Circle Regions

Date: 01/28/2001 at 06:24:07
From: Anonymous
Subject: Regions in a circle

What is the maximum number of regions you can have with n chords in a 
circle?

I've already found:

With 0 chords you have 1 field
With 1 chord you have  2 fields; 1 more
     2                 4         2 more
     3                 7         3  
     4                11         4
        etc.

Now I need a formula and a proof why this is so. Can you help me?

Thanks.

Date: 01/28/2001 at 08:43:45
From: Doctor Anthony
Subject: Re: Regions in a circle

We can find a general formula for the number of regions when the 
interior of a circle is divided by n lines.

Suppose the number of regions is given by f(n) when there are n lines 
drawn in the circle. Now draw one more line cutting all the other n 
lines. There are n points on the additional line, and so this line 
must traverse n+1 of the available f(n) regions, dividing each into 
two parts. It therefore adds n+1 more regions to those present. Thus

      f(n+1) = f(n) + n+1

We can write  f(n+1) - f(n) = n+1  and now form a succession of 
equations as follows:

      f(1) - f(0) =  1
      f(2) - f(1) =  2
      f(3) - f(2) =  3
     ...................
     ...................
  f(n-1) - f(n-2) =  n-1
  f(n)   - f(n-1) =  n
 -------------------------         adding all the equations, 
    f(n) - f(0)  = SUM( 1 to n)   (note cancellation between lines)

   f(n) - 1  = n(n+1)/2

          f(n) = n(n+1)/2  + 1

Check if this is correct  n=0  gives  f(0) = 1
                          n=1  gives  f(1) = 2
                          n=2  gives  f(2) = 4  and these are correct.

With n = 4 we get   4 x 5/2 + 1  =  11  regions.

And so on.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   

Date: 04/19/2003 at 05:00:58
From: Sheila
Subject: Regions in a circle

We drew a circle and then put two points on the line and joined the 
points. The circle is divided into two parts. If there are 4 points 
(not evenly spaced) and they are each joined to all the other points,
the circle is divided up into 8 regions. 

There is a conjecture that says that if the process is repeated, with 
more points along the edge of the circle, not evenly spaced, then a 
rule for the maximum number of regions is 2 to the power of (the 
number of points take one) r=2^(n-1) where ^ = to the power of.

For 2,3,4,5,6 points, they follow the formula, but for 7 points, it 
doesn't; there are 57 regions, and if it followed the rule, there'd 
be 64.  Am I not counting them properly? Is there a simple way of 
understanding how many regions there are, because there must be a 
constant formula, shouldn't there be?

Date: 04/19/2003 at 05:30:48
From: Doctor Anthony
Subject: Re: Regions in a circle

Regions of a Circle Cut by Chords to n Points
----------------------------------------------
n points are distributed round the circumference of a circle and each 
point is joined to every other point by a chord of the circle.  
Assuming that no three chords intersect at a point inside the circle 
we require the number of regions into which the circle is divided.

With no lines the circle has just one region. Now consider any 
collection of lines. If you draw a new line across the circle which 
does not cross any existing lines, then the effect is to increase the 
number of regions by 1. In addition, every time a new line crosses an 
existing line inside the circle the number of regions is increased by 
1 again.

So in any such arrangement

number of regions = 1 + number of lines + number of interior 
intersections

                  = 1 + C(n,2) + C(n,4)

Note that the number of lines is the number of ways 2 points can be 
chosen from n points. Also, the number of interior intersections is 
the number of quadrilaterals that can be formed from n points, since 
each quadrilateral produces just 1 intersection where the diagonals 
of the quadrilateral intersect.

Examples:

        n=4    Number of regions = 1 + C(4,2) + C(4,4) =  8
        n=5    Number of regions = 1 + C(5,2) + C(5,4) = 16
        n=6       "         "    = 1 + C(6,2) + C(6,4) = 31
        n=7       "         "    = 1 + C(7,2) + C(7,4) = 57

Note that formula 2^(n-1) starts to go wrong at n=6

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/