Problem: Let $H$ be the orthocenter and $G$ be the centroid of
acute-angled triangle $ABC$ with $AB \neq AC$.
The line $AG$ intersects the circumcircle
of $ABC$ at $A$ and $P$.
Let $P'$ be the reflection of $P$ in the line $BC$.
Prove that $\angle CAB = 60$ if and only if $HG = GP'$.

\end{enumerate}

Solution: The following complex numbers solution was given by Stefan Tudose.
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
\[ p' = b+c-bc\ol p = \frac{ab+ac-b^2-c^2}{2a-b-c}. \]

Now let $D$ be the midpoint of $\ol{HP'}$.
Then
\begin{align*}
  d = \frac{h+p'}{2} & = \frac{a^2-b^2-c^2+ab+ac-bc}{2a-b-c} \\
  h-p' &= \frac{2(a^2-bc)}{2a-b-c} \\
  g-d &= \frac{2b^2+2c^2-a^2+bc-2ab-2ac}{3(2a-b-c)}
\end{align*}
Then $HG = GP' \iff \ol{GD} \perp \ol{HP'}$ and we have
\[
  \frac{g-d}{h-p'}
  = \frac{2b^2+2c^2-a^2+bc-2ab-2ac}{6(a^2-bc)}
\]
which we would like to be pure imaginary.
However the negative conjugate equals
\[
  \ol{\left( \frac{g-d}{h-p'} \right)}
  = -\frac{2a^2c^2+2a^2b^2-b^2c^2+a^2bc-2abc^2-2ab^2c}{6bc(bc-a^2)}.
\]
Expanding, the condition we have becomes
\[ b^3c+bc^3+b^2c^2=a^2bc+a^2c^2+a^2b^2
  \iff (b^2+bc+c^2)(a^2-bc) = 0. \]
Now $b^2+bc+c^2=0 \iff \angle A = 60\dg$ as desired.

\begin{remark*}
  One can also proceed by $|g-h| = |g-p'|$
  which is longer but ultimately the same calculation.
\end{remark*}
\begin{remark*}
  The condition $AB \neq AC$ is not cosmetic;
  it cannot be dropped from the problem condition.
  This is reflected in the presence of $a^2-bc$ factor.
\end{remark*}