- 03 Nov '02 11:09this would be the same problem as the other one, except that the

number of lines has to be calculated first: the combination of n

elements in groups of two: n x (n-1) / 2

And we substitute that into the formula of the other problem

number of regions = 1 + [N x (N+1)] / 2 (the other problem)

with N= n x (n-1) / 2

gives : 1 + [ n x (n-1) x ( n² - n + 2)] / 8

lets try with

2 points: 1 line, 2 regions

3 points: 3 lines 7 regions

4 points: 6 lines, 22 regions

etc.....

or am I talking nonsens (again)?

sin. - 03 Nov '02 16:19you are right (nothing to be afraid of, LOL). I was a bit naive. There

are two important differences with the first problem:

1) In the first problem I could imagine the circle outside ALL

intersections. So the area's between intersections laying outside the

circle have to be deducted.

2) in the first problem, all lines could have intersected at different

points; Here there are a lot of coinciding intersections.

Too complicated to answer just like that (at least starting from this

angle). Back to the study .... Anyone else? - 24 Dec '02 10:40for n is odd, the terms follow the pattern

n,n+1,3n+1,5n+1 etc (because you always have that irritating central region)

and for n is even the terms follow the pattern

n, 2n, 4n, 6n, 8n etc.

its a compound series, but thats as far as my memory will get me. i tried to express it in terms of n but i get all fuzzy and irritated. there must be a mathematician or two on this site who see these things when they watch the sky...