XVIII OBM

Let $O^{\prime }$ be the circumcenter of $ABC$ and $M,N,P$ be the respective midpoints of $AB,AC,AD$. Notice that $O^{\prime },M$ and $N$ are fixed points and, by Thales theorem, $P$ is a variable point on the segment $MN$. Then $O_1$ is the intersection of the perpendicular bisectors of $AB$ and $AD$, $O_2$ is the intersection of the perpendicular bisectors of $AD$ and $AC$ and, similarly, $O^{\prime }$ is the intersection of the perpendicular bisectors of $AB$ and $AC$.



Since $\angle AN{O}_2=\angle AP{O}_2=90^{\circ }$, $APN{O}_2$ is cyclic; analogously, $AM{O}_{1}P$ is cyclic as well. So $\angle O_{2}AN=\angle O_{2}PN=\angle O_{1}PM=\angle O_{1}AM$. Hence $\angle O_{1}A{O}_2=\angle BAC$.

Quadrilateral $AM{O}^{\prime }N$ is cyclic as well, so $\angle M{O}^{\prime }N=180^{\circ }-\angle MAN=180^{\circ }-\angle BAC$. Thus $\angle O_1{O}^{\prime }{O}_2+\angle O_{1}A{O}_2=180^{\circ }$, which implies that $A{O}_1{O}^{\prime }{O}_2$ is cyclic. Its circumcircle is $O$, and so $O$ lies on the perpendicular bisector of $A{O}^{\prime }$, which is fixed. So the locus is a segment contained in such perpendicular bisector.

It remains to find the vertices of this segment. When $D$ tends to $B$, the perpendicular bisectors of $AD$ and $AB$ tends to coincide, so $O_2$ and $O^{\prime }$ tends to coincide and $O$ tends to be the circumcenter of $A{O}^{\prime }B$; analogously, when $D$ tends to $C$, $O$ tends to be the circumcenter of $A{O}^{\prime }C$. So the locus is the open segment with vertices on the circumcenters of $A{O}^{\prime }B$ and $A{O}^{\prime }C$.