Posted by Brenda on Saturday, February 2, 2013 at 10:42am.
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
thank you!!