All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Partitioning Space (Posted on 2004-09-13)
From Pizza Cut, we know the formula for maximum partitioning (pieces) of the circle, given n straight lines (cuts).
______________________________________

1. Determine the maximum number of regions of the plane produced by n intersecting circles.

2. Determine the maximum number of regions of the plane produced by n intersecting ellipses.

3. Determine the maximum number of regions of space produced by n intersecting spheres.

 No Solution Yet Submitted by SilverKnight Rating: 4.5000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): Part 3?? | Comment 7 of 13 |
(In reply to re: Part 3?? by Brian Smith)

Drat!

Ok, another assumption. Notice that the differences between 2, 4, 8, and 14 are 2, 4, and 6. Perhaps this means that the number of regions added is the sum of even numbers.

So the formula for the total regions is related to the sum of the sum of the even integers. The sum of the even integers from 0 to 2n is 2*[the sum of integers from 0 to n] = 2[(n)(n+1)/2] = n^2 + n.

The sum of THAT is:
the sum of n^2 + the sum of n
= [(2n^3 + 3n^2 + n)/6] + [n*(n+1)/2]
= [(2n^3 + 3n^2 + n) + 3(n^2 + n)]/6
= [2n^3 + 6n^2 +4n]/6
= [n^3 + 3n^2 + 2n]/3

This puts me in the right ball park, but I’m off. What am I off by? I’m actually off by n*(n+1)/2. I can’t prove it, I just saw the pattern (1, 3, 6, 10, 15…).

(n^3 + 3n^2 + 2n)/3 – (n^2 + 2)/2
= (n^3 + 3n^2 + 2n)2/6 – (n^2 + 2)3/6
= [2n^3 + 6n^2 + 4n – 3n^2 – 3n]/6
= [2n^3 + 3n^2 + n]/6

[2n^3 + 3n^2 + n]/6

 Posted by nikki on 2004-09-14 14:52:43

 Search: Search body:
Forums (0)