+0

# \$X_1, X_2,\ldots,X_9\$ are nine points on the circumference of circle \$O\$. Line segments are drawn connecting each pair of points. What is th

0
3354
4

\$X_1, X_2,\ldots,X_9\$ are nine points on the circumference of circle \$O\$. Line segments are drawn connecting each pair of points. What is the largest number of different points inside the circle at which at least two of these line segments intersect? (Remember that the points are not necessarily evenly spaced around the circle.)

Mar 27, 2015

#2
+7

I'm going to answer this question....but I'm assuming an even spacing of the points  {I actually don't know if this makes a difference, or not  !!!!!  }

When the number of points is even, it is quite complicated to determine the number of intersecting points.

When the number of points is odd.....number theory shows that no more than two diagonals ever intersect at any one point !!!.... and we have the following "formula" for the number of intersections when the number of points is odd:

C(n, 4)  where n is the number of odd points on the circle

For instance, when n = 5, we have  C(5,4) = 5

When n = 7, we have  C(7,4)  = 35

So, when n= 9,  we have C(9,4) =126 points of intersection

BTW......

With 6 points, there are 13 points of intersection

By the "formula," 6 points should give us 15 points of intersection, but note that 3 diagonals actually meet in the center....thus...two "pair-wise" intersections are "lost"

And with 8 points, there are 47 points of intersection .... (if I counted correctly  !!!)   Mar 28, 2015

#1
+6

If you have 3 points on a circle and connect them in all possible ways, you will have no internal points of intersection.

If you have 4 points, then you get 1 internal point of intersection = 4C0.

If you have 5 points, then you get 5 internal points of intersection = 5C1.

If you have 6 points, then you get 15 internal points of intersection = 6C2.

If you have 7 points, then you get 35 internal points of intersection = 7C3.

Etc.

Can you take it from here?

Mar 28, 2015
#2
+7

I'm going to answer this question....but I'm assuming an even spacing of the points  {I actually don't know if this makes a difference, or not  !!!!!  }

When the number of points is even, it is quite complicated to determine the number of intersecting points.

When the number of points is odd.....number theory shows that no more than two diagonals ever intersect at any one point !!!.... and we have the following "formula" for the number of intersections when the number of points is odd:

C(n, 4)  where n is the number of odd points on the circle

For instance, when n = 5, we have  C(5,4) = 5

When n = 7, we have  C(7,4)  = 35

So, when n= 9,  we have C(9,4) =126 points of intersection

BTW......

With 6 points, there are 13 points of intersection

By the "formula," 6 points should give us 15 points of intersection, but note that 3 diagonals actually meet in the center....thus...two "pair-wise" intersections are "lost"

And with 8 points, there are 47 points of intersection .... (if I counted correctly  !!!)   CPhill Mar 28, 2015
#3
0

Does it count if its non or spaced out equally?

Apr 13, 2016
#4
+1

The maximal number of intersection points inside the circle occurs when no three of our segments pass through the same point inside the circle. Now consider one of the intersection points and the two segments which meet at it. These two segments are the diagonals of a quadrilateral which has the endpoints of the two segments as its vertices. This is true for every intersection point inside the circle. Moreover, it is true that if we select any four of the nine given points on the circle, connecting them will form a quadrilateral whose diagonals provide one intersection point inside the circle. Therefore, our problem becomes one of counting the number of ways to pick 4 of the 9 points, which can be done in \({9\choose 4} = \boxed{126}\) ways.

Jul 1, 2016