This question already has an answer here:

This is an extension question of this question.

They have given a plenty nice way to fold a A4 sheet in 3 equal parts.

  • What is the Mathematics behind the foldings?
  • Can we use the same way for $A5,A6,\ldots$?

marked as duplicate by Jack M, user147263, Joel Reyes Noche, Jyrki Lahtonen Nov 3 '15 at 14:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ For each $An$ the rate is $1:\sqrt2$ (so that folding $An$ along its long side results in a double $A(n+1)$). $\endgroup$ – drhab Nov 3 '15 at 11:42
  • $\begingroup$ The "duplicate" is the question linked from this one. I think this is indeed an extension of the previous question rather than a duplicate, in that it asks for details that were not considered necessary in the answer to the previous question. $\endgroup$ – David K Nov 3 '15 at 12:50

The method described works for any rectangular piece of paper. And it can be adapted to any $m$ instead of $3$: Find $n$ with $2^n\ge m$, repeatedly halve one of the sides of the rectangle $n$ times (so that it is divided into $2^n$ equal parts. Then the diagonal fold across $m$ of these parts produces intersections at the preceeding folds at $\frac1m,\frac2m,\ldots$ of their length; this can be transfered to the other side of the rectangle by the last fold.

Geometrically, there is merely the intercept theorem behind it.


You could use the "proportion theorem" in Euclidean Geometry: A line drawn parallel to one side of a triangle divides the other sides proportionally.

Folding the paper in quarters, draws three parallel lines, equal distances apart. Folding the cross section from a corner to the third fold, creates two right angle triangles, with two lines parallel to the third side - the side we want to divide in three.

Consider the triangle whose sides are both on the edge of the paper: the parallel lines are still the same distance apart which means the one right angle side is divided into three equal parts (thirds). Using the proportion theorem, we can conclude that the hypotenuses is now also being divided into thirds by the parallel "lines".

Finally we fold perpendicular lines towards the side we want to divide through these intersection points. These lines are again parallel to a side in this triangle therefor dividing the the other side proportionally, which is again in thirds.

By extension you can then fold a paper into n fold by folding it in half m times so that $2^m>n$ then folding the cross section to the n-th parallel lines and then folding the n perpendiculars that will divide the side as desired.


Not the answer you're looking for? Browse other questions tagged or ask your own question.