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
