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 formula for the maximum number of pieces if you cut a crescent moon (think capital letter 'C') with n straight lines.

2. Determine the formula for the maximum number of pieces of cheesecake (a cylinder) produced by n plane cuts.

Crescent Moon:

Well, I noticed that you could make your first cut so that it was tangent to the "inner" part of the crescent. Then you made 2 new pieces – one is just like you were making your first cut in the pizza, and you get a bonus piece when you make sure it’s tangent. On your next cut, you follow the same rules as the pizza where you make sure your new cut goes through every preceding cut PLUS make sure it is tangent to the inner part of the crescent. This time you make 3 new cuts – again this like the two new cuts in the Pizza problem, plus the bonus one for making it tangent.

Continuing in this way, I believe the formula for the maximum number of pieces with n straight cuts is the sum from 1 to n+1, so that is (n+1)(n+2)/2.

If you’re wondering why I didn’t "add one" for the initial piece, it’s because I "used it" in my formula. The pattern for the new cuts actually started at 2 (not 1), so I used the first piece to complete the series that I was summing.

Cylindrical Cheesecake:

This one is trickier. I started seeing a pattern. The first cut made 1 new piece (so 2 total). The second cut made 2 new pieces (so 4 total). The third cut made 4 new pieces (so 8 total). At first I thought it would be something like n^2. But then on the fourth cut I saw that you could make 7 new pieces (not 8 which would make 16 total pieces like my n^2 idea). But wait, now I see a new pattern. So the additional pieces made are 1, 2, 4, 7. The differences between these numbers are 1, 2, 3… this is like the pattern for the pizza cuts (except offset by one), and now we are summing them. Well, the pattern for the additional pieces seems to be (n-1)n/2+1. The sum of n(n+1)/2 is n(n+1)(n+2)/6. So the sum of (n-1)n/2 should be (n-1)n(n+1)/6 and the sum of 1 is just n.

In this case I do need to add in the first piece, because I didn’t use it in my formula. So I think the maximum number of pieces is (n-1)n(n+1)/6 + n + 1.

Posted by nikki on 2004-08-23 15:20:08