Let $Y$ be the antipode to $O$ on $(OMG)$ and let $X$ be the second intersection of $(AO), (OMG)$ so that $OX$ is the radical axis mentioned. Let $Q = AN \cap (ABC)$, let $A'$ be the antipode to $A$ on $(ABC)$, and let $Z = AY \cap (ABC)$. A homothety centered at $A$ with factor $2$ sends $OX$ to $A'Z$. Thus it suffices to show $N,A',Z$ are collinear. Note that $Y$ lies on $BC$, the polar of $N$ w.r.t. $(ABC)$, so $N$ lies on $\ell$, the polar of $Y$ w.r.t. $(ABC)$. Let $A_1 = QY \cap (ABC)$. By Brokard on quadrilateral $AQA_1Z$ we know that $AQ \cap A_1Z \in \ell$. But then since we know $N$ is also $AQ \cap \ell$, we must have $N=N'$ and $N,A_1,Z$ are collinear, so it suffices to show $A_1=A'$. To do this, we let $P = QZ \cap AA_1$. Then the polar of $P$ w.r.t. $(ABC)$ must be $NY$ by Brokard on quadrilateral $AQA_1Z$, thus $OP \perp NY$ so we know $A,O,P,A_1$ are collinear $\implies AA_1 \perp NY \implies A_1 = A'$ as desired.

Here's a similar solution that avoids using any projective results. We use directed angles. Let $S$ be the intersection of the circle with diameter $AO$ with $(OMG)$ other than $O$, and let $D$ be the antipode of $A$ on $(ABC)$. Then the homothety with scale factor $2$ centered at $A$ brings the midpoint of $AN$ to $N$, $O$ to $D$, and $S$ to a point $T=AS\cap (ABC)$ other than $S$. We deduce that it suffices to prove $N,D,T$ collinear. Lemma. $N$ lies on the radical axis of $(ABC)$ and $(OMG)$. Proof. The power of $N$ with respect to $(ABC)$ is $NO^2-R^2$, where $R$ is the circumradius of $\triangle ABC$, and its power with respect to $(OMG)$ is $NO\cdot NM = NO(NO-MO)$ since $O,M,N$ are collinear. Then it suffices to prove \[ R^2=NO\cdot MO. \]By the Extended Law of Sines on right triangles $OBM$ and $OBN$, \begin{align*} NO\cdot MO &= NO\cdot (R\sin\angle MBO) \\ &= \left(\frac{R}{\sin\angle CBO}\right)(R\sin\angle MBO) \\ &= R^2. \end{align*}The result follows. Let $X$ be the intersection of lines $AT$ and $NG$. Then by Thales' Theorem we get \[ \measuredangle DTX = \measuredangle DTA = 90^{\circ}. \]Similarly $\measuredangle DGX = 90^{\circ}$, so $GXTD$ is cyclic. But since $N$ lies the radical axis of $(ABC)$ and $(OMG)$ and on the radical axis of $(DGX)$ and $(OMG)$, it must be the radical center of $(ABC)$, $(DGX)$, $(OMG)$, and hence $N,D,T$ are indeed collinear. We are done.

Set $X = AA \cap ON$, $G^\ast = AO \cap BC$, $T = G^\ast N \cap AX$. Since $T$ is the inverse of $(OMG) \cap (AO)$ w.r.t $(ABC)$, we just need to show $TO$ bisects $AN$. By Ceva, it's equivalent to show $XG^\ast \parallel AN$. But $XAMG^\ast$ is cyclic, $AN$ is symmedian, and $A$-altitude (isogonal to $AO$) is parallel to $XM$, hence done.

Define $G' \equiv AO\cap BC$, $\{A, F\} \equiv AN \cap \odot(ABC)$, $\{U, V\} \equiv NG'\cap \odot(ABC)$. Construct a circle centered at $N$ through $B$ and $C$; an inversion about this circle gives $$(N, G'; V, U)=(P_{\infty,UV}, M_{UV}; V, U)=-1$$since $O$ and $N$ are antipodes in $\omega$, and so the inverse of $G'$ is the midpoint of $UV$. Now $$-1=(N, G'; V, U)\stackrel{A}{=}(F, D; V, U)\stackrel{N}{=}(A,A',U,V)$$where $\{D, A'\} \equiv ND \cap \odot(ABC)$. Since the tangents at $A$ and $A'$ intersect on $UV$, it follows by inversion about $\odot(ABC)$ that line $UV$ is the radical axis of $\odot(ABC)$ and $\odot(OMG)$ and so the midpoint of $\overline{AA'}$ lies on $\odot(OMG)$. Finally, since $\angle AA'N=\angle AGN=\frac{\pi}{2}$, quadrilateral $AA'GN$ is cyclic and so the centers of the circle with diameter $\overline{AA'}$, $\odot(ABC)$, and the circle with diameter $AN$ are collinear, as desired.

Lemma:Let incircle of $ABC$ meets $BC,CA,AB$ at $D,E,F$ ,a line orthogonal to $BC$ at $D$ meets $EF$ at $X$ then $AX$ meets the midpoint of $BC$ Taking inversion with center at incenter of $ABC$ and with some little argument to this problem .

Here is a really straightforward way to solve the problem with linearity of power of a point. Let $\omega$, the circle with diameter $AO$ intersect $\odot (BOC)$ at $K$. Since, $OA, OK, ON$ are collinear by radical axis, $A, K, N$ are also collinear. Let $Q$ denote the midpoint of $AN$. We wish to show that $\mathbb{P}(Q, \omega, \odot (OMG))=\mathbb{P}(Q, \omega) - \mathbb{P}(Q, \odot (OMG)) = 0$. We use linearity of power of a point to compute \begin{align*} \mathbb{P}(N, \omega, \odot (OMG)) &= \frac{1}{2} \left(\mathbb{P}(A, \omega, \odot (OMG)) + \mathbb{P}(N, \omega, \odot (OMG))\right) \\ &= \frac{1}{2}(0 - AO\cdot AG + NK\cdot AN - NM\cdot NO)\\ &= \frac{1}{2}(- AK\cdot AN + NK\cdot AN - NM\cdot NO)\\ &= \frac{1}{2}(AN(NK-AK) - NM\cdot NO)\\ \end{align*} Let $AN$ intersect $\odot (ABC)$ for a second time at $R$. Then $K$ is the midpoint of $AR$ since $OK\perp AK$. So $$\frac{1}{2}(AN(NK-AK) - NM\cdot NO) = \frac{1}{2} (AN(NR\cdot NA) - NM\cdot NO) = \frac{1}{2}(NB^2 - NM\cdot NO) = 0,$$where the last equality follows since $\angle BOM = \angle MBN \implies NB$ is tangent to $\odot (OBM)$.

v_Enhance wrote: Set $X = AA \cap ON$, $G^\ast = AO \cap BC$, $T = G^\ast N \cap AX$. Since $T$ is the inverse of $(OMG) \cap (AO)$ w.r.t $(ABC)$, we just need to show $TO$ bisects $AN$. By Ceva, it's equivalent to show $XG^\ast \parallel AN$. But $XAMG^\ast$ is cyclic, $AN$ is symmedian, and $A$-altitude (isogonal to $AO$) is parallel to $XM$, hence done. Can you prove why is (AN) the symmedian ? Thank you.

Because $NB$ and $NC$ are tangents to the circumcircle of $\triangle ABC$, as $\angle NOB = \angle NOC = 90^{\circ}$.

Nice problem YadisBeles wrote: Let $O$ be the circumcenter of triangle $ABC$, and $\omega$ be the circumcircle of triangle $BOC$. Line $AO$ intersects with circle $\omega$ again at the point $G$. Let $M$ be the midpoint of side $BC$, and the perpendicular bisector of $BC$ meets circle $\omega$ at the points $O$ and $N$. Prove that the midpoint of the segment $AN$ lies on the radical axis of the circumcircle of triangle $OMG$, and the circle whose diameter is $AO$. Let $\Gamma$ denote $\odot(ABC)$. Let $T=\overline{AO} \cap \overline{BC}$ and $(BC, TT^{*})=-1$ be a harmonic bundle. Let $A'$ be the antipode of $A$ in $\Gamma$ and $P=\overline{NA'} \cap \Gamma$ with $P \ne A'$. Let $X=\overline{AT^{*}} \cap \overline{KO}$. Because $\overline{NB}, \overline{NC}$ are tangents to $\Gamma$ we have $-1=(BC, TT^{*}) \overset{A}{=} (BC, A'P)$ hence $\overline{AT^{*}} \perp \overline{NA'}$. Along with $\overline{KO} \parallel \overline{NA'}$ we get $\overline{KO}$ is the radical axis of $\odot(AO), \odot(OT^{*})$. By Power of Point, $$\overline{TO} \cdot \overline{TG}=\overline{TB} \cdot \overline{TC}=\overline{TM} \cdot \overline{TT^{*}}$$hence $\overline{KO}$ is the radical axis of $\odot(AO)$ and $\odot(OMG)$ as desired.

Since $\angle OMC = \angle OGN = 90^{\circ}$, $BC$ and $NG$ meet $(OMG)$ at a point $T$. Let $AG$ intersect $(ABC)$ at $H$, midpoint of $AN$ be $S$ and $R$ be the radius of $(ABC)$. Claim: $H$ is the orthocenter of triangle $ATN$.

Now since $S$ and $O$ are midpoints of $AN$ and $AH$ respectively, we get that $SO \perp AT$. Since $O$ is on the radical axis and $AT$ is parallel to the line connecting the centers, it follows that $S$ is also on the radical axis.

My solution is almost(or totally?) same as v_Enhance's solution , but as I worked on it for long, I'll definitely post my solution YadisBeles wrote: Let $O$ be the circumcenter of triangle $ABC$, and $\omega$ be the circumcircle of triangle $BOC$. Line $AO$ intersects with circle $\omega$ again at the point $G$. Let $M$ be the midpoint of side $BC$, and the perpendicular bisector of $BC$ meets circle $\omega$ at the points $O$ and $N$. Prove that the midpoint of the segment $AN$ lies on the radical axis of the circumcircle of triangle $OMG$, and the circle whose diameter is $AO$. Perform inversion around $(ABC)$. Let $P=(AO) \cap (OMG)$ other than $O$ and let $Q=AA \cap OM$. Let $X^*$ denote image of $X$ under inversion. We have $A,B,C$ fixed, $M$ and $N$ swap, $G^*=AO \cap BC$ and $P^*=AQ \cap G^*N$. We need to show that $P^*O$ bisects $AN$, but by Ceva's it is equivalent to show $QG^* \parallel AN$. As $\angle QMG^*=QAG^*=90^{\circ} \implies QAMG^*$ is cyclic, and $AN$ is $A-$symmedian, we can complete this with a short angle chase: $$\angle MQG^*=\angle MAG^*=\angle MAC-\angle OAC=\angle BAN+\angle B-90^{\circ}=\angle MNA$$. So, we are done. $\blacksquare$

My solution (idk how I forgot to post it ) Let $AO \cap BC = X, NG \cap BC = Y, AO \cap (O) = Z$. Note that $N$ is the pole of $BC$ wrt $(O)$. $-1 = (B, C; O, N) \stackrel{G}{=} (B, C; X, Y) \stackrel{A}{=} (B,C; Z, AY\cap (O))$, so $N, Z, AY\cap (O))$ are collinear. Thus $Z$ is the orthocenter of $ANY$. Now make $ANY$ the ``reference" triangle. The point where $(OMGY)$ (which has diameter $OY$) meets $(AO)$ is clearly on $AY$, and the radical axis of these two circles is perpendicular to $AY$. But the midline of $ANZ$ is parallel to $NZ$, which is perpendicular to $AY$, so we are done.

it's easy to notice that $AN$ is symmedian first solution (by fouad al-mouoine) $\textbf{Define the points:}$ $E = AN \cap (BOC)$ $F = (AO) \cap (OMG)$ $Y = OF \cap AN$ $T = YT \cap (OMG)$ Firstly, $\angle{OEN} = 90$ so $E \in (AOF)$. Secondly, quadrilateral $AETG$ is cyclic since $Y \in OF$ which is the radical axis of $(AO),(OMG)$. Thirdly, we are going to prove that points $A,T,M$ are collinear, angle chaising gives: $$\angle{YTA}=\angle{YEG}=\angle{NEG}=\angle{NOG}=\angle{MOG}=\angle{GTM} $$As required. $\blacksquare$ Lastly, it suffices to show that $YA=YG$, notice the fact that $OA$ is tangent to $AMN$, because by power of a point: $$ OA^2 = OB^2 = OM.ON $$Thus angle chaising again gives: $$\angle{YAG} = \angle{NAO} = \angle{OMA} = \angle{OMT} = \angle{OGT} = \angle{YGA}$$As desired, hence $YA=YG=YN$ because $\angle{NGA}=90$. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1082305.57120616, xmax = 7424606.789412475, ymin = -224989.70157482213, ymax = 5569707.945595611; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); 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A solution similar to post#14 It suffices to prove that $LO\perp AX$ let $X=BC\cap GN, Z=AO\cap (ABC), Y=AN\cap (ABC)$ Claim $Z$ is the orthocenter of $\triangle AXN$ pf) note that $Z, Y, N, G$ are cyclic since $\angle AYZ=\angle ZGN=90$ and obviously $X$ lies on $(OMG)$ $\therefore$ $X$ is the radical center of $(BOC), (ABC), (ZYNG)$ $AG\perp XN, XY\perp AN$ and we are done. now note that $O$ and $L$ are the midpoints of $AZ$ and $AN$. $NZ\parallel LO\perp AX$ (by claim).

Shorter linearity solution: WLOG let $\angle B>\angle C,$ $O'$ be the circumcenter of $\triangle BOC,$ $X$ be the foot from $O$ to $AN$ (and thus the second intersection of $(AO)$ and $(BOC)$), and $x=NO,$ the diameter of $(BOC).$ Now, observe that \begin{align*} \text{Pow}(N,(AO)) &= NX\cdot AN \\ &= AX\cdot AN - AX^2 + NX^2 \\ &= AO\cdot AG - (AO^2 - OX^2) + (NO^2 - OX^2) \\ &= \text{Pow}(A,(OMG)) - R^2 + x^2. \end{align*}However, we also have \begin{align*} \text{Pow}(N,(OMG)) &= NM\cdot NO \\ &= x(x-R\cos A) \\ &= x^2 - R^2, \end{align*}so because the function $f(\bullet) = \text{Pow}(\bullet, (OMG)) - \text{Pow}(\bullet, (AO))$ is linear, we have \[f(P) = \frac{1}{2}(f(N)+f(A)) = \frac{1}{2}((x^2-R^2-\text{Pow}(A,(OMG)) + R^2 - x^2))+(\text{Pow}(A,(OMG)-0)) = 0,\]which solves the problem. $\blacksquare$

Let us define $$f(\bullet):= \text{Pow}(\bullet, (AO))-\text{Pow}(\bullet, (OMG))$$which is a linear map from the plane to $\mathbb{R}$. Let $K$ be the midpoint of $AN$, $Q=AN\cap (ABC)$ and $D$ be the A--Dumpty point of $\triangle ABC$. Notice that \begin{align*} \text{Pow}(K, (AO))-\text{Pow}(k, (OMG))&=f(K)\\ &= \frac{1}{2}(f(A)+f(N))\\ &= \frac{1}{2}(0-AO\cdot AG+NP\cdot NA-NM\cdot NO ) \\ &= \frac{1}{2}(-AP\cdot AN +NP\cdot NA -NC^2)\\ &= \frac{1}{2}(NA\cdot (NP-PA)-NC^2)\\ &= \frac{1}{2}(NA\cdot NQ-NC^2)\\ &= \frac{1}{2}(\text{Pow}(N, (ABC))-{Pow}(N, (ABC)))\\ &=0 \end{align*}so we are done.

Here's my solution: Let $Y = \odot{(AO)} \cup \omega$, $T = \odot{(AO)} \cup \odot{(OMG)}$, $Z = AT \cup NG$, $A'$ be the reflection of $A$ over $Y$, and finally, let $X$ be the midpoint of $\overline{AN}$. Clearly, $Z$ lies on $\odot{(OMG)}$, since $O$ is the Miquel point $\triangle{ANZ}$. Note that $$\measuredangle{OMZ} = \measuredangle{OTZ} = \measuredangle{OTA} = 90^{\circ} = \measuredangle{OMB}$$means that $Z$ lies on $BC$. Since $\measuredangle{OYA'} = \measuredangle{OYA} = 90^{\circ}$ we have $\overline{OY}$ is the perpendicular bisector of $\overline{AA'}$, therefore $\overline{OA'}=\overline{OA}$, which gives us $A' \in \odot{(ABC)}$. Let $f(\bullet) = \text{Pow}(\bullet, \odot{(OMG))} - \text{Pow}(\bullet, \odot{(AO))}.$ Claim. $f(X) = 0$.

From claim we have $$\text{Pow}(X, \odot{(OMG))} = \text{Pow}(X, \odot{(AO))}$$which means $X$ lies on radical axis of $\odot{(AO)}$ and $\odot{(OMG)}$, as desired. $\blacksquare$

Let $R = BC \cap GN$, the midpoint of $AN$ be $X$, the second intersection between $(AGN)$ and $(ABC)$ be $Y$, the $A$-antipode wrt $(ABC)$ be $A_1$, and the centers of $(OMG), (AO)$ be $O_1, O_2$ respectively. It's easy to see that $ON$ is a diameter of $\omega$. In addition, we have $$Pow_{(ABC)}(R) = RB \cdot RC = Pow_{\omega}(R) = RG \cdot RN = Pow_{(AYGN)}(R)$$so $R$ lies on $AY$. Now, because $$\measuredangle AYN = \measuredangle AGN = \measuredangle OGN = 90^{\circ} = \measuredangle AYA_1$$we know $A_1 \in YN$ and $A_1N \perp AR$. Since $$\measuredangle OGR = \measuredangle OGN = 90^{\circ} = \measuredangle OMR$$we know $OMGR$ is cyclic with diameter $OR$. Now, Thales' implies $O_1O_2$ is the $O$-midline of $AOR$. But it's easy to see that $OX$ is the $A$-midline of $AA_1N$, so $$O_1O_2 \parallel AR \perp A_1N \parallel OX$$which yields the desired result, as $O$ lies on $(OMG)$ and $(AO)$. $\blacksquare$ Remarks: I solved this backwards by first noticing the midlines and reducing the condition to $AA_1N \perp AR$, which is equivalent to proving $Y \in AR$. Also, my solution could've been somewhat shorter if I'd added the midpoint of $AY$, i.e. the second intersection between $(OMG)$ and $(AO)$, and considered a homothety at $A$ with scale factor $\frac{1}{2}$.

Let $F=AN\cap (BOC)$. So since $\angle OFN=\angle OFA=90^\circ$ therefore $F$ lies on $(AO)$. Let \[ f(\bullet)=P(\bullet, (AO))-P(\bullet, (OMG)) \]Clearly this function is linear over the plane to real number. Let the midpoint of $AN$ is $J$. Now by power of point linearity we have $$f(J)=\frac{1}{2}\left(f(A)+f(N)\right)$$We want to show $f(J)=0$. Let $K=AN\cap (ABC)$. Since $\angle AFO=90^\circ$ therefore $F$ is the midpoint of $AK$. So by power of point, $NK\cdot NA=NB^2$. And $NM\cdot NO=NC^2=NB^2$ because $NB$ is tangent to $(OMC)$ and $NB=NC$. Now \begin{align*} f(J)&=\frac{1}{2}\left(f(A)+f(N)\right)\\ &=\frac{1}{2}\left(0-AF\cdot NA+NF\cdot NA-NB^2\right)\\ &=\frac{1}{2}\left(NA(NF-AF)-NB^2\right)\\ &=\frac{1}{2}\left(NA\cdot NK-NB^2\right)\\ &=\frac{1}{2}\cdot 0=0 \end{align*}and we are done.

Cute. After inversion wrt $(ABC)$, we can rephrase the problem as follows: if $G'N \cap AA=T$, then prove that $TO$ bisects $AN$. Add $AO \cap (ABC)=A'$; we need $TO \parallel A'N$ because then $TO$ is midline in $\triangle AA'N$ and we will be done. Now, add $BB \cap AA=D, CC \cap AA=R$; notice that now we get an incircle config in $\triangle NRS$ (with $(ABC)$ being the incircle). By a lemma in ISL 2005 G6, $T=G'N \cap RS$ is the midpoint of $RS$ and if $A'N \cap RS=V$ is the $N$-excircle touchpoint, then $T$ is midpoint of $AV$ and $TO$ is midline in $\triangle AA'V$, done.

Ok I forced linpop/pop on this problem, posting because I'm new to the technique. It suffices to prove that\[\mathbb P(D,(AO),(OMG))=1/2\mathbb P(A,(AO),(OMG))+1/2\mathbb P(N,(AO),(OMG))=0\iff AO\cdot(AO+OG)+NM\cdot NO=NO\cdot NH\iff AO\cdot(AO+OG)=NO\cdot(HO+OM)\iff AO^2=OM\cdot NO\iff\measuredangle OMA=\measuredangle OAN,\]which is evident by drawing the A-altitude AI and since AI is isogonal to AO, $\measuredangle OAD=\measuredangle MAI=\measuredangle OMA$. (Wait hold on is it OMA or AMO? AMO goes counterclockwise so it might be that but idk how directed angles work using parallel line angle postulate or whatever.)