The world’s best tortilla puzzle

fold it over itself 3 times. make 1 cut straight down the middle such that it is bisected?

The first, obvious, strategy is to make 4 cuts. Horizontal, vertical, and each of the two diagonals.

However, theoretically, each cut should be able to double the number of slices. If we stack the slices together, we can achieve this effect and get eight slices with 3 cuts. First cut gives us semicircles, which we stack and bisect with the second cut, giving us quarters, which we stack, bisect with the third cut and get eights.

Reducing this further requires folding, I think. So I decided to experiment with a sheet of paper.

Folding a paper in half and cutting at the fold gives us 2 pieces, each half of the paper. Folding in half twice and cutting gives us 3: the two ends of the paper and the middle section. This progresses such that, folding a paper in half x times (folding in the same direction each time) and cutting along that fold yields x+1 pieces.

However, they are not all the same size. Since the ends of the paper aren’t connected, they each form a piece half as small as all the others. Since each fold halves the size of each piece, a paper of length L folded x times will yield x+1 pieces:

x-1 pieces L/x in length
2 pieces L/2x in length

This works out as we would have:
2*(L/2x) + (x-1)(L/x)
L/x + L(x-1)/x
L/x + Lx/x – L/x
L/x + L – L/x
L

How do we compensate for the smaller end pieces? By not cutting in half. Without going into the math, if we position the first fold correctly (with the end pieces being longer), and then continue to fold on that line, the end result will have all the pieces being the same.

How does this translate to a tortilla? We can still fold a tortilla, in half even, it’d just look different. So I think the same principles apply, we’d just have to work out where the fold would have to be, but a single cut along that fold should yield eight equal pieces, if positioned correctly.

Only if the paper is square.

In any event, an assumption of mine is wrong. I stated that:

“Folding a paper in half and cutting at the fold gives us 2 pieces, each half of the paper. Folding in half twice and cutting gives us 3: the two ends of the paper and the middle section. This progresses such that, folding a paper in half x times (folding in the same direction each time) and cutting along that fold yields x+1 pieces.”

This is not true. In testing I discovered the following:

No folds = 1 piece
1 fold = 2 pieces
2 folds = 3 pieces
3 folds = 5 pieces
4 folds = 9 pieces

What does appear to hold is that each of the smaller “end” pieces halves it’s area with each fold, and the remaining folds are always twice in size.

Thus, the area of the end pieces is A/(2^n), where A is the area of the sheet of paper and n is the number of half-folds.

The area of the rest of the pieces, then, is 2A(2^n). Since the total area of all pieces must equal A, and since there will always be 2 end pieces, we get:

2*(A/(2^n)) + x*(2A/(2^n)) = A

Solving for x:

x*(2A/(2^n)) = A – 2*(A/(2^n))
x*2A = A*(2^n) – 2*(A/2^n)*(2^n)
x*2A = A*(2^n) – 2*A
x = A*(2^n)/2A – 2A/2A
x = (2^n)/2 – 1
x = 2^(n-1) – 1

Unfortunately, two get 8 total pieces (2 ends, and 6 remaining) x would equal 6 which does not produce a whole number for n.

So I’ll have to abandon that strategy.

@Travis:

Looks like you’re right. It’s hard to visualize, let alone to describe the visualization. It looks like after 3 such folds (circle -> semicircle, semicircle -> 90-degree sector, 90-degree sector -> 45-degree sector) if you cut in half you’d get 8 pieces.

Makes my previous thinking seem a bit silly.

What I find infuriating is having folds in my tortilla slices.

Well, what Steve says is pretty impressive. As a theoretical question for a paper, its easy to make folds and cuts, but for a thing like tortilla, making 4 cuts is easier. So Presh!!! You posed a wonderful question, and Steve gave a wonderful answer.