Denote by $\triangle DEF$ the orthic triangle. Note that $\overline{AG}$, $\overline{EF}$, $\overline{BC}$ are concurrent at $R$ by radical axis, and that $\overline{PA}$ and $\overline{PG}$ are tangents to $\gamma$. Now, consider circles $(PAGM)$, $(MFDNE)$, and $(MBC)$. They intersect at $M$ but have radical center $R$, so are coaxial; assume they meet again at $T \in \overline{RM}$, say. Then $\angle PTM$ and $\angle MTN$ are both right angles, hence $T$ lies on $\overline{PN}$. Finally $H$ is the orthocenter of $\triangle ARN$, and thus the circle with diameter $\overline{RN}$ passes through $G$, $Q$, $N$.

My solutions: We have $\overline{G, H, N}$ (well-known). Let $AG\cap BC=S, AH\cap BC=D$, and let $MK\perp PN at K$. We have $AG, KM, DN$ are concurent at point $S$ is the radical center of ${\odot (APGKM), \odot (MKDN), \odot (AGDN)}$. Becaues $AH\perp SN, NH\perp AS$ $\Rightarrow H$ is othorcenter of $\triangle ANS$ $\Rightarrow Q\in SH$. We have $SK.SM=SG.SA=SB.SC$ $\Rightarrow K\in \odot (BCM)$. We have $K\in \odot (SN)\equiv (GQN)$. So, $(GNQ)\cap (MBC)\cap PN=K$. DONE

I always overcomplicate Evan's medium geometry proposals lol

Let $U,V$ be the feet of altitudes from $B,C$ respectively. Let $UV\cap BC=X$ and let $XM \cap \mathcal{N} =T$ where $\mathcal{N}$ is the nine-point circle of $\triangle{ABC}$. Considering radical axes of $\mathcal{N},(ABC),(MBC)$ we found that $X$ is the radical center and so $T \in (MBC)$. Considering those of $(PAG),(ABC),(MBC)$ we found that $X$ is the radical center and so $T\in (PAG)$. Then, we have $\angle{PTM}=90^{\circ}$ and $\angle{NTM}=\angle{NUM}=90^{\circ}$. So $T\in PN$. From $\angle{XTN}=90^{\circ}=\angle{XQN}=\angle{XGN}=90^{\circ}$ we get $T\in (GQN)$. Hence $(GQN)\cap(MBC)=T\in PN$.

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Let $DEF$ be the orthic triangle, $V$ be the harmonic conjugate of $D$ with respect to segment $\overline{BC}$, and $T'$ be the harmonic conjugate of $D$ with respect to $EF$ in the nine-point circle. Projecting from $M$ onto line $EF$ implies that $M,T',V$ are collinear, so $$VT'\cdot VM=VF\cdot VE=VB\cdot VC$$so $T'\in\odot(MBC)$, and since $\angle MT'N=\pi-\angle VT'N=\frac{\pi}{2}$, $T'\in \odot(GQN)$, hence $T'\equiv T$. Finally, $$VT\cdot VM=VB\cdot VC=VG\cdot VA$$so pentagon $APGTM$ is cyclic, and $\angle PTM=\pi-\angle PAM=\frac{\pi}{2}$.

V_Enhance, how did you come up with this problem?

Let $ S= AG\cap BC $ we know that $NG\cap SQ=H$ is the orthocenter of $ASN$ if J is the center of $NSGQ$ then $\widehat{JGM}=\widehat{GNS}=\widehat{GQS}=\widehat{GQH} $ thus $GJ$ is tangent to $(M)$ i.e. $\overline{JGP} \perp GM $ besides $ OM \perp AG,NG\perp AG $ then $OM \parallel NG$ implies $ \widehat{GPM}=\widehat{JGN}=\widehat{GQH}=\widehat{GAH} =\widehat{GAM} $ hence $GPMA$ s cyclic but $GJ\perp GM$ thus $PA \perp AM=AH$ implies $PA\parallel BC $ , applying the converse of Reim's to $N-R-P , S-G-A$ where $ R=NP\cap (GMAP) $ yields $ GRSB$ is cyclic .In the other hand $RM\perp PR, SR\perp NR $ we deduce $S,R,M $ are collinear ; further $GA,RM,BC$ are concurrent so by radical axis 's we conclude $RMBC$ is cyclic. R HAS.

Nice problem! Let $\triangle DEF$ be the orthic triangle. First, $\angle AQH=\angle AQA' = 90$ where $A'$ is the antipode of $A$. Thus $Q, H, N, A'$ are collinear. An easy angle chase gives $\angle MFN =\angle MEN =90$, so $QH$, the tangent to $(AEF)$ at $E$, and the tangent at $F$ concur. Thus $QFHE$ is harmonic. A perspectivity at $A$ gives $XBDC$ harmonic where $X=AQ\cap BC$. Thus $X\in EF$ by a well known lemma. On the other hand, since $NH\perp AQ$ and $AH\perp XN$, $XH\perp AG$ and $X, H, G$ are collinear. Moreover, $(XQGN)$ is cyclic. Let $T$ be the intersection of circles $(XQGN)$ and $(MFDNE)$ other than $N$. Note $XT\cdot XM = XF\cdot XE=XB\cdot XC$, so $T$ lies on $(MBC)$. By Radical Axis on $(AQFHGE), (XQGN), (MFDNE)$, $GQ, EF, TN$ concur. Let $R$ be the point of concurrence. On the other hand, by Radical Axis on $(MQDG), (XQGN), (MFDNE)$, we have $MD, GQ, TN$ concur. Thus $R$ lies on $AH$ as well. We claim that $P, R, N$ are collinear. Note $MO\parallel QHN\perp QA$, so $PM\perp QA$. Thus $PQ=PA$, so $P$ is the intersection of the tangents to $(AQFHGE)$ at $A, Q$. But recall $N$ was the intersection of the tangents at $E, F$. Thus by Pascal's on $AQHQAG$ we deduce $P, R, N$ are collinear, as desired. $\square$

Niche ne ponel no pooral

First wrote: V_Enhance, how did you come up with this problem? Well, the original proposal was to show that the nine-point circle and $(MBC)$ intersected on line $PN$. The points $G$ and $Q$ were added after the fact, to prevent people from using straight coordinates. Here is also my original solution: Let $L$ be diametrically opposite $A$ on the circumcircle. Denote by $\triangle DEF$ the orthic triangle. Let $X = \overline{AH} \cap \overline{EF}$. Finally, let $T$ be the second intersection of $(MFDNE)$ and $(MBC)$. We begin with a few easy observations. First, points $H$, $G$, $N$, $L$ are collinear and $\angle AGL = 90^\circ$. Also, $Q$ is the foot from $H$ to $\overline{AN}$. Consequently, lines $AG$, $EF$, $HQ$, $BC$, $TM$ concur at a point $R$ (radical axis). Moreover, we already know $\angle MTN = 90^\circ$. This implies $T$ lies on the circle with diameter $\overline{RN}$, which is exactly the circumcircle of $\triangle GQN$. Note by Brokard's Theorem on $AFHE$, the point $X$ is the orthocenter of $\triangle MBC$. But $\angle MTN = 90^\circ$ already, and $N$ is the midpoint of $\overline{BC}$. Consequently, points $T$, $X$, $N$ are collinear. Finally, we claim $P$, $X$, $N$ are collinear, which solves the problem. Note $P = \overline{GG} \cap \overline{AA}$. Set $K = \overline{HNL} \cap \overline{AP}$. Then by noting \[ -1 = (D,X;A,H) \overset{N}{=} (\infty, \overline{NX} \cap \overline{AK}; A, K) \]we see that $\overline{NX}$ bisects segment $\overline{AK}$, as desired. (A more projective finish is to show that $\overline{PXN}$ is the polar of $R$ to $\gamma$).

Let $A^\star$ be the antipode of $A$ in $(ABC)$ and let $QN\cap (ABC)=\{A,Q^\star\}$. It is well known that $A^\star$ is the reflection of $H$ over $N$. As $HQ\parallel Q^\star A^\star$ (they are both perpendicular to $AN$) and $NH=NA^\star$, we infer that $HQA^\star Q^\star$ is a parallelogram, so $NQ=NQ^\star$. Also note that $G-H-N$ are collinear. Let $\{T\}=AG\cap HQ,\ \{R\}=AH\cap GQ$. Then $T$ lies on the polar of $P$ wrt $\gamma$, so $P$ lies on the polar of $T$ wrt to $\gamma$, i.e. $P\in NR$. Observe that by Power of Point we have $$ NG\cdot NH=NA\cdot NQ=NA\cdot NQ^\star=NB\cdot NC$$ The inversion $\mathcal{I} (N,NB^2)$ swaps $G$ with $H$, $Q$ with $A$, and leaves $B$ and $C$ invariant. Let $\mathcal{I}(M)=M^\star$. Under inversion, the concurrency of $(NGQ),(MBC)$ and $PN$ is equivalent to the concurrency of $AH,(M^\star BC)$ and $NP$, i.e. to proving that $R \in (M^\star BC)$.

shows us that $BR\perp MC$, whence $R$ is the orthocenter of $\triangle{MBC}$, so $$\widehat{BRC}=180^\circ-\widehat{BMC}=\widehat{BM^\star C}\Rightarrow R \in (M^\star BC)$$

Very beautiful problem! My solution: Lemma:The circumcircle of triangle $BHC$,$k$, also passes through $Q$ Proof:construct parallelogram $ACRB$ by taking $R$ to be the symmetric of $A$ with respect to $N$.Then $\angle BRC=\angle BAC=180^{\circ} -\angle BHC$ so $R$ lies on $k$.We also have $BR\parallel AC,BH\perp AC\Rightarrow BH\perp BR$ and $HQ\perp QR$ so $HQRB$ is cyclic which proves the lemma. Now the radical axes of $\gamma$,$k$ and the circumcircle of $ABC$ concur, so $AG,HQ,BC$ all pass through the same point $S$.Since $SH\perp AN,AH\perp SN$ so $H$ is also the orthocenter of triangle $ASN$ so $\angle AGN=90^{\circ} $.But we know that since $G$ lies on $\gamma$ that $\angle AGH=90^{\circ} $ so $G,H,N$ are collinear. Now consider $T$ to be the intersection point of $MS$ and the circumcircle of $MBC$.We will prove that the circumcircle of $NGQ$ passes through $T$ and that $P,T,N$ are collinear, finishing the problem. We have that $ST\cdot SM=SB\cdot SC=SG\cdot SA=SH\cdot SQ$ which yields that $AMGT$,$MTHQ$ are cyclic.So $\angle AGT=\angle TMH=\angle TQS$ and hence $GTQS$ is cyclic.Now we have $\angle GTS=\angle GQS=\angle GNS$ so $GTNS$ is cyclic and $\angle STN=\angle SGN=90^{\circ} $ (1).We also obtain that $TQNS$ is cyclic so $\angle QTN=\angle QSN=\angle QGN$ proving that the circumcircle of $NGQ$ passes through $T$. We also easily obtain that $PA$ is also tangent to $\gamma$ so $PAGM$ is cyclic which, since we have shown that $AMTG$ is also cyclic, means that $PMTG$ is cyclic so $\angle PTM=\angle PGM=90^{\circ} $ (2). From (1) and (2) we obtain that $\angle PTM=\angle STN$ which means that $P,T,N$ are collinear and the proof is complete.

Well known that: 1. $G, H, N$ colinear. 2. $BHQC$ cyclic. 3. $AGBC, AGHQ$ cyclic. Now, radical axis theorem $(AGBC), (BHQC), (AGHQ)$ means $AG, HQ, BC$ concur at a point say $Z$. So $\angle ZQN=\angle ZGN=90^{\circ}$ means $H$ is orthocenter of $\triangle AZN$. Redefine $T$ be intersection of $(GQN)$ and $MZ$. Circles $(AGHQ), (ZGQN)$ are orthogonal, thus $MA^2=MG^2=MT\cdot MZ$, so $\angle MAT=\angle MZG=\angle MGT$, so $MTGA$ is cyclic. Clearly $PGMA$ is cyclic, so $PGTMA$ is cyclic. So $\angle PTM=90^{\circ}$. Now, radical axis theorem $(AGBC), (AMTG), (BTMG)$ means $Z\in TM$. Let $D=AH\cap BC$. Now since $-1=(B, C;D, Z)$, then $ZD\cdot ZN= ZB\cdot ZC= ZT\cdot ZM$, so $MTDN$ cyclic, so $\angle MTN=90^{\circ}$. Combine above means $P, T, N$ colinear, as desred.

Radical axis on $\gamma, (ABC), (BFEC)$ gives $AG, BC, EF$ concurrent on $R$. Since $HQND$ is cyclic, we have $\angle QNR = \angle QND = \angle AHQ = \angle AGQ = 180-\angle RGQ$ so $R, G, Q, N$ cyclic. $O, M$ lies on the perpendicular bisector of $AG$, so $P$ lies on it as well, so $PA$ is also a tangent to $\gamma$. Now since $(PAMG), (MFDNE), (MBC)$ meet at $M$ and $R$ has equal powers to three of them, we can see that these three circles are coaxial with radical axis $MR$. Therefore, there must me $T \in MR$ such that $T$ lies on both $(MBC)$ and $(MFDNE)$. Easily check that $\angle PTM = \angle NTM = 90$, so $T \in PN$. By Brokard on $(BFEC)$, we see that $H$ is the orthocenter of $\triangle ARN$. Therefore, $\angle RTN = \angle PTM = 90 = \angle RGN$, which gives us $T, G, N, Q$ cyclic.

Let $K$ denote the intersection of the nine point circle of $\triangle ABC$ with $PN$. We claim that $K$ lies on $(GNQ)$ and $(MBC)$. Since $OM$ bisects $AG$, $P$ is also tangent to $PA$. Therefore, $PA\parallel BC$. Let $D, E$ denote the feet of the $B$ and $C$ altitudes. Let the radical axis of $(ABC), (BCDE), (AGH)$ be concurrent at $L$. Then $K, L, M$ are collinear by taking radical axis of $(APG), (NDME), (AGH)$. Therefore $LB\cdot LC = LD\cdot LE = LK\cdot LM \implies BCKM$ is cyclic. Since $AG\cap BC = L$ and $PK\cap BC = N$, Reim's theorem implies $LGKN$ is cyclic. Since $AG\cap BC = L$ and $AQ\cap BC=N$, Reim's theorem implies $LGQN$ is cyclic. Hence $K$ lies on $(GNQ)$. $\square$

My solution: Lemma Let $A'$ be foot of $A-altitude$. Circles $\odot (MA'N),\cdot (GNQ),\odot (MCB)$ have common point. Proof: Let $R$ be radical center for circles $\odot (MA'N),\odot (AGHB'),\odot (ABC)$. $BC,B'C',AG$ are concurrent at $R$ ($B',C'$ defined analogously). Let $X,Y=\odot(MBC) \cap \odot (AGH)$. Let $p(X,k)$ be power of $X$ WRT $k$. $p(R,\odot(ABC))=p(R,\odot(A'MN))=p(R,\odot (AGH))$ and $p(R,\odot(ABC))=p(R,\odot (MBC))=RB\cdot RC$ so it implies that $X,Y,R$ are collinear. Let $\psi$ be inversion WRT $\odot(AGH)$. We have that $\psi(\odot(MBC))\cap \psi(\odot(MA'N))=R$ and from $\triangle RAA'$ we have that $\angle ARA'=\angle GNA$ so $R\in \odot( GQN)$ so it implies that circles $\odot (MA'N),\odot (GNQ),\odot (MCB)$ are concurrent at $\psi(R)$. Back to main problem : Let $T_1=\odot(GQN)\cap PN$. We have that($\angle GAA'=a$) $\angle GQN=\angle HQN+\angle GQH=90+\angle AGN=90+a$($H$ is on $GN$) it implies that $GT_1N=90+a$ which implies $\angle GT_1P=90-a$ which implies $T\in\odot(PAMG)$. (because $P$ is on $OM$ we have that $OM\perp AG$ which implies that $PG=PA$) $\angle PT_1M=90$ which implies $\angle MT_1N=90$ so $T\in \odot(MA'N)$. Now from Lemma we have that $T_1\in \odot (MBC)$ Therefore we have $T_1\equiv T$ and it implies that $P,T,N$ are collinear.

Solved with mathogenie1211

Since $OM\perp AG$, $P$ also lies on the tangent at $A$ to $\gamma$. Now let $T=(NQG)\cap (NHA)$ be the Miquel point of $AG\to HQ$. Then $T$ lies on the line through $N=AQ\cap GH$ and $AH\cap GQ$ by Brocard, which passes through $P$ by Pascal. Moreover, the inverse $T'$ of $T$ wrt $\gamma$ is $AG\cap QH$ which lies on $BC$ and satisfies $$T'T\cdot T'M = \text{pow}_{\gamma}(T') = T'G\cdot T'A = T'B\cdot T'C$$ so $T$ also lies on $(MBC)$ and we're done.

Might have overcomplicated it just a bit (Still solved it pretty quickly for myself so yay!) 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Define $X \neq M$ by $(BMC) \cap (DEF)$. Now we claim that $(BHQC)$ is cyclic. To see this note that $\angle HBC = 90 - \angle C$ so it suffices to show that $\angle NQC = \angle C$. To show this we will show that $\triangle NQC \sim \triangle NCA$. Indeed note, \begin{align*} NQ \cdot NA = NE^2 = NC^2 \end{align*}as desired. Then using this note that by radical axis on $(BHQC)$, $(ABC)$ and $(AEF)$ we have the lines $\overline{AG}$, $\overline{BC}$ and $\overline{HQ}$ are concurrent at a point say $Y$. We claim that $\overline{EF}$ is also concurrent at the same point. This follows as $G$ is the Miquel point of cyclic $BCEF$, so by Brokards we must have $\overline{EF} \cap \overline{BC}$, $G$ and $A$ collinear. Thus the lines $\overline{AG}$, $\overline{BC}$, $\overline{HQ}$ and $\overline{EF}$ are all concurrent at a point $Y$. Now we claim that $Y$ lies on the radical axis of $(DEF)$ and $(BMC)$ which follows from radical center on $(DEF)$, $(BMC)$ and $(BFEC)$. Thus $X$, $Y$, and $M$ are all collinear. Moving on, we claim that $YGNQ$ is cyclic. Indeed note that we have $\angle YGN = 180 - \angle AGN = 90$, so it suffices to show that $\angle YQN = 90$. This is equivalent to showing $H$, $Q$ and $Y$ are collinear which we have already shown so we're done. Now we claim that $X \in (YGNQ)$. This follows as $\angle YXN = 180 - \angle MXN = 90$ due to the homothety at $H$ that sends $M \mapsto A$ and $N \mapsto A'$, where $A'$ is the $A$-antipode. Thus we have $X \in (GNQ)$ and $X \in (BMC)$. We then just need to show $\overline{PN}$ passes through $X$. To do this we will show $P$, $X$ and $N$ collinear, or equivalently $\angle PXM = 90$. To do this note, $PA$ is tangent to $(AEF)$. To show this first note, $\overline{OM} \perp \overline{AG}$ due to the homothety at $A$ that maps $\overline{OM} \mapsto \overline{HA'}$. However we have $H$, $A'$ and $G$ collinear as $\angle AGH = \angle AGA' = 90$. Hence $\overline{OM}$ is the perpendicular bisector of $\overline{AG}$. Then the conclusion is obvious. $PAGMX$ is cyclic, which would finish. This follows as by converse of radical axis on $(DEF)$ and $(AEF)$ which implies $AGMX$ is cyclic, and the fact that $\angle MGP = \angle MAP = 90$ which implies $PAGM$ is cyclic. Together these imply the claim. Using these two facts we find that $\angle PXM = 90$ as $\overline{PM}$ is a diameter in $(PAGMX)$. Thus the collinearity follows and we're done.

Observe that by symmetry $\overline{AP}$ is tangent to $\gamma$ as well. Let $E$ and $F$ be the feet of the altitudes from $B$ and $C$ respectively, and let the well-known concurrency point of the four (!) lines $\overline{AG},\overline{EF},\overline{HQ},\overline{BC}$ be $X$. Let $T$ be the foot of the perpendicular from $M$ to $\overline{PN}$. By radical center on $(PAGMT),(NEFMT),\gamma$—since $\overline{NE}$ and $\overline{NF}$ are well-known to be tangent to $\gamma$—we find that $\overline{MT}$ passes through $X$ as well. Hence $\measuredangle XQN=\measuredangle XGN=\measuredangle XTN=90^\circ$, so $TGNQX$ is cyclic. Furthermore, we have $XT\cdot XM=XE\cdot XF=XB\cdot XC$, hence $TMBC$ is cyclic, finishing the problem. $\blacksquare$

We have that $Q$ is the $A-$Humpty Point and $G$ is the $A-$Queue Point. Let $X=EF\cap BC$ and let points $T,N$ be the intersection of $(GNQ)$ and $(DEF)$. It is well-known that $H$ is the orthocenter of $AXN$ and $AD,NG,XQ$ are the altitudes. Hence, $NGQX$ is cyclic with diameter $\overline{NX}$, which gives $\angle XTN=90^\circ$. Moreover, $\overline{MN}$ is the diameter of $(DEF)$ (nine point circle). So, $\angle MTN=90^\circ=\angle XTN$. So, $M,T,X$ are collinear. Note that $MG=MA$ and $OG=OA$, as $M$ is the center of $(AGH)$ and $O$ is center of $(ABC)$. So, $OM$ is the perpendicular bisector of $\overline{AG}$. It follows that $PA=PG$, which means that $\overline{PA}$ is tangent to $(AGH)$ at $A$. Hence, $P,A,M,G$ are concyclic. Claim: $T\in(MBC)$ Proof. Let $K,G$ be the intersection of $(GNQ)$ and $(ABC)$. It is well-known that $K$ is the reflection of $Q$ in $\overline{BC}$ and that $AK$ is the $A-$symmedian. Let $A,H'$ be the intersection of line $AH$ with $(ABC)$. It is well-known that $H'$ is the reflection of $H$ in $\overline{BC}$. Finally, note that $\angle XQN=\angle HQN=\angle HDN=90^\circ$, which means $H,Q,N,D$ are cyclic. Reflecting about $\overline{BC}$, we get that $H',K,N,D$ are cyclic. We now perform $\sqrt{-HA\cdot HD}$ inversion, which is the inversion about the circle with center $H$ and radius $\sqrt{HA\cdot HD}$, and then reflection about $H$. Under this inversion, the nine-point circle $(DEF)$ goes to $(ABC)$ and the circle $(GNQ)$ goes to itself as $G$ goes to $N$, $Q$ goes to $X$, $N$ goes to $G$ and $X$ goes to $Q$. This can be proved using the fact that $H$ is the orthocenter of $AXN$ with altitudes $AD,XQ,NG$. So, $T$ goes to an intersection of $(GNQ)$ and $(ABC)$. So, $T$ goes to either $G$ or $K$. However, $N$ goes to $G$, so $T$ must go to $K$. Finally, note that $D$ goes to $A$ and $H'$ goes to $M$, as $HM=\frac{HA}{2}$ and $HH'=2HD$, while $N$ goes to $G$. So, circle $(H'KND)$ goes to $(MTGA)$. Hence, $T\in(PAMG)$. So, $XG\cdot XA=XT\cdot XM$. But since $A,G,B,C$ are concyclic, $XG\cdot XA=XB\cdot XC$. So, $XT\cdot XM=XB\cdot XC$ which means $T\in(MBC)$. $\blacksquare$ To conclude, note that $\overline{MP}$ is the diameter of $(PMT)$, so $\angle MTP=\angle MTN=90^\circ$, which means $T\in\overline{PN}$. So, $T\in\overline{PN}$, $T\in(GNQ)$ and $T\in(MBC)$. Hence, $T$ is the required point of intersection. $\blacksquare$

Since $\overline{GH} \perp \overline{GA}$ and $\overline{OM} \parallel \overline{NH}$, it follows that $OM$ is the perpendicular bisector of $\overline{AG}$. So, $\overline{AP}$ is also tangent to $\gamma$; this means $\overline{AP} \parallel \overline{BC}$. Let $X$ be the intersection of lines $AG$ and $BC$. Then, it's well known that $X$, $H$ and $Q$ are collinear, so $XGQN$ is cyclic. Claim: lines $PN$ and $XM$ are perpendicular. Proof: Let $P'$ be the reflection of $A$ over $P$, so that $P'$ lies on line $NHG$. We will in fact show that $\triangle AXH \sim \triangle P'NA$, which implies our claim (since the segments we're trying to show perpendicular are the medians of the respective triangles). We already have $\angle XAH = \angle NP'A$ since $\overline{XA} \perp \overline{HP'}$; furthermore, \[ \angle AHX = 180^{\circ} - \angle ANX = \angle P'AN,\]where the step is because $H$ is the orthocenter of $\triangle AXN$. So, our claim is proved. Let $U$ be the intersection of $\overline{PN}$ and $\overline{XM}$. As stated earlier, $PM$ is the perpendicular bisector of $\overline{AG}$, so $(PAMG)$ is cyclic with diameter $PM$. Since $\angle PUM = 90^{\circ}$, it follows that $GUMA$ is cyclic. So, by the radical axis theorem, $BUMC$ is cyclic. Since $\angle XUN = 90^{\circ}$, we also have $XQUN$ cyclic, finishing.

Note that $G$ is the $A$-Queue point, $Q$ is the $A$-Humpty point. Let $X$ be the $A$-Ex point. Reflecting about line $OM$, we see that $G$ goes to $A$, so $PA$ is tangent to $(AH)$. Thus, $P,A,M,G$ are concyclic. Let $T' \neq M$ be the intersection between $(PAMG)$ and $(MBC)$. By radax on $(PAMG)$, $(ABC)$, $(MBC)$, we get that $AG, BC, T'M$ are concurrent. Thus $T'$ lies on $XM$. Let $D$ be the foot of the $A$-altitude. It is well-known that we can take a suitable inversion about $X$ that swaps the following pair of points: $(A,G), (M,T'), (H,Q), (D,N)$. Thus the $A$-altitude $AMHD$ is taken to the circle $(XGT'QN)$. Now, since $\measuredangle MT'N = \measuredangle XT'N = \measuredangle XGN = \measuredangle AGN = 90^{\circ} = \measuredangle MAP = \measuredangle MT'P$, we get that $P,N,T'$ are collinear.

For some reason I love overcomplicating things. Humpty point properties imply that $B,H,Q,C$ are concyclic. Thus applying the radical center theorem on $\gamma, (BHQC),$ and $(ABC)$ gives that $AG, QH,$ and $BC$ are concurrent, say at point $X.$ Claim 1: $XGQN$ is cyclic with diameter $XN.$ Proof: It is well-known that $G,H,N$ are collinear, so we see that $\angle XGN = \angle AGH = \angle AQH = \angle XQN = 90^\circ,$ as claimed. We now get rid of the point $O$ from the problem: Claim 2: $PA$ is tangent to $\gamma$ as well. Proof: Note that $AM = MH$ and $AO = OG,$ so $MO$ is none other than the perpendicular bisector of $AG.$ Since $PG$ is tangent to $\gamma,$ this implies that $PA$ is as well. Thus $P$ is just the intersection of the tangents to $\gamma$ at $A$ and $G.$ Now let $Z$ be the intersection of the circumcircle of $GQN$ with $PN.$ By Claim 1, we see that $Z$ is just the foot of the altitude from $X$ to $PN.$ However, Claim 3: $XM \perp PN.$ Proof: We in fact show the stronger statement that $PM$ is the polar of $X$ with respect to $\gamma.$ First, we show that $P$ lies on the polar of $X.$ By La-Hire this is equivalent to showing that $X$ lies on the polar of $P,$ which is just $AG,$ and this holds by the definition of $X.$ Now we show that $N$ lies on the polar of $X,$ which proves the claim. Again by La-Hire this is equivalent to showing that $X$ lies on the polar of $N,$ which is just $EF$ where $E$ the foot of the altitude from $B$ to $AC$ and $F$ is the foot of the altitude from $C$ to $AB.$ But by the radical center theorem on $\gamma, (BFEC),$ and $(ABC),$ we see that $X,E,F$ are collinear. Therefore, $X$ lies on the polar of $N,$ which proves our claim. Hence $Z$ is the foot of the altitude from $N$ to $XM.$ Now, let $AH$ intersect $BC$ at point $D.$ Then $\angle MZN = \angle MDN = 90^\circ,$ so $MZDN$ is cyclic. Thus by power of a point at $X,$ we get $XZ \cdot XM = XD \cdot XN.$ Since $\angle AGN = \angle ADN = 90^\circ,$ we get that $AGDN$ is cyclic, so $XD \cdot XN = XG \cdot XA = XB \cdot XC$ by power of a point on $(ABC).$ Combining everything together, we get \[ XZ \cdot XM = XD \cdot XN = XG \cdot XA = XB \cdot XC, \]which implies that $ZMBC$ is cyclic, and we are done.

Given that the intersection of $QG$ with $BC$ is at $K$, and $N$ is the midpoint of the arc $BC$ that does not contain $A$, we start with the known result that $ND$ and $IP$ intersect at $Q$. Given that $DP \parallel AN$, by Reim's theorem, $Q$, $P$, $D$, and $G$ are concyclic. It's apparent that they all lie on the $D$-Apollonius circle with respect to $BC$, leading to the relation $QB \cdot CG = QC \cdot BG$. This implies the cross-ratio $-1 = (Q, G; B, C)$, indicating that $GK$ must align with the $G$-symmedian of triangle $BGC$. Therefore, $DG$ bisects the angle $\angle QGM$. Furthermore, $QK$ aligns with the $Q$-symmedian of triangle $QBC$. Consequently, the tangents to $(ABC)$ at $B$ and $C$ intersect at $T$ on line $QK$, establishing that the cross-ratio $(Q, G; T, K) = -1$. This conclusion leads to the result that $DM$ bisects $\angle GMQ$. Hence, it follows that $D$ serves as the incenter of triangle $GQM$, as required.

v_Enhance wrote: Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$. Denote by $M$, $N$ the midpoints of $\overline{AH}$, $\overline{BC}$. Suppose the circle $\gamma$ with diameter $\overline{AH}$ meets the circumcircle of $ABC$ at $G \neq A$, and meets line $AN$ at a point $Q \neq A$. The tangent to $\gamma$ at $G$ meets line $OM$ at $P$. Show that the circumcircles of $\triangle GNQ$ and $\triangle MBC$ intersect at a point $T$ on $\overline{PN}$. Proposed by Evan Chen Let $D$, $E$, $F$ be the foot of altitude from vertex $A$, $B$, $C$ in triangle $ABC$ respectively $\hspace{0.5cm}$$S$ be the intersection of lines $EF$ and $BC$; $T$ be the intersection of lines $PN$ and $MS$ Examine radical axis of circles $\gamma$, circle with diameter $BC$, $(ABC)$ we see that lines $EF$, $BC$, $HQ$ concur at $S$, so $\angle SQN = \angle HQA = 90^\circ$ Cuz $OM \perp AG$ so $P$ is the pole of $AG$ respect to $\gamma$, and we ez to see that $N$ is the pole of $EF$ respect to $\gamma$ Therefore, $PN$ is the polar of $S$ respect to $\gamma$ (Follow La Hire theorem) Lead to $SM \perp PN$ at $T$ and $\overline{ST}.\overline{SM} = \overline{SE}.\overline{SF} = \overline{SB}.\overline{SC} $ Combine with $\angle SPN = \angle SQN = 90^\circ$, we see that $T$ lie on circles $(TQN)$, circle $(BMC)$, done

Let $D,E,F$ be the altitudes from $A,B,C$ to $BC,CA,AB$. $EF\cap BC=T$. Let $M^*$ be the reflection of $A$ with respect to $BC$ and $K$ be the point on $TM^*$ such that $GK\perp BC$. $L$ is on $TM^*$ such that $GL\perp AD$. $OM$ is the perpendicular bisector of $AG$ hence $PA$ is tangent to $(AGH)$ which means $AP\parallel BC$. Invert from $A$ with radius $\sqrt{AH.AD}$. $PN\leftrightarrow (AQP^*)$ where $TP^*=TP$ and $P^*$ is on $AP$. $(MBC)\leftrightarrow (M^*EF)$ and $(TQGN)\leftrightarrow (TQNG)$. When we invert from $T$ with radius $\sqrt{TB.TC},$ $A,Q,K,P^*$ swap with $G,H,M^*,L$ which are cyclic since $\angle GLM^*=\angle AP^*M^*=\angle TAP^*=\angle GHA$. Thus, $(AQP^*)$ pass through $K$. Also $TK.TM^*=TG.TA=TE.TF$ hence $K$ is on $(M^*EF)$. $\angle TKG=\angle KGT=\angle TNG$ so $K$ also lies on $(TNGQ)$ which gives that $(AQP^*),(M^*EF),(TNQG)$ are concurrent as desired.$\blacksquare$

Let $X$ be the $A$ expoint. We make a few observations:- $P$ is the pole of $AG$ in $\gamma$ since it is the intersection of one tangent and perpendicular bisector of $AG$ $\implies (PAGM)$ is cyclic. $Q$ is the $A$ humpty point and it lies on $HX$ so $(GXNQ)$ is cyclic and $\Delta AXN$ has orthocenter $H$. Now, by radax on $(ABC)$, $(PAMG)$ and $(MBC)$, $X$ lies on the radical axis of $(MBC)$ and $(PAMG)$. By radical axis on the nine point circle, $(MBC)$ and $(PAMG)$ , we get that they are coaxial and that the second intersection $T$ lies on $PN$ as well as the circle with diamter $XN$ which is what we needed to prove.

Finally got time to latex this! v_Enhance wrote: Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$. Denote by $M$, $N$ the midpoints of $\overline{AH}$, $\overline{BC}$. Suppose the circle $\gamma$ with diameter $\overline{AH}$ meets the circumcircle of $ABC$ at $G \neq A$, and meets line $AN$ at a point $Q \neq A$. The tangent to $\gamma$ at $G$ meets line $OM$ at $P$. Show that the circumcircles of $\triangle GNQ$ and $\triangle MBC$ intersect at a point $T$ on $\overline{PN}$. Proposed by Evan Chen Let $PN\cap RM=T$. Note that $G$ is the $A-$Queue Point, so we have the following claims Claim I: ${G-H-N}$ is collinear. [Coxeter] Claim II: $\odot(GAND)$ [a**] Claim III: $EF, BC, AG$ concur at $R$, the radical center of $\odot(AGHQ) \odot(AGBC), \odot(BHQC)$. Claim IV: $\odot(RGQN)$ with $RN$ as the diameter. Proof: We need to show $\angle RQN=90^{\circ}$, which is true since $\overline{R-H-Q}$ by Brocard's Theorem on $BCEF$, we get $RH \perp AN$. Claim V: $\odot(AMGP)$ Proof: $P$ is the pole of $AG$ w.r.t $\gamma$ and it lies on the perpendicular bisector of $AG$. So we also have $PN$ is the polar of $R$ w.r.t $\gamma\implies PN\perp RM$ where $PN\cap RM=T$. Claim VI: $\overline{R-T-M}\iff \angle MTN=90^{\circ}\iff T \in (AMGP)$ Proof: $\angle APT=\angle RNT=\angle TGA\implies \boxed{T \in \odot (RGTQN)}$. Claim VII: $\boxed{T \in \odot(MBC)}$ Proof: $RM \cdot RT=RE\cdot RF=RC\cdot RB$

Let $S,E,F$ be feel of altitudes from $A,B,C$, respectively. Denote $X : EF \cap AH$ and $I$ be midpoint of $EF$. Also, let $\gamma_1$ be circle with diameter $MX$. Note that $MIN$ collinear and $E,F$ lies on $\gamma$ with their tangent intersect at $N$. By Brokard, $X$ is orthocenter of $\triangle MBC$. Note that $\angle MIX = 90^{\circ}$, so $ I \in \gamma_1$.But $I$ also lies on the median of $\triangle MBC$, so $I$ is humpty point of $\triangle MBC$. Let $Y$ be midpoint of $GA$, $R$ be $EF \cap BC$. It is well-known that $AGR$ are collinear. and $PA$ is tangent to the circumcircle of $ABC$( the line $OM$ is coaxis of $\gamma$ and the circumcircle $ABC$, so reflecting $G$ across perpendicular bisector of this line will give point $A$.) Note that $\angle MRY = 180^{\circ} - \angle ASR$, so $MYR$ is cyclic. Also, it is well known that $(R,S;B,C) = -1 \rightarrow NR \times NS = NB^2 = NI \times NM$ Thus, $SRYMI$ is cyclic. Inverting this circle about $\gamma$ gives $PXN$ collinear. Let $T’$ be $PN \cap (MBC)$, so $T’$ is queue point of $\triangle MBC$, that is $NX \times NT = NB^2 = NI \times NM$. Inverting about $(BCEF)$ map $(MBC)$ to $(IBC)$ and $(GNQ)$ to $AH$($Q$ is humpty point so we have $NQ \times NA = NB^2$). Also, it maps $T’$ to $X$. Since $X$ already lies on $AH$, it is suffice to show that $X \in (IBC)$, which is a well known property of $M$ humpty point that it lies on the circle with orthocenter and the remaining vertex of triangle $MBC$. 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It's a really nice problem! Evan Chen's medium geometry problems are the best! Let $D,E,F$ be the foot of altitudes in $\triangle{ABC}$. Firstly, note that $G$ is the $A$-queue point and $Q$ is the $A$-humpty point. It's well known that $\overline{G,H,N}$ and $BHQC$ is cyclic. Radical Axis Theorem at $(ABC),(BCEF),(AGFHQE),(BHQC)$ tells us that lines $AG,EF,QH,BC$ are concurrent, say the intersection point to be $R$. Notice that $BDHF$ and $AGFH$ cyclic easily implies $RGFB$ is cyclic by Miquel Point. Moreover, $P$ lies on the perpendicular bisector of $AG$ which is $OM$ and $PG$ is tangent to $(AGFHQE)$ at $G$, so $PA$ must be tangent to the same circle at $A$ and therefore is parallel to $BC$. One can show that $R$ lies on $(GQN)$ easily. Let $RM$ hits $(MBC)$ at $J$. Observe that $AD,BE,CF$ are concurrent at $H$ and $R=EF\cap BC$, so by Ceva-Menelause, $(R,D;B,C)=-1$. Since that $N$ is the midpoint of $BC$, we have $RD\times RN=RB\times RC$. But by Power of Point's Theorem, we have $RJ\times RN=RB\times RC=RD\times RN$. This results in $J$ lying on the nine-point circle of $\triangle{ABC}$, so $\angle{RJN}=180^\circ-\angle{MJN}=180^\circ-\angle{MDN}=90^\circ$, so $J$ lies on $(RGQN)$. Lastly, we claim that $J$ lies on $(APGM)$. Indeed, \begin{align*} \angle{GJM}&=180^\circ-\angle{GJR}\\ &=180^\circ-\angle{GNR}\\ &=180^\circ-\angle{HND}\\ &=90^\circ+\angle{GHA}\\ &=90^\circ+\frac{180^\circ-2\angle{GPM}}{2}\\ &=180^\circ-\angle{GPM} \end{align*} For the final act, simply see that $\angle{MJN}+\angle{PJM}=\angle{MDN}+\angle{PGM}=90^\circ+90^\circ=180^\circ$, so $\overline{P,J,N}$, as desired. $\blacksquare$ P.S. I was about to invert when I suddenly realized that synthetic approach is more than enough for this problem