# How to divide a set of $n$ points of the plane into two subsets so that $MA_1+ MA_2+ ...+MA_k= MB_1+ MB_2+ ...+MB_{n-k}$ for some point $M$?

Definition. If a some set of $$n$$ points of the plane can be divided into two subsets $$\{ A_1, \; A_2, \; ..., \; A_k\}$$ and $$\{ B_1, \; B_2, \; ..., \; B_{n-k}\}$$ so that there is a point $$M$$ of the plane such that $$MA_1+ MA_2+ ...+MA_k= MB_1+ MB_2+ ...+MB_{n-k}$$, then we shall say that there exists a good partition of the set of these $$n$$ points of the plane.

a) Prove that for every set of $$n \ge 2$$ points of the plane there is at least one good partition.

b) Prove that if $$n$$ is even then for every set of $$n \ge 2$$ points of the plane there is at least one good partition such that both subsets of points contain $$\frac{n}{2}$$ points each.

My work. I solved the problem for cases when $$n=2$$ and $$n=4$$. If $$n = 2$$ then everything is easy: the point $$M$$ is an arbitrary point of the perpendicular bisector of a line segment $$A_1B_1$$. If $$n = 4$$ then we randomly denote four points by letters $$A_1,\; A_2, \;B_1, \; B_2$$. Let $$a$$ be a real number such that $$a>A_1B_1+A_1B_2+A_1A_2$$. We construct an ellipse $$w_1$$ with foci at points $$A_1$$ and $$A_2$$ and with semi-major axis $$a$$. Then all two points $$B_1, \; B_2$$ are inside this ellipse. We construct an ellipse $$w_2$$ with foci at points $$B_1$$ and $$B_2$$ and with semi-major axis $$a$$. These ellipses $$w_1$$ and $$w_2$$ will intersect at two points. Let $$M$$ be a intersection point of ellipses $$w_1$$ and $$w_2$$. Then $$MA_1+ MA_2=2a= MB_1+ MB_2$$. I have no idea how to solve the problem in general.

Let your points be $$p_1, \ldots, p_n$$. If $$n$$ is even, take a vector $$v$$ such that the dot products $$p_i \cdot v$$ are all distinct, sort your points according to these dot products, and take the least $$n/2$$ values for one subset ($$A$$) and the greatest $$n/2$$ for the other ($$B$$). Note that if $$t$$ is sufficiently large and positive, all points in $$A$$ are farther from $$tv$$ than are all points in $$B$$; the reverse is true for $$-tv$$. Use the Intermediate Value Theorem to show that you can take $$M = t v$$ for some real $$t$$.
EDIT: To clarify, the function to use the Intermediate Value Theorem on is the sum of distances from $$tv$$ to the points in $$A$$ minus the sum of distances from $$tv$$ to the points in $$B$$. It is easy to prove that this is continuous, using the fact that the sum or difference of continuous functions is continuous.
If $$n$$ is odd, let $$A$$ consist of $$p_1$$ and the $$(n-1)/2$$ closest points to $$p_1$$ (breaking ties arbitrarily), $$B$$ the farthest $$(n-1)/2$$. For $$M = p_1$$ the sum of distances to points in $$A$$ is $$\le$$ the sum of distances to points in $$B$$. For $$M$$ very far away the sum of distances to points in $$A$$ is greater. Again use the Intermediate Value Theorem.