parmenides51 wrote: Let P be an arbitrary point on the arc $AC$ of the circumcircle of a fixed triangle $ABC$, not containing $B$. The bisector of angle $APB$ meets the bisector of angle $BAC$ at point $P_a$ the bisector of angle $CPB$ meets the bisector of angle $BCA$ at point $P_c$. Prove that for all points $P$, the circumcenters of triangles $PP_aP_c$ are collinear. by I. Dmitriev A circumcenter cannot collinear

@Above, the problem is saying the locus of all possible circumcenters of a certain triangle is a line. P is not a constant point. The other two points depend on P as well.

My solution. $I$ is incenter, $M_C$ is the midpoint of the minor arc of $AB$, and $M_B$ is the midpoint of the minor arc of $BC$ First, by simple angle chasing, we could know that $(I P_A P_C P)$ is concyclic. Also let we point the intersection of circle $(I P_A P_C)$ and $(ABC)$ by $X$ Then the circumcenter will lye on the perpendicular bisector of $IX$ FIrst, $I$ is the constant point. Then by showing $X$ is the constant point will prove that the locus of circumcenter of $P P_A P_C$ is line. We could know that $\triangle X P_A M_C \sim \triangle X P_C M_A$, so $\frac{M_C X}{M_B X}=\frac{M_C P_A}{M_A P_C}=\frac{I M_C}{I M_B}$ (by sine law in triangle $\triangle M_C I P_A, \triangle M_A I P_C$) so the point $X$ is not moving because it is the intersection of circumcircle of $ABC$ and the Apollonius circle.