Let $A'$ be the antipode of $A$ with respect to the circumcircle of $ABC$. It is well known that $A'$ is the reflection of $H$ about $M$. Since $MO = NO$ and $AH = 2\times MO$, $AH = NM$ Here, $AH \parallel NM$ so $\square AHMN$ is a parallelogram. This implies that $MO = NO = AN = HM = MA'$, or, $M$ is the circumcenter of $\triangle HOA'$, which is equivalent to $\angle AOG = 90^{\circ}$.

The Main observation is ANMH is a parallelogram. If NO=NA , angle chasing will do. If AOG=90, This implies AOH=90 Let K be midpoint of AH. Therefore KOMH is a parallelogram. Now just angle chase and we are done

Has a nice complex bash soln as well. Let ($ABC$) be the unit circle with $A = 1$ $G = \frac{1+b+c}{3}$ If $\angle AOG = 90^{\circ}$ we get $G$ is imaginary or $Re(b+c) = -1$ $M = \frac{b+c}{2}$ $N = -M$ Hence $Re(N) = \frac{1}{2}$ and N lies on the perpendicular bisector of $AO$ which is equivalent to $NO = NA$ all statements are equivalent hence showing both directions

Let $X$ be midpoint of $AH$. We know $AH \parallel OM$ and $HM = XO$ and we get $ANOX , XOMH$ as parallelograms If $AN=NO$ then $AN=XO$ and $NO=XA=XH$ give us $\angle AOH = 90 \implies \angle AOG = 90$ If $\angle AOG = \angle AOH = 90$ then $OX=AX=HX$ As $ANOX$ is parallelogram $AN=XO=XA=ON$ $\blacksquare$

Okay, my first ever complex bash! Very doable in contest in my opinion. Solution: Introduce the orthocenter $H$ of $\triangle ABC$ and let $A'$ be the antipode of $A$ in $\odot(ABC)$. [asy][asy] import olympiad; size(240); defaultpen(fontsize(12pt)); pair A = dir(130); pair B = dir(210); pair C = dir(330); pair O = circumcenter(A,B,C); pair G = (A+B+C)/3; pair M = (B+C)/2; pair N = 2*O-M; pair H = orthocenter(A,B,C); pair A_ = 2*O - A; draw(A--B--C--A, red); draw(circumcircle(A,B,C), magenta); draw(H--A_, orange); draw(M--N, orange); draw(N--A, orange); draw(A--H, orange); draw(A--A_, fuchsia); draw(N--A_, fuchsia); draw(A--M, fuchsia); draw(H--O, blue); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$O$", O, dir(150)); dot("$G$", G, dir(140)); dot("$M$", M, dir(M)); dot("$N$", N, dir(N)); dot("$H$", H, dir(H)); dot("$A'$", A_, dir(A_)); [/asy][/asy] Toss the system on the complex plane with $|a| = |b| = |c| = 1$. We can easily compute \[m = \frac{b+c}{2} \qquad n = -\frac{b+c}{2}.\]Now observe that \begin{align*} \angle AOG = 90^\circ & \iff \Re\left(\frac{g}{a}\right) = 0 \\ \iff \Re\left(\frac{a+b+c}{3a}\right) = 0 & \iff \Re\left(\frac{1}{3} \left(\frac{b+c}{a} + 1\right)\right) = 0 \\ \iff \Re\left(\overline{a}(b + c)\right) = -1 \end{align*}Now, finally for the length condition, we have \begin{align*} \left| \frac{b+c}{2} + a\right| = \left| \frac{b+c}{2}\right| \iff |b + c + 2a|^2 = |b + c|^2 \\ \iff (b+c+2a)\overline{(b+c+2a)} = (b+c)\overline{(b+c)} \\ \iff 2 + b\overline{c} + c\overline{b} + 2\overline{a}(b+c) + 2a(\overline{b} + \overline{c}) + 4 = 2 + b\overline{c} + c\overline{b} \\ \iff a\overline{(b+c)} + \overline{a}(b+c) = -2 \iff \Re\left(\overline{a}(b + c)\right) = -1 \end{align*}which means, we're done! $\blacksquare$

Sol:- Let $A'$ be reflection of $A$ across $O$ and $G'$ be reflection of $G$ across $M$. We have $GG'=AG$ so $OGG'A'$ is a trapezoid. $NA=NO \iff MA'=MO \iff $ trapezoid $OGG'A'$ is right angled i.e. $\angle AOG=90^\circ$.

Attachments:

starchan wrote: Let $ABC$ be a triangle with circumcentre $O$ and centroid $G$. Let $M$ be the midpoint of $BC$ and $N$ the reflection of $M$ across $O$. Prove that $NO = NA$ if and only if $\angle AGO = 90^{\circ}$. Proposed by Pranjal Srivastava Shouldn't this be true?

Mainly same solution as aboves, but posting for storage purposes. Since there is a circumcenter and centroid in the statement, we are motivated to tie in the orthocenter using the Euler line. So, in view of this, let $H$ be the orthocenter of $\triangle ABC$. Note that since $AN = NO = OM = \frac{AH}{2}$, so $NM = AH$, and clearly $NM \parallel AH$, so $ANMH$ is parallelogram. Let $X$ denote reflection of $A$ about $O$. But, since $X$ is on circumcircle, diamatrically opposite point to $A$, we have that $H, M, X$ are also collinear. But, by our earlier result, we have $AN = NO = OM = MH = MX$ (from parallelogram and definition of $X$), but since $M$ lies on $HX$, and $MH = MO = MX$, $\triangle HOX$ must be a right angled triangle, so $\angle AOH = 180^{\circ} - \angle HOX = 90^{\circ},$ but $\angle AOH = \angle AOG$, by Euler line theorem, so $\angle AOG = 90^{\circ}$. (Also, everything above was reversible steps, so we are done )

Complex numbers work really well here. I'll show only one part for now cus I'm kind of tired First, we assume $A,B,C$ lie on the unit circle with $a=1$. Now, we first assume that $NA=NO$. This gives $\left|{1+\frac{b+c}{2}}\right|=\left|{\frac{b+c}{2}}\right|$ which gives $Re(b+c)=-1$. Now $G=\frac{1+b+c}{3}$. But clearly, $G$ is purely imaginary. So $AO \perp OG$.

I think this is much simpler than a lot of other solutions here. Say you have that $\angle AOG = 90^{\circ}$ and suppose the midpoint of $AG$ is $X$. This means $NX \parallel OG$ since $AX = XG = GM$ which implies $NX \perp AO$ and $NX$ bisects $AO$, thus $\bigtriangleup AON$ must be isosceles. The other direction follows similarly. $\blacksquare$

Trivial. Let $J$ be the midpoint of $AG$, and let $NJ$ intersect $AO$ at point $K$. Since $AG = 2GM$, it follows that $MG = GJ$. Given that $MO = ON$, it can be concluded that $OG \parallel NJ$. Since $J$ is the midpoint of $AG$, it implies that $K$ is the midpoint of $AO$. Therefore, $NO = NA \Leftrightarrow NJ \perp AO \Leftrightarrow OG \perp AO$. Q.E.D.

$\textcolor{blue}{\text{Coordinates FTW!}}$ Employ Cartesian coordinates. Set $B$ as the origin $(0,0)$ and $C$ as $(1,0)$. Let $A \equiv (a,b)$. Clearly we have: $M \equiv \left(\dfrac{1}{2}.0\right)$ $G \equiv \left(\dfrac{a+1}{3},\dfrac{b}{3}\right)$ $O \equiv \left(\dfrac{1}{2},\dfrac{a^{2}+b^{2}-a}{2b}\right)$ and $N \equiv \left(\dfrac{1}{2},\dfrac{a^{2}+b^{2}-a}{b}\right)$. Using distance formula we get the lengths $\overline{NO},\overline{NA}$ as follows: $\overline{NO}=\dfrac{b^{2}-a^{2}+a}{2ab-b}$ $\overline{NA}=\sqrt{\left(a-\dfrac{1}{2}\right)^{2}+\left(\dfrac{a^{2}-a}{b}\right)^{2}}$ Also slopes of $\overline{AO},\overline{OG}$ can also be found as: Slope of $\overline{AO}$=$\left(\dfrac{b^{2}-a^{2}+a}{2ab-b}\right)$ Slope of $\overline{OG}$=$\left(\dfrac{3a^{2}+b^{2}-3a}{b-2ab}\right)$. Now, $NO=NA$, $\iff \left(\dfrac{b^{2}-a^{2}+a}{2ab-b}\right)^{2}=\left(a-\dfrac{1}{2}\right)^{2}+\left(\dfrac{a^{2}-a}{b}\right)^{2}.$ Expanding the above terms we get, $3a^{4}+3a^{2}-6a^{3}-2ab^{2}+2a^{2}b^{2}+b^{2}-b^{4}=0$ $\iff 3a^{2}b^{2}+b^{4}-3ab^{2}-3a^{4}-a^{2}b^{2}+3a^{3}+3a^{3}+ab^{2}-3a^{2}=b^{2}-4ab^{2}+4a^{2}b^{2}$ $\iff (b^{2}-a^{2}+a)(3a^{2}+b^{2}-3a)={(b-2ab)}^{2}$ $\iff \left(\dfrac{b^{2}-a^{2}+a}{2ab-b}\right)\left(\dfrac{3a^{2}+b^{2}-3a}{b-2ab}\right)=-1$ $\iff \angle{AOG}=90^{\circ}$. $\blacksquare$

Cool problem. Lemma: Let $ABCD$ be a parallelogram with $P$ the midpoint of $BC$; then $AB = BP$ iff $\angle APD = \frac{\pi}{2}$. Proof sketch: Join $AP, PD$; compute $\angle APB + \angle DPC$. Now let $H$ be the orthocentre of $\Delta ABC$. Then applying the above lemma on parallelogram $ANMH$ gives us the result (after a lot of Euler-Line justification and blah blah blah)

Let $H$ denote the orthocenter of $\triangle ABC$ and let $A'$ denote the $A-$antipode. We start off by making the following important observation. Claim : Quadrilateral $AHMN$ is a parallelogram. Proof : We know that lines $\overline{AO}$ and $\overline{HM}$ intersect at the $A-$antipode $A'$. Now since $AH \parallel MO$, consider the homothety centered at $A'$ which scale factor $2$. This clearly maps $O$ to $A$ so it must map $M$ to $H$, so $AH = 2MO$. But then, \[AH = 2MO = MN\]so $AHMN$ is indeed a parallelogram as desired. Now, we know that the circumcenter $O$ , centroid $G$ and orthocenter $H$ of $\triangle ABC$ all lie on the Euler line of $\triangle ABC$. Thus, the condition that $\angle AOG = 90^\circ$ can be more naturally interpreted as $\angle AOH =90^\circ$. We show the argument only for one direction below, but it holds for the opposite direction as well. If $\measuredangle HOA = 90^\circ$, line $\overline{HO}$ which is clearly the $H-$median of $\triangle AHA'$ must in fact be the $AA'$ perpendicular bisector. Thus, \[\measuredangle OHM = \measuredangle AHO = \measuredangle MOH\]so $\triangle HOM$ is isosceles. Further, \[AN = HM = MO = NO\]

First let's prove that, $\angle AOG = 90^{\circ} \rightarrow NA = NO $. Let $\omega$ be the circumcircle of $\triangle ABC$. Define, $D$ to be a point on $\omega$ such that $A,O,D$ are collinear ($AD$ is a diameter), and $H$ to be the orthocenter of $\triangle ABC$. $$AO=OD, NO=OM \rightarrow ANDM \text{ is a parallelogram.}$$Hence, $NA=DM$. We know that $H$ is the reflection of $D$ across $M$, and that $O,G,H$ are collinear. $\therefore AH//OM$. Using the Midpoint Theorem for $\triangle ADH$, $$OM = \frac{1}{2} OM $$As $\angle AOG = 90^{\circ}$, $$AH=HD$$$$\rightarrow NO=OM=\frac{1}{2}HD = MD=NA.$$ Now let's prove that $NA=NO \rightarrow \angle AOG = 90^{\circ}$. $$NA=NO\rightarrow \frac{1}{2}AH=OM=MD=MH=\frac{1}{2} HD $$$$AH=HD\rightarrow \angle AOG = 90^{\circ}$$