Problem 1523

### Part I: Prove the existence of a unique point $P$

1. Symmetry and Rotation:
- Given an equilateral triangle $ABC$ and a point $M$ in the plane, let $A', B', C'$ be the reflections of $A, B, C$ through $M$.
- The transformation from $ABC$ to $A'B'C'$ can be seen as a composition of a rotation of $\frac{2\pi}{3}$ centered at some point $O'$ (the center of $A'B'C'$) and a reflection through $M$.

2. Rotation Analysis:
- This composition results in a rotation of $-\frac{\pi}{3}$ around some point $P$.
- Point $P$ is the center of this rotation, which means $P$ is equidistant from $A$ and $B'$, $B$ and $C'$, and $C$ and $A'$.

3. **Uniqueness of $P$:**
- Since the rotation is uniquely defined by the angle $-\frac{\pi}{3}$ and the points involved, $P$ is uniquely determined.

Thus, we have shown that there exists a unique point $P$ equidistant from $A$ and $B'$, from $B$ and $C'$, and from $C$ and $A'$.

### Part II: Prove the circumcircle of $\triangle MNP$ passes through a fixed point

1. Equilateral Triangle and Midpoint:
- Let $D$ be the midpoint of side $AB$.
- When $M$ varies, $M$ does not coincide with $D$.

2. **Properties of $P$:**
- If $O$ is the center of $\triangle ABC$, then $POO'$ forms an equilateral triangle.
- $M$ is the midpoint of $OO'$, implying $\angle POM = \frac{\pi}{3}$.

3. Concyclic Points:
- Since $PAB'$ is an equilateral triangle and $DM \parallel AB'$, we have $\angle PNM = \frac{\pi}{3}$.
- From the above, $P, O, M, N$ are concyclic, meaning they lie on the same circle.

4. Fixed Point:
- The circumcircle of $\triangle MNP$ always passes through $O$, the center of $\triangle ABC$.

Thus, the circumcircle of $\triangle MNP$ passes through the fixed point $O$.

$\blacksquare$