Folding into Three Parts
You have a rectangular sheet of paper.
You are only allowed to fold the paper and create creases. For example, we can fold the paper into half to get 2 congruent rectangles, and fold it into half again to get 4 congruent rectangles.
Can you create 2 parallel creases along which to fold the paper into 3 congruent rectangles?
Details and Assumptions:
- You are not allowed to use a ruler, compass, scissors or any other geometric aids.
- You are not allowed to guess/estimate any angle or length by the naked eye.
- You may only fold the paper, create creases, and use the edges and corners of the paper and the creases to make new folds.
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1 solution
Nice! How does one show that these lines are indeed a third of a page apart?
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I have added that explanation to the solution.
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How do we get the fold through D (parallel to the side)?
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@Calvin Lin – Yes. You also have point E to line up to. It's still not as clean as some origami. There is a room for a better solution.
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@Marta Reece – Hm, if we can fold the other diagonal G C to obtain the points D ′ , E ′ , that makes it easier to hunt down the line D D ′ , E E ′ . However, I'm not certain if origami allows us to "fold between 2 points", esp if we do not have a ruler to guide us.
(Though, if so, the question then becomes how to obtain B G )
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@Calvin Lin – If rules do not allow folding between points, maybe I should remove the not-quite-solution.
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@Marta Reece – @Agnishom Chattopadhyay Can you clarify what kind of operations are allowed?
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@Calvin Lin – Line BG can be produced in five foldings if placing a corner onto a known point is permitted, and final lines can be obtained if placing an edge on two points is permitted. The solution will be more complicated than described, but still possible. If either of those moves is ruled illegal, then I am no longer sure that the division into thirds could be done.
@Calvin Lin – I wanted to formalize this, but it turns out it is kinda hard to describe. I added some points at the end of the problem, but it still does not capture my idea very rigorously.
In my opinion, Marta's solution is valid because one can reproduce this on any rectangular sheet accurately without any aids.
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@Agnishom Chattopadhyay – I have updated the solution. It is much cleaner, but it still requires point to point folding, which I am pretty sure is allowed, and line to two points folding, which is more questionable.
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@Marta Reece – That's much clearer :)
I agree that "point to point" folding is valid (IE the perpendicular bisector).
Do you think the following "point to line, centered on point" is allowed? I've seen it in Origami books. Specifically, given point X (centered point), point A and line L, I want to mark point B on L such that ∣ X A ∣ = ∣ X B ∣ . In a sense, this gives us a compass at point X, though we can only tell where the circle intersects a line (as opposed to using the paper as a compass itself to trace out a circle).
If so, suppose we want to connect points X and Y.
Then, we fold X to Y creating line L. L is the perpendicular bisector.
Using a center of X, point Y to a point A on line L. AXY is an equilateral triangle.
Using a center of Y, point X to a point A on line L. BXY is an equilateral triangle, on the other side.
Fold A to B creating line K, which passes through X, Y.
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@Calvin Lin – I updated the solution.
There are still problems. The solution works only for a square, not "regular sheet of paper" as stated in the wording of the problem.
Also, I feel that the exact description of permitted types of moves should be provided.
@Calvin Lin – Are you aware of the Huzita-Hatori Axioms ?
@Calvin Lin – It is possible to get folds parallel to the edges of the paper through some point, since you know that it is parallel when it perfectly lines up with the other two perpendicular edges of the paper.
(1) Start with two creases - vertical mid line and a diagonal.
(2) Match corner A to center O to form the next crease.
(3) Fold out both creases D E and B C to the same side and match them to create a crease half way between them.
(4) Match corner C with point F and crease.
The intersection of creases generated in steps (3) and (4), marked X , is one third off the bottom of the square.
(5) Repeat steps (1) through (4) on the right with the other diagonal, point C folded to the center, line half way between those two parallel lines and finally point A matched with point F to generate point X ′ .
(6) Fold the upper edge down so that it lines up with X X ′ . Left and right edges will also line up. The crease generated will be one third of the way from the top.
(7) Fold the lower edge to meet the crease generated in step (6) for the final division.
What's left is to show that the distance between X and the lower edge is indeed one third of the side of the square. Let's set the side equal to 8 and place the origin of the coordinate system at the point A with x-axis in the direction of point C . The equations of the creases will then be:
B C : y = 8 − x , D E : y = 4 − x , G H : y = 6 − x , K L : y = 2 x + 1
The last one is derived as a perpendicular bisector of F C : y = 1 6 − 2 x . The slope for K L has to be 2 1 and it has to go through point ( 6 , 4 ) .
y = 2 x + b , 4 = 2 6 + b , b = 1 , y = 2 x + 1
Intersection of lines y = 6 − x , y = 2 x + 1 is point X : ( 3 1 0 , 3 8 ) .