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Recently I came through a book of Arthur Engel which mentioned a problem called Sylvester Problem which states that-

A finite set $S$ of $n$ points in the Euclidean Plane has the property that any line through two of them passes through a third. Show that all the points lie on a line.

In Engel's book, the problem has been solved using the so called Extremal Principle (for details of the proof see Problem Solving Strategies by Arthur Engel).

But I think that the problem is trivial. My solution is as follows.

Let $A_i$ denote the points in the plane $1 \leq i \leq n$. Now choose any three points, say $A_1$ , $A_ 2$ and $A_3$. By hypothesis there is a straight line through these three points, let it be $L_1$ . Now choose the points $A_2$ , $A_3$ and $A_4$. Again a straight line passes through the three points. Call this line $L_2$. Now since two lines cannot intersect in more than one point, $L_1$ and $L_2$ must coincide. Now applying induction the result is proved.

The result is still valid in planes other than Euclidean Planes in which two straight lines cannot intersect in more than one point.

Now my question is if the theorem is that much trivial (if my proof is correct), why even bother about calling it as a significant theorem?

William Hilbert
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1 Answers1

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The proposed solution starts like this: Now choose any three points, say $A_1$ , $A_ 2$ and $A_3$. By hypothesis there is a straight line through these three points.

The hypothesis of the problem does not say that for any three points $A_1,A_2,A_3$ in $S$ there is a line through the three points. It says that for any two points, say $A_1$ and $A_2$ in $S$, there exists a third point $Q$ in our set $S$ such that $A_1$, $A_2$, and $Q$ are collinear.

Remark: Let $S$ be the set of points of any finite projective plane. Then any two distinct lines have exactly $1$ point in common. Every line contains the same number $n\ge 3$ of points, and certainly not all points in such a plane are collinear.

André Nicolas
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  • But the statement that I have borrowed from Engel doesn't seem equivalent to what you have told. – William Hilbert Apr 15 '14 at 06:33
  • The hypothesis of the problem just says that every line through two of our points contains a third point in our set. If you are familiar with the Fano Plane, or any other finite projective plane, you will know that (in the Fano Plane case) every line contains exactly $3$ points. – André Nicolas Apr 15 '14 at 06:39
  • In that case let me state the argument like the following. Let $A_1$ and $A_2$ be two points on the line $L$. Now by hypothesis this $L$ passes through some $A_i \in S$. Call this $A_i$ as $A_3$ then proceed with my argument. Is there anything which I am still missing? – William Hilbert Apr 15 '14 at 06:46
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    The phrase "any line through two of them passes through a third" means that given any two points, **there is a third such that $\dots$. – André Nicolas Apr 15 '14 at 06:46
  • And I think I have not contradicted that hypothesis in the last modification. Have I? – William Hilbert Apr 15 '14 at 06:49
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    About the comment in which you write "call this $A_i$ as $A_3$," let $A_4$ be a fourth point. The condition then says that there is an additional point of $S$ on the line $A_1A_4$, also an additional point of $S$ on $A_2A_4$, also one on $A_3A_4$. This does not get us towards all points being collinear. – André Nicolas Apr 15 '14 at 06:51
  • @AWertheim: Exactly. Thank you very much. – William Hilbert Apr 15 '14 at 06:55