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Oct 4, 2020 at 18:39 comment added Paul Panzer If I understand the "rules" of folding correctly once you have one of the four main folds the others can be made without any auxiliary folds because folding through a given point perpendicular to a given crease or edge is an allowed operation.
Oct 4, 2020 at 18:30 comment added Retudin I did not show but need the 1/3 and 2/3 fold horizontally and vertically..
Oct 4, 2020 at 18:22 comment added Paul Panzer Generalization is a good point. Mine does generalize in principle but not as straight forward as yours. Re number of folds I count $9$ ($5$ auxiliary $4$ money folds) in both cases.
Oct 4, 2020 at 18:10 comment added Retudin @PaulPanzer My solution needs more folds than yours, so in that sense yours is better. I think my solution is easier to understand, but if one understands Thales's theorem one probably understands your use of the golden ratio, so I might be wrong. I also like about my solution that it is easy to to generalize to any 1/Nth part of total, although that may also be true for yours..
Oct 4, 2020 at 17:11 comment added Paul Panzer I'd say this is up to a minor detail the same as my construction. Which is more elegant is very much a matter of taste. Number of auxiliary folds is the same. Btw. the thing you do with the right angled triangle is called Thales's Theorem.
Oct 4, 2020 at 16:07 comment added Retudin Seems that adds something to the existing answers, so please post.
Oct 4, 2020 at 16:04 comment added Bubbler I thought, by you mentioning the 1/5 problem, you wanted a solution where you can prove the 1/3-ness by dividing the paper into multiple areas and reassembling into three same-sized squares. I do have such a solution right now, though it doesn't place the square at the center. If you're interested, I'll post it when I get to my PC.
Oct 4, 2020 at 15:47 history edited Retudin CC BY-SA 4.0
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Oct 4, 2020 at 15:38 history answered Retudin CC BY-SA 4.0