Problem: Let $ABC$ be an acute scalene triangle.
The reflections of the centroid $G$ and the circumcenter $O$ of $ABC$ in its sides $BC$, $CA$, $AB$
are denoted by $G_1$, $G_2$, $G_3$ and $O_1$, $O_2$, $O_3$, respectively.
Show that 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.

\end{enumerate}

Solution: 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 $p_B = a+c-ac\ol p$, $p_C = a+b-ab\ol p$.
To compute $q_B$ note that
\begin{align*}
  a+q_B &= p_B + a q_B \ol{p_B} \\
  &= a+c-ac\ol p + aq_B \left( 1/a + 1/c - p/ac \right) \\
  \implies ac\ol p - c &= (a/c-p/c) q_B \\
  \implies q_B &= c^2 \frac{a\ol p -1}{a-p}.
\end{align*}
Similarly $q_C = b^2 \frac{a \ol p - 1}{a-p}$.
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 \ol p -1}{a-p} \right)
  \left( b^2(a+c-ac \ol p) - c^2 (a+b-ab\ol p) \right)}%
  {(b-c)(a \ol p-1) + (b^2-c^2) \cdot \frac{a \ol p - 1}{a-p}} \\
  &= \frac{b^2(a+c-ac \ol p) - c^2(a+b-ab\ol p)}%
  {(b-c)(a-p) + (b^2-c^2)} \\
  &= \frac{(b-c)(a(b+c)+bc) - (b-c) abc \ol p}{(a-p)+(b+c)} \\
  &= \frac{ab+bc+ca - abc \ol 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).