9.4. Ramsey theorem in Geometry

The problem: We have 5 point on the plane and any 3 of them are not collinear. Can we prove that 4 point of the 5 makes always a convex polihedron?

Let the 5 points be A, B, C, D and E.

Case 1. The convex hull of the 5 point is a convex rectangle. The we are ready.

Case 2. The convex hull is a convex pentagon. Then joining any 4 points in consecutive way we get a convex rectangle.

Case 3. The convex hull is triangle. Then the convex hull contains 2 points. Let these points be D and E.

If we connect these point with a line e, then e will cross the triangle in a such a way that 2 point will be on the same side (A and B).

Now, A, B, D and E form a convex polihedron.

What about more points?

Theorem 9.9. (Klein, Szekeres, Erdős, 1933.): If 9 points are in the plain in "general position", then there are always 5 points among them which form a convex polyhedron.

More general?

There are given points in the plain, where , integer, Are there always k points among them which form a convex polyhedron?

Erdős-Szekeres Conjecture (1934): The answer is affirmative.

It has been proven that the conjecture is true if .

Theorem 9.10. (Erdős, Szekeres): Let be the minimum cardinality of a set of general position points in the plain, in which always exist k points which form a convex polyhedron, then