How does folding paper affect the geometry angles?

Question

This is how you create equilateral triangles.

But I have noticed that if you start of with a flawed fold, you can still end up with equilateral triangles, how come?

Instructions:

Need: Rectangular strip paper.

Step one make an arbitrary crease. Take the right side of the paper and align in to the crease, now fold, you should have another crease. Now alternate, fold up vertically then fold down.

Code: Down, down, up, down, up, down...

Why does any arbitrary crease lead to equilateral triangles.

Answer 1

Problem setup

Consider the strip laid horizontally as in the question, with the first crease made:

   ,---\-------------------------------
   |   \ 2 = 180º-                 ...
   '-----\-----------------------------

Let $\theta=\theta_1$ be the angle on the left of the first crease as measured along the bottom edge. (In the above diagram, $\theta<90º.$) Let $$\theta_n=\text{left angle of the $n$th crease along the } \begin{cases}\text{bottom edge} &\text{if } n \text{ odd}\\ \text{top edge} &\text{if } n \text{ even}\end{cases}$$ Making the second crease creates a line that bisects the angle $2\alpha=180º-\theta$. A little bit of geometry about alternate angles (see proposition 29) shows that $\alpha$ is in fact $\theta_2$. In general $$ \theta_n = 90º-\frac{\theta_{n-1}}2$$

Limit computation

You could quote a theorem about contractive sequences to see that $\theta_n$ has a limit, which must then be $60º$. Alternatively, observe that we can rewrite the above as $$ \theta_n-60º = \frac{-1}{2}( \theta_{n-1} - 60º)$$ so $T_n:=\theta_n-60º$ is a geometric sequence with common ratio $-1/2$ whose modulus is less than $1$. We deduce $T_n\to 0$ and hence $\theta_n\to 60º$, and moreover we obtain the explicit formula $$ \theta_n =60+ T_n = 60º + \left(\frac{-1}2\right)^{n-1}(\theta_1-60º).$$

Verification of example

In particular if $\theta_1 = 18º$ then the theoretically correct angles are \begin{align} \theta_1 &= 18º\\ \theta_2 &= 81º \\ \theta_3 &= 49.5º \\ \theta_4 &= 65.25º \\ \theta_5 &= 57.375º \\ \theta_6 &= 61.3125º \\ \theta_7 &= 59.34375º \\ \theta_8 &= 60.328125º \end{align} which is quite close to your example.