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Need help finding equation

  1. Jun 11, 2010 #1
    Hello all, I feel a bit out of my league here :) if I've posted in the wrong section I'm so sorry...

    I am looking for help finding an equation or formula. When I was in school years ago (before internet and search engines) one of the lessons in geometry was drawing a certain number of dots in a circle, then connecting the dots with all possible lines to get a number. On tests, this is how we were taught to get the answer. Draw the dots and connect them! I was in love with numbers, and figured there HAD to be a better way. I remember hiding away in my bedroom and after a little while and some scribbled paper I came up with a simple formula to get the number of lines.

    I brought this up in class that Monday morning, and my teacher was amazed. She drew example after example on the chalkboard, and asked me what the highest number I tested it out on was. Which I believe was 20-something. She then gathered the other math teachers at the high school and they spent days pouring through books and updating the class on their progess. Then one day she simply told us they found the formula and mine was just a simplified version of it. I was crushed!

    Since the birth of the internet, I figured I'd search for this formula she had never shown me to learn more about when it was created and see the "unsimplified" version. I have come up with nothing. Since my job (sadly) is not math related, and I don't have any idea how to find this or what it would even be called to search for, I thought I'd seek help here. If anyone reading my post is knowledgable in this - I would love to hear from you!

  2. jcsd
  3. Jun 12, 2010 #2
    Hi pinkshoegirl,

    If you have n points in the plane, no three of which are in a straight line, and connect each pair of points with a line, then there are

    [tex]\frac{(n-1) n}{2}[/tex]


    Is this what you mean?
  4. Jun 12, 2010 #3
    Yes exactly! Any idea what that is called? Thanks for the reply :)

    In school I had come up with (x²-x)/2, but same thing. My teacher had told me that the formula they found was not as simplified as mine but never showed it to us... I still find it amazing this simple equation wasn't taught to us, and we all sat there during exams drawing out the lines!
  5. Jun 12, 2010 #4
    One way to look at this result is that it is a binomial cofficient, i.e. the coefficient of [tex]x^2[/tex] in the expansion of [tex](1+x)^n[/tex], usually written [tex]\binom{n}{2}[/tex].

    Another way to look at the same result is as the number of combinations of n objects taken 2 at a time, or the number of handshakes among a group of n people.
  6. Jun 13, 2010 #5
    Impressive figuring that out at school! Another way to write it is

    [tex]\sum_{k=1}^{n-1}k = \left ( \sum_{k=1}^{n}k \right ) - n[/tex]

    where n is a whole number greater than 1. That's just means: add up all the whole numbers from 1 till you come to one less than n, e.g. if n is 4,


    Maybe you noticed that along the way. A nice way to picture how this relates to the


    formula is to draw an n by n square of "dots" like this. With n = 4 again,

    x o o o
    x x o o
    x x x o
    * * * *

    [tex]1+2+3 = \frac{4^2-4}{2} = 6[/tex]
  7. Jun 13, 2010 #6


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    Here's an associated problem which may be what you are remembering. Mark n points around the circumference of a circle and draw all lines connecting the points. What is the maximum number of regions those lines can divide the disk into?

    If you mark the points equally spaced, you will get many lines intersecting in the center but the point is to move them slightly so you never get more than two line intersecting in one point in order to get the maximum number of regions.

    For example, if there is 1 point, there are, of course, no lines and you just have one region- the entire disk.

    If there are two points, there is only one line between them and you have 2 regions.

    If there are three points, there are the three lines connecting them and so 4 regions.

    If there are four points you get a quadrilateral with the two diagonals also drawn- 8 regions.

    It is easy to see that if you have 5 points you get 16 regions and if you have 6 points, you get 32 regions.

    The is often used as an example of the danger of relying on a few "data points" because that looks like powers of two and that the next number should be 64.

    It is not. If you use 7 points, you get 63 regions, not 64.

    The actual formula is a part of the binomial sum:

    [tex]\sum_{i= 0}^5\begin{pmatrix}n-1 \\ i\end{pmatrix}[/tex]
    where [tex]\begin{pmatrix}n \\ i\end{pmatrix}[/tex] is taken to be 0 if i> n.

    As long as n is less than or equal to 6, that is the entire "binomial sum":
    [tex](1+ 1)^{n-1}= \sum_{i=0}^{n-1}\begin{pmatrix}n-1 \\i \end{pmatrix}[/tex]
    so we get a power of 2. If n= 7, we are missing the last, i= 6, term and so get 1 less than [itex]2^6= 64[/itex].
  8. Jun 13, 2010 #7


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    A simple way to see n(n-1)/2 is the number of handshakes of a group of n people:

    Pick (as outsider) one of the n people. This can obviously be done in n ways. Given that person, choose a second, different, person. This can be done in n-1 ways. So n(n-1) is the number of pairs consisting of two distinct people. Divide by 2 to compensate for the fact that the pairs (a,b) and (b,a) are counted as different 'handshakes'.

    In fact, wikipedia has an article about this: triangle number!
  9. Jun 13, 2010 #8
    Wow, thank you all so much for the feedback! I really appreciate it! We never counted the segments only the lines, but that's interesting as well :) can't wait to read up more on this all!

    Thanks again,
  10. Jun 13, 2010 #9
    The same sort of thing happened to me at 9 or 10. I called them 'triangle numbers'.

    n(n-1)/2 is the number of pennies you have arranged in a triagle with n on each side. Look up Pascal Triangle. (a different sort of triangle). You will find your numbers arranged along a diagonal row.
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