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### Counting Regions Formed by Straight Lines

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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

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

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

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