Let $ABC$ be a triangle. Find the locus of points $P$ such that the triangle formed by the orthogonal projections $A^{\prime}, B^{\prime}, C^{\prime}$ of $P$ on the sides $(BC),(CA),(AB)$ is isosceles at $A^{\prime}$.
Let's use the Law of Sines in triangle $B A^{\prime} C^{\prime}$, noting that its circumcircle is the circle with diameter $[B P]: P B=\frac{A^{\prime} C^{\prime}}{\sin \hat{B}}$. Similarly, we find $P C=\frac{A^{\prime} B^{\prime}}{\sin C}$. Thus, triangle $A^{\prime} B^{\prime} C^{\prime}$ is isosceles at $A^{\prime}$ if and only if $\frac{P B}{P C}=\frac{\sin \widehat{B}}{\sin \hat{C}^{\prime}}$, but the Law of Sines in $A B C$ gives us that $\frac{\sin \widehat{B}}{\sin \hat{C}}=\frac{A B}{A C}$, so this happens only if $P$ lies on the $A$-Apollonian circle of $A B C$.

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