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.

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.
  • $\begingroup$ we will use A0 or other measure to solve the problem? $\endgroup$
    – Oray
    Commented Oct 29, 2017 at 20:40
  • $\begingroup$ @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. $\endgroup$
    – A. P.
    Commented Oct 29, 2017 at 20:43
  • $\begingroup$ 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. $\endgroup$
    – boboquack
    Commented Oct 29, 2017 at 21:58
  • $\begingroup$ @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. $\endgroup$
    – A. P.
    Commented Oct 29, 2017 at 22:12
  • $\begingroup$ Does noting a point on a line that something folds onto count as another fold? $\endgroup$
    – boboquack
    Commented Oct 29, 2017 at 22:20

2 Answers 2


I can do it in just:

4 folds

Initial configuration:

Step 0


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

We have:

Step 1


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:

Step 2


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:

Step 3

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

  • $\begingroup$ Very neat! Note that since AFC' and ABF are similar, you have that AF is the arithmetic mean between AB and AC', leading to the result immediately. $\endgroup$
    – Florian F
    Commented Nov 7, 2021 at 11:31
  • $\begingroup$ Sorry for unaccepting your answer, but I just realized that there is a solution with less folds. $\endgroup$
    – A. P.
    Commented Nov 7, 2021 at 21:03

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 $EF$.

enter image description here

  • 1
    $\begingroup$ I get $EF = \sqrt{\frac{17}{12}} 2^{-\frac{1}{4}} \approx 1.00087$. But as boboquack said, it's not exact. $\endgroup$
    – A. P.
    Commented Oct 29, 2017 at 22:53
  • $\begingroup$ @A.P.That is because I used the millimeter dimensions. $\endgroup$
    – Seyed
    Commented Oct 29, 2017 at 22:55
  • 1
    $\begingroup$ @Seyed Then I still get $EF = \sqrt{\frac{18025153}{18}} \, \text{mm} \approx 1000.698 \, \text{mm}$. $\endgroup$
    – A. P.
    Commented Oct 29, 2017 at 23:09
  • $\begingroup$ @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. $\endgroup$
    – Seyed
    Commented Oct 29, 2017 at 23:16

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