# Create a 1 meter measure

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:

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 have a polygon with at least one edge that is exactly $$1 \, \text{m}$$ long. 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.

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 very odd egde lengths.

• we will use A0 or other measure to solve the problem? – Oray Oct 29 '17 at 20:40
• @Oray Either you use A0 paper to create a polygon with at least one edge that is exactly $1 \, \text{m}$ or A4 paper to create a polygon with a $25 \, \text{cm}$ edge. – A. P. Oct 29 '17 at 20:43
• Isn't folding a corner onto a corner a special case of folding a corner onto a known point, and isn't this a special case of folding a known point onto another known point which also should be possible? Also, egde should be edge. – boboquack Oct 29 '17 at 21:58
• @boboquack Yes, you're right. I've edited it to be more precise. I still left the example with folding a corner onto another corner, so that the examples are more or less ordered in ascending complexity. – A. P. Oct 29 '17 at 22:12
• Does noting a point on a line that something folds onto count as another fold? – boboquack Oct 29 '17 at 22:20

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 $AF\geq0$, $AF=1$.

• This is the shortest solution I found so far. Please correct the errors in the last spoilerbox (AH=AC', AH=AF, ∠ABC'=90°, AB=√√2, and AC'=1/√√2) and use LaTeX, then I'll accept it. – A. P. Oct 29 '17 at 23:00
• @A.P. what's the error I need to correct? – boboquack Oct 30 '17 at 1:34
• You corrected most of them already. The remaining ones are: AH = AC' $\to$ AH = HC', AH = AF $\to$ AH = HF, AB = $\sqrt[4]{2}$ $\to$ AB = $\frac{1}{\sqrt[4]{2}}$, AC' = $\frac{1}{\sqrt[4]{2}}$ $\to$ AC' = $\sqrt[4]{2}$ – A. P. Oct 30 '17 at 7:54
• @A.P.  done!  – boboquack Oct 30 '17 at 10:36

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 $EF$.

• I get $EF = \sqrt{\frac{17}{12}} 2^{-\frac{1}{4}} \approx 1.00087$. But as boboquack said, it's not exact. – A. P. Oct 29 '17 at 22:53
• @boboquack, That is because I used the millimeter dimensions. – Seyed Oct 29 '17 at 22:54
• @A.P.That is because I used the millimeter dimensions. – Seyed Oct 29 '17 at 22:55
• @Seyed Then I still get $EF = \sqrt{\frac{18025153}{18}} \, \text{mm} \approx 1000.698 \, \text{mm}$. – A. P. Oct 29 '17 at 23:09
• @A.P., You are right and I am not insisting that my solution is correct, but it is less than one millimeter and "I think" it might be due to the fact that the dimensions of the iso paper sizes are not rational. – Seyed Oct 29 '17 at 23:16