
Problem 9a.2 Given an acute-angled triangle $A B C$ and let $\alpha, \beta$ and $\gamma$ be its angles respectively to $A, B$ and $C$. For an arbitrary interior point $M$ denote by $A_{1}, B_{1}$ and $C_{1}$ respectively the feet of the perpendiculars from $M$ to $B C, C A$ and $A B$. Find the locus of $M$ for which the triangle $A_{1} B_{1} C_{1}$ is a right angle triangle.


Solution: Let $M$ be a point such that $\Varangle A_{1} C_{1} B_{1}=90^{\circ}$. Since the quadrilateral $A C_{1} M B_{1}$ is inscribed we have $\Varangle B_{1} C_{1} M=\Varangle B_{1} A M=$ $x$. Analogously $\Varangle A_{1} C_{1} M=\Varangle A_{1} B M=y$. Denote $z=\Varangle A M B$. It follows now from the quadrilateral $A M B C$ that $x+\gamma+y+\left(360^{\circ}-\right.$
$z)=360^{\circ}$. Since $x+y=\Varangle A_{1} C_{1} B_{1}=90^{\circ}$ we get $z=90^{\circ}+\gamma$. Therefore $\Varangle A M B=90^{\circ}+\gamma$. Thus, the locus of $M$ such that $\Varangle A_{1} C_{1} B_{1}=90^{\circ}$ is an arc $G_{1}$ in the interior of $\triangle A B C$ for which the segment $A B$ is seen under angle $90^{\circ}+\gamma$. Analogously one can prove that the locus of $M$ such that $\Varangle C_{1} B_{1} A_{1}=90^{\circ}$ is an arc $G_{2}$ in the interior of $\triangle A B C$, for which $A C$ is seen under angle $90^{\circ}+\beta$. Further, the locus of $M$ such that $\Varangle B_{1} A_{1} C_{1}=90^{\circ}$ is an arc $G_{3}$ in the interior of $\triangle A B C$ for which $B C$ is seen under angle $90^{\circ}+\alpha$. The desired locus is the union of the three arcs $G_{1} \cup G_{2} \cup G_{3}$.

