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March 6, 2015

March 6, 2015

Posted by **Brenda** on Saturday, February 2, 2013 at 10:42am.

- math -
**Reiny**, Saturday, February 2, 2013 at 12:21pmThe 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 -
**Brenda**, Saturday, February 2, 2013 at 12:42pmthank you!!

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