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Partitioning Space (Posted on 2004-09-13) Difficulty: 5 of 5
From Pizza Cut, we know the formula for maximum partitioning (pieces) of the circle, given n straight lines (cuts).
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  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)

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Solution 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 Im off. What am I off by? Im actually off by n*(n+1)/2. I cant prove it, I just saw the pattern (1, 3, 6, 10, 15).

So my final answer/guess is:

(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
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