Locus of the centres of equilateral triangles (contest problem) Given a triangle $A_0A_1A_2$ determine the locus of the centres of the equilateral triangles $X_0X_1X_2$ satisfying the condition that each of the lines $X_kX_{k+1}$, $k=0,1,2$ passes through $A_{k+2}$ (all indices are reduced modulo 3).
All I was thinking about are the Napoleon triangle associated to the triangle $A_0A_1A_2$.
What do you think? Thanks! (This problem was proposed at the Romanian TST 2014 for Seniors).

This is an algebraic approach which shows the locus of the center lies on a circle centered at centroid. The algebra is not as horrible as I originally thought. I hope this can inspire someone to construct a more geometrical proof for this interesting problem.`
WOLOG, we will choose a coordinate system where the centroid $C$ of $\triangle A_0A_1A_2$ is the origin. We will assume $A_0, A_1, A_2$ surrounds $C$ in counterclockwise manner. 
We then proceed to identify any point $(u,v)$ in this coordinate system with the complex number $u+iv$. In addition,  for any complex valued expression $???$, we will use the notation $\overrightarrow{???}$ to represent the corresponding vector in $\mathbb{R}^2$. We will abuse the notation and use the same symbol to denote a geometric point and its corresponding complex number. e.g $$C = \frac13(A_0+A_1+A_2) = 0
\quad\iff\quad \vec{C} = \frac13\left(\vec{A}_0+\vec{A}_1+\vec{A}_2\right) = \vec{0}
$$ since the centroid has been chosen as the origin.
Let $P$ be the center of $\triangle X_0X_1X_2$. 
Let $\omega = e^{i2\pi/3}$ be the cubic root of unity. 
Let $\eta$ be a complex number on the unit circle such that $\vec{\eta}$ is the outward pointing normal of the side $X_1X_2$ of $\triangle X_0X_1X_2$. When $\triangle X_0X_1X_2$ is an equilateral triangle, the outward pointing normals for the sides $X_2X_0$ and $X_0X_1$ will be $\overrightarrow{\omega\eta}$ and $\overrightarrow{\omega^2\eta}$ respectively. It is easy to see $\triangle X_0X_1X_2$ is an equilateral triangle when and only when
$$\begin{align}
&\overrightarrow{\eta}\cdot\left(\vec{A_0} - \vec{P}\right)
= \overrightarrow{\omega\eta}\cdot\left(\vec{A_1} - \vec{P}\right)
= \overrightarrow{\omega^2\eta}\cdot\left(\vec{A_2}- \vec{P}\right)\\
\iff &
\begin{cases}
\overrightarrow{(1-\omega)\eta}\cdot\vec{P} &= \overrightarrow{\eta}\cdot\vec{A}_0 - \overrightarrow{\omega\eta}\cdot\vec{A}_1\\
\overrightarrow{(\omega-\omega^2)\eta}\cdot\vec{P} &= \overrightarrow{\omega\eta}\cdot\vec{A}_1 - \overrightarrow{\omega^2\eta}\cdot\vec{A}_2\\
\overrightarrow{(\omega^2-1)\eta}\cdot\vec{P} &= \overrightarrow{\omega^2\eta}\cdot\vec{A}_2 - \overrightarrow{\eta}\cdot\vec{A}_0\\
\end{cases}
\end{align}
$$
Translate this to complex numbers, we have
$$\begin{array}{rcrcrcrcrcr}
(1-\omega)\eta\bar{P} &+& (1-\omega^2)\bar{\eta} P
&=& \eta\bar{A}_0 &+& \bar{\eta}A_0 
&-& \omega\eta\bar{A}_1 &-& \omega^2\bar{\eta}A_1\\
(\omega-\omega^2)\eta\bar{P} &+& (\omega^2-\omega)\bar{\eta}P
&=& \omega\eta\bar{A}_1 &+& \omega^2\bar{\eta}A_1 
&-& \omega^2\eta\bar{A}_2 &-& \omega\bar{\eta}A_2\\
(\omega^2-1)\eta\bar{P} &+& (\omega-1)\bar{\eta}P
&=& \omega^2\eta\bar{A}_2 &+& \omega\bar{\eta}A_2 
&-& \eta\bar{A}_0 &-& \bar{\eta}A_0
\end{array}
$$
Multiply the $2^{nd}$ equation by $\omega$, the $3^{rd}$ equation by $\omega^2$ and sum the three equations, we obtain:
$$3(1-\omega^2)\bar{\eta}P =
(1-\omega^2)\eta\left(\bar{A}_0 + \bar{A}_1\omega^2 + \bar{A}_2\omega\right)
+(1-\omega^2)\bar{\eta}\left(A_0 + A_1 + A_2\right)$$
Since $A_0 + A_1 + A_2 = 3 C = 0$, this gives us
$$P = \frac13 \eta^2 \left(\bar{A}_0 + \bar{A}_1\omega^2 + \bar{A}_2\omega\right)
$$
and hence the locus of $P$ lies on a circle centered at origin with radius
$\frac13 \left|A_0 + A_1\omega + A_2\omega^2\right|$.
Geometrically, this means if we rotate $A_1$ with respect to $C$ for $120^{\circ}$ to get
$\omega A_1$, rotate $A_2$ with respect to $C$ for $240^{\circ}$ to get $\omega^2 A_2$. The centroid of the triangle formed by $A_0, \omega A_1$ and $\omega^2 A_2$ lies on the locus.
This allow us to construct the locus using compass and ruler.
