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It's easy to make a 30° angle on a piece of graph paper or an angle of any degree with a protractor, but if you're looking for a way to make angles and all you've got is a plain ol' piece of paper, there is still hope. Jean Pedersen and Peter Hilton have developed a process by which any rational angle can be created with an infinitely long piece of paper, and on the off chance that you don't have an infinitely long piece of paper, a good approximation is possible with a moderately-sized piece of paper.

The first step is to figure out the angle-folding algorithm for your angle. Pick an acute angle α0=a0π/n (if you wanted, say, a 36° angle, multiply it by π/180 to put it in this form: 36°=π/5), where a0 and n are relatively prime and n is an odd number (this is not a serious restriction, which we'll see later).

 \ _) α0

If a0 is even, we fold the paper so that crease 0 lines up with the top of the paper. α0 is bisected, such that α1=a1π/n where a1=a0/2 (note: this is an integer because a0 is even):

 \–  _) α1
  \      –
   0           1
     \                  –

If a0 is odd, we switchfold the paper so that crease 0 is in line with the bottom of the paper. α0's supplement is bisected, such that α1=a1π/n where a1=(n-a0)/2 (note: this is an integer because both n and a0 are odd):

 \                      /
  \                 /
   0          1
     \      /

Repeat this process with respect to α1, α2, ... until αm0. Then we have our algorithm. Some sample algorithms:

π/5: switchfold — fold — switchfold — fold — switchfold — fold — ...

 \                      /                        –   \ –                                                                         /
   \                 /                  –              \           –                                                         /
   0          1            2                         3                     4                                         5
     \      /      –                                       \                                 –                       /              –

π/7: switchfold — switchold — fold — switchfold — switchfold — fold — ...

 \                      / \ –                                                                  / \ –
  \                 /         \          –                                                   /       \          –
   0          1                 2                 3                                   4            5                  6
     \      /                           \                           –                   /                     \                             –

Now here's the trick: if you start with any angle β0 and apply the angle-folding algorithm for a specific degree measure b0π/n, the angle βim will approach b0π/n as i goes to infinity (m is the period of the algorithm; proof below). After you have made the angle b0π/n, you can halve that angle by applying folds, yielding b0π/(2rn) (This is why the odd n restriction isn't limiting: if you want to find some a0π/n, rewrite n=2rm, where m is odd, find a0π/m, and halve it r times). In English, it doesn't matter what angle you start with: if you use the sequence for π/5, you'll get π/5, and rather quickly. Then can fold it in half to get π/10, π/20, etc.

The next time you are bored, grab a piece of lined paper and rip off the margin. See what interesting angles you can fit on it. If you fold along the right creases, you can turn your strip of paper into a regular (star) {n/a}-gon. To do this, pick a pair of creases that meet at the top edge such that the angle between them is aπ/n. Find n such pairs of creases that are equally spaced along the piece of paper, and fold and twist until your polygon/star is complete.

Try and think of a better way to impress your math buddies than by making a perfect 17-pointed star.

Quasi-Proof: Let α0=a0π/n be acute. Let's say n=1 and 0<a0<1/2 is rational. If αl+1 is created by a fold of αl, al+1=al/2. If αl+1 is created by a switchfold of αl, al+1=(1-al)/2. Let's say our algorithm consists of j1-1 folds, followed by a switchfold, then j2-1 folds, followed by a switchfold, ..., then a switchfold, then jk-1folds:

The more switchfolds you do, the larger the non-a0 term in the numerator grows, until a0 is negligible. Therefore, any algorithm with switchfolds, repeated enough times, will yield a result regardless of the original choice of a0. In the case of an algorithm composed of all folds would yield an angle of 0, which is not terribly interesting, nor is it included in the algorithm. QED

Polster, Burkard. Variations on a Theme in Paper Folding The American Mathematical Monthly. Jan 2004, Vol 111, #1 pp 39-47.

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