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 unheard of and complex concept of the so called convex hull is involved, the proof could be unintelligible to non-mathematicians, high school students inclusive. So, here is the simplest proof of mine, in which you can meet the familiar points and lines only.
Let us think about the five points as of two lines and an extra point. Line l1 is defined by P1 and P2, line l2 is defined by P3 and P4, and P5 is the extra point. There are only 3 cases to consider:
Case 1. Neither line intersects the segment between the points on the other line. One can build a convex quadrilateral by connecting the points on l1 and l2.
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 unheard of and complex concept of the so called convex hull is involved, the proof could be unintelligible to non-mathematicians, high school students inclusive. So, here is the simplest proof of mine, in which you can meet the familiar points and lines only.
Let us think about the five points as of two lines and an extra point. Line l1 is defined by P1 and P2, line l2 is defined by P3 and P4, and P5 is the extra point. There are only 3 cases to consider:
Case 1. Neither line intersects the segment between the points on the other line. One can build a convex quadrilateral by connecting the points on l1 and l2.
Case 2. Each line intersects the segment between the points of the other line. One can build a convex quadrilateral by connecting the points on l1 and l2.
Case 3. Only one of the lines intersects the segment on the other line. Let us assume without loss of generality that l1 intersects the P3P4 segment of l2. P5 can be in any of the 18 parts of the plane, but fortunately those parts are of only 2 types, O (coming from Other) and S (coming from Same). O means there exists a line dividing the plane so that P5 is in one half of the plane and a triangle consisting of three of the other points exists in the other half such that when P5 is connected to two of its vertices a convex quadrilateral is built (see below the line P2P4 allowing us to build the convex quadrilaterals P2P3P4P5 and P1P2P5P4).
S means there exists a line dividing the plane so that P5 is in one half of the plane and a triangle consisting of three of the other points exists in the same half such that when P5 is connected to two of its vertices a convex quadrilateral is built (see below the line P1P2 allowing us to build the convex quadrilateral P1P2P5P4).
We have just exhausted all possible arrangements of two lines and a point 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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