Brazilian Mathematical Olympiad

Let $O'$ be the circumcenter of $ABC$ and $M, N, P$ be the respective midpoints of $AB, AC, AD$. Notice that $O', M$ 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, $O'$ is the intersection of the perpendicular bisectors of $AB$ and $AC$.

Since $\angle ANO_2 = \angle APO_2 = 90^\circ$, $APNO_2$ is cyclic; analogously, $AMO_1P$ is cyclic as well. So $\angle O_2AN = \angle O_2PN = \angle O_1PM = \angle O_1AM$. Hence $\angle O_1AO_2 = \angle BAC$.

Quadrilateral $AMO'N$ is cyclic as well, so $\angle MO'N = 180^\circ - \angle MAN = 180^\circ - \angle BAC$. Thus $\angle O_1O'O_2 + \angle O_1AO_2 = 180^\circ$, which implies that $AO_1O'O_2$ is cyclic. Its circumcircle is $O$, and so $O$ lies on the perpendicular bisector of $AO'$, 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 $O'$ tends to coincide and $O$ tends to be the circumcenter of $AO'B$; analogously, when $D$ tends to $C$, $O$ tends to be the circumcenter of $AO'C$. So the locus is the open segment with vertices on the circumcenters of $AO'B$ and $AO'C$.