Triangle $A_0B_0C_0$ is given in the plane. Consider all triangles $ABC$ such that:
(i) The lines $AB,BC,CA$ pass through $C_0,A_0,B_0$, respectvely,
(ii) The triangles $ABC$ and $A_0B_0C_0$ are similar.
Find the possible positions of the circumcenter of triangle $ABC$.


Austrian-Polish 2000
1. **Understanding the problem**: We are given a triangle \( A_0B_0C_0 \) and need to find the possible positions of the circumcenter of another triangle \( ABC \) such that:
   - The lines \( AB, BC, CA \) pass through \( C_0, A_0, B_0 \) respectively.
   - The triangles \( ABC \) and \( A_0B_0C_0 \) are similar.

2. **Homothety and similarity**: Since \( \triangle ABC \) is similar to \( \triangle A_0B_0C_0 \), there exists a similarity transformation (which can be a combination of rotation, reflection, translation, and homothety) that maps \( \triangle A_0B_0C_0 \) to \( \triangle ABC \).

3. **Homothety**: Specifically, there exists a unique homothety (scaling transformation centered at a point) that maps \( \triangle A_0B_0C_0 \) to \( \triangle ABC \). This is because the lines \( AB, BC, CA \) passing through \( C_0, A_0, B_0 \) respectively, implies that the corresponding sides of the triangles are parallel and proportional.

4. **Circumcenter under homothety**: The circumcenter of a triangle is the point where the perpendicular bisectors of the sides intersect. Under a homothety, the circumcenter of \( \triangle A_0B_0C_0 \) will map to the circumcenter of \( \triangle ABC \). This is because homothety preserves the angles and the relative distances between points.

5. **Conclusion**: Therefore, the circumcenter of \( \triangle ABC \) is the image of the circumcenter of \( \triangle A_0B_0C_0 \) under the homothety that maps \( \triangle A_0B_0C_0 \) to \( \triangle ABC \).

\[
\boxed{\text{The circumcenter of } \triangle ABC \text{ is the image of the circumcenter of } \triangle A_0B_0C_0 \text{ under the homothety.}}
\]