Problem: Let $ABC$ be an acute scalene triangle
and let $P$ be a point in its interior.
Let $A_1$, $B_1$, $C_1$ be projections of $P$ onto
triangle sides $BC$, $CA$, $AB$, respectively.
Find the locus of points $P$ such that
$AA_1$, $BB_1$, $CC_1$ are concurrent
and $\angle PAB + \angle PBC + \angle PCA = 90\dg$.

\end{enumerate}

Solution: In complex numbers with $ABC$ the unit circle,
it is equivalent to solving the following two cubic equations
in $p$ and $q = \ol p$:
\begin{align*}
 (p-a)(p-b)(p-c) &= (abc)^2 (q -1/a)(q - 1/b)(q - 1/c) \\
 0 &= \prod_{\text{cyc}} (p+c-b-bcq) + \prod_{\text{cyc}} (p+b-c-bcq).
\end{align*}
Viewing this as two cubic curves in $(p,q) \in \CC^2$,
by \emph{B\'ezout's Theorem} it follows there are at most nine solutions
(unless both curves are not irreducible,
but it's easy to check the first one cannot be factored).
Moreover it is easy to name nine solutions (for $ABC$ scalene):
the three vertices, the three excenters, and $I$, $O$, $H$.
Hence the answer is just those three triangle centers $I$, $O$ and $H$.

\begin{remark*}
  On the other hand it is not easy to solve the cubics by hand;
  I tried for an hour without success.
  So I think this solution is only feasible
  with knowledge of algebraic geometry.
\end{remark*}
\begin{remark*}
  These two cubics have names:
  \begin{itemize}
    \ii The locus of $\angle PAB + \angle PBC + \angle PCA = 90\dg$
    is the \textbf{McCay cubic},
    which is the locus of points $P$
    for which $P$, $P^\ast$, $O$ are collinear.
    \ii The locus of the pedal condition
    is the \textbf{Darboux cubic},
    which is the locus of points $P$
    for which $P$, $P^\ast$, $L$ are collinear,
    $L$ denoting the de Longchamps point.
  \end{itemize}
  Assuming $P \neq P^\ast$,
  this implies $P$ and $P^\ast$
  both lie on the Euler line of $\triangle ABC$,
  which is possible only if $P=O$ or $P=H$.

  Some other synthetic solutions are posted at
  \url{https://aops.com/community/c6h1243902p6368189}.
\end{remark*}