A simple and interesting proof question

Recently, in algorithm class, the teacher told an interesting proof question.

A finite point set of n points on a plane. Has the following properties: The third point Z can be found on the straight line determined by any two points, X, Y. Try to prove that all points in a are in the same line.

For the proof problem, the most common and systematic method is nothing more than two kinds: Inductive method and disprove method. Other methods, such as synthesis and analysis, are of great relevance to specific problems. If there is no train of thought to solve the problem, these two methods will be a good choice. The following will try to solve this problem in both ways.

One, inductive method.

　　Believe that the person who has studied high school mathematics, no one does not know this famous, but also simple and effective proof method. We will not repeat it here. A proof process is given below.

(1) When |a|≤4, it is clearly established; (|a| indicates the number of midpoint in the collection)

(2) Assuming that when the |a|=k, the above conclusion is established, that is, all points in the same line;

(3) then only need to prove when |a|=k+1, the conclusion is also established;

Take a subset of a ', |a ' |=k, so that the remaining point is X. According to (2) there, a ' is a bit in the same line. Take a ' midpoint y, then there must be another point z in the straight xy. And according to a ' definition, Z can only be in a '. Since two points determine a straight line, A is a bit in the same line.

Is there a problem with this seemingly perfect proof? The landlord has not reacted at first. In fact, there is a big bug in this process. Now let's see where the problem is.

As we all know, the third step of the induction process requires the assumption of the second step, which is also the most difficult part of the mathematical induction method. Here, we assume that all points are in the same line when |a|=k. So he took the conclusion to the third step in a straightforward way. We ignore the hypothetical condition of the topic: The third Point Z can be found in the straight line determined by any two points, X, Y. We write the second step complete should be: When |a|=k, if any two points in A in the line determined by X, Y can find a third point Z, then all points in the same line. Now the problem is clear, a satisfies this nature, does not mean a ' satisfies this nature, only first proves that a ' also has the nature above the proof is established. Although for |a|≥4, we can guess that's right, but it's not easy to prove it.

Next, let's try contradiction to prove it.

Second, to disprove the law.

　　As a matter of fact, many people will think of the absurdity of the problem. Because it's proof of a "all" problem. So, specifically, how to do it?

First, we assume that the points in a are not on the same line. Then for a line L in a, the presence of point D is no longer on the line. According to the nature of a satisfies, there must be three points on l, assuming a, B, C.

Let us first prove a nature.

Over D, perpendicular to L. Obviously, A, B and C must have two points on the same side, assuming A and B. Well, there is distance (d,l) >distance (B,lad). That is, if there is a point not on the line L, we can always find another point, so that the vertical distance of the point to another line is less than a point to the vertical distance of the line L.

With this nature, the proof is basically complete. Since a is a finite point set, and not all points are in a straight line (assuming). Then obviously, the distance to the line must have a non-0 minimum value, that is, there is a point and a straight line so that point to the shortest distance. This is clearly contradictory to the nature of the above. Because we are sure to find another smaller non-0 value.

The certificate is completed.

　　

A simple and interesting proof question