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Pts of Intersection
By Anonymous on Tuesday, January 23, 2001 - 02:55 am:

Hi all,

I am amazed at the level of intelligence at this one gathering! I'd like to tap your expertise please. I came across this question to which I could not answer and hope you can help:

How many points of intersection would there be for 2000 straight lines drawn on the same plane?

Thanks
Ian

By James Lingard (Jchl2) on Tuesday, January 23, 2001 - 11:36 am:

Hi Ian,

The answer to your question really depends on which lines you're talking about. Think about the following different situations:

(i) If all the lines are parallel then there won't be any points of intersection.

(ii) If all the lines are different lines through one point, then there will be precisely one point of intersection.

(iii) If 1999 of the lines are parallel and the other line isn't, then there will be 1999 points of intersection.

(iv) If no two of the lines are parallel, and the lines are arranged so that never more than two lines meet at any particular point (this can be done) then there will be 1999000 points of intersection. (Can you see why this is?)

(v) If any two of the lines are the same line, then you might say that there are infinitely many points of intersection (although I suppose you might think that this was not really 2000 lines).

If all the lines are different, then there can't be any more than 1999000 points of intersection (can you see why this is the case?), but I can't see how to work out exactly which possibilities there are between 1 and 1999000 - for instance, it's impossible to arrange the lines so that there are exacly n points of intersection if n is any number between 2 and 1998. Maybe someone else knows how to do this?

Anyway, I hope that helps,
James.

By Anonymous on Tuesday, January 23, 2001 - 11:40 am:

Of course if you allow one line to lie on top of another there will be an infinite number of intersection points

By Anonymous on Thursday, January 25, 2001 - 06:41 am:

Hi James,

Thanks so much. Your treatment of the solution is great. I think that although the question was worded as such, they might just be alluding to your fourth answer. However, I cannot see how to get that; I know the number is large but why 1999000? Thanks

Ian

By James Lingard (Jchl2) on Friday, January 26, 2001 - 04:46 pm:

OK, the way we'll work it out is to draw the lines (in a hypothetical sense) one by one, and count the number of new points of intersection we get each time we draw a new line.

First we draw one line. Obviously there aren't any points of intersection yet.
Now we draw the second line so that it crosses the first one, and we get one point of intersection.
When we draw the second line, so that it crosses both of the first two, we get two new points of intersection (in addition to the one we already had).
When we draw the third line, so that it crosses all of the first three, we get three new points of intersection (giving us a total now of 6 points of intersection).

And so it continues like this. When we draw the nth line, there are n-1 lines already down, so that the nth line crosses each of these n-1 lines, giving us n-1 new points of intersection.

When we draw the final (2000th) line there are already 1999 lines down, so we get 1999 new points of intersection.

So, altogether we have 1 + 2 + 3 + ... + 1999 points of intersection.

Now to work out what this sum is, we can use the formula n(n + 1)/2 which gives us the sum of the numbers 1 up to n. In our case, we need to use this formula with n = 1999, so we get

1 + 2 + 3 + ... + 1999 = 1999×(1999 + 1)/2 = 1999×1000 = 1999000.

Hope that helps,
James.

By Anonymous on Monday, January 29, 2001 - 01:01 am:

Thanks so much James. Yeah, got it.

Ian