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Again, this is from io9's site: http://io9.com/this-weeks-puzzle-is-as-easy-as-pie-1687313362

Sunday Puzzle #21: Slicing Pie

*One straight slice through one whole pie divides the pie into two pieces. Simple enough. A second straight cut, crossing the first, will produce four pieces. A third cut, directed through the intersection of the first two, will make six pieces –but a third slice, strategically placed, can actually create as many as seven pieces. What is the maximum number of pieces you can produce with six straight cuts? With N straight cuts?*

I would think N straight cuts would end in infinity. As far as 6, would it be something like (2n +1) = y. Hmm, no that can't be right. What do you guys think?

7/3 = X/6 | X= 14?

Seems too simple though...

N straight cuts either matters on the size of the pie or just goes to infinite.

I'm thinking 22? Or for n: ½(n² + n + 2). That is, if I'm correct in assuming the maximum number of pieces for four slices is 11, five is 16, and so on.

1 + Σn as a partial sum formula is 1 + (½n(n + 1)). The square comes from the n being multiplied by n + 1.

Thinking in only two dimensions, three cuts could get you seven slices; but thinking three dimensionally you can get eight. I believe that @BradinDvorak's solution might be the correct answer to this puzzle, but the more slicing planes you use and the more dimensions to which you have access the more pieces you can get.

It's all in how you slice it...

Ugh, that looks like crap on my screen, the sigma is not displaying properly.

EDIT:

1 + sigma n = 22 (for 1 through 6)

So for n straight cuts the answer is ---> 1 + sigma n = y (for 1 through n)

1 + (6 + 5 + 4 + 3 + 2 + 1) = 22 = 6 cuts = 1/2 (6^2 + 6 +2)

1 + (5 + 4 + 3 + 2 + 1) = 16 = 5 cuts = 1/2 (5^2 + 5 +2)

1 + (4 + 3 + 2 + 1) = 11 = 4 cuts = 1/2 (4^2 + 4 +2)

1 + (3 + 2 + 1) = 7 = 3 cuts = 1/2 (3^2 + 3 +2)

1 + (2 +1) = 4 = 2 cuts = 1/2 (2^2 + 2 +2)

1 + (1) = 2 = 1 cut = 1/2 (1^2 + 1 +2)

1 + (n1 + n2 + n3...) = 1/2 (n^2 + n + 2)

I thought this was quite interesting.

I kind of dislike this puzzle, as it is not really a puzzle or a riddle. It's just an advanced math test, and I seem to remember something similar actually BEING on a math test I did once.

The previous ones required logical thinking and deduction. This simple requires math, so if you are great with advanced math it is easy, if you are not it is impossible.

The answer is basically the n+1 in that, yes, the answer would be 22. But the total could be infinity. EDIT: No, it's not. That's why n+1 is wrong.

Mathematician in the comments arguing about infinity:

The real question is 'what is a quantum slice of pizza?' Mathematically, you can do this to infinity but given a normal 14" pizza at some n your smallest slice will be smaller than the plank length and non-measurable. Can you slice a pizza so small that God cannot make it? Anyway ...

f(1)=2, f(2)=4, f(3)=7, f(4)=11, f(5)=16, f(6)=22, f(7)=29.

It it technically possible, though aesthetically ugly to use each new straight cut to cut through all the other straight cuts. That's how you get 7 slices with 3 cuts, the third goes through the 2 other lines. You get 11 slices with a fourth cut that goes through the previous 3 cuts.

So I think that's

F(n)=(Σ(1,n) n) +1

Here's the explanation:

http://io9.com/can-you-solve-one-of-archimedes-most-challenging-puzzle-1688790863

Let n be the number of straight cuts to be made on a 2D pie, then the number of slices can be given by:

Slices=No.of touchpoints+(No.of intersections-previous lines) where

No.of touchpoints=No.of cuts*2 (points where the lines touch the circumference)

No.of intersections=No.of points within the pie where 2 lines cross each other

No.of previous lines=n-1

and No.of intersections=(n-1*(n))/2

The closed form function= (n*2)+((n*(n-1)/2)-(n-1))

This thread is a month old, but... I suppose we can let this necro slide because this sounds like it's right.

EDIT: Nvm, madglee posted the answer or something a while back..