Dividing a Line Segment into Odd Number of Parts of Equal Size and a Fujimoto Approximation

Dividing a segment into even numbered parts of equal size is easy, just fold the ends together and get half, repeat to get a quarter, again to get an eighth, another time to get a sixteenth, and so on.  This does not rely on the squareness of the paper, just a straight segment of a line.

There is a kind of "cheat":  to make seven exactly equal segments, fold a segment three times over, and you have eight equal parts. Fold one of them over (or cut it, although cutting is frowned upon among origami purists) and you have seven equal parts! When I showed this to second graders at Lynch School for Japan Day one year, the chorus of boos and "noooo!"s clearly revealed their position that they were not impressed. But it is easy, direct, and rather precisely achieved without guessing.

Making odd numbers of parts of equal size requires a more analytical approach.  If you don't happen to memorize the exact methods in Lang's summary http://www.langorigami.com/wp-content/uploads/2017/09/origami_constructions.pdf, a quick way to get there is by a very efficient approximation method.  Fujimoto's approximation is explained in http://www.teachersofindia.org/sites/default/files/5_how_do_you_divide_a_strip_into_equal_fifths.pdf, and every fold divides the "error" in half, so by the 4th fold (5th approximation), the "error" is 1/16th of the deviation from your original guess. If your guesstimate were anywhere close to where it should be, the error between your guess and the exact 1/5 is very small. 

Extra credit: use Fujimoto's approximation to find sevenths of a segment.  Remember, this has no dependency on a square of paper, all you need is a line segment.

Extra extra credit: would this work on dividing an angle into odd numbers of equal angles?

One Eighth Along the Edge of a Square, and Three Sevenths

One eighth.  1/8th

The easy way to find 1/8th of the length of a line segment is to divide 3 times.  On a square, fold two adjacent corners together to get half, fold the corner to the center that you just found and get a quarter, and finally one more time, fold paper so the corner lands on the one-quarter mark.  That's three steps.

An interesting shortcut requires only two steps if you have a square.  Make a pinch or mark halfway between two adjacent corners, let's say on the right side:  upper right corner and lower right corner are lined up and the spot halfway between them is marked with a pinch. Take a corner on the opposite side of the square, say the upper left corner, and put it on top of the midpoint of the right side that you just marked.   You don't have to make a full fold, but just find the place (this will be on the bottom of the square if we are using the above landmarks) where the fold intersects the bottom edge.  Pinch this spot and you will have marked 1/8th of the distance along the bottom of the square.  That's two steps.

Challenge: prove this using the equation of a line going through two points where the slope of the line is expressed by the equation:
m = ( y0 - y )/( x0 - x )
where (x0, y0) is a specific point and x and y are variables on the x-y coordinate system. For a straight line, m a constant.

Hints:  Using the above references, and starting with a unit square (length 1 on each side), the midpoint between the top right and bottom right is (1, .5)(let's call that point B) and the upper left corner of the square is (0,1)(let's call this point A). The midpoint between those is -- (.5, .75)(let's call this C).    The line that forms the fold is the perpendicular bisector of the line between the upper left and midpoint of the right side.

The slope of the line through A and B is m = -1/2.  Its perpendicular is -1/m which would be m1 = 2.  The perpendicular bisector of segment AB would have a slope of 2, going through point C (.5, .75).  The equation for that line would be m = (y - .75)/(x - .5) = 2.

Sooo,  2x -1 = y - 3/4

y = 2x -1/4 = 0 where the square is on the x axis.

2x = 1/4

x = 1/8

Voila, the fold is exactly 1/8th of the length of the side of the square, in two folds, if you like, or two pinches, since there is no need to actually fold the crease, just pinch and mark the spot.

More:  if you mark two of these on the same square, and put one of the 1/8th marks on top of the other (that's five pinch marks), the part where the last fold intersects a side, is 3/7th of the length of the side of the square. This five-pinch procedure yields a particularly interesting artifact, that the side of a square can directly and exactly arrive at an odd numbered subdivision of a square. Other exact odd-numbered divisions from folding are demonstrated in http://www.langorigami.com/wp-content/uploads/2017/09/origami_constructions.pdf. 

Extra credit:  prove with algebraic equivalents that the last fold is actually 3/7 of the distance along one of the edges of the square.

Triangle

Triangle

In origami it is natural that the triangle is equilateral, just as the natural base is a square, where the sides are the same length.  Triangles are the first planar object that can be made with the fewest number of line segments (two line segments just are not up to the task).

There is an easy way to make a lot of equilateral triangles, starting with a strip of paper of uniform width.  Even if the first fold is only approximate, the following folds can be made "perfect" as long as the corner is tight and the edge aligns with the existing edge.    

https://www.cutoutfoldup.com/125-fold-a-series-of-equilateral-triangles.php.

Making an equilateral triangle (or equivalently, 60 degree and 30 degree angles) is as easy as making two folds on a rectangular piece of paper. Say we start with a letter size piece of paper, long side up and down (sometimes this is called "portrait mode") and fold it left to right in half:  the resulting fold is vertical. If you take say the upper left corner and put it on the centerline, while adjusting its location so the crease ends up including the upper right corner, and set the crease, you have a 30 degree angle from the top. Folding the paper over where the top edge has landed, makes a 60 degree angle.  Using the point on the centerline as a reference point, complete the equilateral triangle.  Voila!   Constructing 30 and 60 degree angles with two folds can be done almost anywhere with a bit of planning, and it's good to know this shortcut to make an exact 30 or 60 degree angle.

Strips of paper folded into connected equilateral triangles in this manner are very nearly ready to make a basic hexaflexagon.  https://en.wikipedia.org/wiki/Flexagon shows how to make a hexahexaflexagon.

I've used a hexaflexagon to write brief messages that are hard to decipher until the flexagon is returned to its state where the message was originally written.  

See also:

Triangle unit origami

Flexagon, Hexaflexagon, Hexahexaflexagon

2000 doves

 https://cathedral.org/visit-us/doves/

Maybe it should have been 2021 doves. I wonder if there is a video of them moving with the air currents.

Origami parabolas!?

             Last time I mentioned a fold that was equivalent to the shared tangent of two parabolas. In this article, I’d like to talk more about parabolas and origami - finding a parabola’s tangent is actually quite common in origami.

You might know that a parabola is the shape of a graph of a quadratic function; that is, the graph of something in the form of y=ax^2+bx+c.

You might also have heard of parabolas in the context of parabolic reflectors, where rays of light or sound emanating from one point are reflected parallel to each other or vice versa.

Parabolas in origami occur because of another definition of a parabola: the set of all points equidistant from a point (called the parabola’s “focus”) and a line (called the parabola’s “directrix”).

Okay, okay, maybe this is too obtuse and it seems like it’s not going anywhere, but let me explain. When you fold two points so they lie on top of each other, you get the perpendicular bisector of the segment containing the two points, that is, the set of all points equidistant from the two original points. Now, if you keep one of the original points fixed and move the other point around in a line, it stands to reason that the lines will all have a point that’s on a parabola - the parabola whose focus is the fixed point and directrix is the line on which the other point is moving.

Perhaps it’s time for the algorithm. Draw a point close-ish to one of the sides of a piece of paper. Then, take that side and fold it so that when it lands, it goes through the point. Make many folds of this type. Eventually, you should see a parabola emerge - there’ll be a heavily creased region, and a region without any creases, and the boundary between the two is a parabola.

How to trisect an angle with origami

There are a couple things that can’t be done with a compass and straightedge - squaring a circle, doubling a cube, finding the distance from a point to an ellipse.

Among these is the task of trisecting a given angle. However, despite not being able to be done using classical methods, origami has an algorithm for this construction.

It begins with the angle drawn from the corner (I will, WLOG, refer to the example with the angle in the bottom right corner) of the piece of paper with one ray lying on the paper’s edge and the other ray marked or folded. A fold is made parallel to the edge of the paper with the angle’s ray on it, close to the top. Then, the bottom half of the paper is folded in half.

Next, the point at the intersection of the top fold and the left edge of the paper and the vertex of the angle are folded in such a way that they meet the angle’s ray and the bottom horizontal fold, respectively.

Finally, if you fold from the corner to the point where the diagonal fold intersected the bottom horizontal fold and the point where the angle’s vertex landed, on the bottom horizontal fold, you will get two more lines that perfectly trisect your original angle.

Now, here’s the crucial question: What did using origami add that a straightedge and compass couldn’t? Well, it was that diagonal fold. It added what would amount by straightedge and compass to an infinite amount of information. That fold maps to finding the shared tangent of two parabolas, which requires a cubic equation, which cannot be done using Euclidean methods. Indeed, origami is even more powerful than straightedge and compass.

Origami, Geometry, and Math

Origami, Geometry, and Math have many shared concepts and uses.  Other subjects sometimes taught separately such as Algebra and its advanced topics blend right into Calculus.  Math in its many manifestations are representations of the same underlying concepts; for example, (a+b) squared, can be worked out with algebra, but it can also be represented in squares and rectangles of a square with a side of length (a+b). https://www.mathdoubts.com/a-plus-b-whole-square-geometric-proof/   Using either the geometry or the algebra, the answer turns out to be the same.  Why? it's the same underlying idea, we are looking at and solving the same problem through either an algebra viewing lens, or a geometry viewing lens.

Origami of course is closest to geometry in physical appearance. There are, however, things that can be easily done with origami that are hard to do with conventional tools of geometry (straightedge, compass), such as trisecting an arbitrary angle.  See https://plus.maths.org/content/trisecting-angle-origami for details. 

Starting with basic constructions, it is possible to arrive at either exact 30 and 60 degree angles, or usable approximations of a 72 degree angle (a whole 360 degree circle divided into fifths), and those useful approximations can be used in origami construction that are sometimes clumsy using a protractor and ruler.

Coming back to the original point, Origami, Geometry, Algebra, Calculus and other aspects of Mathematics are simply different viewing lenses, or windows with which we look at a set of concepts and principles.  Sometimes Origami can give us an angle that is useful and pleasing to look at.