You have a straight line $l$ in an infinite plane.

  • You can fold the plane along any straight line so the line $l$ becomes two rays with a common starting point. In the picture we fold along line $a_1$ to get the shape that is composed of the red ray $AB$ and the blue ray $AE$
  • You can fold the plane a second time so the two rays becomes some other shape. In the picture we fold along line $a_2$ to get the shape that is composed of the red ray $BF$, the green segments $BD$ and $DE$, and the blue ray $EC$.

enter image description here

After the second fold, the shape has one cross point $C$, where the green segment and the blue ray cross each other.

Question: If you can fold the plane four times, what is the maximum number of cross points you can get?


1 Answer 1


I think the answer is


because the maximal increments at the four foldings are

0,1,6,28 which are the binomial coefficients $\begin{pmatrix}2^0\\2\end{pmatrix},\begin{pmatrix}2^1\\2\end{pmatrix},\begin{pmatrix}2^2\\2\end{pmatrix},\begin{pmatrix}2^3\\2\end{pmatrix}$.

Here the powers $2^k$ arise as the

maximum number of lines/rays/line segments after k foldings.

The binomial coefficients count

The number of pairs that can be formed among them. Indeed, each segment ("mother segment") will be folded into at best two sub segments ("children") and new intersections can only be formed between children not on the same side of the fold. Further, siblings cannot cross because their intersection lies on the fold. And finally of the children of any two mothers at most one new pair can be formed. Because the two candidates pairs (the two non sibling, opposite sides of fold pairs)---being mirror images of each other---intersect on opposite sides of the fold, hence at most one of them can be manifest.

This concludes proof (well, sketch of proof) that the number I give is an upper bound.

To construct a maximal example simply

make the first fold almost perpendicular to the given line. And the consecutive folds almost parallel to the first. We can then make sure the maximum number of segments is created by always bisecting the corridor left by the two preceding folds, We can ensure the maximum number of intersections by always very slightly deviating from the new fold being exactly parallel instead rotating it such the all the intersections with the previous fold can be separated from all the intersections with the fold before that by a line parallel to the given line.


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