_a'$, $X_b'$ and $X_c'$ lie on the $A$-, $B$- and $C$-Apollonian circles respectively, and since there are at most two intersections of a circle and a line, we know that $X_a = X_a'$, $X_b = X_b'$ and $X_c = X_c'$. Thus $HG$ is the diameter of $(X_a X_b X_c)$ where $G$ is the centroid of triangle $ABC$, so the circumcenter of triangle $X_a X_b X_c$ is the midpoint of segment $\overline{HG}$, as desired.

$(AK_aL_a)$ is the $A$-Apollonius circle of $\triangle ABC$, which makes $X_a$ the $A$-Humpty point. It's well known that $AX_a \perp X_aH$, and by definition $AX_a$, $BX_b$, $CX_c$ concur at $G$. This implies that $GH$ is the diameter of $(X_aX_bX_c)$, so the center is the midpoint of $GH$, as desired.

Let $\omega$ be the orthocentroidal circle (circle with diameter $GH$, where $G$ is the centroid and $H$ is the orthocentre). Its centre clearly lies on the Euler line, so it suffices to show that $X_a,X_b,X_c$ lies on $\omega$. Note that $(AK_aL_a)$ is the $A$-Apollonius circle, so $X_a=(AK_aL_a) \cap AM_a$ is the $A$-Humpty point. It is well-known that $HX_a \perp AM$, so $\angle HX_aG = 90^{\circ}$, from which it follows that $X_a$ lies on $\omega$. Similarly, $X_b,X_c$ lie on $\omega$, so we are done.

very beautiful problem. here is my solution: Claim: Let $H_a$ be the $A$ humpty point of a circle and $O,N$ denote the circumcenter and 9 point center respectively then $\frac{H_{a}O}{H_{a}N} = 2$. Proof: observe that $\frac{HO}{HN} = \frac{GO}{GN} = 2$ where $H,G$ are the orthocenter and centroid respectively.We also know that $\angle HH_{a}N = 90^{\circ}$. So $H_a$ lies on $(HG)$ and $(HG)$ is the Apollonian circle of $O,N$ with ratio 2. Hence the assertion is true. PS: we actually only need the concylic part but the result itself is cool so i added it Claim : the circumcenter of $(H_{a}H_{b}H_{c})$ lies on the euler line. proof: $(H_{a}H_{b}H_{c}) \equiv (HG)$ from the above claim so the circumcenter is on $HG$ which is the euler line. Claim: $X_i \equiv H_i$ where $i= a,b,c$. proof: as $(BC;L_{a}K_{a})=-1$ so from P1 we have $M_{a}A^2 = ML_{a} \cdot MK_{a} = MX_{a} \cdot A$. Done. This completes the proof.

It's clear that $(B, C; K_A, L_A) = -1. \implies M_AB^2 = M_AC^2 = M_AK_A \cdot M_AL_A = M_AX_A \cdot M_AA$. Thus, $(ABX_A), (ACX_A)$ are tangent to $BC \implies X_A$ is the $A$-Humpty point. Analogously, $X_B, X_C$ are $B$-Humpty and $C$-Humpty points, respectively. Since $X_A$ is $A$-Humpty point, we deduce that $HX_A$ is perpendicular to $AM_A$, where $H$ is the orthocenter of $ABC$. Therefore, $\angle HX_AG = 90^{\circ} \implies X_A$ lies on the circle with diameter $HG$, where $G$ is the centroid of $ABC$. Similarly, $\angle HX_BG = \angle HX_CG = 90^{\circ}$. Hence, $HGX_AX_BX_C$ is cyclic and its diameter is $HG \implies$ center of $(X_AX_BX_C)$ lies on $HG$. $\blacksquare$

From the internal external angle bisectors, $(L_a, K_a; B, C)=-1,$ so we know $M_a B^2 = M_aK_a \cdot M_aL_a.$ By PoaP, $$M_a X_a \cdot M_a A = M_aK_a \cdot M_aL_a= M_a B^2,$$which implies $A$ and $X_a$ are inverses wrt the circle with diameter $BC,$ say $w_{bc}.$ Letting $H_c = w_{bc} \cap AB$ and $H_b = w_{bc} \cap AC,$ we notice $\angle BH_cC = \angle BH_bC = 90,$ so $CH_c \cap BH_b = H,$ the orthocenter. By Brocard, $H \in polar(A).$ However we know $polar(A)$ also passes through the inverse of $A,$ which is $X_a$. Thus, $HX_a=polar(A)$ and $\angle HX_aA=90.$ Notice that $G \in AM_a,$ so $\angle HX_aG=90.$ Hence $X_a, X_b, X_c$ all lie on circle with diameter $HG,$ and we are done.

It is well known that $(AK_aL_a)$ is the $A$ apollonius circle so it cuts the $A$ median at the $A$ humpty point which is $X_a$ and similarly $X_b$ and $X_c$ are humpty points. It is trivial to see that $X_i \in (HG)$ where $i \in$ $\{a,b,c\}$ and $G$ is centroid of $\Delta ABC$ because the humpty point is defined as the perpendicular from $H$ to the corresponding median so we get the desired result.

Only consider $X_a$ initially. Notice that $(AK_aL_a)$ forms the exact configuration of an Apollonian circle; that is, we have \[\frac{XB}{XC} = \frac{AB}{AC},\] for all points $X$ on $\overline{BC}$. Note that the $A$-Humpty point $X_a'$ in $\triangle ABC$ can be defined as the point on the $A$-median satsifying $\tfrac{X_a'B}{X_a'C} = \tfrac{AB}{AC}$. Hence, $X_a' = X_a$, meaning that $HX_a \perp GX_a$ (where $H$ is the orthocenter and $G$ is the centroid). An analogous argument for $X_b$ and $X_c$ implies that $(X_aX_bX_c)$ has diameter $\overline{HG}$. $\square$

On my way to another one-line solution. Point on the median? It has to be humpty! The intersection of $(AK_a L_a)$ (the Apollonian circle) with the median gives the Humpty point $X_a$, which by definition lies on the perpendicular from $AH$. Therefore $X_aX_bX_c$ lie on the circle with diameter $HG$. $\blacksquare$

What? Used the 10\% hint on ARCH. It turns out that $X$ is exactly the $A$-Humpty point, so $\angle HXG=90^{\circ}$ therefore the circumcenter of $X_aX_bX_c$ is exactly the midpoint of $GH$. But note that $X$ lies on $AM$ and the $A$-apollonius circle of $ABC$, so we are done. $\blacksquare$

Re-solved because I somehow forgot that I solved this. Note that $(AKL)$ is the circle through $A$ that has a constant $PB/PC$. Thus $X_A$ is exactly the $A$-Humpty point. Thus the reflection of $H$ over $X$, say $Y$, has $GH=GY$. Similarly for the other two, so $(Y_aY_bY_c)$ has circumcenter $G$, so $(X_aX_bX_c)$ has circumcenter $(G+H)/2$. $\blacksquare$ Remark. A few hours later, I discovered that the $A$-, $B$-, and $C$-Ex Points are collinear by some harmonic stuff. Now the negative inversion centered at $H$ swapping $A$ and the foot of the $A$-altitude and similar maps the Ex points to the Humpty points, so indeed we get that the Humpty points are also concyclic with $H$. Thus the three perpendicular bisectors involving $H$ must concur, and at that point it's easy to see what's happening.