Given equilateral triangle $ABC$, let $P$ be a point that varies inside triangle $ABC$ so that $\angle APC = 120^\circ$. Let $M$ be the intersection of lines $CP$ and $AB$, and let $N$ be the intersection of lines $AP$ and $BC$. Find the locus of the circumcenter of triangle $BMN$.
So far, I have only found the range of P and I can't seem to generalize the circumcircle. Any help would be greatly appreciated!
Notice first of all that triangles $ACN$ and $BCM$ are congruent: $\angle CAN\cong\angle BCM$ (inscribed angles subtending the same arc $CP$), $\angle NCA\cong\angle MBC$ and $AC\cong BC$ by hypothesis. In an analogous way, one proves that triangles $ACM$ and $ABN$ are congruent.
Let now $D$ be the point on $AC$ such that $CD\cong BN\cong AM$ and thus $AD\cong BM\cong CN$. Triangle $DMN$ is by construction equilateral and has the same center $G$ as $ABC$.
If $O$ is the symmetric of $D$ with respect to $G$, we have $ON\cong OM$ and $\angle NOM=120°$. It follows that $O$ is the circumcenter of $PNBM$ and the requested locus is then the symmetric of side $AC$ with respect to $G$.