1. DonationAcolyte
    Now With Added BA
    04 Jul '02
    30 Oct '02 19:36
    For this you will need a pen and paper (a ruler helps as well).

    Draw a circle. Draw n lines across it. How many regions can you divide the circle into? Try
    it for small n, and see if you can spot the pattern.
  2. Donationbelgianfreak
    stitching you up
    08 Apr '02
    30 Oct '02 20:24
    I've already expressed my lack of knowledge in maths, so this has
    got to be wrong. But isn't it just n*2??
  3. DonationAcolyte
    Now With Added BA
    04 Jul '02
    30 Oct '02 23:13
  4. Joined
    01 Dec '01
    30 Oct '02 20:48
    just trying: 1 + nx(n-1)/2 ?
  5. Joined
    01 Dec '01
    30 Oct '02 20:49
    sorry I meant: 1 + nx (n+1)/2
  6. Asheville
    20 Sep '02
    30 Oct '02 21:07

    I'll call 1 + n(n+1)/2 = f(n) for simplicity

    f(3) = 1 + 3(4/2) = 7
    f(4) = 1 + 4(5/2) = 10

    Looks good so far! Nicely done! What was your reasoning? Or did you
    just try to fit an equation to your observations?

  7. Joined
    01 Dec '01
    30 Oct '02 21:17
    sorry: f(4) = 11
    and I can make 11 area's with 4 lines, I believe...
  8. Asheville
    20 Sep '02
    31 Oct '02 17:32
  9. Joined
    29 Jul '01
    30 Oct '02 22:06
    Gilbert, do you have a proof? Can you prove your formula to be true
    for when n = 100 say?

    The Squirrel Lover

    PS. There is more than one way of doing it. There is a particularly
    attractive way of doing it however that does in fact kind of tie in with
    the other thread on arithmetic progressions!!!!
  10. Joined
    01 Dec '01
    30 Oct '02 23:21
    Oh Mark, that has been years ago since I tried things like this. I
    cannot do that just like that. At most try to 'plan' a proof on intuitive
    basis. The formal proof would have to follow. Let me see:

    If I can proof that when it is true for in-1, then it holds also for n, AND
    knowing it is true for say n=1, then it is ok isn't it? That is another
    way of expressing the progression, I think.

    I think the point is to prove that with n-1 non-parallel lines, the n-th
    line creates n NEW regions, because at any (of the n-1) intersection
    (s) the line splitses a region in two, but except from one extreme, the
    others are counted twice (keft and right).

    That's as far as I get right now. Saved by the bedtime-bell. lol.

  11. Joined
    01 Dec '01
    03 Nov '02 11:20
    my math is too rusty. Why don't you give the correct proof (or the
    correct answer should mine be wrong)?

    I decided to stick to my intuition. As Einstein said once: "as soon as
    the mathematicians got involved I didn't understand relativity
  12. Joined
    29 Jul '01
    07 Nov '02 12:30

    f(n) denotes the number of regions in the circle with n lines in it.

    As you say, by drawing one more line there will be n points on that
    line. In other words, of the f(n) regions the line will traverse n+1 of
    them. Each region gets split into two so n+1 regions are added as a
    result of this extra line.

    So before we had f(n), now we have f(n+1) and this extra line has
    given us n+1 extra regions.

    Therefore, f(n+1) = f(n) + n+1

    Now the cunning bit...

    Rearranging the above line will give f(n+1) - f(n) = n+1

    Consider what this is saying (sorry for all the dots, they are necessary
    to line up the columns, if I leave spaces the forum software scrunches
    everything up).

    f(1) - f(0) = 1 (since here n = 0 so n+1 = 0+1 = 1)
    f(2) - f(1) = 2 (since here n = 1 so n+1 = 1+1 = 2)
    f(3) - f(2) = 3 (since...etc)
    f(4) - f(5) = 4

    and so on up to:-

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

    Add up the columns (note how the top left to bottom right diagonal on
    the left hand side of the equals sign cancels, this is what Acolyte
    mentioned in one of his posts where all the terms cancel leaving just
    two, (usually) just the first one and the last one).

    We get f(n) - f(0) = 1+2+3+4+...+(n-1)+n

    The right hand side is simply the sum of the first n integers!!! This is
    where it ties in with the other thread involving Gauss and arithmetic
    progressions!! I love maths! We can use the formula Hotpawn and
    Gilbert gave in that thread (the proof of which follows the same
    pattern as the explanation I gave Jon with n replacing 100).

    So, we have:-

    f(n) - f(0) = n(n+1)/2

    We know f(0) to be 1 so...

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

    Which, adding one to both sides, is the formula that Gilbert arrived at
    by intuition and testing the first few cases.

    f(n) = n(n+1)/2 + 1
  13. Donationbbarr
    Chief Justice
    Center of Contention
    14 Jun '02
    03 Nov '02 11:26
    sintubin's formula can be proven via mathematical induction, if you
    are looking for a proof in the rigorous sense of the term.
  14. Asheville
    20 Sep '02
    30 Oct '02 20:53
    How are we drawing the lines? Through a central point? If so, then the
    figure would have:

    1 lines = 2 regions
    2 lines = 4 regions
    3 lines = 6 regions
    4 lines = 8 regions
    Which would wake the pattern regions = 2n

    Or more simply, just noticing that every line divides 2 regions into 4
    regions (double), we get the same result.

    However, if you can make the like anywhere, the circle with three lines
    in it can have 7 regions, the circle with 4 can have 10, the circle with 5
    can have 14 regions (am I correct) and I fail to see a pattern.

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