We claim that all the circles pass through the Anti-Steiner point $T$ of the Euler line. Indeed, $\measuredangle TG_2A=\measuredangle AGO=\measuredangle TG_3A$, and similarly for the other five circles.

Here is an approach with complex numbers. Let $P$ be an arbitrary point. Let $P_B$ and $P_C$ be the reflections of $P$ across $AB$ and $AC$ and let $Q_B$ and $Q_C$ be the second intersections of lines $AP_B$ and $AP_C$ with the circumcircle. Then we will compute the intersection of $(A P_B P_C)$ and $(A Q_B Q_C) \equiv (ABC)$. We have $p_B = a+b-ab\overline p$, $p_C = a+c-ac\overline p$. To compute $q_B$ note that \begin{align*} a+q_B &= p_B + a q_B \overline{p_B} \\ &= a+b-ab\overline p + aq_B \left( 1/a + 1/b - p/ab \right) \\ \implies ab\overline p - b &= (a/b-p/b) q_B \\ \implies q_B &= b^2 \frac{a\overline p -1}{a-p}. \end{align*}Then, the desired intersection is \begin{align*} \frac{p_B q_C - p_C q_B}{p_B - p_C + q_C - q_B} &= \frac{ \left( \frac{a \overline p -1}{a-p} \right) \left( c^2 (a+b-ab\overline p) - b^2(a+c-ac \overline p) \right)} {(b-c)(1-a \overline p) + (c^2-b^2) \cdot \frac{a \overline p - 1}{a-p}} \\ &= \frac{c^2(a+b-ab\overline p)-b^2(a+c-ac \overline p)} {-(b-c)(a-p) + (c^2-b^2)} \\ &= \frac{a(c+b) + cb - abc \overline p}{a-p+c+b} \\ &= \frac{ab+bc+ca - abc \overline p}{a+b+c-p} \end{align*}which is in any case symmetric in $a$, $b$, $c$. Moreover taking $p = \frac13(a+b+c)$ and $p=0$ give the same numbers (and indeed any $p$ on the Euler line).

Angle chasing shows that $\odot (ABC)$, $\odot (AO_3O_2)$, $\odot (BO_1O_3)$ and $\odot (CO_1O_2)$ concur. Also $\odot (ABC)$, $\odot (AG_3G_2)$, $\odot (BG_1G_3)$ and $\odot (CG_1G_2)$ concur. Let $H$ be the orthocenter of $\triangle ABC$ and let $H_1$, $H_2$ and $H_3$ be the reflections of $H$ WRT $BC$, $CA$ and $AB$ respectively. We know $H_i$, $G_i$ and $O_i$ are collinear for $i=1,$ $2$, $3$. Let $P \equiv G_1O_1 \cap G_2O_2$. Since $CO_1=CO_2$, $CG_1=CG_2$ and $\measuredangle O_1CO_2 = \measuredangle G_1OG_2$ we get $C$ is the center of spiral similarity that sends $G_1O_1$ to $G_2O_2$ thus $C \in \odot (PO_1O_2)$ and $C \in \odot (PG_1G_2)$. Furthermore $\measuredangle O_1PO_2 = \measuredangle O_1CO_2 = 2 \measuredangle ACB = \measuredangle H_1PH_2$. Thus $P \in \odot (ABC)$. We conclude $\odot (ABC)$, $\odot (AO_3O_2)$, $\odot (BO_1O_3)$, $\odot (CO_1O_2)$, $\odot (AG_3G_2)$, $\odot (BG_1G_3)$ and $\odot (CG_1G_2)$ concur at $P$.

A more straight-forward solution using complex numbers(without using spiral similarity in complex numbers as v_Enchance did). As v_Enchance did we will find the coordinates of the intersection between the circle $(ABC)$ and the circle $(AP_bP_c)$, where $P$ is a random point inside $\triangle{ABC}$ and $P_b$ , $P_c$ are his reflections across lines $AC$, respectively $AB$. Let this intersection be $X$ . We have that $\overline{x}=\frac{1}{x}$ and $\frac{(p_b-a)(p_c-x)}{(p_c-a)(p_b-x)}$ is real, where $p_b=a+b-ab\overline p$ and $p_c=a+c-ac\overline p$ $\Longleftrightarrow$ $$\frac{bx+ax-px-ab}{cx+ax-px-ac}=\frac{ac+bc-abc\overline{p}-cx}{ab+bc-abc\overline{p}-bx} \Longleftrightarrow$$$$x^2(c-b)(a+b+c-p)+x(c-b)(abc\overline{p}+ap-a^2-2ab-2ac-bc)+a(c-b)(ab+bc+ac-abc\overline{p})=0 \Longleftrightarrow$$$$x^2(a+b+c-p)+x(abc\overline{p}+ap-a^2-2ab-2ac-bc)+a(ab+bc+ac-abc\overline{p})=0 \Longleftrightarrow$$$$x_{1,2}=\frac{-(abc\overline{p}+ap-a^2-2ab-2ac-bc)\pm\sqrt{(abc\overline{p}+ap-a^2-2ab-2ac-bc)^2-4a(ab+bc+ac-abc\overline{p})(a+b+c-p)}}{2(a+b+c-p)} \Longleftrightarrow$$$$x_{1,2}=\frac{-(abc\overline{p}+ap-a^2-2ab-2ac-bc)\pm\sqrt{((ab+bc+ac-abc\overline{p})+a(a+b+c-p))^2-4a(ab+bc+ac-abc\overline{p})(a+b+c-p)}}{2(a+b+c-p)} \Longleftrightarrow$$$$x_{1,2}=\frac{-(abc\overline{p}+ap-a^2-2ab-2ac-bc)\pm\sqrt{((ab+bc+ac-abc\overline{p})-a(a+b+c-p))^2}}{2(a+b+c-p)} \Longleftrightarrow$$$$x_{1,2}=\frac{-(abc\overline{p}+ap-a^2-2ab-2ac-bc)\pm(bc+ap-a^2-abc\overline{p})}{2(a+b+c-p)} \Longleftrightarrow$$$$x=\frac{ab+bc+ac-abc\overline{p}}{a+b+c-p}$$The last one we derive because $a\neq x$. We have that $\frac{ab+bc+ac-abc\overline{\frac{a+b+c}{3}}}{\frac{2}{3}(a+b+c)}=\frac{ab+bc+ac}{a+b+c}$ And because the relation is symmetric in $a,b,c$ we are done.

I'm not quite sure, but here it goes: Let $H_1$, $H_2$ and $H_3$ be the reflection points of the orthocenter $H$ of $ABC$ across $a=BC$, $b=CA$ and $c=AB$. It is well known that all three are on the circum circle $\mathcal{C}$ of the triangle $ABC$. A composition $\mathcal{Z}_b \circ \mathcal{Z}_a$ of reflection across $a$ and $b$ is a rotation $\mathcal{R}$ around $C$ and an angle $2\gamma$ which takes $G_1$ to $G_2$, $O_1$ to $O_2$ and $H_1$ to $H_2$. Say $F$ is intersection point of the lines $G_1O_1$ and $G_2O_2$. Since $\mathcal{R}$ takes $G_1O_1$ to $G_2O_2$ we have \begin{eqnarray*} \angle O_1FO_2 &=& \angle O_1CO_2 \\ \angle G_1FG_2 &=& \angle G_1CG_2 \\ \angle H_1FH_2 &=& \angle H_1CH_2 \\ \end{eqnarray*}So $F$ is on $\mathcal{C}$ and $FO_1CO_2$ and $FG_1CG_2$ are concyclic. Now we can define $F'$ as intersection point of the lines $G_1O_1$ and $G_3O_3$. But then $F' \in G_1O_1\cap \mathcal{C} = \{F,H_1\}$ and since $F'\ne H_1$ we have $F'=F$ and thus conclusion.

Lemma: Let $P$ be a point on a fixed line $\ell$, and let $P_b$ and $P_c$ be the reflections of $P$ over $AB$ and $AC$. Then, the circumcircle of $AP_bP_c$ passes through a fixed point. Proof: Invert at $A$, and $\ell$ becomes a fixed circle $\ell'$ passing through $A$ that intersects $AB$ and $AC$ at $X$ and $Y$. $P_b'$ and $P_c'$ are still the reflections of $P'$ over $AB$ and $AC$ as $AP=AP_b=AP_c$ and the inversion fixes $AB$ and $AC$. But now note that the circumcircle of $AP_bP_c$ inverts to line $P_b'P_c'$, which is known to pass through the orthocenter of $AXY$, as desired. Let $H_1, H_2, H_3$ be the reflections of the orthocenter $H$ of $ABC$. Applying this lemma with $\ell$ as the Euler line of $ABC$ gives us that $(AG_2G_3)$, $(AO_2O_3)$, and $(AH_2H_3)=(ABC)$ concur. Similarly, $(BG_3G_1),(BO_3O_1),(ABC)$ and $(CG_1G_2),(CO_1O_2),(ABC)$ concur. Hence, it remains to show that $(BO_3O_1),(CO_1O_2),(ABC)$ concur. But $\angle BO_3O_1=\angle(BO_3,AC)=90-B$ and $\angle CO_2O_1=\angle (CO_2,AB)=90-C$, so if $(BO_3O_1)$ and $(CO_1O_2)$ intersect at $P$ then $\angle BPC=90-B+90-C=A$, as desired.

Wait, isn't that obvious? The concurrency point is the Euler reflection point of $\triangle ABC$.

I think to compute the values of $o_1,o_2,o_3,g_1,g_2,g_3$ is easier than computing the intersection coordinate for general point.

rmtf1111 wrote: A more straight-forward solution using complex numbers(without using spiral similarity in complex numbers as v_Enchance did). As v_Enchance did we will find the coordinates of the intersection between the circle $(ABC)$ and the circle $(AP_bP_c)$, where $P$ is a random point inside $\triangle{ABC}$ and $P_b$ , $P_c$ are his reflections across lines $AC$, respectively $AB$. Let this intersection be $X$ . We have that $\overline{x}=\frac{1}{x}$ and $\frac{(p_b-a)(p_c-x)}{(p_c-a)(p_b-x)}$ is real, where $p_b=a+b-ab\overline p$ and $p_c=a+c-ac\overline p$ $\Longleftrightarrow$ $$\frac{bx+ax-px-ab}{cx+ax-px-ac}=\frac{ac+bc-abc\overline{p}-cx}{ab+bc-abc\overline{p}-bx} \Longleftrightarrow$$$$x^2(c-b)(a+b+c-p)+x(c-b)(abc\overline{p}+ap-a^2-2ab-2ac-bc)+a(c-b)(ab+bc+ac-abc\overline{p})=0 \Longleftrightarrow$$$$x^2(a+b+c-p)+x(abc\overline{p}+ap-a^2-2ab-2ac-bc)+a(ab+bc+ac-abc\overline{p})=0 \Longleftrightarrow$$$$x_{1,2}=\frac{-(abc\overline{p}+ap-a^2-2ab-2ac-bc)\pm\sqrt{(abc\overline{p}+ap-a^2-2ab-2ac-bc)^2-4a(ab+bc+ac-abc\overline{p})(a+b+c-p)}}{2(a+b+c-p)} \Longleftrightarrow$$$$x_{1,2}=\frac{-(abc\overline{p}+ap-a^2-2ab-2ac-bc)\pm\sqrt{((ab+bc+ac-abc\overline{p})+a(a+b+c-p))^2-4a(ab+bc+ac-abc\overline{p})(a+b+c-p)}}{2(a+b+c-p)} \Longleftrightarrow$$$$x_{1,2}=\frac{-(abc\overline{p}+ap-a^2-2ab-2ac-bc)\pm\sqrt{((ab+bc+ac-abc\overline{p})-a(a+b+c-p))^2}}{2(a+b+c-p)} \Longleftrightarrow$$$$x_{1,2}=\frac{-(abc\overline{p}+ap-a^2-2ab-2ac-bc)\pm(bc+ap-a^2-abc\overline{p})}{2(a+b+c-p)} \Longleftrightarrow$$$$x=\frac{ab+bc+ac-abc\overline{p}}{a+b+c-p}$$The last one we derive because $a\neq x$. We have that $\frac{ab+bc+ac-abc\overline{\frac{a+b+c}{3}}}{\frac{2}{3}(a+b+c)}=\frac{ab+bc+ac}{a+b+c}$ And because the relation is symmetric in $a,b,c$ we are done. To avoid solving quadratics. We can use Vieta jumping. We begin from the following equation (which rmtf1111 derived). $$\frac{bx+ax-px-ab}{cx+ax-px-ac}=\frac{ac+bc-abc\overline{p}-cx}{ab+bc-abc\overline{p}-bx}$$Or , $$(bx+ax-px-ab)(ab+bc-abc\overline{p}-bx)-(cx+ax-px-ac)(ac+bc-abc\overline{p}-cx)=0$$Observe that $x=a$ satisfies above equation (via inspection). Therefore by vieta, the other root is \begin{align*}\frac{[1]}{a[x^2]}&=\dfrac{-ab(ab+bc-abc\overline{p})+ac(ac+bc-abc\overline{p})}{a(-b(b+a-p)+c(c+a-p))}\\ &=\dfrac{a(ac^2-ab^2+c^2b-b^2c-c(abc\overline{p})+b(abc\overline{p}))}{a(c^2-b^2+ac-ab-pc+pb)}\\ &=\dfrac{a((ac+ab)(c-b)+bc(c-b)-abc\overline{p}(c-b)}{a((c+b)(c-b)+a(c-b)-p(c-b))}\\ &=\dfrac{a(c-b)(ab+bc+ca-abc\overline{p})}{a(c-b)(a+b+c-p)}\\ &=\dfrac{ab+bc+ca-abc\overline{p}}{a+b+c-p}\end{align*}Where $[1]$ and $[x^2]$ denote the coefficients of $1$ and $x^2$ in the quadratic equation. As this is symmetric with $a,b,c$, the rest is plugging in $p=0$ and $p=\dfrac{a+b+c}{3}$ to check that this yield equal value. Which is very easy. PS. Am I the only one who think that this is the easier than P4?

In fact, this problem is well-known. The point of intersection is the Anti-Steiner point. O-G-H is Euler line and H1 the reflection of H to BC and similar for H2 (CA) and H3 (AB). Let X be the point of intersection between H1G1 and H2G2. Easy to show that BH1XH2 is insciptible quadrilater so X is on the circumcircle of ABC. Easy to show that G1XCG2 is inscriptible quadrilater (∢G1XG2=∢G1CG2=2C) so X on the circumcircle of G1CG2. Similar for the rest of the triangles.

It's an easy problem about Anti-Steiner point.

tenplusten wrote: Then $\frac{(p-a)(b-c)}{(p-c-a)b}=\frac{(\frac{1}{p}-\frac{1}{a})(\frac{1}{b}-\frac{1}{c})}{\frac{1}{x}-\frac{1}{c}-\frac{1}{a})\cdot \frac{1}{b}}$ from here we easily get $x=\frac{ab+bc+ac}{a+b+c}$ I just tried to complex bash this problem, but wasn't very successful. This seems the 'easiest' solution, but I don't understand this step. Does this mean that the 'Anti-Steiner' point, as discussed above, of $ABC$ in the complex plane (with circumcircle the unit circle ... ) is $\frac{ab+bc+ac}{a+b+c}$ ? EDIT: Apologies for the general and open-ended bump. The post was because of two unfortunate reasons: i) lack of internet resources at olympiad level for the Anti-Steiner point aside from this, and ii) no complex number support in Geogebra

Please everybody i have a question I just want to know what was your intuition ti think about anti-steiner point In fact me i don't know this particular point so how can i do it ?!

liberator wrote: We claim that all the circles pass through the Anti-Steiner point $T$ of the Euler line. Indeed, $\measuredangle TG_2A=\measuredangle AGO=\measuredangle TG_3A$, and similarly for the other five circles.

We will complex bash with $(ABC)$ as the unit circle. Let $P$ denote either $O$ or $G$. Then, \begin{align*} p_1 &= b+c-bc\overline{p} \\ p_2 &= a+c - ac\overline{p} \\ p_3 &= a+b - ab\overline{p}, \end{align*}and \begin{align*} \overline{p}_1 &= 1/b+1/c-\frac{1}{bc}p \\ \overline{p}_2 &= 1/a+1/c - \frac{1}{ac}p \\ \overline{p}_3 &= 1/a+1/b - \frac{1}{ab}p. \end{align*}Let $Z=(P_1P_2C)\cap(ABC)$. We have that \[\frac{z-p_1}{z-p_2}\div\frac{c-p_1}{c-p_2}\in\mathbb{R}.\]Note that \[\frac{c-p_1}{c-p_2} = \frac{bc\overline{p}-b}{ac\overline{p}-a}=\frac{b}{a},\]so \[\frac{z-p_1}{z-p_2}\cdot\frac{a}{b} = \frac{1/z-\overline{p}_1}{1/z-\overline{p}_2}\cdot\frac{b}{a}.\]Thus \begin{align*} &a^2(z-b-c+bc\overline{p})(z(1/a+1/c-p/ac)-1)\\=&b^2(z-a-c+ac\overline{p})(z(1/b+1/c-p/bc)-1), \end{align*}or \begin{align*} f(z):=&a(z-(b+c-bc\overline{p}))(z(c+a-p)-ac)\\-&b(z-(a+c-ac\overline{p}))(z(c+b-p)-bc)=0. \end{align*}The $z^2$ coefficient of $f(z)$ is \[a(c+a-p)-b(c+b-p)=c(a-b)+(a-b)(a+b)-p(a-b)=(a-b)(a+b+c-p),\]the $z$ coefficient is \begin{align*} &b(a+c-ac\overline{p})(c+b-p)-a(b+c-bc\overline{p})(c+a-p)+b^2c-a^2c\\ =&(b^2a-a^2b)+c^2(b-a)+c(b^2-a^2)-pc(b-a)-c(ab^2-a^2b)\overline{p}+c(b^2-a^2)\\ =&(b-a)(ab+c^2+bc+ac-pc-abc\overline{p}+bc+ac) \\ =&-(a-b)(ab+2bc+2ca+c^2-pc-abc\overline{p}), \end{align*}and the constant term is \begin{align*} &a^2c(b+c-bc\overline{p})-b^2c(a+c-ac\overline{p})\\ =&abc(a-b)+c^2(a-b)(a+b)-abc^2\overline{p}(a-b) \\ =&(a-b)c(ab+c(a+b)-abc\overline{p}) \\ =&(a-b)c(ab+bc+ca-abc\overline{p}). \end{align*}Since $a\not=b$, we can write the equation $f(z)=0$ as \[(a+b+c-p)z^2-(ab+2bc+2ca+c^2-pc-abc\overline{p})z+c(ab+bc+ca-abc\overline{p})=0.\]Note that the point $C$ is clearly on both circles, so we expect a factor of $z-c$ to come out here. We see the equation can be written as \[(z-c)((a+b+c-p)z-(ab+bc+ca-abc\overline{p}))=0.\]As noted, we want the other intersection, so we finall arrive at \[(P_1P_2C)\cap(ABC) = Z = \frac{ab+bc+ca-abc\overline{p}}{a+b+c-p}.\]This is symmetric in $a,b,c$, so all $(P_iP_jC)$ pass through $Z$, and it is straightforward to verify that plugging in $p=0$ and $p=\frac{1}{3}(a+b+c)$ gives the same value for $Z$, namely $\frac{ab+bc+ca}{a+b+c}$ (in fact its true for any $p=k(a+b+c)$ whre $k\in\mathbb{R}$). Thus, $(G_1G_2C)$, $(G_1G_3B)$, $(G_2G_3A)$, $(O_1O_2C)$, $(O_1O_3B)$, $(O_2O_3A)$ and $(ABC)$ all pass through $Z$, as desired. EDIT (August 6th, 2019): Here's the solution using Steiner lines (see @below for a detailed discussion).

[asy][asy] unitsize(1inch); pair A, B, C, O, G, H, X, Y, Z, D, E, F, I, J, K, L, M, N, P, Q, R, S, T, U, V, W, u, v, w, h, i; A = dir(110); B = dir(195); C = dir(345); O = (0,0); G = centroid(A,B,C); H = orthocenter(A,B,C); X = foot(H,B,C); Y = foot(H,C,A); Z = foot(H,A,B); D = 2X - H; E = 2Y - H; F = 2Z - H; I = foot(G, B, C); J = foot(G, A, C); K = foot(G, A, B); L = 2I - G; M = 2J -G; N = 2K - G; P = B+C-O; Q = C+A-O; R = A+B-O; T = M + (0,.05); S = extension(D, L, E, M); U = foot(S, B, C); V = foot(S, C, A); W = foot(S, A, B); w = 2U - S; h = foot(H, B ,C); i = 2h - H; u = orthocenter(S, B, C); v = 2U - u; draw(S--w, dotted); draw(S--V, dotted); draw(S--W, dotted); draw(A--v, deepgreen); draw(H--w, deepgreen); draw(A--i, dotted); draw(circumcircle(A,B,C), dashed+heavygreen); draw(A--B--C--cycle, deepblue); draw(A--W, deepblue); draw(U--V--W, dashed+deepgreen); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$H$", H, E+.5); dot("$D$", S, dir(S)); dot("$X$", U, (-.3,-.5)); dot("$Y$", V, (-.3,-.5)); dot("$Z$", W, (.5,1.5)); dot("$U$", v, dir(v)); dot("$D'$", w, dir(w)); dot("$H'$", i, dir(i)); [/asy][/asy] Given a point $D$ on $(ABC)$, the Steiner line of $D$ (with respect to $ABC$) is the line passing through the reflections of $D$ over the sides of $\triangle ABC$. We will prove $H$ lies on the Steiner line. Let $X,Y$ be the feet of $D$ onto $BC$ and $AC$, and let $U$ be the second intersection of line $DX$ with $(ABC)$. Then $\measuredangle DUA = \measuredangle DCA = \measuredangle DCY = \measuredangle DXY$ so $AU$ is parallel to the Simson line of $D$. Let the reflections of $D,H$ over $\overline BC$ be $D',H'$ . Then $HD'\parallel AU$ because $\measuredangle H'HD' = \measuredangle DH'H = \measuredangle DH'A = \measuredangle DUA$, meaning that $H$ lies on the Steiner line of $D$. Now given a line $\ell$ passing through $H$ of $\triangle ABC$, we can construct the anti-Steiner point $S$ by reflecting $\ell$ over $AB$, $BC$, or $CA$ and taking the second intersection with $(ABC)$. In particular, this means that the reflections of the Euler line over $AB,BC,CA$ concur at some $S$ on $(ABC)$. Let $H_1,H_2,H_3$ be the reflections of $H$ over $BC,CA,AB$. Then $$\measuredangle O_2AO_3 = \measuredangle G_2AG_3 = \measuredangle H_2AH_3 = \measuredangle H_2SH_3 = \measuredangle G_2SG_3 = \measuredangle O_2SO_3$$$$\measuredangle O_3BO_1 = \measuredangle G_3BG_1 = \measuredangle H_3BH_1 = \measuredangle H_3SH_1 = \measuredangle G_3SG_1 = \measuredangle O_3SO_1$$$$\measuredangle O_1CO_2 = \measuredangle G_1CG_2 = \measuredangle H_1CH_2 = \measuredangle H_1SH_2 = \measuredangle G_1SG_2 = \measuredangle O_1SO_2,$$so all seven circumcircles have a common point $S$. [asy][asy] unitsize(1inch); pair A, B, C, O, G, H, X, Y, Z, D, E, F, I, J, K, L, M, N, P, Q, R, S, T, h, i; A = dir(110); B = dir(195); C = dir(345); O = (0,0); G = centroid(A,B,C); H = orthocenter(A,B,C); X = foot(H,B,C); Y = foot(H,C,A); Z = foot(H,A,B); D = 2X - H; E = 2Y - H; F = 2Z - H; I = foot(G, B, C); J = foot(G, A, C); K = foot(G, A, B); L = 2I - G; M = 2J -G; N = 2K - G; P = B+C-O; Q = C+A-O; R = A+B-O; T = M + (0,.05); h = foot(H, B, C); i = 2h - H; S = extension(D, L, E, M); draw(circumcircle(A,B,C), dashed+heavygreen); draw(A--B--C--cycle, deepblue); draw(H--G--O, blue); draw(D--L--P, lightblue); draw(E--M--Q, lightblue); draw(F--N--R, lightblue); draw(F--S, dotted+lightblue); draw(M--S, dotted+lightblue); draw(L--S, dotted+lightblue); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$O$", O, NE); dot("$G$", G, E+.5); dot("$H$", H, E+.5); dot("$H_1$", D, dir(D)); dot("$H_2$", E, dir(E)); dot("$H_3$", F, dir(F)+.3); dot("$G_1$", L, SW+1); dot(M); dot("$G_3$", N, NW+.5); dot("$O_1$", P, dir(P)+.7); dot("$O_2$", Q, NE+1.5); dot("$O_3$", R, dir(R)); dot("$S$", S, dir(S)); label("$G_2$", T, dir(T)); [/asy][/asy]

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -18.09, xmax = 10.77, ymin = -8.05, ymax = 13.69; /* image dimensions */ pen wwzzqq = rgb(0.4,0.6,0); pen qqqqcc = rgb(0,0,0.8); /* draw figures */ draw(circle((-2.17,2.29), 7.098394184602601), linewidth(0.8)); draw((-2.6276499906343615,9.37362594199273)--(-9.08930704538719,0.7055789664185856), linewidth(0.8)); draw((-9.08930704538719,0.7055789664185856)--(3.90106249194755,-1.3882333013102403), linewidth(0.8)); draw((3.90106249194755,-1.3882333013102403)--(-2.6276499906343615,9.37362594199273), linewidth(0.8)); draw((xmin, -1.3944246994249494*xmin-0.7359015977521396)--(xmax, -1.3944246994249494*xmax-0.7359015977521396), linewidth(1.2) + linetype("4 4") + wwzzqq); /* line */ draw((xmin, 0.7276506843553079*xmin-0.7764266200296193)--(xmax, 0.7276506843553079*xmax-0.7764266200296193), linewidth(0.8) + dotted + blue); /* line */ draw((xmin, -1.9761875054618554*xmin + 12.500221401928765)--(xmax, -1.9761875054618554*xmax + 12.500221401928765), linewidth(1.2) + dotted + blue); /* line */ draw((xmin, -0.3453388004307663*xmin + 4.492270217827564)--(xmax, -0.3453388004307663*xmax + 4.492270217827564), linewidth(0.8) + dotted + blue); /* line */ draw(circle((-2.4085858391711774,3.8873920509447912), 7.399727720669627), linewidth(0.8) + linetype("0 3 4 3") + red); draw((-8.84092976835196,7.545386298722886)--(-9.08930704538719,0.7055789664185856), linewidth(1.2) + qqqqcc); draw((-9.08930704538719,0.7055789664185856)--(-2.605298181358001,2.8969905357003584), linewidth(1.2) + qqqqcc); draw((-9.08930704538719,0.7055789664185856)--(-3.622214717553847,-3.4121336381395437), linewidth(1.2) + qqqqcc); /* dots and labels */ dot((-2.17,2.29),dotstyle); label("$O$", (-2.09,2.49), NE * labelscalefactor); dot((-2.6276499906343615,9.37362594199273),dotstyle); label("$A$", (-2.55,9.57), NE * labelscalefactor); dot((-9.08930704538719,0.7055789664185856),dotstyle); label("$B$", (-9.69,0.39), NE * labelscalefactor); dot((3.90106249194755,-1.3882333013102403),dotstyle); label("$C$", (4.19,-1.61), NE * labelscalefactor); dot((-3.475894544074004,4.1109716071010744),linewidth(4pt) + dotstyle); label("$H$", (-3.39,4.27), NE * labelscalefactor); dot((-2.605298181358001,2.8969905357003584),linewidth(4pt) + dotstyle); label("$G$", (-2.53,3.05), NE * labelscalefactor); dot((4.910296803987434,2.7965542097795186),linewidth(4pt) + dotstyle); label("$AntiSteinerPoint$", (4.99,2.95), NE * labelscalefactor); dot((-3.0182445534396413,-2.9726543348916543),dotstyle); label("$O_1$", (-2.93,-2.77), NE * labelscalefactor); dot((3.4434125013131895,5.69539264068249),dotstyle); label("$O_2$", (3.53,5.89), NE * labelscalefactor); dot((-9.546957036021554,7.789204908411315),dotstyle); label("$O_3$", (-9.47,7.99), NE * labelscalefactor); dot((-3.622214717553847,-3.4121336381395437),linewidth(4pt) + dotstyle); label("$G_1$", (-3.55,-3.25), NE * labelscalefactor); dot((3.1061615305824292,6.361863795245499),linewidth(4pt) + dotstyle); label("$G_2$", (3.19,6.53), NE * labelscalefactor); dot((-8.84092976835196,7.545386298722886),linewidth(4pt) + dotstyle); label("$G_3$", (-8.77,7.71), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Notation: Let $X$ be the AntiSteinerPoint of $HG$. One line proof : Thus, $\angle XG_3B =\angle BGH = 180^\circ - \angle BGO = 180^\circ -\angle BG_1X$ due to reflection and hence, $BG_3XG_2$ is cyclic and as $X$ only depends on $HG$, the result follows.

Solution by complex numbers: Suppose $X=O_2G_2\cap O_3G_3$. Then a unique spiral similarity centered at $A'=(XO_2O_3)\cap (XG_2G_3)$ sends $O_2G_2$ to $O_3G_3$. But since $A$ is also such a center, we have $A'=A$, and so $X=(AO_2O_3)\cap (AG_2G_3)$. Let $(ABC)$ be the unit circle, then $o_a=b+c$, $g=\frac13(a+b+c)$, and $$g_a=b+c-bc\overline{g_a}=b+c-\frac13\left(b+c+\frac{bc}{a}\right)=\frac13\left(2b+2c-\frac{bc}{a}\right),$$etc. We can compute $x$: $$ x =\frac{(\overline{o_c}g_c-o_c\overline{g_c})(o_b-g_b)-(\overline{o_b}g_b-o_b\overline{g_b})(o_c-g_c)}{(\overline{o_c}-\overline{g_c})(o_b-g_b)-(\overline{o_b}-\overline{g_b})(o_c-g_c)} $$The denominator: \begin{align*}(\overline{o_c}-\overline{g_c})(o_b-g_b)&=\left(\frac1a+\frac1b-\frac13\left(\frac2a+\frac2b+\frac{c}{ab}\right)\right)\left(a+c-\frac13\left(2a+2c-\frac{ac}{b}\right)\right)\\&=\frac{1}{9}\cdot \frac{(a+b+c)(ab+bc+ca)}{ab^2},\end{align*}So by symmetry \begin{align*}&(\overline{o_c}-\overline{g_c})(o_b-g_b)-(\overline{o_b}-\overline{g_b})(o_c-g_c) \\=&\ \frac19(a+b+c)(ab+bc+ca)\left(\frac{1}{ab^2}-\frac{1}{ac^2}\right)\\=&\ \frac19\cdot\frac{(a+b+c)(ab+bc+ca)(c+b)(c-b)}{ab^2c^2}.\end{align*}The numerator: \begin{align*} &(\overline{o_c}g_c-o_c\overline{g_c})(o_b-g_b)\\ =&\ \left(\frac13\left(\frac1a+\frac1b\right)\left(2a+2b-\frac{ab}{c}\right)-\frac13(a+b)\left(\frac2b+\frac2a-\frac{c}{ba}\right)\right)\left(a+c-\frac13\left(2a+2c-\frac{ac}{b}\right)\right)\\ =&\ \frac19\left(\frac{c}{a}+\frac{c}{b}-\frac{a}{c}-\frac{b}{c}\right)\left(a+c+\frac{ac}{b}\right), \end{align*}So by symmetry \begin{align*} &(\overline{o_c}g_c-o_c\overline{g_c})(o_b-g_b)-(\overline{o_b}g_b-o_b\overline{g_b})(o_c-g_c)\\ =&\ \frac19(ab+bc+ca)\left(\frac{c}{ab}+\frac{c}{b^2}-\frac1c-\frac{a}{bc}-\frac{b}{ac}-\frac{b}{c^2}+\frac{a}{bc}+\frac1b\right). \end{align*}Hence we can simplify: \begin{align*} x &= \frac{bc^3+ac^3-ab^2c-a^2bc-b^3c-b^3a+abc^2+a^2bc}{(a+b+c)(c+b)(c-b)}\\ &= \frac{(ab+bc+ca)(c+b)(c-b)}{(a+b+c)(c+b)(c-b)}\\ &= \frac{ab+bc+ca}{a+b+c}. \end{align*}This is symmetric in $a,b,c$, so what's left is to prove that this lies on unit circle. But since $$|a+b+c|=\Big|\frac1a+\frac1b+\frac1c\Big|=|ab+bc+ca|,$$$X$ does lie on the unit circle. Synthetic solution: Same as above, first notice that $X=(AO_2O_3)\cap (AG_2G_3)$. On the other hand, since $OG$ passes through the orthocenter, $X$ is the Anti-Steiner point of line $OG$, which lies on $ABC$. Hence $X$ is the desired intersection.

Unfortunately, when I first solved the problem, I accidentally took $H$ instead of $G$, and the complex bash was a lot cleaner, but it somehow still worked! (See the comments in above posts about it working for any point on the Euler Line.) But the bash for centroid is a... little worse .

Set $(ABC)$ as the unit circle. We see that $\triangle{CO_AO_B}~\triangle{CG_AG_B}$, so $C$ is the spiral center between $O_AO_B$ and $G_AG_B$, meaning that $(O_AO_BC), (G_AG_BC)$ intersect again at $X=O_AG_A\cap O_BG_B$. Because $X\in O_AG_A$ is equivalent to the reflection of $X$ over $BC$ being on the Euler line, we have $\frac{b+c-bc\bar{x}}{a+b+c}$ is real; similarly, from $X\in O_BG_B$ we have $\frac{a+c-ac\bar{x}}{a+b+c}$ is real. From here it easy to solve for $x$ to get $x=\frac{ab+bc+ca}{a+b+c}$. It is trivial to check $|x|=1$, so $X\in (ABC)$. And since $x$ is symmetric in $a, b, c$, we have that $X$ lies on all of the other six circles, so we are done. Remark: We never needed to use the specific values of $O, G$; they can be replaced by any two points on the Euler line.

Set $\triangle ABC$ unit circle, then $g = \frac{a+b+c}{3}$, $O = 0$. By formula plug and chug or homothety, $o_3 = a + b$ and cyclic, and by formula plug and chug, $g_3 = a + b - \frac{ab}{3} \left( \frac 1 a + \frac 1 b + \frac 1 c \right)$. Instead of intersecting this nasty thing, we reflect orthocenters instead to get $h_3 = - \frac{ab}{c}$. Let $z$ denote the desired point, then $z \overline z = 1$ because circumcenter, giving us the matrix \begin{align*} \begin{pmatrix} z^2 & 1 & z \\ - \frac{ab}{c} & - \frac{c}{ab} & 1 \\ a + b & \frac 1 a + \frac 1 b & 1 \end{pmatrix} = 0. \end{align*}This gives us a quadratic in $z$, which we can cheese using Vieta's(since we already know one of the intersects with the circumcircle is the reflection of the orthocenter). Expanding and using vieta, we find $z = \frac{ab + bc + ca}{a + b + c}$. Now the verifications are pretty easy. We evaluate \begin{align*} \frac{o_1 - c}{o_2 - c} \cdot \frac{o_2 - z}{o_1 - z} &= \frac{b}{a} \cdot \frac{c + a - \frac{ab + bc + ca}{a + b + c}}{b + c - \frac{ab + bc + ca}{a + b + c}} \\ &= \frac{b}{a} \cdot \frac{a^2 + ac + c^2}{b^2 + bc + c^2}, \end{align*}which we can conjugate to get $\frac{a}{b} \cdot \frac{b^2c^2 + ab^2c + a^2b^2}{a^2c^2 + a^2bc + a^2b^2}$ which is equal to the original. Thus $Z$ lies on all of the circumcentery circles. To finish, note that \begin{align*} \measuredangle AO_3G_3 = \measuredangle AO_3Z = \measuredangle AO_2Z = \measuredangle AO_2G_2, \end{align*}while properties of reflections instantly give $\measuredangle O_3AG_3 = \measuredangle O_2AG_2$ and $O_3G_3 = O_2G_2$, thus giving us a spiralsim that's actually spiral congruency. Now the finish to show that Z lies on the centroidy circles is clear by angle chase. Actually the only thing we used about $G$ was that it lies on the Euler line, so this should actually work for any point on it. Nice.

Lemma. Let $Q$ be any point and let $P$ be the Anti-Steiner Point of line $QH$. Let $Q_B$, $Q_C$ be the reflections of $Q$ w.r.t. $\overline{CA}$, $\overline{AB}$, respectively. Then $P \in (AQ_BQ_C)$. Proof of the lemma. Notice that $$\angle Q_CAH_C=\angle Q_CAQ-\angle QAH-\angle H_CAH$$$$\angle Q_BAH_B=\angle H_BAH-\angle HAQ-\angle Q_BAQ$$Since $\angle Q_CAQ+\angle QAQ_B=\angle H_CAH+\angle HAH_B=2\angle BAC$, we have that $\angle Q_CAH_C=\angle Q_BAH_B \implies =\angle Q_BAQ_C=\angle H_BAH_C=180^{\circ}-\angle H_BPH_C=180^{\circ}-\angle Q_BPQ_C \implies PAQ_BQ_C$ is cyclic, as desired. $\square$ Now since $G$, $O$, $H$ are collinear we are immediately done by the lemma. $\square$

We start by showing that the three lines $O_1G_1, O_2G_2, O_3G_3$ concur. Routinely, we have $$o_1 = b+1,o_2 = a+c,g_1 = \frac{2ac+2ab-bc}{3a},g_2 = \frac{2ab+2bc-ac}{3b}$$Then using the intersection formula, we compute $$d = \frac{(o_2\overline{g_2}-\overline{o_2}g_2)(o_1-g_1)-(o_1\overline{g_1}-\overline{o_1}g_1)(o_2-g_2)}{(\overline{o_1}-\overline{g_1})(o_2-g_2)-(\overline{o_2}-\overline{g_2})(o_1-g_1)}$$$$ = \frac{\frac{(a+c)(ab+ac+bc)(ac-b^2)}{9a^2bc}-\frac{(b+c)(ab+ac+bc)(bc-a^2)}{9ab^2c}}{\frac{(a+b+c)(ab+ac+bc)}{9b^2c}-\frac{(a+b+c)(ab+ac+bc)}{9a^2c}}$$$$ = \frac{b(a+c)(ac-b^2)-a(b+c)(bc-a^2)}{(a+b+c)(a^2-b^2)} = \frac{ab+bc+ac}{a+b+c}$$Now notice that $$(ab+ac+bc)(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}) = (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$$$\Rightarrow |ab+bc+ac|^2 =|a+b+c|^2 \Rightarrow \frac{|ab+ac+bc|}{|a+b+c|} = 1$$which implies that $$O_1G_1\cap O_2G_2\in(ABC)$$Now since the value is symmetric wrt $a,b,c$, we have that the three lines $O_1G_1,O_2G_2, O_3G_3$ concur. To finish, we present a lemma, \begin{lemma} For two similarly oriented triangle $\Delta ABC$, and $\Delta ADE$, let $X = BC\cap DE$. We have that $D\in(ABD)$, $D\in(ACE)$. \end{lemma} \begin{proof} Angle Chasing (Spiral sim) \end{proof} Thus we have that the point $\frac{ab+bc+ac}{a+b+c}$ is a common point of the seven circumcircles, as desired.

Notice that the orthocenter $H$ lies on $OG$ by the Euler line. Lemma: If $\ell$ is an arbitrary line passing through $H$ then the reflections of $\ell$ about $AB$, $BC$, and $CA$ concur on $(ABC)$. Let $P$ be a point on $(ABC)$ where the simson line $\ell_E$ wrt $ABC$ is parallel to $\ell$. Now, consider the homethety $\mathcal{H}$ centered at $P$ taking $\ell_E$ to $\ell$. Recall, $\ell_E$ bisects $PH$ so $\mathcal{H}$ has a dilation factor of $2$. Thus, if we let $X$, $Y$, and $Z$ be the feet of the simson line to $BC, CA,$ and $AB$ and $X'$, $Y'$, and $Z'$ be their respective points under $\mathcal{H}$. We see $X',Y',Z' \in \ell$ and $X'$, $Y'$, and $Z'$ are the reflections of $P$ across $AB$, $BC$, and $CA$. So, the reflections of $\ell$ across $AB$, $BC$, and $CA$ all pass through $P$ as desired. $\square$ Claim: If given some point $Q \in \ell_E$ with $Q_2$ and $Q_3$ being its reflections across $AC$ and $AB$, respectively. Then, $Q_2AQ_3P$ is cyclic. Notice that $AQ_2=AQ=AQ_3$ so it is sufficient to show $\angle Q_2PA = \angle Q_3PA$. Indeed, we have $\angle Q_2PA = \angle QY'A$. We see that $AY'=AP=AZ'$ so $\angle QY'A= \angle QZ'A= \angle Q_3PA$. Proving the claim. $\square$ Thus, setting $\ell_E$ to be the euler line yieds $(G_1G_2C)$, $(G_1G_3B)$, $(G_2G_3A)$, $(O_1O_2C)$, $(O_1O_3B)$, $(O_2O_3A)$ to all pass through $P$ which lies on $(ABC)$ as desired. $\blacksquare$

We claim that the concurrency point is the Anti-Steiner point of the line $HO$. Let $H$ be the orthocenter and $H_1,H_2$ be the reflection of $H$ over $BC, CA$ on $(ABC)$. Clearly $OH$ and $O_1H_1$ are the reflection with respect to $BC$. It is well known that $H_1O_1$ hits $(ABC)$ at the Anti-Steiner point of $HO$, suppose it is $S$. Now we have to prove $S, C, O_1,O_2$ are cyclic. For the proof, Obviosuly $\angle O_1CO_2=2\angle C$. Now \begin{eqnarray*} \angle O_1SO_2=180^\circ -\angle H_1BH_2=180^\circ -(90^\circ -\angle C+90^\circ -\angle C)=2\angle C. \end{eqnarray*}So points $S, O_1, O_2, C$ are concyclic. Similarly $S, O_1, O_3,B$ and $S, O_2, O_3, A$ are also concyclic. Since $G$ lies on the line $HO$ therefore it is same as before. Hence all the circles passes through the point $S$. $\square$

neba42 wrote: Please everybody i have a question I just want to know what was your intuition ti think about anti-steiner point In fact me i don't know this particular point so how can i do it ?! We know $O,G,H$ are collinear and the reflection of $O, G$ are are given. And we know Anti-Steiner point is about the reflection of line through $H$ about three sides of a triangle. Therefore Anti-Steiner point comes. If you don't know that point and its property then you have to discover the collinearity of $H_1, G_1, O_1$(not hard). Then step by step$\cdots$

Let $X$ be the Anti-Steiner Point of the Euler line. Notice $$180-\angle AO_2X=\angle G_2O_2A=\angle GOA=\angle G_3O_3A$$by reflection so $X$ lies on $(AO_2O_3).$ Similarly, it lies on the other circles. $\square$

Prolly the same as some of the solutions above, but I will still post it for storage. Let $S$ be the Anti-Steiner point wrt Euler Line. $$\angle G_2SG_3=\angle (H_BG_2, H_CG_3)=\angle (AC, HP)+\angle (AC, AB)+ \angle (HP, AB)=2 \angle BAC=\angle (G_2AG_3) \implies S, A, G_2, G_3$$Exactly same angle chase can be done for $O$

My solution is same as rmtf1111 so I m not gonna rewrite but I will give my handmade diagram.

We will first do the circumcenter portion of the problem. It is much more tame. We have $$O_2=a+c,O_3=a+b$$clearly. Now, we will use the spiral center trick. Let $AO_2$ intersect $(ABC)$ again at $P_2$, and let $AO_3$ intersect $(ABC)$ again at $P_3$. Now, we have $$P_2=\frac{-c}{a(\frac{1}{a}+\frac{1}{c})-1}=\frac{-c^2}{a+c-c}=\frac{-c^2}{a}.$$Similarly, we have $$P_3=\frac{-b^2}{a}.$$Thus, the spiral center, the intersection of $(ABC)$ and $(AO_2O_3)$, is $$\frac{(a+b)\frac{-c^2}{a}+(a+c)\frac{b^2}{a}}{a+b-\frac{c^2}{a}-a-c+\frac{b^2}{a}}$$$$=\frac{-c^2(a+b)+b^2(a+c)}{a(b-c)+b^2-c^2}$$$$=\frac{ab^2+cb^2-ac^2-bc^2}{(b-c)(a+b+c)}$$$$=\frac{(b-c)(ab+ac+bc)}{(b-c)(a+b+c)}=\frac{ab+ac+bc}{a+b+c}.$$Since this is symmetric in $a,b,c$, clearly $(BO_1O_3)$ and $(CO_1O_2)$ also pass through this point. Now, we embrace the true bashing spirit and do the centroid portion of the problem. This part is far worse than the previous part. Fortunately, we already know the common point, so we do not have to use the spiral center to find it again. We first compute $$G_2=\frac{2}{3}a+\frac{2}{3}c-\frac{ac}{3b},G_3=\frac{2}{3}a+\frac{2}{3}b-\frac{ab}{3c}.$$Now, it suffices to show that $$a,\frac{2}{3}a+\frac{2}{3}c-\frac{ac}{3b},\frac{2}{3}a+\frac{2}{3}b-\frac{ab}{3c},\frac{ab+ac+bc}{a+b+c}$$are concyclic. First, in order to expedite the remaining calculation, we will make the substitution $$u=\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}).$$Thus, the points that we wish to show are concyclic are $$a,a+c-acu,a+b-abu,\frac{ab+ac+bc}{a+b+c}.$$This is equivalent to $$\frac{\frac{ab+ac+bc}{a+b+c}-a}{\frac{ab+ac+bc}{a+b+c}-a-c-acu}\div \frac{b-abu}{b-c-abu+acu}\in R.$$Multiplying the left fraction by $a+b+c$ makes this become $$\frac{bc-a^2}{bc-a^2+(acu-c)(a+b+c)}\cdot\frac{(b-c)(1-au)}{b(1-au)}.$$Of course, we cancel to get $$\frac{bc-a^2}{bc-a^2+c(au-1)(a+b+c)}\cdot\frac{b-c}{b}.$$The conjugate of this expression is $$\frac{\frac{1}{bc}-\frac{1}{a^2}}{\frac{1}{bc}-\frac{1}{a^2}+(\frac{1}{ac}(\frac{1}{3}(a+b+c))-\frac{1}{c})(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}\cdot\frac{\frac{1}{b}-\frac{1}{c}}{\frac{1}{b}}$$which simplifies to $$\frac{a^2-bc}{a^2-bc+ab(\frac{1}{3}b+\frac{1}{3}c-\frac{2}{3}a)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}\cdot \frac{c-b}{c}.$$Comparing this to the original expression, the numerators are actually the same (both ones are flipped signs), so we can actually just compare the denominators and try to show that $$b(bc-a^2+c(au-1)(a+b+c))=c(a^2-bc+ab(\frac{1}{3}b+\frac{1}{3}c-\frac{2}{3}a)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})).$$We can manipulate the left side to get $$b(bc-a^2+c(au-1)(a+b+c))$$$$=b^2c-a^2b+bc(au-1)(a+b+c).$$Plugging in the definition of $u$, we get $$b^2c-a^2b+bc(\frac{a}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-1)(a+b+c)$$$$b^2c-a^2b+\frac{abc}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{3}{a})(a+b+c)$$$$b^2c-a^2b+\frac{1}{3}(-2bc+ab+ac)(a+b+c) (*).$$We can also rewrite the left side as $$c(a^2-bc+ab(\frac{1}{3}b+\frac{1}{3}c-\frac{2}{3}a)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}))$$$$=ca^2-bc^2+\frac{1}{3}(b+c-2a)(ab+ac+bc) (**).$$Thus, it suffices to show that $(*)=(**)$, which multiplying by 3 becomes $$3b^2c-3a^2b+(-2bc+ab+ac)(a+b+c)=3ca^2-3bc^2+(b+c-2a)(ab+ac+bc),$$which is evident upon expansion.

Let $(ABC)$ be the unit circle. Then it follows that $o = 0$, and $G = \frac{a + b + c}{3}$ Then \[o_1 = b + c\]\[o_2 = a + c\]\[o_3 = a + b\]And \[g_1 = b + c - bc\overline{g}\]\[g_2 = a + c - ac\overline{g}\]\[g_3 = a + b - ab\overline{g}\] Then let the intersection of $(ABC)$ and $(O_1O_2C)(\neq C)$ be $X$. By the concyclic formula, we have \[\frac{x - o_1}{x - o_2} \div \frac{c - o_1}{c - o_2} \in \mathbb R\]\[\frac{x - b - c}{x - a - c} \cdot \frac{a}{b} = \overline{\frac{x - b - c}{x - a - c}} \cdot \overline{\frac{a}{b}} = \frac{\frac{1}{x} - \frac{1}{b} - \frac{1}{c}}{\frac{1}{x} - \frac{1}{a} - \frac{1}{c}} \cdot \frac{b}{a}\]$\newline$ \[\frac{\frac{bc - bx - cx}{bcx}}{\frac{ac - ax - cx}{acx}} = \frac{bc - cx - bx}{ac - ax - cx} = \frac{x - b - c}{x - a - c} \cdot \frac{a}{b}\]$\newline$ \[(bc - cx - bx)(bx - ab - bc) = (ac - ax - cx)(ax - ab - ac)\]\[(-cx - b(x - c)(-ab + b(x - c) = (-cx - a(x - c)(-ab + a(x - c)\]\[(b(x - c) + cx)(b(x - c) - ab) = (a(x - c) + cx)(a(x - c) - ab)\]\[b^2(x - c)^2 - ab^2(x - c) + bcx(x - c) = a^2(x - c)^2 - a^2b(x - c) + acx(x - c)\]Dividing by $x - c$, we get \[b^2(x - c) - ab^2 + bcx = a^2(x - c) - a^2b + acx\]\[(b - a)(b + a)(x - c) - ab(b - a) + cx(b - a) = 0\]Dividing by $b - a$, we get \[(a + b)(x - c) - ab + cx = 0\]\[ax - ac + bx - bc - ab + cx = 0\]\[(a + b + c)x - ab - ac - bc\]\[x = \frac{ab + ac + bc}{a + b + c}\] Since this is symmetric in terms of $a$, $b$, and $c$, $O_1O_3B$ and $O_2O_3A$ also coincide on this point. Now we need to prove that it lies on $G_1G_2C$. Similarly, we need to prove that \[\frac{x - g_1}{x - g_2} \div \frac{c - g_1}{c - g_2} \in \mathbb R\]\[\frac{\frac{ab + ac + bc}{a + b + c} - b - c + bc\overline{g}}{\frac{ab + ac + bc}{a + b + c} - a - c + ac\overline{g}} \div \frac{b(c\overline{g} - 1)}{a(c\overline{g} - 1)} \]\[\frac{ab + ac + bc + (a + b + c)(-b - c + \frac{ab + ac + bc)}{3a})}{ab + ac + bc + (a + b + c)(-a - c +\frac{ab + ac + bc}{3b})} \div \frac{b}{a}\]\[\frac{3a(ab + ac + bc) + (a + b + c)(-2ab - 2ac + bc)}{3b(ab + ac + bc) + (a + b + c)(-2ab - 2bc + ac)}\]\[\frac{3a^2b + 3a^2c + 3abc -2a^2b - 2a^2c + abc - 2ab^2 - 2abc + b^2c - 2abc - 2ac^2 + bc^2}{3ab^2 + 3abc + 3b^2c - 2a^2b - 2abc + a^2c - 2ab^2 - 2b^2c + abc = 2abc - 2b^2c + ac^2}\]\[\frac{a^2b + a^2c - 2ab^2 + b^2c - 2ac^2 + bc^2}{ab^2 + ac^2 - 2bc^2 - 2a^2b + a^2c + ac^2}\]\[\frac{(\frac{1}{a^2b} + \frac{1}{a^2c} - \frac{2}{ab^2} + \frac{1}{b^2c} - \frac{2}{ac^2} + \frac{1}{bc^2}) \cdot a^2b^2c^2}{(\frac{1}{ab^2} + \frac{1}{ac^2} - \frac{2}{bc^2} - \frac{2}{a^2b} + \frac{1}{a^2c} + \frac{1}{ac^2}) \cdot a^2b^2c^2} = \frac{a^2b + a^2c - 2ab^2 + b^2c - 2ac^2 + bc^2}{ab^2 + ac^2 - 2bc^2 - 2a^2b + a^2c + ac^2}\]So $x$ lies on $(G_1G_2C)$, and by symmetry, it lies on $(G_1G_3B)$ and $(G_2G_3A)$, so we are done. : D

The fact that this is very trivially and directly complex bashable is slightly disturbing. Here goes, Set $ABC$ be the unit circle as usual with center $O$. Then, \[o_1=b+c-bc\overline{o}=b+c\]Similarly, \begin{align*} o_2 &= c+a\\ o_3 &= a+b \end{align*}Also, \begin{align*} g_1 &= b+c-bc\overline{g}\\ &= b+c -bc\left(\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3} \right)\\ &= \frac{3ab+3ac-ab-bc-ca}{3a}\\ &= \frac{2ab+2ca-bc}{3a} \end{align*}Similarly, \begin{align*} g_2 &= \frac{2ab+2bc-ca}{3b}\\ g_3 &= \frac{2bc+2ca-ab}{3c} \end{align*}Now, consider the point represented by the complex number $x=\frac{ab+bc+ca}{a+b+c}$. Note that, \begin{align*} \overline{x} &= \frac{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\\ &= \frac{a+b+c}{ab+bc+ca}\\ &= \frac{1}{x} \end{align*}This means that $X$ lies on $(ABC)$. Next, \begin{align*} \frac{(c-x)(o_2-o_1)}{(c-o_1)(o_2-x)} & = \frac{\left(c-\frac{ab+bc+ca}{a+b+c}\right)\left(a+c-b-c\right)}{(c-b-c)\left( a+c - \frac{ab+bc+ca}{a+b+c} \right)}\\ &= \frac{\frac{(ac+bc+c^2-ab-bc-ca)(a-b)}{a+b+c}}{\frac{b(ab+bc+ca-a^2-ab-ac-ac-bc-c^2)}{a+b+c}}\\ &= \frac{(c^2-ab)(b-a)}{b(a^2+ac+c^2)} \end{align*}Then, we also see that, \begin{align*} \overline{\left(\frac{(c-x)(o_2-o_1)}{(c-o_1)(o_2-x)}\right)} &= \overline{\frac{(c^2-ab)(b-a)}{b(a^2+ac+c^2)}}\\ &= \frac{\left(\frac{1}{c^2}-\frac{1}{ab}\right)\left(\frac{1}{b}-\frac{1}{a}\right)}{\frac{1}{b}\left(\frac{1}{c^2}+\frac{1}{ac}+\frac{1}{a^2}\right)}\\ &= \frac{\left(\frac{ab-c^2}{abc^2}\right)\left(\frac{a-b}{ab}\right)}{\frac{1}{b}\left(\frac{a^2+ac+c^2}{a^2c^2}\right)}\\ &= \frac{(c^2-ab)(b-a)}{b(a^2+ac+c^2)}\\ &= \left(\frac{(c-x)(o_2-o_1)}{(c-o_1)(o_2-x)}\right) \end{align*}Thus, $\left(\frac{(c-x)(o_2-o_1)}{(c-o_1)(o_2-x)}\right) \in \mathbb{R}$ which implies that $X$ lies on $(O_1O_2C)$. Similarly, $X$ also lies on $A_1O_3B$ and $O_2O_3A$ ($x$ is symmetric in $a,b,c$ so it suffices to verify one). Next, note that \begin{align*} \frac{(c-x)(g_2-g_1)}{(c-g_1)(g_2-x)} &= \frac{\left(c-\frac{ab+bc+ca}{a=b+c}\right) \left(\frac{2ba+2bc-ac}{3b}-\frac{2ab+2ac-bc}{3a}\right)}{\left(c-\frac{2ab+2ca-bc}{3a}\right)\left(\frac{2ba+2bc-ca}{3b}-\frac{ab+bc+ca}{a=b+c}\right)}\\ &=\frac{\frac{(ac+bc+c^2-ab-bc-ca)(2a^2b+2abc-a^2c-2ab^2-2abc+b^2c)}{3ab(a+b+c)}}{\frac{(3ac-2ab-2ca+bc)(2a62b+2ab^2+2abc+2abc+2b^2c+2bc^2-a^2c-abc-ac^2-3ab^2-3b^2c-3abc)}{9ab(a+b+c)}}\\ &= \frac{9ab(a+b+c)(a-b)(c^2-ab)(2ab-ac-bc)}{3ab(a+b+c)(ac+bc-2ab)(2a^2b-ab^2+2bc^2-b^2c-a^2c-ac^2)}\\ &= \frac{3(c^2-ab)(b-a)}{(2a^2b-ab^2+2bc^2-b^2c-ac^2-a^2c)} \end{align*}Now, we check that \begin{align*} \overline{\left(\frac{(c-x)(g_2-g_1)}{(c-g_1)(g_2-x)}\right)} &= \overline{\left(\frac{3(c^2-ab)(b-a)}{(2a^2b-ab^2+2bc^2-b^2c-ac^2-a^2c)}\right)}\\ &= \frac{3\left(\frac{1}{c^2}-\frac{1}{ab}\right)\left(\frac{1}{b}-\frac{1}{a}\right)}{\left(\frac{2}{a^2b}-\frac{1}{ab^2}+\frac{2}{bc^2}-\frac{1}{b^2c}-\frac{1}{ac^2}-\frac{1}{a^2c}\right)}\\ &= \frac{3a^2b^2c^2(ab-c^2)(a-b)}{a^2b^2c^2(2bc62-ac^2+2a^2b-a^2c-ab^2-b^2c)}\\ &= \frac{3(c^2-ab)(b-a)}{(2a^2b-ab^2+2bc^2-b^2c-ac^2-a^2c)}\\ &= \left(\frac{(c-x)(g_2-g_1)}{(c-g_1)(g_2-x)}\right) \end{align*}Thus, $\left(\frac{(c-x)(g_2-g_1)}{(c-g_1)(g_2-x)}\right) \in \mathbb{R}$ which implies that $X$ must lie on $(G_1G_2C)$. Similarly, $X$ also lies on the circles $(G_1G_3B)$ and $(G_2G_3A)$. This means that indeed the circumcircles of triangles $G_1G_2C$ , $G_1G_3B$ , $G_2G_3A$ $O_1O_2C$ , $O_1O_3B$ , $O_2O_3A$ and $ABC$ have a common point (which is the previously described point $X$).

From Anti-Steiner we find the $\overline{O_1G_1}$, $\overline{O_2G_2}$ and $\overline{O_3G_3}$ concur at $T$ on $(ABC)$. We claim $T$ is the desired point of concurrency. First we demonstrate there exists a spiral similarity at $B$ mapping $\overline{O_1G_1} \mapsto \overline{O_3G_3}$. The proof is through complex numbers. Set $(ABC)$ as the unit circle, and observe that, \begin{align*} o_1 &= b + c + o - bc\overline{o} = b + c\\ o_3 &= a + b + o - ab\overline{o} = a + b\\ g_1 &= b + c + g - bc\overline{g} = \frac{2ab + 2ca - bc}{3a}\\ g_3 &= a + b + g - ab\overline{g} = \frac{2ca + 2bc - ab}{3c} \end{align*}Now we compute their spiral center; \begin{align*} \frac{g_1o_3 - g_3o_1}{g_1 + o_3 - g_3 - o_1} &= \frac{\frac{2ab + 2ac - bc}{3a}(a + b) - \frac{2ca + 2bc - ab}{3c}(b + c)}{\frac{2ab + 2ac - bc}{3a} + a + b - \frac{2ca + 2bc - ab}{3c} - b - c}\\ &= \frac{\frac{c(a+b)(2ab + 2ac - bc) - a(b + c)(2ca + 2bc - ab)}{3ac}}{\frac{c(2ab + 2ac - bc) - a(2ca + 2bc - ab) + 3ac(a - c)}{3ac}}\\ &= \frac{c(2a^2b + 2a^2c + 2ab^2 + abc - b^2c) - a(abc + 2b^2c - ab^2 + 2c^2a + 2bc^2)}{2abc + 2ac^2 - bc^2 - 2ca^2 - 2abc + a^2b + 3a^2c - 3ac^2}\\ &= \frac{2a^2bc + 2a^2c^2 + 2ab^2c + abc^2 - b^2c^2 - a^2bc - 2ab^2c + a^2b^2 - 2c^2a^2 - 2abc^2}{(ab + ac + bc)(a - c)}\\ &= \frac{b(ab + bc + ca)(a - c)}{(ab + ac + bc )(a - c)}\\ &= b \end{align*}which implies that $T = \overline{O_1G_1} \cap \overline{O_3G_3}$ is the second intersection of the circles $(O_1O_3B)$ and $(G_1G_3B)$. The result follows by symmetry.

We use complex numbers with $(ABC)$ being the unit circle. Let lowercase letters denote the complex numbers representing their corresponding uppercase letter points, as usual. Note that $BOCO_1$ is a rhombus and therefore a parallelogram, so $o_1 = b+c-0 = b+c.$ Similarly, $o_2 = c+a$ and $o_3 = a + b.$ Now we will find the second intersection if $(ABC)$ and $(CO_1 O_2)$ and show that it is symmetric in $a,b,c.$ Let $Z_1 = CO_1 \cap (ABC) \ne C$ and $Z_2 = CO_2 \cap (ABC) \ne C.$ By a well-known formula, we have \[ z_1 = \frac{c-o_1}{c\overline{o_1} - 1} = \frac{c-(b+c)}{c\left(\frac{1}{b}+\frac{1}{c}\right) - 1} = -\frac{b^2}{c} \]and similarly \[ z_2 = -\frac{a^2}{c}. \]Now, if $P$ is the intersection of our two circles, then there is a spiral similarity centered at $P$ sending $\overrightarrow{Z_1 Z_2}$ to $\overrightarrow{O_1 O_2}$, so \begin{align*} p &= \frac{z_1 o_2 - z_2 o_1}{z_1 + o_2 - z_2 - o_1} \\ &= \frac{-\frac{b^2}{c} (a+c) + \frac{a^2}{c} (b+c)}{-\frac{b^2}{c} + a + c + \frac{a^2}{c} - b - c} \cdot \frac{c}{c} \\ &= \frac{a^2 b + a^2 c - b^2 a - b^2 c}{a^2 - b^2 + c(a-b)} \\ &= \frac{(a-b)(ab+bc+ca)}{(a-b)(a+b+c)} \\ &= \frac{ab+bc+ca}{a+b+c}, \end{align*}which is indeed symmetric in $a,b,c.$ Therefore, $(ABC),(CO_1O_2),(AO_2 O_3),(BO_3 O_1)$ share point $P.$ Now we show that $P$ lies on $(CG_1 G_2);$ since $p$ is symmetric in $a,b,c,$ this will finish the problem. First, we clearly have $g = \frac{a+b+c}{3},$ so by complex reflection, \[ g_1 = b+c-bc\left(\frac{1}{3} \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \right) = \frac{2b}{3} + \frac{2c}{3} - \frac{bc}{3a} \]and similarly \[ g_2 = \frac{2c}{3} + \frac{2a}{3} - \frac{ca}{3b}. \]We would thus like to show that \[ \frac{c-g_1}{c-g_2} \cdot \frac{p-g_2}{p-g_1} \in \mathbb{R}. \]To find the first fraction, we blindly plug in note that $\measuredangle G_1 C G_2 = 2 \measuredangle BCA,$ so the inscribed angle theorem implies that this is equal to $\frac{b}{a}$ since $CG_1 = CG_2 = CG.$ On the other hand, we have no choice but to plug in for the second fraction. We get \begin{align*} \frac{c-g_1}{c-g_2} \cdot \frac{p-g_2}{p-g_1} &= \frac{b}{a} \cdot \frac{\frac{ab+bc+ca}{a+b+c} - \left(\frac{2c}{3} + \frac{2a}{3} - \frac{ca}{3b}\right)}{\frac{ab+bc+ca}{a+b+c} - \left(\frac{2b}{3} + \frac{2c}{3} - \frac{bc}{3a}\right)} \\ &= \frac{b}{a} \cdot \frac{a}{b} \cdot \frac{3b(ab+bc+ca)-(a+b+c)(2bc+2ab-ca)}{3a(ab+bc+ca)-(a+b+c)(2ac+2ab-bc)} \\ &= \frac{(3ab^2+3b^2c+3abc-(2abc+2a^2b - a^2 c + 2b^2 c + 2ab^2 - abc + 2bc^2 + 2abc - ac^2)}{3a^2 b + 3abc + 3a^2 c - (2a^2 c + 2a^2 b - abc + 2abc + 2ab^2 - b^2 c + 2ac^2 + 2abc - bc^2)} \\ &= \frac{ab^2 + b^2 c - 2a^2 b + a^2 c + ac^2 - 2bc^2}{a^2 b + a^2 c - 2ab^2 + bc^2 + b^2 c - 2ac^2}. \end{align*}The conjugate of this expression is equal to \[ \frac{\frac{1}{ab^2} + \frac{1}{b^2 c} - \frac{2}{a^2 b} + \frac{1}{a^2 c} + \frac{1}{ac^2} - \frac{2}{bc^2}}{\frac{1}{a^2 b} + \frac{1}{a^2 c} - \frac{2}{ab^2} + \frac{1}{bc^2} + \frac{1}{b^2 c} - \frac{2}{ac^2}} \cdot \frac{a^2 b^2 c^2}{a^2 b^2 c^2} = \frac{ab^2 + b^2 c - 2a^2 b + a^2 c + ac^2 - 2bc^2}{a^2 b + a^2 c - 2ab^2 + bc^2 + b^2 c - 2ac^2}, \]thus the expression is real. Hence $C,G_1,G_2,P$ are concyclic, and we conclude.

Notice $G_1O_1,G_2O_2,G_3O_3$ concur at $X$ on $(ABC)$ by antistiner. Then $\measuredangle AG_2X=\measuredangle AG_2O_2=-\measuredangle AGO=\measuredangle AG_3O_3=\measuredangle AG_3X$ and $\measuredangle AO_2X=\measuredangle AO_2G_2=-\measuredangle AOG=\measuredangle AO_3G_3=\measuredangle AO_3X$

What a coincidence coming across this problem, just after a few days after reading about anti-steiner points and some related lemmas. Observe: $G_1O_1,G_2O_2,G_3O_3$ concur at the anti-steiner point of $OG$. Let that point be $X$ Step I: Consider the orthocenter of $\triangle ABC$ to be $H$, then consider the reflection of the point over sides $BC, CA, AB$ to be $H_1,H_2,H_3$. Step II: Since $AO_1=AO_2$, $AG_1=AG_2$ and $\measuredangle O_1AO_2 = \measuredangle G_1OG_2$, $A$ is the center of spiral similarity that sends $G_1O_1 \mapsto G_2O_2$ similarly, $G_2O_2\mapsto G_3O_3$. Step III: Angle Chasing: $\measuredangle AO_3X=\measuredangle AOG=-\measuredangle AO_2G_2=\measuredangle AO_3G_3=\measuredangle AO_2X$ and $\measuredangle AG_3X=\measuredangle AG_3O_3=-\measuredangle AGO=\measuredangle AG_2O_2 =\measuredangle AG_2X $. Thus the intersection of lines $G_2O_2$ and $G_3O_3$ intersect at $X$. Doing this for other $3$ lines gets us our desired result. $\square$

Let $(ABC)$ be the unit circle. We compute the intersection $X$(different from $A$) of $(ABC)$ with $(AO_2O_3)$ and show it lies on all circles. We have $o_2=a+c$, $o_3=a+b$, $\overline x=1/x$ and \[\frac{a-o_2}{a-o_3}\cdot\frac{x-o_3}{x-o_2}\in\mathbb R\]After expanding and canceling a factor of $b-c$, this come out to \[x\cdot\frac{a+b+c}{a}+\frac{ab+bc+ca}{x}-a-2b-2c-\frac{bc}{a}=0\]Notice that $a$ is a solution of this equation so by Vieta we get $x=\frac{ab+bc+ca}{a+b+c}$. Since this is symmetric, it lies on $AO_3O_1$ and $AO_1O_2$. Therefore it suffices to check that it lies on $AG_2G_3$ and the others follow from symmetry. We have $g_2=a+c-ac\overline{g}$, $g_3=a+b-ab\overline g$ and we need to prove that \[\frac{a-g_3}{a-g_2}\cdot\frac{x-g_2}{x-g_3}\in\mathbb R\] This comes out to \[\frac{c^2a+c^2b-2a^2c-2b^2c+a^2b+b^2a}{b^2a+b^2c-2a^2b-2c^2b+a^2c+c^2a}\in\mathbb R\]Which is clear. $\blacksquare$

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