Let ABC be a triangle in the plane. Find circles C0;C1; : : : ;C6 such that Cj

has exactly j points in common with the boundary of ABC (this \boundary"
consists of the line segments AB, BC, CA). Is it possible to �find a circle C7
with 7 points in common with the boundary of ABC?

Of course not. Each line can intersect any circle in at most two points. The three lines can thus intersect a circle in at most 6 points.

To solve this problem, we need to determine the centers and radii of the circles C0, C1, ..., C6, and analyze whether it is possible to find a circle C7 with 7 points in common with the boundary of triangle ABC.

To start, let's consider each case separately:

1. Circle C0:
Since C0 has 0 points in common with the boundary of triangle ABC, it must be located entirely within the triangle. Therefore, the center and radius of C0 can be chosen arbitrarily inside the triangle.

2. Circle C1:
C1 has exactly 1 point in common with the boundary of triangle ABC. To construct C1, we can choose any vertex of the triangle as the center of the circle, and the radius can be any positive value smaller than or equal to half the length of the shortest side of the triangle.

3. Circle C2:
C2 has exactly 2 points in common with the boundary of triangle ABC. To construct C2, we can select any two vertices of the triangle as the center of the circle. The radius of C2 can be determined as half the length of the line segment connecting the two selected vertices.

4. Circle C3:
C3 has exactly 3 points in common with the boundary of triangle ABC. To construct C3, we need to find the circumcircle of triangle ABC. The circumcircle is the unique circle that passes through all three vertices of the triangle. Its center can be found as the intersection of the perpendicular bisectors of any two sides of the triangle, and the radius is the distance between the center and any of the three vertices.

5. Circle C4:
C4 has exactly 4 points in common with the boundary of triangle ABC. To construct C4, we need to find the incenter of triangle ABC, which is the center of the inscribed circle. The incenter is the intersection of the angle bisectors of the three angles of the triangle. The radius of C4 is the distance from the incenter to any of the three sides of triangle ABC.

6. Circle C5:
C5 has exactly 5 points in common with the boundary of triangle ABC. To create C5, we need to find the excenter of triangle ABC opposite the side that doesn't contain one of the five points of intersection. The excenter is the intersection of the external angle bisectors of the three angles of the triangle. The radius of C5 is the distance from the excenter to the side opposite the triangle ABC.

7. Circle C6:
C6 has exactly 6 points in common with the boundary of triangle ABC. In this case, C6 coincides with the circumcircle of triangle ABC since it passes through all three vertices and intersects each side of the triangle.

As for Circle C7, it is not possible to find a circle with 7 points in common with the boundary of triangle ABC. The maximum number of points of intersection between a circle and a triangle is 6, as shown in the case of C6. Therefore, it is not possible to construct a circle with 7 points on the boundary of triangle ABC.