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Topic: Collinear
Replies: 9   Last Post: May 8, 2013 1:12 PM

 Messages: [ Previous | Next ]
 William Elliot Posts: 2,383 Registered: 1/8/12
Re: Collinear
Posted: May 7, 2013 4:39 AM

On Tue, 7 May 2013, David Bernier wrote:
> On 05/07/2013 03:48 AM, William Elliot wrote:

> > Let C be a collection of n points with the property that
> > any line L with two points of C on it, has a third point
> > of C on it.
> >
> > How is it that C is collinear, ie all points of C are
> > on a single line?

>
> I think this can be proved by induction on 'n' the
> number of points.
>
> For n >= 3, let P(n) denote the statement:
> "In any collection C of n points in the plane,
> if C is such that any line L with two points
> of C on it has a third point on it (i.e., L),
> then the set C consists of collinear points."
>
> P(3) is obvious.
>
> Then, for any n >= 3, we want to prove:
> P(n) implies P(n+1).
>
> If we can do that, we're done.
>
> Sketch: assume P(n) and let C be any collection
> of n+1 points in the plane. Let Q be some point
> in C and let C' = C \ {Q}.
>
> Then C' has n points. By the induction hypothesis,
> P(n) is assumed.

> xxxx I dunno ...

Why whould C' have the desired property?

> I would try to show that
> P(3) implies P(4), so try to show P(4) using
> the property (assumed) P(3) ...

P(4) is easy without induction. Pick a,b from S and thus there's
some third c in S with collinear { a,b,c }. Now for d, the fourth
point, the line ad must contain either b or c.
Thus the lines ad and abc coincide and S is collinear.

With more to-do, P(5) can proved directly without induction.

Date Subject Author
5/7/13 William Elliot
5/7/13 David Bernier
5/7/13 William Elliot
5/7/13 David Bernier
5/7/13 LudovicoVan
5/7/13 Robin Chapman
5/7/13 Barry Schwarz
5/8/13 David C. Ullrich
5/7/13 fom
5/7/13 fom