# Folding paper into corners

Here is a piece of paper:

Fold it once, and you can get a shape with 9 corners:

Starting with a rectangular sheet of paper and folding twice (along any line), what is the largest possible number of corners that may result?

Rule for counting corners: both concave and convex corners count (as you can see in the example), but they must not be "covered" by the rest of the paper. In other words, you are counting the corners of the silhouette.

Credit: Alex C Weiner

• Nice puzzle - you may have to ask to fold it thrice. So far 0, 1, 2 folds give 4, 9, 16 corners - looks interesting!
– Tom
Jun 28, 2016 at 10:26
• May need to say if can just fold the rectangle or along any line.
– Tom
Jun 28, 2016 at 20:30
• If you make it even thinner, you can fold the 16-solution like this nup.pw/QhvFL5.png, giving 28 corners. So, (x+2)^2 is not optimal. Jun 28, 2016 at 21:18
• Carl you have missed two concave corners in your diagram, so the answer for three folds must be at least 30. Jun 29, 2016 at 4:31
• @CarlLöndahl: The 5-pointed star made from a thin strip folded 4 times has: 5 inner triangles, a central pentagon, 5 exterior angles, 2x4 "fold" corners, plus the 9 angles from crossing the two strip ends, for a total of 15 + 5 + 5 +8 +9 = 42, so the "rule" does not hold. Jun 30, 2016 at 0:14

Using a long and thin rectangular strip, you can get up to

$17$ corners.

Here's how:

• The last three corners come from the triangular hole inside the shape. Feb 26, 2018 at 20:21
• Looks like a new winner!
– Owen
Feb 26, 2018 at 21:35

I managed to make

$16$ corners with a thin strip, or $15$ with any rectangle including a square.

Like so:

For a strip:

For a square:

- fold along the red lines
(the black lines on the back are the same as the front except rotated to have the 16th lines in the bottom quarter rather than the top one.)

For a rectangle with less width than height do the same but move the first red line's ends nearer the centre and the second red line's ends nearer the quarter marks (at some point you can make $16$ corners instead though).

• Is this always possible for an arbitrary n x m rectangle? I feel like it isn't but I wouldn't know how to go about proving that. Jun 28, 2016 at 13:40
• @RowlandB It needs to be pretty thin to allow both the hole and the two ends to cross (try it with an A4 sheet!); I have not calculated the limits. Jun 28, 2016 at 14:08
• @RowlandB - I added a case for one less corner that works for a square. Jun 28, 2016 at 17:38
• The strip thing didn't convince me (as not following the spirit of this challenge), but the second part deserves a +1. Jun 29, 2016 at 8:41
• i.imgur.com/ph8DPCb.png <-- diagram. Showing the limits as the hole shrinks to a point and the sticking out corners reduce to touching an edge. I am assuming that this is minimal when using symmetric equilateral triangles, but you could do some calculus on varying the angles to find out. +1 to JA for considering identifying a case outside the general case. Jun 29, 2016 at 22:54

I have managed to come up with

14 corners with two folds.

Following are the images of the paper with each fold.

Unfolded

One fold

Two folds

Corners highlighted

14

by

take a strip of paper (still technically a rectangle) and fold it so the ends cross leaving a triangular hole.