# Fold a standard paper to get a Rhombus

A follow up to the "Create a 3 inch measurement" post.

Can you create a Rhombus ( a parallelogram with all sides equal AND which is not a square-- my restriction) using a standard letter size paper ( 8.5 x 11 inches) merely by folding, in 7 steps or less?

The solution must be a folded paper that is a Rhombus and no paper sticking out.

No marking or cutting. No use of any tools (like rulers). Only folding. No external help like computers or books etc.

• I think fold a paper then unfold it is "marking" the papper – Jamal Senjaya Oct 27 '17 at 3:42

I can do it in just three folds and a single unfold.

1. Fold the paper so that two opposite corners (call them A and C) meet. This creates a fold going from one long side to the other; call the points where this fold meets the long edges B and D so that A is on the same side as D and B is on the same side as C. Or see the image.
2. Unfold.
3, 4. Fold along AB and CD.
Image below:

Proof that this is a rhombus:

Note that the fold in the first step, BD, is the perpendicular bisector of AC. Therefore, AB = BC, CD = DA. By symmetry (since BD passes through the centre of the paper), we rotate 180 degrees to find that BC = DA. So ABCD is a rhombus.

• Ooh, nice!   – boboquack Sep 26 '17 at 19:23
• ...and is also always the largest possible rhombus that can be folded from any specific rectangle!!! – Penguino Sep 26 '17 at 22:23
• @Peter Tailor ...but then it would be a parallelogram but not a rhombus (unless I am misunderstanding what you mean by bisecting DE)... – Penguino Sep 27 '17 at 20:00

Here is a 6-step way:

Fold the paper vertically and horizontally in half, and unfold
Call the points where the folds meet the edges A, B, C and D clockwise
Fold the shape ABCD, by symmetry, this is a rhombus (all side lengths are the same), and not a square (one diagonal is 8 in and the other 11.5 in)
[Note: it is well-known that any specified convex polygon contained within a piece of paper can be folded in finitely many moves, and all rhombi are convex actually, don't need this since folding the sides in doesn't pass through the exterior of the rhombus (and it also doesn't bound the number of moves)]
This takes 2+4=6 moves

• Maybe could someone edit an image in? Thanks! – boboquack Sep 23 '17 at 21:35
• you can fold the diagonals while at the quarter-size stage for 3 less folds (or is it???) – JMP Sep 23 '17 at 23:07
• Is this on a 8.5 x 11 inch paper? Can you explain the geometry. May be I dont understand it. – DEEM Sep 23 '17 at 23:24
• @DEEM: It will work on any rectangular piece of paper. Think of it as drawing a line between the mid points of adjacent sides. Its obvious that each line will be the same length and that it is not a square (unless the original piece of paper is a square). – Chris Sep 24 '17 at 0:27
• @JonMarkPerry they need to fold in different directions to be able to open up again So you could do them with the quarter sized to prove they are all equal, but it would be the same number of folds and a little more awkward. – user19641 Sep 24 '17 at 4:41