My solution: Let $ N, K $ be the 9-point center, symmedian point of $ \triangle ABC $, respectively . Let $ O $ be the circumcenter of $ \triangle ABC $ and $ O' $ be the reflection of $ O $ in $ BC $ . Let $ T $ be the reflection of $ G $ in $ N $ . Easy to see $ N $ is the midpoint of $ AO' $ . ... $ (\star) $ Since the reflection of $ H $ in $ BC $ lie on $ \odot (O) $ (well-known) , so $ B, C, H, P' $ lie on a circle with center $ O' \Longrightarrow O'H=O'P' $ . Since $ HP' $ is the Steiner line of $ P $ WRT $ \triangle ABC $ , so $ HP' $ is perpendicular to the isogonal conjugate $ AK $ of $ AP \equiv AG $ WRT $ \angle BAC $ . ____________________________________________________________ If $ GH=GP' : $ Since $ O'G $ is the perpendicular bisector of $ HP' $ , so $ O'G \perp HP' \Longrightarrow OG' \parallel AK \Longrightarrow T \in AK $ ( from $ (\star) $ ) , hence from $ \angle HAT=\angle OAG $ and $ HT=OG $ we get $ AO=AH \Longrightarrow \angle BAC=60^{\circ} $ . If $ \angle BAC=60^{\circ} : $ From $ AO=AH $ and $ HT=GO $ we get $ \angle HAT=\angle OAG \Longrightarrow T \in AK $ , so from $ (\star) $ we get $ O'G \perp HP' \Longrightarrow G $ lie on the perpendicular bisector of $ HP' \Longrightarrow GH=GP' $ . Q.E.D

This problem is also approachable with complex numbers. WLOG let the circumcircle of $ \triangle{ABC} $ be the unit circle and denote the coordinates of points by lowercase letters. It's well-known that $ g = \frac{a + b + c}{3} $ so we have that $ \frac{p - a}{\overline{p} - \overline{a}} = \frac{g - a}{\overline{g} - \overline{a}} \Longrightarrow p = \frac{-bc(b + c - 2a)}{ab + ac - 2bc}. $ Therefore $ p' = b + c + \frac{ab + ac - 2bc}{b + c - 2a} = \frac{b^2 + c^2 - ab - ac}{b + c - 2a}. $ Now since $ h = a + b + c $ we have that $ 3HG = 3|h - g| = 2|a + b + c|. $ We also have that $ 3GP' = |3g - 3p'| = \left|\frac{(a + b + c)(b + c - 2a) - 3(b^2 + c^2 - ab - ac)}{b + c - 2a}\right| = 2\left|\frac{ab + ac + bc - a^2 - b^2 - c^2}{b + c - 2a}\right|. $ Therefore, setting the two expressions equal and squaring, we have that $ (a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) = \left(\frac{ab + ac + bc - a^2 - b^2 - c^2}{b + c - 2a}\right)\left(\frac{abc(a + b + c) - a^2b^2 - a^2c^2 - b^2c^2}{abc(ab + ac - 2bc)}\right). $ Since $ abc(b + c - 2a)(ab + ac - 2bc) \ne 0 $ we have that this is equivalent to: \[(a + b + c)(ab + ac + bc)(b + c - 2a)(ab + ac - 2bc) - (ab + ac + bc - a^2 - b^2 - c^2)(abc(a + b + c) - a^2b^2 - a^2c^2 - b^2c^2) = 0 \Longleftrightarrow\] \[3(a^2 - bc)^2(b^2 + bc + c^2) = 0\] But $ AB \ne AC \Longrightarrow a^2 \ne bc $ which implies that $ GH = GP' \Longleftrightarrow b^2 + bc + c^2 = 0 \Longleftrightarrow \angle{A} = 60 $ as desired. How did I do that crazy last step factorization? Well, we want to factor $ b^2 + bc + c^2 $ out of the expression since this is equivalent to $ \angle{A} = 60. $ We also want to use the fact that $ AB \ne AC, $ which in complex numbers is equivalent to $ a^2 = bc $ so I knew $ a^2 - bc $ factored into the expression as well. At this point I guessed, but really with enough time just dividing would probably be possible.

I felt bad about my last proof (the computations are kind of nasty) so here's a REALLY easy synthetic proof. I will only prove the "only if" direction but the "if" direction can be easily recovered. Assume WLOG that $ AB < AC. $ Assume that $ \angle{A} = 60 $ and let $ O $ be the circumcenter of $ \triangle{ABC}. $ Also let $ M $ be the midpoint of $ BC. $ Also, denote the circumradius of $ \triangle{ABC} $ by $ R. $ Let $ O' $ be the reflection of $ O $ over $ BC. $ Since $ OO' = 2OM = 2R\cos{A} = R $ we have that $ O' $ lies on the circumcircle of $ \triangle{ABC} $ so if we let $ \omega $ be the circle with center $ O' $ and radius $ R, $ it's clear that this circle passes through $ B, C, H, O, $ and $ P'. $ Since $ O'P' = O'H $ it suffices to show that ray $ O'G $ bisects $ \angle{P'OH}. $ Now we have that $ AH = 2R\cos{A} = R = O'H. $ Moreover since $ \frac{AG}{GM} = 2 $ and since $ M $ is the midpoint of segment $ OO', $ we have that $ G $ is the centroid of $ \triangle{OO'A} $ which since $ OO' = OA $ means that $ O'G = AG. $ Therefore quadrilateral $ AGO'H $ is a kite which implies that $ \angle{GO'H} = \angle{GAH} = \angle{GPP'} $ since $ PP' \parallel AH. $ Therefore it suffices to show that quadrilateral $ P'PO'G $ is cyclic. But since quadrilateral $ AOO'G $ is a kite and since $ \triangle{AOP} $ is isosceles we have that $ \angle{P'PG} = \angle{P'PO} + \angle{GPO} = \angle{P'O'O} + \angle{GAO} = \angle{P'O'O} + \angle{GO'O} = \angle{GO'P'} $ so quadrilateral $ P'PO'G $ is cyclic as desired so we're done.

Ok, I've done it using barycentric coordinates on $ABC$ It's easy to see that $P=(-a^2,b^2+c^2,b^2+c^2)$ . Call $x= \frac{b^2+c^2-a^2}{2}$ e cyc. Let $X=(0,b^2+c^2-z,b^2+c^2-y)$ , it's easy to see that $PX \perp BC$ so $X$ is the midpoint of $PP'$ and so $P'=(a^2,2c^2-a^2,2b^2-a^2)=(y+z,2x+y-z,2x+z-y)$ . It's well known that $H=(yz,xz,xy)$ . So the midpoint of $P'H$ (call it $M$) is $$M=(6xyz+2y^2z+2yz^2+xy^2+xz^2,(2x+y-z)(xy+xz+yz)+xz(4x+y+z),(2x+z-y)(xy+xz+yz)+xy(4x+y+z))$$ Now $BCHP'$ is cyclic and it's center is $O'$ , the reflection of $O$ in the line $BC$ . We have $O'=(-xy-xz,2xz+xy+yz,2xy+xz+yz)$. Finally $HG=GP'$ iff $M,G,O'$ are on the same line. It's well known that $G=(1,1,1)$ so $HG=GP'$ iff $$ \left[\begin{matrix} 1 & 1 & 1 \\ -xy-xz & 2xz+xy+yz & 2xy+xz+yz \\ 6xyz+2y^2z+2yz^2+xy^2+xz^2 &(2x+y-z)(xy+xz+yz)+xz(4x+y+z) & (2x+z-y)(xy+xz+yz)+xy(4x+y+z) \end{matrix}\right] = 0 $$ Expanding we get $$(y-z)(xy+xz+yz)(3x^2-xy-xz-yz)=0$$ But $y=z \rightarrow b=c$ and that's not ok! And $xy+xz+yz$ can't be $0$. So $GH=GP'$ iff $3x^2=xy+xz+yz$ or iff $(2x)^2 = (x+z)(x+y)$. So iff $(b^2+c^2-a^2)^2=(bc)^2 $ and since $ABC$ is acute we have $b^2+c^2 > a^2 $ and so $a^2=b^2+c^2-bc$. So because of Carnot we get $GH=GP' $ iff $ \angle{BAC}= \frac{\pi}{3}$.

Yeah, as above complex numbers works nicely... it even lets you handle the condition $AB \neq AC$, which cannot be dropped. First \[ \frac{pa(b+c)-bc(p+a)}{pa-bc} = \frac{b+c}{2} \implies p = -\frac{2bc-ab-ac}{bc(2a-b-c)}. \]Then \[ p' = b+c-bc\overline p = \frac{ab+ac-b^2-c^2}{2a-b-c}. \]Then \[ p' - \frac{a+b+c}{3} = \frac23 \frac{ab+bc+ca-a^2-b^2-c^2}{2a-b-c} \]and the condition that this has length equal to $\frac23 \left\lvert a+b+c \right\rvert$ amounts to \begin{align*} 0 &= \left( ab+bc+ca-a^2-b^2-c^2 \right)\left( \frac{a+b+c}{abc} - a^{-2}-b^{-2}-c^{-2} \right) \\ &- \left( a+b+c \right)\left( \frac1a+\frac1b+\frac1c \right)\left( 2a-b-c \right)\left( \frac2a-\frac1b-\frac1c \right). \end{align*}This factors as \[ 0 = 3(abc)^{-2}(b^2+bc+c^2)(a^2-bc)^2 \]which is good game. You know a priori based on the problem statement that $b^2+bc+c^2$ and $a^2-bc$ are both factors (since they respectively mean "$BC = \sqrt 3 \iff \angle A = 60^{\circ}$ and $AB=AC$) so this factoring requires no ingenuity at all. In fact, one can (and should) do the above expansion by taking "modulo $(a^2-bc)(b^2+bc+c^2)$", which cleans up the computation by reducing the number of terms which have to be expanded.

Well, the complex bash can be simplified even more

Ah, that's good! The $GD \perp HP'$ trick reduces the degree of the polynomial in question to degree $4$ (since the denominator of $(bc-a^2)$ goes away), and thus the factorization $(a^2-bc)(b^2+bc+c^2)$ can even be reverse-engineered. I'll remember that, thanks!

Let $X$ be the pole of $BC$ WRT $\odot (ABC)$, $Y \equiv \odot (ABC) \cap AX (Y \not=A), M\equiv BC \cap AG$, and let $Z$ be the midpoint of $AY$ Since $AX$ is $A$-Symmedian of $\triangle ABC \Rightarrow YP \parallel BC \Rightarrow Y,M$ and $P'$ are collinear. Let $Q$ the reflection of $A$ with respect to $G \Rightarrow YQ=GP'$ and $AG=GQ \Rightarrow GZ \parallel YQ \Rightarrow ZG=\frac{GP'}{2}$, and $T \equiv HO \cap AX$ $HG=GP' \Longleftrightarrow ZG=GO \Longleftrightarrow TG=GO \Longleftrightarrow TO=2TH \Longleftrightarrow OX=2AH \Longleftrightarrow 3OM=MX$ $\Longleftrightarrow \angle BXO=30^{\circ} \Longleftrightarrow \angle BAC=60^{\circ}$

Did anyone try a metric approach to bashing? It seems that this would be nearly as easy as complex bashing considering that we have expressions for $GH$, can easily find $GP'$, and can reduce $\angle CAB=60$ to $a^2=b^2-bc+c^2$.

pi37 wrote: Did anyone try a metric approach to bashing? It seems that this would be nearly as easy as complex bashing considering that we have expressions for $GH$, can easily find $GP'$, and can reduce $\angle CAB=60$ to $a^2=b^2-bc+c^2$. I found an expression of $GP'$ in terms of $a,b$ and $c$ but it is hard for the equality $GH=GP'$ to be reduced to $a^2=b^2-bc+c^2$. The main steps are using : 1)$GP'^2=GP^2+PP'^2-2GP.PP'\cos(\widehat{APP'})$ 2)$GP=GM+MP$ and $MP.MA=MB.MC$ and $GM=\frac{1}{3}AM$ 3)$PP'=2PK$ ($K$ is the intersection of $PP'$ and $BC$) and $\sin(\widehat{AMB})=\frac{PK}{MP}$ 4)$\widehat{APP'}=\frac{\pi}{2}-\widehat{KMP}=\frac{\pi}{2}-\widehat{AMB}$

Let $M$ be the midpoint of $BC$ and let the tangents at $B$ and $C$ at $\odot(ABC)$ meet at $N$. Let $O$ be the circumcenter of $\triangle ABC$ and let $D$ be the reflection of $O$ with respect to $M$. Since the reflections of $H$ and $P'$ with respect to $BC$ lie on $\odot(ABC)$, it follows that $B,C,H,P'$ lie on a circle $\omega$ with center $D$. We know that $AN$ and $AP$ are isogonal conjugates with respect to $\widehat{BAC}$, so the Steiner line of $P$ with respect to $\triangle ABC$, which is $HP'$, is perpendicular to $AN$. Another well known result is that $B,C,H,O$ are concyclic if and only if $\widehat{BAC}=60^\circ$. $(*)$ If $GH=GP'$, since $DH=DP'$, we get that $GD\perp HP'$ and therefore $GD\parallel AN$. But now from the fundamental theorem of similarity we obtain $ND=2DM=DO$, hence $D$ is the midpoint of $NO$. This means that $D$ is the center of the circle $\odot(BOC)$ or $O\in \omega$. and we are done by $(*)$. If $\widehat{BAC}=60^\circ$, then $O\in \omega$. We just do the reverse of the above, which is: See that $D$ is the midpoint of $NO$ and deduce that $ND=2MD$, which gives us $GD\parallel AN$. Now notice that this implies $GD\perp HP'$ and conclude that $GH=GP'$.

Is only an simple observation

Here's an alternative way to use complex numbers (the bashing only comes at the very end): Let $(ABC) = (O)$ be the unit circle, and let $H'$ and $G'$ be the reflections of $H$ and $G$ over $BC$. Let $N$ and $Q$ be the projections of $G $and $H$ onto $BC$, and let $M$ be the midpoint of $BC$. Then $n = \tfrac{2m+q}{3} = \tfrac{a+3b+3c-\frac{bc}{a}}{6}$ by the 1-2 property of the Euler line. Also\[h' = -\frac{bc}{a} \quad \text{and} \quad g' = 2n - g = \frac{2b+2c-\frac{bc}{a}}{3}.\]$HG = GP'$ is equivalent to $H'G' = G'P$, or that $G'O$ is perpendicular to $H'P'$. This is equivalent to $\tfrac{g'}{\overline{g'}} = h'p$ because $h'$ and $p$ lie on $(O)$ (and thus $\tfrac{h' - p}{\overline{h'} - \overline{p} } = -h'p$). So we do a straightforward bash now and we obtain \[\left(2b+2c-\frac{bc}{a}\right)\left(2-\frac{a}{b}-\frac{a}{c}\right) = -\frac{bc}{a}(b+c-2a)\left(\frac{2}{b}+\frac{2}{c}-\frac{a}{bc}\right) \iff \frac{2(b^2+bc+c^2)(a^2-bc)}{abc} = 0.\] $ABC$ is acute, so $a^2 - bc \neq 0$. So if $HG = GP'$, then $b^2 + bc + c^2 = 0 \implies b^3 = c^3 \implies \angle CAB = 60^\circ$, and if $\angle CAB = 60^\circ$, then $HG = GP'$, as desired.

Let $H',G',O'$be the reflection of $H,G,O$ in the line $BC.$ $H'$ lie on $\odot(ABC).$ Since $\angle{H'OP}=2\angle{H'AP},$ $HG=GP'\Longleftrightarrow H'G'=G'P\Longleftrightarrow\angle{H'OG'}=\angle{POG'}\Longleftrightarrow\angle{POG'}=\angle{H'AP}.$ Also, $\angle{G'OO'}=\angle{GO'O},$ so $\angle{G'OP}=\angle{G'OO'}+\angle{O'OP}=\angle{GO'O}+\angle{O'OP}=\angle{G'OO'}+\angle{O'MP}-\angle{GPO}.$ From $OM//AH,$ we get $\angle{HAP'}=\angle{GMO}=\angle{O'MP}.$ $\angle{G'OP}=\angle{G'OO'}+\angle{O'MP}-\angle{O'MP}=\angle{HAP'}+\angle{O'MP}-\angle{GPO}.$ $\angle{POG'}=\angle{H'AP}\Longleftrightarrow\angle{HAP'}+\angle{O'MP}-\angle{GPO}=\angle{H'AP}$ $\Longleftrightarrow\angle{O'MP}=\angle{GPO}\Longleftrightarrow G,O',O,P$ is cyclic $\Longleftrightarrow OM\times O'M=GM\times MP=$ $\frac{1}{3} AM\times MP=\frac{1}{3} BM\times CM$ $\Longleftrightarrow OM^2=\frac{1}{3} BM^2 \Longleftrightarrow \angle{BOM} = 60^{\circ} \Longleftrightarrow \angle{BAC} = 60^{\circ}$ So, $HG=GP'\Longleftrightarrow \angle{BAC} = 60^{\circ}$

We will prove a more general result: Problem. Given an acute triangle $ABC$ inscribed in $(O)$ ($AB\neq AC$) with orthocenter $H$. Let $P$ be an arbitrary point on arc $BC$, $P'$ be the reflection of $P$ wrt $BC$. $(OPP')$ intersects $AP$ again at $G$. Then $GH=GP'$. Proof. Let $OP'$ intersect $AH$ at $J$ then $\angle JAG=\angle GPP'=\angle GOJ$ therefore $AJGO$ is a cyclic quadrilateral. $(AGH) $ meets $HP'$ again at $K$ then $G$ is Miquel point of triangle $JHP$ wrt $(A,O,K).$ We get $KGOP' $ is cyclic. Let $GH$ cut $(O_a)$-the circumcircle of $(BHC)$- again at $M$. We have $\angle GMP'=\angle HMP'=\angle HAP=\angle HKG$, then $M\in (OPP')$. Let $O_aH$ cut $(OPP')$ at $L$. Since $O_aH=O_aM=O_aP',$ we obtain $H$ is incenter of triangle $ LMP'$. Then $GL=GH=GP'.$ Back to our problem. Let $PP' $ intersect $(O)$ at $U$ then $HP' \parallel AU'$. Since $PU$ is not the diameter of $(O)$ then $\angle (HP',AP)\neq 90^\circ$. Therefore the perpendicular bisector of $HP' $ meets $AP$ at one point. Applying problem above we get $GH=GP'$ iff $G\in (OPP')$ $\Leftrightarrow MG\cdot MP=MO\cdot MO_a=MO^2 \Leftrightarrow \dfrac{1}{3}MA\cdot MP=MO^2 \Leftrightarrow \dfrac{1}{3}MB^2=MO^2 \Leftrightarrow \angle BAC=60^\circ.$

Let $K$ be the intersection of the $A$-symmedian with the circumcircle of $\odot(ABC)$. It is well-known that $K$ is the reflection of $AG\cap(BHC)$ on the same side of $BC$ as $A$ over $BC$. Since $\odot(BHC)$ is the reflection of $\odot(ABC)$ about $BC$, it follows that $P'$ lies on $\odot(BHC)$. However, since $\odot(BHC)$ is also the reflection of $\odot(ABC)$ about $M$, the midpoint of $BC$, we get that $M$ is the midpoint of $P'K$. Let $A'$ be the point such that $ABA'C$ is a parallelogram; reflecting the entire diagram about $M$ sends $G$ to the centroid of $\triangle A'BC$, $P'$ to $K$, and $H$ to the antipode of $A$ in $\odot(ABC)$. Another dilation at $A$ with ratio $\tfrac{1}{2}$ sends $K$ to the midpoint of the symmedian chord $K'$, the centroid of $\triangle A'BC$ to $G$, and the antipode of $A$ to $O$, the circumcenter of $\odot(ABC)$. Thus, $G$ is the center of negative homothety with ratio $\tfrac{1}{2}$ sending $P'$ to $K'$ and $H$ to $O$. It is also well-known that $K'$ lies on $\odot(BOC)$. If $GP'=GH$, this implies that $K'OP'H$ is an isosceles trapezoid, hence it is cyclic. Therefore $K'O$ and $PH$ share a perpendicular bisector; since $K'$ lies on $\odot(BOC)$, this implies that the circumcenters of $\odot(BOC), \odot(BHC)$ lie on this perpendicular bisector, along with $G$. Since the line formed by the circumcenters is the perpendicular bisector of $BC$, these three points are collinear if and only if $AB=AC$, or the two circumcenters are the same point; since we can rule out the former case, the latter must be true, implying that $\odot(BOC)\equiv \odot(BHC)$, or that $O$ lies on $\odot(BHC)$, whence $$2\angle A=180-\angle A\Longrightarrow \angle A=60^{\circ}$$ Conversely, if $\angle A=60^{\circ}$, this implies that hexagon $BHK'OP'C$ is cyclic. Additionally, since $G\equiv OH\cap K'P'$, $2GK'=GP'$, and $2GO=GH$, this implies that lines $K'O$ and $HP'$ are homothetic with respect to $G$, hence they are parallel, and since $K'OHP$ is cyclic, we get that it is an isosceles trapezoid, or $GH=GP'$, as desired.

Not really contributing anything, but here's a complex bash. It really is pretty clean aside from one grind at the end. Set $\odot(ABC)$ as the unit circle, so the circumcenter of $ABC$ is the origin, and let the complex number of each point be its corresponding lowercase letter. Solving for $p$, we have \[ \frac{p-a}{\frac{1}{p}-\frac{1}{a}}=\frac{g-a}{\overline{g}-\frac{1}{a}} \implies p = \frac{(2a-b-c)bc}{ab+ac-2bc} \]By the projection formula, \[ p' = b+c-bc \overline{p} = \frac{ab+ac-b^2-c^2}{2a-b-c} \]One can easily show that $HG = \frac{2}{3}|a+b+c|$. Also, $P'G = \frac{2}{3}|\frac{ab+bc+ca-a^2-b^2-c^2}{2a-b-c}|$. $|w|^2=w \overline{w}$ for complex $w$, so setting $HG = P'G$ and squaring, we have \[ (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(2a-b-c)(\frac{2}{a}-\frac{1}{b}-\frac{1}{c})=(ab+bc+ca-a^2-b^2-c^2)(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}-\frac{1}{a^2}-\frac{1}{b^2}-\frac{1}{c^2}) \]Now this simplifies to \[ ((ab+bc+ca)-(a^2+b^2+c^2))((a^2bc+ab^2c+abc^2)-(a^2b^2+b^2c^2+c^2a^2)) = (a+b+c)(ab+bc+ca)(2a-b-c)(2bc-ab-ac) \]After some lengthy computations, this simplifies to \[ 3(a^4b^2+a^4c^2+b^4c^2+b^2c^4+a^4bc-2a^2bc(b^2+c^2+bc)) = 0 \]Factoring, this $3(a^2-bc)^2(b^2+bc+c^2) = 0$. I claim that $a=bc$ is impossible. Indeed, since $AB \ne AC, (a-b)(\frac{1}{a}-\frac{1}{b}) \ne (a-c)(\frac{1}{a}-\frac{1}{c})$. This further simplifies as $a^2 \ne bc$, thus $a^2=bc \leftrightarrow AB = AC$. Hence, $b^2+bc+c^2=0 \implies b = \omega c$ or $b=\omega^2 c$ where $\omega = e^{i \frac{2\pi}{3}}$. Considering this as rotation about origin, we have $\angle CAB = 60^{\circ}$. The if direction can be easily recovered so we are done.

Remarkably, introducing a bunch of points to tame the "rather metric condition $HG=GP'$," basically solves both the directions simultaneously. I will try to attach a diagram asap.

Wolstenholme wrote: \[(a + b + c)(ab + ac + bc)(b + c - 2a)(ab + ac - 2bc) - (ab + ac + bc - a^2 - b^2 - c^2)(abc(a + b + c) - a^2b^2 - a^2c^2 - b^2c^2) = 0 \Longleftrightarrow\]\[3(a^2 - bc)^2(b^2 + bc + c^2) = 0\] I got stuck at this place while complex bashing...just wondering, how much time it should normally take to expand and factorize ? By the way, the bash can be reduced a bit further by letting $C = \bar{b}$ (and letting $2k := \frac{b+\bar{b}}{2}$, you get everything as a function of $a$ and $k$), and $P$ can be calculated in less than 10 seconds using the formula that the second intersection of line $AZ$ with unit circle (Where $A$ is a point in the unit circle and $Z$ is an arbitrary complex number) is given by $\frac{Z - A}{1-A\bar{Z}}$

You know that $b^2+bc+c^2=0$ is needed, so $b^2+bc+c^2$ must be a factor of the expression, which makes the factorization doable

EDIT: So actually Aiscrim got the same reduction as me just realized. Basically its a way to reduce a 100+ term factorization to about ~10 terms. Aiscrim wrote: Well, the complex bash can be simplified even more

So I found a way to avoid having super big expressions at a time in the complex bash. The idea is that $|x/y|=1\iff x-y\perp x+y$, and with the appropriate $x$ and $y$, we will get $x-y$ to have two terms. This is great as the condition $\frac{x+y}{x-y}\in i\mathbb{R}$ will have the $x-y$ term cancel out when written out with conjugates. Note that I use a stupid (albeit pretty cool ) method to derive $P'$, one could just as well do it directly. As usual, set $(ABC)$ to the unit circle. The first step is to compute $p'$. Here is a really nice way to do it that avoids using the complicated arbitrary intersection formula. Let $D$ denote the intersection of the $A$-symmedian with $(ABC)$. We know that $(BC;AD)=-1$, so \[\frac{d-b}{d-c}=-\frac{a-b}{a-c},\]or $(c-a)(d-b)=(a-b)(d-c)$, or $(b+c-2a)d=2bc-ac-bc$, or \[d=\frac{2bc-ac-ab}{b+c-2a}.\]Since $AD$ and $AP$ are isogonal in $\angle A$, we have $p=bc/d$. Thus, \begin{align*} p'&=b+c-bc\overline{p}\\&=b+c-d\\ &= b+c-\frac{2bc-ac-ab}{b+c-2a} \\ &= \frac{(b+c)^2-2a(b+c)-2bc+ac+ab}{b+c-2a} \\ &= \frac{b^2+c^2-ab-ac}{b+c-2a}. \end{align*}I just realized that doing this directly as $AG\cap(ABC)$ would probably have been faster . Let's move on. Note that \[\frac{g-p'}{g-h}=\frac{3g-3p'}{3g-3h}=\frac{3p'-h}{2h},\]so letting $\mathcal{P}$ be the condition that $HG=GP'$, we see that \[\mathcal{P}\iff \left|\frac{3p'-h}{2h}\right|=1.\]We compute \begin{align*} (b+c-2a)(3p'-h) &= 3(b^2+c^2-ab-ac)-(b+c-2a)(a+b+c) \\ &= 2(a^2+b^2+c^2-ab-bc-ca), \end{align*}so \[\mathcal{P}\iff \left|\frac{a^2+b^2+c^2-ab-bc-ca}{(a+b+c)(b+c-2a)}\right|=1.\]Let \[x=a^2+b^2+c^2-ab-bc-ca\]and \[y=(a+b+c)(b+c-2a)=-2a^2+b^2+c^2-ab-ac+2bc.\]We see that $\mathcal{P}\iff|x|=|y|$. However, it is easy to see geometrically that $|x|=|y|\iff x-y\perp x+y$, so we have \[\mathcal{P}\iff 3(a^2-bc)\perp bc-a^2+2(b^2+c^2-ab-ac).\]This is ABSOLUTELY WONDERFUL, as $\overline{(a^2-bc)}\propto a^2-bc$, so we get nice cancellation. We see that \begin{align*} \mathcal{P} &\iff a^2-bc\perp bc-a^2+2(b^2+c^2-ab-ac) \\ &\iff \frac{bc-a^2+2(b^2+c^2-ab-ac)}{a^2-bc}\in i\mathbb{R} \\ &\iff bc-a^2+2(b^2+c^2-ab-ac) = a^2bc\left(\frac{1}{bc}-\frac{1}{a^2}+2\left(\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}-\frac{1}{ac}\right)\right) \\ &\iff 2(bc-a^2) = 2\left(\frac{a^2c}{b}+\frac{a^2b}{c}-ac-ab-(b^2+c^2-ab-ac)\right) \\ &\iff bc(bc-a^2) = a^2(b^2+c^2)-bc(b^2+c^2) \\ &\iff (bc-a^2)(b^2+c^2+bc)=0. \end{align*}But as $AB\not=AC$, we have $bc\not=a^2$, so therefore $\mathcal{P}\iff b^2+bc+c^2=0$. It is well known that $b^2+bc+c^2\iff \angle CAB=60^\circ$, so we have $\mathcal{P}\iff\angle CAB=60^\circ$, as desired.

Let $M$ be the midpoint of $\overline{BC}$, $L$ be the reflection of $O$ over $\overline{BC}$, and $A'$ the antipode of $A$, so that $M$ is the midpoint of $\overline{HA'}$. Let the $A$-symmedian meet $(ABC)$ again at $D$. Since $\overline{BC}$ bisects $\angle AMD$, $D$ and $P$ are reflections across $\overline{OL}$. Note the following series of equivalences, in which all steps are reversible, whence our conclusion goes both ways. $\angle CAB=60^\circ$, one side of the condition.

that $\overline{AL}$ is the perpendicular bisector of $\overline{OH}$. $\measuredangle DAL=\measuredangle LAG=\measuredangle GLA$, in which the first equality comes from $\overline{AD}$ and $\overline{AG}$ being isogonal. $\overline{AD}\parallel\overline{GL}$, which is immediate. $\overline{DA'}\perp\overline{GL}$, since $\angle ADA'=90^\circ$. $\overline{HP'}\perp\overline{GL}$, by homothety at $M$ with scale factor $-1$. $GH=GP'$, since $OH'=OP$ implies $LH=LP'$. Hence, we are done. $\square$

Beautiful problem. We give a proof by using the fact $\measuredangle BAC=60^{\circ}$.The other direction is reversible using phantom points. Now we should add in more points.Let $H'$ be the reflection of $H$ over $\overline{BC}$ which lies on $(ABC)$.Let $G',L$ be the reflections of $G,O$ wrt $\overline{BC}$.Now note that $H',G',L$ are obviously collinear because $\overline{HGO}$ is the euler line of $\triangle ABC$. Now check that it suffices to show $H'G'=G'P$. Let $H''$ be the antipode of $H'$ which is clearly the reflection of $A$ wrt $\overline{OL}$.Finally let $G''$ be the reflection of $G'$ wrt $L$.Also note that $H'G'=G'G''$ due to euler line ratios. [asy][asy] size(8cm); pair A=dir(140); pair B=dir(210); pair C=dir(330); pair O=(0, 0); pair H=orthocenter(A, B, C); pair G=(A+B+C)/3; pair Hp=2*foot(A, B, C)-H; pair L=dir(270); pair M=(B+C)/2; pair P=intersectionpoint(M -- (M+100*(M-A)), circle(O, 1)); pair X=foot(G,B,C); pair Gp=2*X-G; pair Ho=2*O-Hp; pair Go=2*L-Gp; draw(A -- B -- C -- A,linewidth(1.3)); draw(circumcircle(A, B, C)); draw(Hp -- A -- L, dashed); draw(G -- L,gray); draw(H -- O, dashed); draw(A--M--P,gray); draw(Hp--P,linewidth(1.3)); draw(Hp--Go,linewidth(1.3)); draw(Gp--M--Ho,dashed); draw(O--M--L,dashed); draw(Go--P,linewidth(1.3)); draw(P--Ho,dashed); draw(Hp--Ho,gray); dot("$A$", A, A); dot("$B$", B, W); dot("$C$", C, E); dot("$H$", H, N); dot("$G$", G, 2*dir(80)); dot("$H'$", Hp, SW); dot("$L$", L, 2*S); dot("$P$", P, 2*dir(90)); dot("$M$", M, NE); dot("$O$", O, N); dot("$G'$", Gp, 2*dir(240)); dot("$H''$", Ho, dir(0)); dot("$G''$", Go, 2*dir(270)); [/asy][/asy] If $\measuredangle BAC = 60^{\circ}$ then note that $\measuredangle BOC = 120^{\circ}$.Whence $L$ lies on $(ABC)$.Now a crucial claim: Claim: The points $G',M,H''$ are collinear. Proof: Trivial.Menelaus on $\triangle H'OL$ and noting that $\tfrac{H'G'}{G'L}=2$ due to euler line ratios. Now note that $\overline{G'MH''}$ and $\overline{AMP}$ are reflections of each other wrt $\overline{OL}$..Whence $\overline{LH''}$ bisects $\measuredangle G'H''P$ which readily shows that $G''$ should clearly lie on $\overline{H''P}$.But $\overline{HH''}$ is the diameter.Whence $\measuredangle H'PG'' = 90^{\circ}$ and so $H'G'=G'P$ due to $H'G'=G'G''$ and we are done.$\blacksquare$

Let $\Omega=(ABC)$ with center $O$. Let $H',G'$ are the reflections of $H,G$ over $\overline{BC}$. Reflecting over $\overline{BC}$, notice \[ GH=GP' \iff G'H'=G'P\iff \overline{OG'} \perp \overline{H'P}, \]where the last step follows since $H',P\in \Omega$. Use complex numbers with $\Omega$ unit circle. Let $M$ be the midpoint of $\overline{BC}$. Since $P=\overline{AM}\cap \Omega$, we have \[ \overline{a}+\overline{p} = \overline{m} +\overline{a}\overline{p}m \Rightarrow \overline{p} = \frac{1/a - \frac{1/b+1/c}{2}}{\frac{b+c}{2a}-1} = \frac{2bc-a(b+c)}{bc(b+c-2a)}. \]And since $G=\tfrac{a+b+c}{3}$, we have \[g' = b+c-bc\left( \tfrac{1/a+1/b+1/c}{3}\right)=\tfrac13\left(2b+2c-\tfrac{bc}{a}\right) \Rightarrow \overline{g'} =\frac{2b+2c-a}{3bc} \]by conjugating. It is easy to see $h'=-bc/a$, so $\overline{h'} = -\tfrac{a}{bc}$. We have \begin{align*} \frac{\overline{p} - \overline{h'}}{\overline{g'}} &=\frac{\frac{2bc-a(b+c)}{bc(b+c-2a)} +\frac{a}{bc}}{\frac{2b+2c-a}{3bc}} = \frac{(2bc-ab-ac) + a(b+c-2a)}{\tfrac13(2b+2c-a)(b+c-2a)} \\ &= \frac{6(bc-a^2)}{(2b+2c-a)(b+c-2a)}. \end{align*}Hence \begin{align*} \overline{OG'} \perp \overline{H'P} &\iff \frac{p-h'}{g'} \in i\mathbb{R} \iff \frac{\overline{p} - \overline{h'}}{\overline{g'}} \in i\mathbb{R} \\ &\iff \frac{bc-a^2}{(2b+2c-a)(b+c-2a)} = -\frac{\tfrac{1}{bc} - \tfrac{1}{a^2}}{(2/b+2/c-1/a)(1/b+1/c-2/a)} \\ &\iff (2b+2c-a)(b+c-2a) = a^2bc(\tfrac2b+\tfrac2c-\tfrac1a)(\tfrac1b+\tfrac1c-\tfrac2a) \\ &\iff bc(2b+2c-a)(b+c-2a) = (2ac+2ab-bc)(ac+ab-2bc) \\ &\iff -2 a^2 b^2 - 2 a^2 b c + 2 b^3 c - 2 a^2 c^2 + 2 b^2 c^2 + 2 b c^3 = 0 \\ &\iff 2 (-a^2 + b c) (b^2 + b c + c^2) = 0. \end{align*}Since $AB\not = AC$, the above is equivalent to $b^2+bc+c^2=0$, i.e. $b=c\omega$ for some $\omega^3=1$, $\omega\not = 1$, i.e. $\angle BOC=120^\circ$, i.e. $\angle BAC=60^\circ$.

Let $M$ be the midpoint of $BC$ and $O$ be the circumcenter of $ABC$. We will use the standard notations for sides, radii and angles. Firstly, by Leibniz' Theorem we have $OG^2 = R^2 - \frac{a^2+b^2+c^2}{9}$, so from the Euler line we obtain $HG^2 = 4R^2 - \frac{4}{9}(a^2+b^2+c^2)$. Having in mind $R^2 = \left(\frac{abc}{4S}\right)^2 = \frac{a^2b^2c^2}{(a+b+c)(a+b-c)(b+c-a)(a+c-b)}$, we deduce $$HG^2 = \frac{4}{9}\frac{-a^6 + a^4 b^2 + a^4 c^2 + a^2 b^4 - 3 a^2 b^2 c^2 + a^2 c^4 - b^6 + b^4 c^2 + b^2 c^4 - c^6}{2a^2b^2 + 2b^2c^2 + 2c^2a^2 - a^4 - b^4 - c^4}.$$ Now let us focus on $GP'^2$. Denoting $\angle AMB = \varphi$, we have $\angle GMP' = 180^{\circ} - 2\varphi$ and so by the Cosine Law $GP'^2 = MG^2 + MP'^2 - 2 \cdot MG \cdot MP' \cos(180^{\circ} - 2\varphi) = MG^2 + MP^2 + 2 \cdot MG \cdot MP \cdot \cos(2\varphi)$. Now $MG = \frac{AM}{3}$ and Power of a Point imply $MP = \frac{3a^2}{4MA}$ and since $AM^2 = \frac{1}{4}(2b^2 + 2c^2 - a^2)$ we obtain $GP'^2 = \frac{2b^2+2c^2-a^2}{36} + \frac{9a^2}{4(2b^2+2c^2-a^2)} + \frac{a^2\cos(2\varphi)}{6}$. To compute $\cos(2\varphi) = 2\cos^2\varphi - 1$, let $AK \perp BC (K\in BC)$ - then $\cos^2\varphi = \frac{KM^2}{AM^2} = \frac{4(\frac{a}{2} - c\cos\beta)^2}{2b^2+2c^2-a^2} = \frac{(b^2-c^2)^2}{a^2(2b^2+2c^2-a^2)}$ and so $\cos(2\varphi) = \frac{2(b^2-c^2)^2}{a^2(2b^2+2c^2-a^2)} - 1$. To sum up, we have $$ GP'^2 = \frac{7a^4 - 16a^2b^2 -16a^2c^2 + 81a^2 + 16b^4 - 16b^2c^2 + 16c^4}{36(2b^2+2c^2-a^2)}. $$Finally, by clearing denominators in $HG^2 = GP'^2$ and factoring out multiples of $b-c$ and $a^2 - b^2 - bc - c^2$ gives the result.

WLOG let $(ABC)$ be the unit circle. We know $$\frac{p-a}{\overline{p}-\overline{a}}=\frac{a-g}{\overline{a}-\overline{g}}$$so $p=\frac{(2a-b-c)bc}{ab+ac-2bc}.$ Hence, $$p'=b+c-bc\overline{p}=b+c+\cdot\frac{ab+ac-2bc}{2a-b-c}=\frac{ab+ac-b^2-c^2}{2a-b-c}.$$Therefore, $$GP'=\frac{2}{3}\left|\left(\frac{ab+ac+bc-a^2-b^2-c^2}{2a-b-c}\right)\left(\frac{a^2bc+b^2ac+c^2ab-b^2c^2-a^2c^2-a^2b^2}{(2bc-ac-ab)(abc)}\right)\right|.$$Since $HG=\frac{2}{3}\left|a+b+c\right|,$ we see that $HG=GP'$ if and only if $$\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{ab+ac+bc-a^2-b^2-c^2}{2a-b-c}\cdot\frac{a^2bc+b^2ac+c^2ab-b^2c^2-a^2c^2-a^2b^2}{(2bc-ac-ab)(abc)}$$which is equivalent to \begin{align*}&(ab+ac+bc-a^2-b^2-c^2)(a^2bc+b^2ac+c^2ab-b^2c^2-a^2c^2-a^2b^2)\\&-(2a-b-c)(2bc-ac-ab)(a+b+c)(ab+ac+bc)=0\end{align*}which is equivalent to $$3(a^2-bc)^2(b^2+bc+c^2)=0.\quad(*)$$Since $AB\neq AC,$ we know $$(*)\iff b^2+bc+c^2=0\iff\angle CAB=60.$$$\square$

Very nice problem! Let $R$ the reflexion of $H$ on $G$, $M$ be the midpoint of $BC$ and $H_A$ be the $A$-Humpty point so it's well-known that $\angle HH_AM = 90$ and $H_AM=MP$ this implies that $ \angle H_AP'P = 90$ and $P'H_A \parallel BC$. Let $O$ the center of $\odot (ABC)$ and $O'$ the center of $\odot (BHC)$ so $O'$ is the reflexion of $O$ on $BC$. And also that $H_A,P' \in \odot (HBC)$. Let $A'$ the reflexion of $A$ on $M$ so $A' \in \odot (HBC)$. Case 1: If $\angle CAB= 60 \implies HG=GP'$ Proof: Then $O \in \odot (HBC)$ then it's enough to prove that $\angle RP'H = 90$ the it's enough to prove that $R \in P'A'$ let $K$ a point on $\odot (HBC)$ such that $A'K \parallel HG$ but $O \in HG$ so $K$ is the reflexion of $O$ on $O'$ then it's enough prove that $P'H_AHO'$ is harmonic but watching from $A'$ and projecting on the line $OO'$ we have: $$(P',H;H_A,K) \stackrel{A'}{=} (L,O';M,K)$$With $L= A'P' \cap OO'$ be the point of intersection of the tangents in $B$ and $C$ respect $\odot (HBC)$ and this is true because watching from $B$ and using that $\angle CAB = 60$ this give us the result. $\square$ Case 2: If $HG = GP' \implies \angle CAB = 60$ Proof: Then $\angle RP'H = 90 \implies R \in P'A'$. Let $K$ a point on $\odot (HBC)$ such that $KA' \parallel HR$ and let $K' = OO' \cap AK$ so $HO'=O'A' \implies OO'=O'K'$ Then: $$ -1 = (R,H;G,K) \stackrel{A'}{=} (L,O';M,K')$$With $L= A'R \cap OO'$ be the point of intersection of the tangents in $B$ and $C$ respect $\odot (HBC)$. Then if $OM=l \implies O'M=l$ and $O'K'=2l$ but $-1= (L,O';M,K')$ then $ML=3l$ then $BM^2 = O'M.ML = 3l^2 \implies BM= \sqrt{3}l$ and $O'M=l$ so $\angle O'BM= 30 \implies \angle OBM =30 \implies \angle CAB = 60$. $\blacksquare$

Here is my (bad) solution:

Let the tangents at $B, C$ to $(ABC)$ meet at $N$. $OBNC$ cyclic implies $MN \cdot MO = MB \cdot MC = MB^2$. Let $O', A', K$ be the reflections of $O, H, P'$ in midpoint $M$ of $BC$. $K, A'$ lie on $(ABC)$ with diameter $AA'$, and $O'$ is the centre of $(CP'HB)$. A homothety centre $M$ scale factor $\frac{1}{3}$ maps $A \to G$, reflection means $HP' \parallel A'K$. Since $K, P$ are symmetric in $OM$, $BK = CP$ thus $AK$ is isogonal to $AP$ hence is the $A$-symmedian, i.e. $A-K-N$. Finally, $\triangle ABC$ acute guarantees that $O', N$ are on the same side of $M$. $\angle CAB = 60 \iff \angle OBM = 30 \iff \frac{MB}{MO} = \sqrt{3} \iff \frac{MO}{MN} = \frac{1}{3} \iff \frac{MO'}{MN} = \frac{1}{3} \iff O'G \parallel AN \iff O'G \perp A'K \iff O'G \perp HP' \iff GH = GP'$.