I think the incenters of $\triangle MNP$ and $\triangle ABC$ should coincide, not the circumcenters. Here is a proof of this fact. It is well-known that $I$ lies on $(AYZ), (BXZ), (CXY)$ (just add the incircle touchpoints and find congruent triangles using the length conditions). Thus $IX=IY=IZ$ since $I$ should be midpoint of the three arcs $XY, YZ, ZX$. To finish, notice that the distances from $I$ to $MN, NP, PM$ are equal since $IX/2=IY/2=IZ/2$ and $MN$ is the perpendicular bisector of $IX$ and similarly for the other two sides, so we are done.

Yes, the incenters are supposed to coincide. Thanks for pointing it out.