The Happy Ending Problem (named so by Paul Erdős as it led to the marriage of his friends Esther Klein and George Szekeres) was originally stated by Esther Klein in the following manner:
There are five points on a flat surface, no two of which are coinciding and no three of which are on a straight line. Prove that four of these points are vertices of a convex quadrilateral.
This simple theorem has a simple proof but, as the complex concept called "convex hull" is involved, the proof could be unintelligible to non-mathematicians, high school students inclusive. So, here is the latest proof of mine, in which you can meet the familiar points and lines only.
Let us start with the observation that any three of those five points form a triangle and let us call the other two points extra points. Where can the extra points be? There are five cases to consider.
Case 1. There exists at least one extra point in an OK part of the plane (see Fig. 1 below). The convex quadrilateral is easy to construct.
There are five points on a flat surface, no two of which are coinciding and no three of which are on a straight line. Prove that four of these points are vertices of a convex quadrilateral.
This simple theorem has a simple proof but, as the complex concept called "convex hull" is involved, the proof could be unintelligible to non-mathematicians, high school students inclusive. So, here is the latest proof of mine, in which you can meet the familiar points and lines only.
Let us start with the observation that any three of those five points form a triangle and let us call the other two points extra points. Where can the extra points be? There are five cases to consider.
Case 1. There exists at least one extra point in an OK part of the plane (see Fig. 1 below). The convex quadrilateral is easy to construct.
Fig. 1
Case 2. There exist two extra points in two ? parts of the plane. (see Fig. 2 below). The convex quadrilateral is easy to construct.
Fig. 2
Case 3. There exist two extra points in one ? part of the plane (see Fig. 3 below). The line connecting those points may intersect zero or two sides of the triangle. Connect those extra points to an unintersected side of the triangle and there it is, our convex quadrilateral.
Fig. 3
Case 4. There exists two extra points inside the triangle. There is no need to draw here, the reasoning is the same as in Case 3. The only difference in Case 4 is that the line between the extra points always intersects two sides of the triangle.
Case 5. There exist one extra point inside the triangle and one in ? part of the plane (see Fig. 4 below). The line connecting those points will always intersect two sides of the trianble. Start your "walk" from the extra point outside the triangle to the one inside it. Take the points of the side you cross first, connect them to the extra points and there you have it, our convex quadrilateral.
Case 5. There exist one extra point inside the triangle and one in ? part of the plane (see Fig. 4 below). The line connecting those points will always intersect two sides of the trianble. Start your "walk" from the extra point outside the triangle to the one inside it. Take the points of the side you cross first, connect them to the extra points and there you have it, our convex quadrilateral.
Fig. 4
We have just exhausted all possible arrangements of a triangle and two points in a plane and proved that a convex quadrilateral can be built in each and every one of them. Q.E.D.
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