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

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David Bernier

Posts: 3,892
Registered: 12/13/04
Re: Collinear
Posted: May 7, 2013 6:24 AM
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On 05/07/2013 04:39 AM, William Elliot wrote:
> 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?

You're right. I'm nowhere near solving this right now.

>> 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.

Jesus is an Anarchist. -- J.R.

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