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#1 2008-07-03 04:51:04
- Daniel123
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- Registered: 2007-05-23
- Posts: 663
Points of intersection
How many points of intersection are there on a plane containing n straight lines? (assume none of the lines are parallel, and that no more than two lines intersect at any one point)
Last edited by Daniel123 (2008-07-03 04:53:42)
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#2 2008-07-03 05:04:29
- ZHero
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- Registered: 2008-06-08
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Re: Points of intersection
Total number of points of intersection is nC2
i.e. n!/[2!(n-2)!]
If two or more thoughts intersect, there has to be a point!
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#3 2008-07-03 05:17:41
- Daniel123
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- Registered: 2007-05-23
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Re: Points of intersection
Total number of points of intersection is nC2
i.e. n!/[2!(n-2)!]
Can you prove that is the case?
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#4 2008-07-03 05:35:32
- mathsyperson
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- Registered: 2005-06-22
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Re: Points of intersection
To put it another way, when you have n lines, then you will have the (n-1)th triangular number of intersections.
None of the lines are parallel, so any two lines are guaranteed to intersect exactly once.
None of the intersections involve more than 2 lines, so any two lines will have a unique intersection point.
Knowing that, you can think of it in the same way as the handshake problem.
- With one line, there's nothing for it to intersect with, so there are no intersections.
- When the second line gets added it intersects with the first, which makes one.
- The third line intersects with the other two (and in different places), making three intersection when you count the one we already have.
- The fourth line intersects with the three we already have, bringing the total up to 6.
...
- The nth line intersects with the (n-1) already drawn, which adds n-1 to the total.
So to be technical, n lines will create
intersections.But since 1-1=0, this is the same as
(ie. the (n-1)th triangular number)Why did the vector cross the road?
It wanted to be normal.
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#5 2008-07-03 05:43:36
- Daniel123
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- Registered: 2007-05-23
- Posts: 663
Re: Points of intersection
That's what I was looking for. When you add an extra line, it increases the number of intersections by the number of lines there were before the extra line was added, as the new line intersects each of the other lines exactly once (as you said). If N(x) is the number of intersections of x lines, N(n) = N(n-1) + n-1. If the intitial n is replaced by n-1, n-2 etc it becomes clear that N(n) = 1 + 2 + 3 + ... n-1, which is just an arithmetic sequence.
Last edited by Daniel123 (2008-07-03 05:49:40)
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#6 2008-07-03 06:12:27
- ZHero
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- Registered: 2008-06-08
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Re: Points of intersection
Yo man..
1+2+3+...+(n-1)
= (n-1)n/2
= n(n-1)(n-2)(n-3)...(1)/[2(n-2)(n-3)...(1)]
= n!/[2!(n-)!]
= nC2 !!
And you know that an extra line contributes to one extra point of intersection to the previous no. of points of intersection!!
Last edited by ZHero (2008-07-03 06:16:55)
If two or more thoughts intersect, there has to be a point!
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