My problem

Let $D,E,F$ be the altitudes from $A,B,C$ to $BC,CA,AB$ respectively. Let $S$ be the $A-$humpty point. Take the inversion centered at $M$ with radius $MB=MC=ME=MF$. We want to show that $(MPQ)\leftrightarrow AH,(MAH)\leftrightarrow SP,(DEFM)\leftrightarrow EF$ are concurrent. Since $ME,MF$ are tangent to $(APFHSE),$ we get that $(E,F;AH\cap EF,Q)=-1=(E,F;S,A)=(PE,PF;PS,PA)=(E,F;PS\cap EF,Q)$ thus, $PS,AH,EF$ are concurrent as desired.$\blacksquare$

Let $K$ be the midpoint of $AH$. Since $P$ is the $A$-queue point, it is known that $P$, $H$ and $M$ are collinear. Therefore, define $T$ such that $HT\bot PM$ and also let $L=KQ\cap N_9$ be the phantom point of $G$. We´re going prove that $L \in (MPQ)$ and $L \in (MHA)$, thus, $L=G$. Initially, note that, by rotation, $\angle PMQ=\angle AHT$. Since $Q$ belongs to the radical axis of both $(APH)$ and $N_9$: $QL\cdot QK=QP\cdot QA \Rightarrow PLKA$ is cyclic Hence $\angle PLQ=\angle PAK=\angle PAH=\angle AHT= \angle PMQ \Rightarrow L \in (PMQ)$ However, since $KM$ is the diameter of $N_9$, $\angle KLM=90$ and, from $KA=KP$, $\angle KAP= \angle KPA=\angle KLA$ Then, $\angle ALM = \angle ALK + \angle KLM = \angle KAP + 90 = \angle AHT + 90 = \angle AHM \Rightarrow L \in (AHM)$ Q.E.D $\blacksquare$

Very standard configurational geometry here. Let $D$ , $E$ and $F$ be the feet of the altitudes of $\triangle ABC$. We rename $P$ to $Q_A$ - the $A-$Queue point and $Q$ to $X_A$- the $A$-Ex point respectively. Further note by $H_A$ the $A-$Humpty Point and by $N$ the midpoint of $AH$. Denote by $\omega_9$ the nine point circle of $\triangle ABC$. First, it is well known that $N$ lies on $\omega$ and that $N$ is the $M-$antipode on this circle (since both these points lie on the $EF$ perpendicular bisector). Now, let $R = \overline{X_AN} \cap \omega_9$. Our claim is that this point coincides with $G$. To see why, first note that $H_A$ and $R$ clearly lies on $(MQ_AX_A)$ which is well known to have diameter $X_AM$ (since $\angle AX_AM = \frac{\pi}{2}$) the latter since, \[\measuredangle MRX_A = \measuredangle MRN = \frac{\pi}{2}\]Further, note that by the converse of the Radical Center Theorem, quadrilateral $RNH_AH$ is cyclic. Thus, \[\measuredangle HRX_A = \measuredangle HH_AN = \measuredangle NHH_A = \measuredangle HRH_A\]from which it follows that $\overline{X_AN}$ is simply the external $\angle H_ARH$-bisector. Then, since $MR \perp X_AN$ we have that $\overline{MR}$ is the internal $\angle H_ARH$-bisector. Now, \[\measuredangle MRH= \measuredangle H_ARM = \measuredangle H_AX_AD = \measuredangle H_AAD = \measuredangle MAH\]implying that $R$ indeed lies on $(AHM)$ as well. Thus, $R$ lies on all the circles we need and indeed $R\equiv G$ which finishes the proof of the problem.

Let $D, E, F$ be the feet of the altitudes, $K=PD\cap (MPQ)$ and $L=(MPQ)\cap(AHP)$. Define $\psi$ as an inversion of center $H$ and radius $\sqrt{HA\cdot HD}$ followed by a reflection over $H$. Lemma 1: $\psi(Q)=L$ and $L\in AM$. Proof: It's well-known that $\psi(P)=M$, thus $(MPQ)$ "inverts" to itself. Clearly $\psi(BC)=(AHP)\Rightarrow\psi(Q)=L$. Notice that $\psi((HDM))=AP\Rightarrow L\in(HDM)\Rightarrow\angle HLM=\angle HDM=90^{\circ}$. But $\angle ALH=90^{\circ}\Rightarrow L\in AM \ _{\blacksquare}$ Lemma 2: $K\in(ABC)$.

Proof: Notice that $DP\cdot DK=DQ\cdot DM=DH\cdot DA=DB\cdot DC$, Click to reveal hidden text

, therefore $K\in(ABC) \ _{\blacksquare}$ See that $\psi(PD)=(AHM)$ and $\psi((MPQ))=(MPQ)\Rightarrow \psi(K)=G$. But $\psi((ABC))=(DEF)\Rightarrow G\in(DEF)$, as we wanted. $_{\blacksquare}$

Official Solution: Since \( M \) belongs to \( (MPQ) \), \( (AHM) \), and the nine-point circle, we only need to prove that the centers of these three circles are collinear, as this would imply that they are coaxial and thus also intersect at \( G \). Let \( D \) be the foot of the altitude relative to the side \( BC \), \( J \) the center of \( (AHM) \), \( L \) of \( (MPQ) \), \( N_9 \) of the nine-point circle, and \( K \) the midpoint of \( AH \). Notice that \( \angle MPQ = 90^\circ \), because \( M \), \( H \), and \( P \) are collinear, therefore \( L \) is the midpoint of \( MQ \). Also, observe that \( K \), \( M \), and \( D \) belong to the nine-point circle, but \( \angle MDK = 90^\circ \), therefore \( N_9 \) is the midpoint of \( MK \). Thus, to prove that \( L \), \( N_9 \), and \( J \) are collinear, we will show that \( KJML \) is a parallelogram, whose diagonals intersect at the midpoint, solving the problem. Let \( I \) be the circumcenter of triangle \( AMQ \). Note that \( H \) is the orthocenter of \( AMQ \), so \( J \) (the circumcenter of \( AHM \)) is the reflection of \( I \) across \( AM \), hence \( AIJM \) is a parallelogram. Moreover, it is a known property of the orthocenter that \( AKLI \) is a parallelogram (proven by the homothety centered at the centroid that maps the medial triangle to the original triangle, and the circumcenter to the orthocenter), thus we can conclude that \( KLMJ \) is also a parallelogram, completing the problem.