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n points are arranged on a cirlce and all the chords drawn. Let I(n) be the number of intersection points inside the circle if no 3 chords are concurrent. Find a formula for I(n).

  • math -

    The condition should have been
    " ... if no 2 or more chords are parallel or concurrent."

    The number of chords possible = C(n,2)
    = n!/(2!(n-2)!) = n(n-1)/2
    but all chords that join adjacent point will not result in any intersection
    so we have to subract n
    number of usable chords or lines = n(n-1)/2 - n
    = (n(n-1) - 2n)/2
    = (n^2 - n - 2n)/2 = n(n-3)/2 , where n > 3

    if no lines are parallel or concurrent:
    1 line intersects in 0 points
    2 lines intersect in 1 point
    3 lines intersect in 3 points
    4 lines intersect in 6 points
    getting a bit messy to sketch
    but let's think about it.
    Every time we add a new line it would intersect each of the previous lines once, adding that to the total
    e.g
    if we add a 5th line, we would add 4 new points to the 6 we already have , so
    5 lines intersect in 10 points
    6 lines intersect in 15 points (10+5)
    7 lines intersect in 21 points (15+6)
    etc

    So in n lines we would have .?.?... points

    lets investigate
    0 1 3 6 10 15 21 ... perhaps you recognize these as the triangular numbers. (billiard balls are racked up in that fashion)
    they are produced by the formula n(n-1)/2
    check: if n=5 , 5(4)/2 = 10

    now be careful:
    for every n chords we get n(n-1)/2 points of intersection
    but for every n points on the circle we get n(n-3)/2 lines that intersect

    so the number of intersection points caused by n points on the circle

    = ( [n(n-3)/2] [ n(n-3)/2 - 1] ) /2
    = ...
    = n(n-3)(n(n-3) - 2)/8

    let's test this;
    We know by investigating that
    4 points result in 2 usable lines which result in 1 points
    so when n = 4
    I get 4(1)(4(1)-2)/8 = 1
    Well, how about that ????
    5 points on the circle ---> 5(2)( 5(2) - 2)/8 = 10

    YEahhh it works
    I enjoyed that

  • math -

    thank you!!

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