Let $A B C$ 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 $B C, C A, A B$, respectively. Find the locus of points $P$ such that $A A_{1}, B B_{1}, C C_{1}$ are concurrent and $\angle P A B+\angle P B C+$ $\angle P C A=90^{\circ}$.
In complex numbers with $A B C$ on the unit circle, it is equivalent to solving the following two cubic equations in $p$ and $q=\bar{p}$: 
$$ \begin{aligned} (p-a)(p-b)(p-c) & =(a b c)^{2}(q-1 / a)(q-1 / b)(q-1 / c) \\ 0 & =\prod_{\text {cyc }}(p+c-b-b c q)+\prod_{\text {cyc }}(p+b-c-b c q) \end{aligned} $$
Viewing this as two cubic curves in $(p, q) \in \mathbb{C}^{2}$, by Bézout'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 $A B C$ scalene): the three vertices, the three excenters, and $I, O, H$. Hence the answer is just those three triangle centers $I, O$ and $H$.  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.  Remark. These two cubics have names: 
- The locus of $\angle P A B+\angle P B C+\angle P C A=90^{\circ}$ is the McCay cubic, which is the locus of points $P$ for which $P, P^{*}, O$ are collinear.
- The locus of the pedal condition is the Darboux cubic, which is the locus of points $P$ for which $P, P^{*}, L$ are collinear, $L$ denoting the de Longchamps point. Assuming $P \neq P^{*}$, this implies $P$ and $P^{*}$ both lie on the Euler line of $\triangle A B C$, which is possible only if $P=O$ or $P=H$.