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/
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