That is, without any gaps or overlap? This is of course trivially simple for squares, rectangles, rhombuses and kites. As for parallelograms, some time ago I posted here about an alternate way of folding rectangles in half and the same method generalizes to all parallelograms. For a trapezoid one folds the shorter parallel edge to the longer one and then fold in the two remaining corners.
That leaves quadrilaterals not fitting any other category. I suspect it’s possible but I’m not sure how one would demonstrate that the solution was generalized to all quadrilaterals.
My wife was a math prof / teacher & engineer. I had a bachelors-level minor in math back when we wrote our proofs on papyrus using squid ink and crushed sheep shit. You totally hijacked our Friday night and destroyed a very large stack of Post-its. Thank you. You bastard!
Our tentative but unproven conclusion in “no”. For the same reasons you can’t succeed in folding a scalene triangle in half, but generalized up one vertex.
Our collective argument rests on angles. A convex quadrilateral has 360 degrees of internal angle. In essence, your folds result in points meeting in the interior that also must meet the 360 degree criteria. And must either meet in all 4 together (picture folding in the corners of a square along creases connecting the side midpoints), or in two sets of two (picture folding a 4x4 square into a 2x1 rectangle and the generated angles and half-angles at the creases). Or meeting along a pair of sides.
For each of the above approaches given a scalene trapezoid / trapezium it’s trivial to construct a counterexample. It’s easier the more pathological your quadrilateral’s sides. An almost square is arguable, but once the sides are of length 1, 10, 100, 1000 it’s real obvious that won’t work.
FYI, the wiki on trapezoid was interesting in that it exposed a total difference in my wife’s and my’s notion of the terminology for the less-regular quadrilaterals: trapezoids, trapeziums, etc. Her and my texts had taken different sides in the terminology wars and it took us awhile to notice we were talking past one another. Gee thanks. For your entertainment:
Scalene triangles are easy. Fold it parallel to the longest side until that vertex ends up on the side. Then fold the other two corners in to the same point.
All it would take is to say that every surface is actually a double sheet.
In fact, I have a solution in mind for a very loose version of the problem: suppose one side of the paper is red (the other is white). Can you fold or roll the paper so that every red surface touches another red surface?
The gotcha is (probably) the terminology issue I described. The terms “trapezoid” and “trapezium” have different and incompatible definitions in different lexicons. So what I wrote and what you read were /are (probably) using different definitions. Read that wiki I cited for the details.
Or else I flat used the wrong words. It was kinda late for a good write-up of my/our hand-waving.
In any case, I was trying to refer to convex quadrilaterals with zero parallel sides. As you say, given two parallel sides the problem is trivial regardless of the similarity or dissimilarity of the other side’s angles.
Sorry to confuse. But it was a fun problem to chew on.
That won’t be possible, because that shape can’t exist in the first place. In any closed polygon, the length of the longest side must be less than the sums of the lengths of all of the other sides.
D’oh!! Thank you. I realized that example was bogus as you say about 30 minutes after I’d posted it.
My believe my underlying point stands however that the more lopsided and asymmetrical the figure, the more easily it intuitively forms a counterexample.
My overall thoughts are admittedly a long, long way short of rigorous here.
Actually, I don’t think it matters how extreme it is. In the general case, there won’t be any two corners with supplementary angles, and so the only possibility would be all four corners meeting at a point. But the only way for that to happen would be for the endpoints of the folds to be the midpoints of the sides. And we also have the constraint that the folds must be perpendicular to each other, because after folding, each corner of the folded shape must have two layers that were originally 180º. Take any elongated almost-rectangle, where the elongation is relatively large and the deviations from right angles are relatively small, and you can’t satisfy both of those conditions at once.
That’s not a very rigorous argument, and I think there might be some holes in it if, for instance, you allow an infinite number of folds, but that’s the basic idea.
My understanding of the implicit limitations of the problem were you could only fold to create something 2 plies thick anywhere and therefore by virtue of the half-area constraint 2 plies thick everywhere. I’m not seeing how large numbers (much less infinite numbers) of folds could even be used in that case.
Large finite numbers of folds won’t help. But you might end up with a situation where, say, a sharp corner gets folded around itself repeatedly in a spiral.
Just off the top of my head- I would take the example diagram above and make the two sides not parallel. The folding two sides together will create two uncovered triangles as a result of the difference between the two sides’ lengths, usually scalene. Unlike the parallel sides example, these will not be isoceles, so cannot be folded in half to completely cover themselves. As a result, you have an uncovered triangle with only one edge not exposed. Unless you’re allowed double coverage, you can’t fold a triangle with only one exposed side. If you are allowed to fold over and over several layers, then must the folds meet at an edge at all?
