Let $D$ be the midpoint of $BC.$ The requested locus is the circle $\omega$ with center $D$ and radius $BD \sqrt{3},$ excluding those two points that lie on $BC$ itself. Let $A'$ be the reflection of $A$ across $D.$ The given condition is equivalent to $BGCA'$ being cyclic, so by Power of a Point, we need $DG \cdot DA' = DB \cdot DC.$ Noting that $DA' = DA = 3DG,$ we get that $$\frac{AD^2}{3} = \frac{BC^2}{4}.$$Therefore, $$AD^2 = \frac{3 \cdot BC^2}{4},$$so $AD = \frac{\sqrt{3}}{2} BC = \sqrt{3} BD.$ Thus $A$ must lie on $\omega.$ It is easy to show that all points $A$ on $\omega$ work except for those that lie on line $BC$ (otherwise the triangle is degenerate), so we are done.

My solution differs only in the fact that I decided to reflect $G$ and not $A$ across $D$ I decided however to pursue also the Ptolemy's way to see how doable it was:

Doable but too much grind... Whoever chose this way had much less time for beautiful problem 6.

you can also bash it. let wlog $b = 1$ and $c=-1$ the points $B$ and $C$ on the complex plane, and $A=a$ variable. We want points $a, \frac{a-1}{2}, \frac{a+1}{2}, \frac{a}{3}$ concyclic, i.e., $$ \dfrac{1-a}{2} : \dfrac{-4a}{3-a} \in \mathbb{R} \iff \dfrac{(1-a)(3-a)}{-8a} = \dfrac{(1-\Bar{a})(3-\Bar{a})}{-8\Bar{a}} \iff \Bar{a}(a^2-4a+3) = a(\Bar{a}^2 - 4\Bar{a} +3) \iff a\Bar{a} = 3 \iff |a| = \sqrt{3}$$. that means that $A$ must lie on the circle centered in the midpoint of $BC$ and with radius $BC\sqrt{3}$.

Why to complex bash if u can bary bash it? As usual, let $A = [1,0,0], B = [0,1,0], C = [0,0,1], G = [1,1,1], M = [0,1,1]$ Let $G'$ be the reflection of $G$ on $M$. Obviously, $G' = [-1,2,2]$ and all we want is that it is at the circumcircle of $ABC$. So we get $2a^2 -b^2 -c^2 =0$ By cosine law at $AMB$ and $AMC$ and the above observation we get that $AM = \sqrt{3}MB$ What a nice problem to bary.

You can also finish the barybash with a coordbash. After obtaining $2a^2-b^2-c^2=0$, let $A=(x,y)$, $B=(-1,0)$ and $C=(1,0)$. Then $a^2=4$, $b^2=(x-1)^2+y^2$ and $c^2=(x+1)^2+y^2$, which gives $x^2+y^2=3$, that is, $MA=\sqrt{3}MB$.

We claim that the locus is a circle centered at the midpoint of $BC$ with radius $\frac{\sqrt{3}}{2}BC$. Let $BC=a,AC=b,AB=c$. The main idea of the problem is the following: \begin{claim} The condition is equivalent to $2a^2=b^2+c^2$. \end{claim} Let $D,E,F$ denote the midpoints of $BC,AC,AB$. The condition is equivalent to $AFGE$ being cyclic, which is equivalent to $$BF\times BA=BG\times BE$$$$\frac{1}{2}c^2=(\frac{1}{3}\sqrt{2a^2+2c^2-b^2})(\frac{1}{2}\sqrt{2a^2+2c^2-b^2})$$$$\frac{1}{2}c^2=\frac{1}{6}(2a^2+2c^2-b^2)$$$$3c^2=2a^2+2c^2-b^2$$$$2a^2=b^2+c^2,$$as desired. We promptly make the transition to Cartesian coordinates. WLOG $B=(0,0)$ and $C=(1,0)$, and suppose $A=(x,y)$. The condition is then $$x^2+y^2+(x-1)^2+y^2=2,$$which simplifies to $$(x-\frac{1}{2})^2+y^2=\frac{3}{4},$$which is a circle centered at $(1/2,0)$ with radius $\frac{\sqrt{3}}{2}$, and scaling up, we get the desired locus.

Draw point $G'$ where $BGCG'$ is a parallelogram, let $M$ be the midpoint of $BC$. Note that $ABG'C$ is cyclic hence \[3\cdot MA\cdot MG'=3\cdot MB^2=3\cdot MC^2\]\[MA^2=3\cdot MB^2=3\cdot MC^2\]let $BC=2r$ then locus is a circle centered at $M$ with radius $r\sqrt{3}$ minus the two points on line $BC$.

This is a contest problem from Brazil 2023?

soryn wrote: This is a contest problem from Brazil 2023? As it is stated on the source, this is Iberoamerican math Olympiad P4. The name of the post comes from the fact that this year the Olympiad was hosted by Brazil.

WLOG let $B$ and $C$ be at $-1$ and $1$ respectively. Let $A$ be at the complex number $a$. Then $G$ is at $\frac{a}{3}$. Let the reflection of $G$ over the origin be $H$ at $-\frac{a}{3}$. We want $ABHC$ to be cyclic. By PoP, this means that we need $\frac{|a|^2}{3}=1$ so $|a|=\sqrt3$. This means that for general $B$ and $C$, the locus of points $A$ is a circle of radius $\frac{BC\sqrt3}2$ centered at the midpoint of $BC$.

Nice,thx

Let $M$ be the midpoint of $\overline{BC}$. The locus is the circle centered at $M$ with radius $\sqrt{3}\cdot BM$, excluding the points on $\overline{BC}$. Denote $G'$ as the reflection of $G$ over $M$. Note that \[180^\circ = \angle BAC + \angle BGC = \angle BAC + \angle BG'C,\] which implies that $BACG'$ is a cyclic quadrilateral. Hence, Power of a Point on $M$ yields \[BM \cdot CM = G'M \cdot AM \implies AM^2 = 3BM^2.\] This yields $AM = \sqrt{3} \cdot BM$, which implies the desired locus.

Let the midpoint of $BC$ be $M$. The locus is the circle centered at $M$ with radius $\frac{BC\sqrt{3}}{2}$. To see this, reflect $G$ over $M$ and we need $BACG'$ being cyclic. Power of a point implies $AM \cdot MA' = \frac{BC^2}{4}$ which coupled with median ratios implies the desired.