Create a 1 meter measure

Question

How many folding steps do you need to create a $1 \, \text{m}$ measure from an ideal DIN A0 sheet? This sounds easy, but consider that this paper has edge lengths of $\sqrt[4]{2} \, \text{m}$ and $\frac{1}{\sqrt[4]{2}} \, \text{m}$ respectively and I want a (theoretically) exact $1 \, \text{m}$ measure. That's the whole question – but to make sure everybody understands how the construction should be done, I will show the allowed moves in an example:

In general all moves that don't leave a degree of freedom are allowed. More specifically:
• Folding one corner onto another, for example:
  
  (Unfolding doesn't count as additional step.)
• Folding one edge onto another (or itself), for example:
  
  
• A generalization of the first listed move: Folding a known point onto another known point, for example:
  
  
• Folding around an edge inside the figure, for example:
  
  
• Folding a corner onto an edge. To make sure this has no degree of freedom left, the kink can be fixed by going through a known point, for example:
  
  
• Folding through two known points, for example:
  
  
If somebody comes up with another move that doesn't leave a degree of freedom it is of course also allowed.

In the end you should contruct two points which are exactyl $1 \, \text{m}$ apart. The answer with the least number of moves will be accepted. But to make sure people understand your answer please include a drawing or a photo of how the paper is folded. Also provide a calculation that proves that the edge is really $1 \, \text{m}$. If you prefer to fold a real paper while solving this task you don't need a DIN A0 paper, but instead you can use a DIN A4 paper ($\frac{\sqrt[4]{2}}{4} \, \text{m} \cdot \frac{1}{4 \sqrt[4]{2}} \, \text{m}$) and find a $25 \, \text{cm}$ measure.

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^{Yes, I'm aware of the questions "Create a 3 inch measurement" and "Create a 1 inch measurement". But this question is not so straightforward, because you have to deal with irrational egde lengths to contruct an integer length.}

Answer 1

I can do it in just:

4 folds

Initial configuration:

First:

Folding the paper over a corner so two adjacent sides overlap
In the diagram, this is folding AC over AB

We have:

Now:

Fold between the tip of the paper and the opposite edge, and unfold
In the diagram, this is folding A onto C'

We get the mark:

Finally:

Fold the tip over the side of the mark not on the diagonal onto the line of the paper underneath the main paper, noting the point which the tip folds onto by making another fold through that point not coinciding with the line already passing through that point.
In the diagram, this is folding A over H onto BG

You get:

And the required distance is:

Between the tip and the noted point
In this diagram, this is AF

Why this works:

Because of step 2, $AH=HC'$. But because of step 3, $AH=HF$.
So $H$ is the circumcentre of $\triangle AFC'$, where $AC'$ is a diameter.
A diameter subtends a right angle, so $\angle AFC'=90^\circ$.
$\angle FBC'=90^\circ$ too. Since then $180^\circ-\angle AFC'=\angle FBC'$, with $B$ on the same side of $FC$ as $A$, by alternate segment theorem, $AF$ is tangent to the circumcircle of $\triangle FBC'$.
But then by power of a point, $AF^2=AB\cdot AC'$.
$AB=\frac{1}{\sqrt[4]2}$, and $AC'=\sqrt[4]2$. So $AF^2=1$.
Since $AF\geq0$, $AF=1$.

Answer 2

The $A0$ papers are $1189mm$x$841mm$. Points $B$, $C$, $D$ and $E$ are the midpoints. 1- Fold along $BC$. 2- Fold along $DE$. 3- Fold along $AB$. The $EF$ is $1$ meter. To find the coordinate of point $F$, we write the equation of lines $AB$ and $CD$, then we can use the pythagorean theorem to find the length of $EF$.