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 α_{0}=a_{0}π/n (if you wanted, say, a 36° angle, multiply it by π/180 to put it in this form: 36°=π/5), where a_{0} and n are relatively prime and n is an odd number (this is not a serious restriction, which we'll see later).

_________________________________________________________________

\ ___)__ α_{0}

\

0

\

_____\________________________________________________________________

If a_{0} is even, we fold the paper so that crease 0 lines up with the top of the paper. α_{0} is bisected, such that α_{1}=a_{1}π/n where a_{1}=a_{0}/2 (note: this is an integer because a_{0} is even):

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\– ___)__ α_{1}

\ –

0 1

\ –

_____\______________–___________________________________________________

If a_{0} is odd, we switchfold the paper so that
crease 0 is in line with the bottom of the paper. α_{0}'s supplement is bisected, such that α_{1}=a_{1}π/n where a_{1}=(n-a_{0})/2 (note: this is an integer because both n and a_{0} are odd):

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\ /

\ /

0 1

\ /

_____\/_)α___{1}__________________________________________________________

Repeat this process with respect to α_{1}, α_{2}, ... until α_{m}=α_{0}. Then we have our algorithm. Some sample algorithms:

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

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\ / – \ – /

\ / – \ – /

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 β_{0} and apply the angle-folding algorithm for a specific degree measure b_{0}π/n, the angle β_{im} will approach b_{0}π/n as i goes to infinity (m is the period of the algorithm; proof below). After you have made the angle b_{0}π/n, you can halve that angle by applying folds, yielding b_{0}π/(2^{r}n) (This is why the odd n restriction isn't limiting: if you want to find some a_{0}π/n, rewrite n=2^{r}m, where m is odd, find a_{0}π/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 α_{0}=a_{0}π/n be acute. Let's say n=1 and 0<a_{0}<1/2 is rational. If α_{l+1} is created by a fold of α_{l}, a_{l+1}=a_{l}/2. If α_{l+1} is created by a switchfold of α_{l}, a_{l+1}=(1-a_{l})/2.
Let's say our algorithm consists of j_{1}-1 folds, followed by a switchfold, then j_{2}-1 folds, followed by a switchfold, ..., then a switchfold, then j_{k}-1folds:

a_{j1-1}=a_{0}/2^{j1-1}

a_{j1}=(1-a_{0}/2^{j1-1})/2=(1-a_{0})/2^{j1}

a_{j2+j1-1}=(1-a_{0})/2^{j2+j1-1}

a_{j2+j1}=(1-(1-a_{0})/2^{j2+j1-1})/2=(2^{j2+j1}-1+a_{0})/2^{j2+j1}

.

.

.

The more switchfolds you do, the larger the non-a_{0} term in the numerator grows, until a_{0} is negligible. Therefore, any algorithm with switchfolds, repeated enough times, will yield a result regardless of the original choice of a_{0}. 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

Source:

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