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).
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!!
To find a formula for I(n), we can start by considering smaller values of n and observing any patterns.
Let's start with the smallest possible case when n = 2. In this case, we have 2 points arranged on a circle, and we can draw a single chord between them. Since there are no other chords, there are no intersection points, i.e., I(2) = 0.
Next, let's consider the case when n = 3. We have 3 points arranged on a circle, and we can draw all possible chords between them. By connecting the 3 points, we form a triangle inside the circle. There are no chords inside the triangle, so there are no intersection points, i.e., I(3) = 0.
Now, let's examine the case when n = 4. We have 4 points arranged on a circle, and we can draw all possible chords between them. By connecting the 4 points, we form a convex quadrilateral inside the circle. Let's label the points A, B, C, and D (in clockwise order). If we draw the chords AB and CD, they will intersect at a single point O inside the circle. So, I(4) = 1.
If we increase n to 5, we have 5 points arranged on a circle, and we can draw all possible chords. By connecting the points, we form a pentagon. If we label the points A, B, C, D, and E (in clockwise order), and draw the chords AC and BD, they will intersect at a single point O inside the circle. So, I(5) = 1.
From the above observations, we can notice a pattern. For n = 4 and n = 5, we have only one intersection point inside the circle. We can also observe that the number of intersection points inside the circle is equal to the number of diagonals of a regular (n-sided) polygon minus 2.
Therefore, we can generalize the formula for I(n) as follows:
I(n) = (n choose 4) - 2
This formula holds for n greater than or equal to 4, as for n less than 4, I(n) = 0.