On the plane are marked $2n+1$ points, with no three points collinear and no four on a circle. Prove that there exists a circle passing through three of these points, such that $n-1$ of the remaining points lie inside it, and the other $n-1$ lie outside.
Are there any ideas? Or the solution.
We can assume that one point is the origin so the set of points is $\{0, p_1, \ldots, p_{2n+2}\}$. Let $q_k = p_k/\lVert p_k \rVert^2$ be the inversion of $p_k$ for $k\in\{1, \ldots, 2n+2\}$. Pick two points $q_a, q_b$ such that there are $n$ points from the inverted set on either side of the line through $q_a$ and $q_b$. (For example pick any $q_a$ and sweep a line through this point over the plane to find a matching $q_b$.) Note that this line avoids $0$ since the points are in general position. Now inversion is an involution that turns circles through $0$ into lines avoiding $0$ and vice versa. So $\{0, p_a, p_b\}$ is a set with the required property.
Incomplete answer, response to your "Any ideas?"
First two ideas which come up are combinatorics and induction.
Combinatorics: How many circles are possible? (A lot).
Every three non-collinear points define a circle. So there are $\begin{pmatrix}{2n+1\\ \ \ \ \ 3} \end{pmatrix}$ distinct circles possible = $ \frac{(2n+1)(2n)(2n-1)}{3!}$ = $\frac{8n^3 - 2n}{6}$ distinct circles.
$n = 0 \ \ $ gives only one point and no circles so is not reasonable given the question constraints.
$n = 1 \ \ \rightarrow 2n+ 1 = 3\ $ so there is one circle with zero points inside and zero outside, correct.
$n = 2 \ \ \rightarrow 2n+ 1 = 5$ so there are $\begin{pmatrix}{5\\ 3} \end{pmatrix} = 10 \ $ distinct circles possible. Two points are not on the circle. We want to show that one of these circles has one detached point in the interior and one in the exterior.
I don't have a definitive answer yet but am considering something with triangles: If ABC define a circle they also define a triangle. Anything interior to the triangle is also interior to the circle. Careful, exterior is not always true. But given two line segments AD and BE intersecting at C, anything interior to triangle ABC and circle ABC is exterior to triangle DEC and circle DEC.
With $n = 2$ and five points there are $\begin{pmatrix}{5\\2} \end{pmatrix} = 10 \ $ lines possible. An argument could be made about points either side of each line. To consider.
Induction: The case for $n = 1$ forms a base step and if you can prove the case for $n = 2$ you have a methodology; then work inductively adding two points.
Clearly I can draw a line that separates the points into two groups of any size I choose. Draw a line to separate the group into groups of $n{+}2$ and $n{-}1$ points. The first group is intended to be inside and on the circle (perimeter); the latter group outside.
Turn the line into a circle of very large radius enclosing the $n{+}2$ points. Then reduce the radius until the circle touches a point; this is our first candidate perimeter point. Maintaining the circle through that point, continue reducing radius until you touch a second perimeter point. Continue contracting to touch a third perimeter point.
In that final contraction it is possible that the circle will instead touch a point in the designated outer-points group, in which case contract the circle past the two initial perimeter points found, keeping the new perimeter point until another (fourth, I guess) perimeter point is found, then expand the circle again to retrieve one of the two original perimeter points (leaving the other one switched to being an outer point).