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Counting Regions Formed by Straight Lines


Date: 04/18/98 at 15:00:29
From: Jay Shah
Subject: Infinite divisions of a plane

Dear Doctor Math -

I have been exploring dividing planes into distinct regions with 
straight lines, mainly what happens when more lines are added. The way 
you add them seems to make a big difference in the number of regions 
you come up with. My question is: how many regions are formed by n 
straight lines if no three meet in a single point and no two are 
parallel? 

I have come up with .5n^2+.5n+1, which I think is the maximum number 
of regions possible. Now what happens if parallelism is allowed? What 
happens if concurrence is allowed? What happens if both are allowed? 

I would really appreciate it if you could see what you could come up 
with for this problem. I greatly appreciate any help you can give me. 
Thank you very much.

From a great lover of math,
Jay Shah


Date: 04/18/98 at 15:29:33
From: Doctor Anthony
Subject: Re: Infinite divisions of a plane

Suppose we draw n straight lines on the plane so that every pair of
lines intersects (but no 3 lines intersect at a common point). Into 
how many regions do these n lines divide the plane?

With n = 1 we divide the plane into 2 regions. With n = 2 we have 4 
regions; with n = 3 we get 7 regions. A fourth line will meet the 
other 3 lines in 3 points and so traverse 4 regions, dividing them 
into two parts and adding 4 new regions. In general the nth line will 
add n new regions:

     u(1) = 2
     u(2) = 4
     u(3) = 7
     u(4) = 11   

And so on, where u(n) = number of regions with n lines.

We get the recurrence relationship:

     u(n+1) = u(n) + (n+1)

We get the following chain of equations:

       u(n) - u(n-1) = n
     u(n-1) - u(n-2) = n-1
     u(n-2) - u(n-3) = n-2
     ......................
     ........................
       u(4) - u(3)   =  4
       u(3) - u(2)   =  3 
       u(2) - u(1)   =  2
     -------------------------- 
       u(n) - u(1)   =  2 + 3 + 4 + ..... + (n-1) + n

which we find by summing the equations.

All other terms on the left cancel between rows. We are left with:

     u(n) = u(1) + 2 + 3 + 4 + ....+ n   and   u(1) = 2

Thus:

     u(n) = 1 + [1+2+3+4+...+n]

          = 1 + n(n+1)/2

          = (2 + n^2 + n)/2

So:

     u(n) = (n^2+n+2)/2

If you allow parallel lines and more than 2 lines to intersect at a 
point, the answer becomes undefined.

-Doctor Anthony,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry
High School Sequences, Series

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