Cutting pizza

[rocdocmac]

Suppose you have a pizza being cut vertically with one straight cut, you'll be left with two pieces. The ratio of number of pieces to cuts in such an instance will be 2 (a whole number, i.e.  no remainder).

If you cut a pizza using two straight vertical cuts, you'll have four pieces left (pieces-to-cut ratio = 2, also a whole number).

Cutting the pizza thrice, the maximum pieces possible is seven (pieces/cuts = 7/3 = 2.3333..., i.e. there is a remainder of 1).

See attached image.

Questions ...

(1) What is the maximum number of pieces that one would get by cutting the pizza 500 times using only straight vertical lines (no horizontal cuts allowed)?

(2) How many cuts should one make to get a maximum of at least 5 million pieces?

(3) The minimum piece-to-cut ratio with no remainder is exactly 2 as indicate above. What would be the next higher number of cuts, such that the number of resulting pieces divided by the number of cuts has no remainder?

 

[phil1882]

i dont really understand the question. your picture seem contrary to your description.

your picture seems to suggest you can cut in any direction, but your description suggests all cuts have to be in the same direction.

i'll solve both.

for lines in any direction: http://mathworld.wolfram.com/CircleDivisionbyLines.html

1/2*(n^2 +n +2), n=500; 125251

5000000 = 1/2*(n^2 +n +2) n=3162

for cuts in the same direction:

each cut adds 1 new region.

for 500 cuts, that 501 rejoins

to get 5000000 peices 4999999 cuts are needed

[TimeSpaceLightForce]

Spoiler

The 4th slice can cut across 3 existing cuts..the 5th slice can cut across 4 existing cuts..so on.  But why 500 times? why pizza?

 

[rocdocmac]

Rather change "pizza" to an enormous 2D circle then!

I used pizza as to eliminate any horizontal cut,  but it will be a hell of a mess should one really had to cut a pizza into so many pieces!

[rocdocmac]

All cuts should be vertical and in any direction ... I only emphasized that no horizontal cut is allowed, e.g. one could possibly divide a cake with three cuts to yield 8 pieces ... 2 vertical cuts and 1 horizontal cut!

(1) 125251 pieces after 500 cuts (correct)

(2) 3162 cuts gives 500704 pieces  (also correct)

(3) Never mind the 3rd answer since I reckon that it must be infinity, although 199999 cuts as a first approximation comes very, very close! However, there will always be a remainder of either 1 or a pretty large number.

Therefore, 2 remains the minimum and maximum  pieces-to-cuts ratio. I cannot, though, prove this mathematically (yet)!

I wasn't aware of an entry at "Mathworld"! So, my question should actually have been a straight-forward quickie!

 

[phil1882]

your description still doesn't make sense. how can all cuts be both vertical  and in any direction and no horizontal. which is it, all vertical (same direction) or any direction?

Edited June 26, 2018 by phil1882

[rocdocmac]

Vertical (in any direction) = straight down.

Horizontal as in figure, e.g. a cake. Two vertical cuts give 4 pieces, then the 3rd cut (horizontal ) divides those 4 pieces into two each for a total of 8. Horizontal cuts are therefore excluded.

 

Edited June 26, 2018 by rocdocmac

[DejMar]

It appears what you (rocdocmac) are saying is:
Assume the pizza is a 2-dimensional circle (in a plane of Euclidean space). All cuts are straight lines (i.e., line segments).
The confusion that occurs is with the terms vertical and horizontal in the assumed plane, as they imply all vertical cuts are parallel to the imaginary y-axis and all horizontal cuts, if such were allowed, to be parallel to the imaginary x-axis. Yet, as the cuts are given to be in any direction, there was an implied contradiction to that inference.

[rocdocmac]

Thanks Dejmar for clearing!

[EventHorizon]

Proof for (3)

Spoiler

n cuts result in a maximum of 1+.5*n*(n+1) pieces.

Case 1: n is odd and greater than 2.

Since n is odd, n+1 is even and .5*(n+1) is an integer.  So the number of pieces is 1+n*(some integer).  The remainder by dividing this by n is clearly 1 since the number of pieces is one more than a multiple of n.

Case 2: n is even and greater than 2.

The number of pieces is 1+(n/2)*(n+1).  n is even, so n+1 is odd.  (n/2) times an odd number greater than 2 results in n/2 more than a multiple of n.  Adding the 1 from the formula makes the number of parts 1+(n/2) plus a multiple of n.  Since n is greater than 2, 1+(n/2) is less than n.  So the remainder when dividing by n is 1+(n/2).

Since if n>2 there will always be a remainder of either 1 or 1+(n/2), There is no number of cuts greater than 2 that will produce a maximum number of pieces that is evenly divided by the number of cuts.