# PFG: A pretty spiral!

PFG = Paper Folding Golf

You bought a piece of paper from your local op shop, but your shop is always dodgy. So instead of a normal piece of A4, you got an infinitely long strip, a metre wide.
(Actually, it's a bit strange, it seems to have a flaw that causes it to change size to a yard wide when you cross the border into the US. You live near the border in Canada)

Anyway, you want to fold it some number of times then make a single straight cut across the paper you get a rectangle (you still need to find a way to make the line exactly perpendicular to the edge of the paper).
This rectangle should have the property that if you cut its width from its length, you will get another rectangle that has the same ratio of length to width, which will mean you can get a pretty spiral of squares and nice-looking rectangles.

Technicalities:

• No tools apart from the paper, folding (is that a tool?) and the magical one-shot straight-line cutter allowed.
• $length>width$
• Your paper is really semi-infinite, it has one finite edge.
• Because of magic, your paper doesn't form a black hole under its gravity or anything. It somehow tapers at the end into an infinitely dense roll with weight 50 grams and of diameter 5cm (2" in the US) which you can unroll and re-roll at will without changing its properties. Outside of the roll the paper acts like normal paper.

Quick check of possible minimality:

I got $min(folds)\le4$.

Note (may help?):

It is said that this rectangle looks the nicest.

Googling is prohibited! D: Hence the tag!

• Well, " infinitely long strip" as "line", or as "ray" ? Feb 24, 2017 at 10:00
• Does the infinite strip have an end? One that is orthogonal? Feb 24, 2017 at 12:38
• @JanIvan As in ray. Feb 24, 2017 at 21:00

I'll assume the strip is actually semi-infinite, and its one end has a straight cut orthogonal to the infinite edges. The following solves it in 4 folds. The 4th fold is not needed if the cutting machine can automatically cut perpendicular to the sides of the strip.

Assuming the rectangle has width 1, then it needs to have length $\phi$, the golden ratio $(1+\sqrt5)/2$.

Proof: Let $x$ be the length of the rectangle. The ratio of its sides is $1/x$. If you then cut the width from its length, i.e. cut off a 1x1 square, you are left with a rectangle with length $1$ and width $x-1$. The ratio of its sides is $(x-1)/1$. The equation $1/x = (x-1)/1$ has the solution $x = (1+\sqrt5)/2$, or $\phi$.

Note: It is also possible to cut a short rectangle off, of $1$ by $1/\phi$, which has the same edge ratio, but my solution cuts a $1$ by $\phi$ rectangle. 1. Fold corner A to the opposite long edge, so that the crease goes through the other corner. This diagonal fold marks a point B on the same edge as A but at distance of 1 from it.
2. Fold the short edge over to point B. This marks off point C that lies halfway between A and B.
3. Fold C to the top edge, with the fold going through the other corner. This marks off point D such that A-D has length $\phi$.
4. You need to cut the strip at D. To mark the line through which to cut, simply fold the strip through point D.

Proof that this works:
Clearly $|AB|=1$ so $|AC|=1/2$.
Let O be the other corner, and let $\alpha$ be the angle between OC makes with the long edge at O. Note that OC is not drawn in the pictures, as that line never becomes a fold. When OC is folded onto the long edge, this is the angle that is bisected to get create fold line OD.
We have $\tan\alpha=2$, and Pythagoras gives you $\sin\alpha=2/\sqrt5$, and $\cos\alpha=1/\sqrt5$.
You can then use the half-angle formula for tan: $\tan\frac{\alpha}{2}=\frac{\sin\alpha}{1+\cos\alpha}$
This gives that $\tan\frac{\alpha}{2} = \frac{2}{\sqrt5+1} = \frac{1}{\phi}$.
OD is the diagonal of our rectangle, and its angle with the side has tangent $1/\phi$. That makes the ratio of its sides $1$ to $\phi$, and hence AD has length $\phi$.

It the strip does not end in a straight orthogonal cut, or if it were infinite at both ends, then an extra fold is needed, so then my solution becomes 5 folds.

Simply fold the strip onto itself to create a straight end. Then proceed with the first three folds of the solution above. You need a different 4th fold, however.
To cut the rectangle from the strip with a single cut, you will have to insert it into the cutting machine with the strip folded so that A lies on top of D. Cutting the paper along the orthogonal fold at A then automatically cuts the other layer orthogonally at D, resulting in the rectangle we want.

• Nice! Could you amend why folding C upwards (giving D) makes AD distance $\phi$? Feb 24, 2017 at 10:05
• @BmyGuest I was already working on it, and I have added the proof now. Feb 24, 2017 at 10:32
• @humn Are you assuming the strip of paper is infinite at both ends? I read the question as being about a semi-infinite strip, as it says it "tapers at the end into a roll", not both ends. If it is infinite, then you'd indeed need one extra fold at the start to create the short edge. I'd then also replace the last fold in my answer by a fold bringing A onto D, so that I could cut the rectangle from the strip by a single cut through two layers. Feb 24, 2017 at 13:00
• @humn - Ah, yes, I did assume that the end of the strip was already straight and orthogonal. If that were not the case, the changes from my previous comment about the infinite strip apply in exactly the same way. Feb 24, 2017 at 13:49
• (Took me a while to notice your edit, Jaap Scherphuis, so my comments are at last deleted. Again, a brilliant way to establish $\phi$!) Feb 24, 2017 at 15:42

well, if u have infinite paper with no beginning or end: