Today's post is a little bit different than usual, but it's somewhat math-related, so I figured I'd post it here anyway. Recently, I folded Eric Joisel's origami dwarf

My folded dwarf

(you can see Joisel's rough instructions for the dwarf here). One interesting feature of the dwarf that Joisel points out in his instructions is that the dwarf is folded out of a 28 by 28 grid. As Joisel observes, usually origami models use grids whose dimensions are powers of 2 - it's simple to fold a piece of paper in half repeatedly to obtain an 8 by 8 grid, or a 32 by 32 grid. But 28 by 28 is trickier. In fact, Joisel advises you to use a ruler to form the grid instead of bothering to fold 28ths by hand. But it turns out that it's not so hard to fold 28ths after all. That's what I'm writing about today. But before jumping straight into folding 28ths, we'll start with a slightly easier topic.

Folding a Square in Thirds

Here is a nice little folding sequence to fold a piece of paper in thirds. First, take your square and fold it in half. Unfold, and you are left with a vertical crease cutting the square in half.

Next, fold and unfold the square in half diagonally. Now, fold and unfold a crease from the bottom right corner to the middle of the top edge. And now you're done! The two diagonal creases intersect each other at a point one third of the way across the paper!

Why Does This Work?

There's probably some sort of clever argument you can make using Euclidean geometry and similar triangles and the like to show that this algorithm really does find you one third of the paper. But I think it's easier to use coordinate geometry instead. Let's imagine our square of paper as living in the plane so that its right edge is the $y$ axis and the bottom edge is the $x$ axis.

Note that the two diagonal creases that we folded lie along the lines $y = x + 1$ and $y = -2x$. Now, we can solve for their intersection. \[\begin{aligned} x + 1 &= -2x\\ 3x &= -1\\ x &= -1/3 \end{aligned}\] we find that they intersect at $x = -1/3$. That's one third of the way across the paper!

Generalizing to Arbitrary Fractions

The simple fact that $3 = 2+1$ played a crucial role in our proof above. The $2x$ on the right hand side and the single $x$ on the left hand side combined to give us a factor of $3x$. And that $3$ became the denominator of $1/3$. So what would happen if our right hand side were $-4x$ instead of $-2x$?

\[\begin{aligned} x + 1 &= -4x\\ 5x &= -1\\ x &= -1/5 \end{aligned}\]

Then, instead of finding a point $1/3$ of the way across the page, we would find a point $1/5$ of the way across the page! In general, if we can fold a line with a slope of $-n$, we can then fold the paper into segments of width $1/(n+1)$.

And how do we fold a line of slope $-n$? Earlier we folded a line with slope $-2$ by first folding the paper in half, and then folding a diagonal cutting one of the halves in half. This creates a line of slope $-2$ because half of a square is a $2:1$ rectangle, and its diagonal has slope $-2$. Similarly, we can use an $n:1$ rectangle to fold a diagonal of slope $-n$.

So given a fold $1/n$ of the way across the paper, we can find $1/(n+1)$ as follows: Suppose we start with a square that has a crease $1/n$ of the way across.

Next, fold and unfold the square in half diagonally. Now, fold and unfold a crease from the bottom right corner to the top of our starting crease. And now you're done! The two diagonal creases intersect each other at a point $1/(n+1)$ of the way across the paper!

Folding 28ths

This procedure gives us a straightforward, if tedious method of folding a square into 28ths: First fold it in half, then find $1/3$, then use $1/3$ to find $1/4$, then use $1/4$ to find $1/5$, and so on, until we finally use $1/27$ to find $1/28$. Of course, this is a terrible idea for several reasons. It would take a long time to fold, and would leave countless extra creases on your square. With a little bit of thought, we can fold 28ths with far less effort, and making minimal extra creases.

$28 = 4 \cdot 7$. Folding things in quarters is easy: just fold in half twice. So the only difficult part of folding 28ths is folding 7ths. $7 = 6 + 1$, so we can obtain $1/7$ by first folding $1/6$. And $1/6$ is just half of $1/3$, which we already know how to fold. Here is whole folding sequence:

First, take your square and fold it in half. Unfold, and you are left with a vertical crease cutting the square in half.

Next, fold and unfold the square in half diagonally. You only need to make a strong crease in the top right. Now, fold the diagonal from the bottom right corner to the middle of the top edge. Make a pinch where this crease intersects the other diagonal crease. As we saw earlier, this intersection is $1/3$ across the paper. Now, fold the right edge to the intersection you just made, pinching at the top of the paper. This creates a pinch $1/6$ across the paper. Now, fold the a diagonal from the bottom right corner to the top of the $1/6$ pinch you just made. Pinch where this diagonal intersects your original diagonal. This intersection is $1/7$ across the paper. Finally, you can fold the right edge to your $1/7$ intersection to create $1/14$, and you can fold the right edge to the $1/14$ crease to create $1/28$. Then you're done!