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Lines Partitioning a Plane


cDate: 03/09/98 at 01:32:27
From: Curtis Regimbald
Subject: Problem solving

A plane is partitioned into 2 regions by 1 line, and into 4 regions by 
2 intersecting lines. Into how many disjoint regions do 5 coplanar 
lines partition the plane, if no 2 lines are parallel and no 3 of them 
are congruent?

I have no idea what they want or how to solve it.


Date: 03/09/98 at 11:51:02
From: Doctor Anthony
Subject: Re: Problem solving

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

Suppose the number of regions is given by f(n) when there are n lines 
drawn on the plane. 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 2 
parts. It therefore adds n+1 more regions to those present. Therefore:

    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, we get         
    f(n) - f(0)  = SUM(1 to n)       cancellations 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,
which they are.

With n = 5, we get 5 * 6/2 + 1 = 16 regions.

So 5 lines will divide the plane into 16 regions.

-Doctor Anthony, The Math Forum
 http://mathforum.org/dr.math/   
    
Associated Topics:
High School Equations, Graphs, Translations

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