5. The eight points below are the vertices and midpoints of the sides of a square. We want to draw a number of circles through the points in such a way that each pair of points lies on (at least) one of the circles.

Find the smallest possible number of circles with which this can be done.
5. Of the eight points, we color four black and the other four white, as indicated in the left figure. The circle through the four black points we call $C_{1}$ and the circle through the four white points we call $C_{2}$. If a pair of points lies on a circle $C$, we say that $C$ covers that pair.
![](https://cdn.mathpix.com/cropped/2024_04_17_7c5565a512609fc44047g-4.jpg?height=336&width=782&top_left_y=1329&top_left_x=640)

Circle $C_{1}$ covers all pairs of black points and circle $C_{2}$ covers all pairs of white points. It is easy to verify that each of the $4 \cdot 4=16$ pairs consisting of one white and one black point is covered by one of the four circles in the right figure. Together, these six circles thus form a solution.

We will now prove that there is no solution with five (or fewer) circles. First, we note that a circle passing through more than two black points must be identical to $C_{1}$, and a circle passing through more than two white points must be identical to $C_{2}$. A circle is fixed once we choose three points it must pass through.

A circle passing through at most 2 black points covers at most one of the $\frac{4 \cdot 3}{2}=6$ pairs of black points. A solution with five circles must therefore include circle $C_{1}$ (otherwise, at most 5 pairs of black points would be covered). Similarly, the solution must include circle $C_{2}$.

Each of the remaining three circles passes through at most 2 black and at most 2 white points. Thus, such a circle covers at most $2 \cdot 2=4$ pairs consisting of one white and one black point. In total, the five circles cover at most $0+0+3 \cdot 4=12$ such pairs out of the 16 in total. Therefore, the five circles do not form a correct solution.

We conclude that the smallest possible number of circles in a solution is 6.