Let $A, B, C, D$ and $E$ be five points in the plane in general position (that is, such that no three of them are collinear, and no four of them are concyclic). A separator is defined as a circle passing through three of these points, such that one of the two remaining points is located inside the circle, and the other outside. How many separators are there?
We quickly realize that if we try to tackle the problem naively, we will soon end up with perfectly sordid case studies. So we stop for a minute to think about how to simplify this case study as much as possible. To do this, we would like to get rid of as many circles as possible, which encourages us to invert through one of our points, say $A$. Another way to say the same thing is that these separation questions will be simpler with a point at infinity, as it will always be clear whether this point is separated from the others or not.

Throughout the following, I will call $ABC$ the circle passing through the three points $A, B$, and $C$, and the same for other circles. Let's therefore look at the effect of our inversion on the separators. We check that the inversion sends a separator passing through $A$, say $ABC$, to a line $B^{\prime} C^{\prime}$ separating $D^{\prime}$ and $E^{\prime}$: the segment $DE$, which intersects $ABC$ at exactly one point (distinct from $A$), is sent to an arc of a circle intersecting the line $B^{\prime} C^{\prime}$ at exactly one point. Conversely, we check in the same way that such a line is sent back to a separator.

The second case is that of separators not passing through $A$. Suppose then that $BCD$ separates $A$ and $E$. The segment $AE$ intersects $BCD$ at exactly one point (distinct from $A$), so the image of the segment, which is a ray starting from $E$, intersects the circle $B^{\prime} C^{\prime} D^{\prime}$ at exactly one point: $E^{\prime}$ is therefore inside this circle (whether $E$ is inside or outside $BCD$). Conversely, if $B^{\prime} C^{\prime} D^{\prime}$ contains $E^{\prime}$, we check in the same way that $BCD$ is a separator.

We have therefore reduced the problem to the following: we consider four points in the plane in general position, and we want to count the number of lines separating two of these points, as well as the number of circles passing through 3 points containing the 4th. There are two cases to consider. Case 1: if one of the points, say $D$, is inside the triangle $ABC$, then there are three such lines, $AD, BD$, and $CD$, and one such circle, $ABC$ (it is the only one by an angle argument). Case 2: if the points are the vertices of a convex quadrilateral. Then there are two lines that work, the diagonals, and two circles: there must be two vertices, say $A^{\prime}$ and $C^{\prime}$, such that the sum of the angles at these two vertices is strictly greater than $180^{\circ}$ (since the points are not cocyclic), and by an angle argument the only circles that work are $A^{\prime} B^{\prime} D^{\prime}$ and $C^{\prime} B^{\prime} D^{\prime}$.

Bonus! An exercise not done in class, but already typed before the session, so I leave it here.