Let $ABC$ be a triangle in a plane $\pi$. Find the set of all points $P$ (distinct from $A,B,C$ ) in the plane $\pi$ such that the circumcircles of triangles $ABP$, $BCP$, $CAP$ have the same radii.
To solve the problem, we need to find all points \( P \) in the plane \( \pi \) such that the circumcircles of triangles \( ABP \), \( BCP \), and \( CAP \) have the same radii. We will denote the angles \( \angle BPC \), \( \angle CPA \), and \( \angle APB \) as \( \alpha \), \( \beta \), and \( \gamma \), respectively.

Given that the circumcircles of \( \triangle ABP \), \( \triangle BCP \), and \( \triangle CAP \) have the same radius, we have the following condition:
\[ \sin \alpha = \lambda \sin \hat{A}, \quad \sin \beta = \lambda \sin \hat{B}, \quad \sin \gamma = \lambda \sin \hat{C} \]
where \( \lambda \) is a constant.

We will prove that \( \lambda = 1 \). In that case, all three circles will have radius \( R \), equal to the circumradius of the circumcircle of \( \triangle ABC \). There are exactly four circles of radius \( R \) passing through two of the vertices of \( \triangle ABC \): its circumcircle and the other three intersecting at its orthocenter. Therefore, the solution to the problem is that \( P \) can be any point on the circumcircle or the orthocenter.

### First Case: \( P \) is an inner point of \( \triangle ABC \)

1. Since \( P \) is an inner point of \( \triangle ABC \), we have:
   \[ \alpha + \beta + \gamma = 360^\circ \]
   Using the sine rule for angles, we get:
   \[ \sin \gamma = -\sin \alpha \cos \beta - \cos \alpha \sin \beta \]

2. For the angles of \( \triangle ABC \), we have:
   \[ \hat{A} + \hat{B} + \hat{C} = 180^\circ \]
   Using the sine rule for angles, we get:
   \[ \sin \hat{C} = \sin \hat{A} \cos \hat{B} + \cos \hat{A} \sin \hat{B} \]

3. By manipulating the equations, we obtain:
   \[ \sin \hat{A} (\cos \hat{B} + \cos \beta) + \sin \hat{B} (\cos \hat{A} + \cos \alpha) = 0 \]
   \[ \sin \hat{B} (\cos \hat{C} + \cos \gamma) + \sin \hat{C} (\cos \hat{B} + \cos \beta) = 0 \]
   \[ \sin \hat{C} (\cos \hat{A} + \cos \alpha) + \sin \hat{A} (\cos \hat{C} + \cos \gamma) = 0 \]

4. Suppose \( \lambda < 1 \). Then:
   \[ |\cos \alpha| > |\cos \hat{A}|, \quad |\cos \beta| > |\cos \hat{B}|, \quad |\cos \gamma| > |\cos \hat{C}| \]
   Since \( \alpha + \beta + \gamma = 360^\circ \), two of the angles, w.l.o.g. \( \beta \) and \( \gamma \), have \( \cos \beta < 0 \) and \( \cos \gamma < 0 \). This leads to:
   \[ \sin \hat{B} (\cos \hat{C} + \cos \gamma) + \sin \hat{C} (\cos \hat{B} + \cos \beta) < 0 \]
   which contradicts the second equation above.

5. Suppose \( \lambda > 1 \). Then:
   \[ |\cos \alpha| < |\cos \hat{A}|, \quad |\cos \beta| < |\cos \hat{B}|, \quad |\cos \gamma| < |\cos \hat{C}| \]
   W.l.o.g., we can suppose \( \hat{B}, \hat{C} < 90^\circ \). Then:
   \[ \sin \hat{B} (\cos \hat{C} + \cos \gamma) + \sin \hat{C} (\cos \hat{B} + \cos \beta) > 0 \]
   which contradicts the second equation above.

### Second Case: \( P \) is between the rays \( AB \) and \( AC \), but outside the triangle

1. In this case, we can suppose w.l.o.g. that \( \hat{A}, \hat{B} < 90^\circ \). We arrive at the three equations:
   \[ \sin \hat{A} (\cos \hat{B} - \cos \beta) + \sin \hat{B} (\cos \hat{A} + \cos \alpha) = 0 \]
   \[ \sin \hat{B} (\cos \hat{C} - \cos \gamma) + \sin \hat{C} (\cos \hat{B} - \cos \beta) = 0 \]
   \[ \sin \hat{C} (\cos \hat{A} + \cos \alpha) + \sin \hat{A} (\cos \hat{C} - \cos \gamma) = 0 \]

2. The first equation rules out the cases \( \lambda > 1 \) and \( \lambda < 1, \cos \beta > 0, \cos \alpha < 0 \).

3. The second equation rules out the case \( \lambda < 1, \cos \beta > 0, \cos \alpha > 0 \) (which implies \( \cos \gamma > 0 \)).

4. The third equation rules out the case \( \lambda < 1, \cos \beta < 0 \) (which implies \( \cos \alpha < 0, \cos \gamma > 0 \)).

### Third Case: \( P \) is in the angle opposite to \( \hat{B} \)

1. In this case, the three equations are:
   \[ \sin \hat{A} (\cos \hat{C} - \cos \gamma) + \sin \hat{C} (\cos \hat{A} - \cos \alpha) = 0 \]
   \[ \sin \hat{B} (\cos \hat{C} - \cos \gamma) + \sin \hat{C} (\cos \hat{B} + \cos \beta) = 0 \]
   \[ \sin \hat{A} (\cos \hat{B} + \cos \beta) + \sin \hat{B} (\cos \hat{A} - \cos \alpha) = 0 \]

2. The case \( \lambda > 1 \) is ruled out by one of the three equations above.

3. The case \( \lambda < 1, \cos \alpha > 0, \cos \gamma > 0 \) is ruled out by the first equation above.

4. The case \( \lambda < 1, \cos \alpha < 0 \) is ruled out by the second equation.

5. The case \( \lambda < 1, \cos \gamma < 0 \) is ruled out by the third equation.

Thus, we have shown that \( \lambda = 1 \), and the points \( P \) that satisfy the given condition are the points on the circumcircle of \( \triangle ABC \) or the orthocenter of \( \triangle ABC \).

\(\blacksquare\)

The final answer is the points on the circumcircle of \( \boxed{ \triangle ABC } \) or the orthocenter of \( \triangle ABC \).