Given a triangle $ABC$. Point $M$ moves along the side $BA$ and point $N$ moves along the side $AC$ beyond point $C$ such that $BM=CN$. Find the geometric locus of the centers of the circles circumscribed around the triangle $AMN$.
Answer: The line $\ell$ passing through the circumcentre $O$ parallel to the internal angle bisector of $\angle A$.
Let the circumradius of $ABC$ be $R$ and let $X$ be a point on $\ell$. Let the circle $(X,XA)$ intersect $AB$ and $AC$ again at $M$ and $N$. We want to prove that $BM=CN \Leftrightarrow AM-AN = c-b$. Denote $\theta = \angle (\ell, XA)$. Then
$$XA = \frac{cos(C+A/2)}{\sin \theta} R$$$$AM = 2XA\cos (A/2+\theta) $$$$AN = 2XA\cos (A/2-\theta) $$$$\therefore AM-AN = -4XA\sin A/2 \sin \theta = -4R \sin A/2 \cos (C+A/2) = 2R(\sin C - \sin (A+C)) = c-b.$$
Note that by similarity, the midpoint $X$ of arc $BAC$ of $(ABC)$ is the center of spiral similarity sending $BM$ to $CN$, so $(AMN)$ passes through $X$. Thus the circumcenter of $(AMN)$ lies on the perpendicular bisector of $AX$, which is a fixed line. It is easy to see that all points on that line can be the centers of such circles.