# Geometry haberdasher problem - square to equilateral triangle variation

Let me remind the haberdasher's problem, proposed in 1907 by the puzzle composer Henry Dudeney. Dissect an equilateral triangle to a square, with only three cuts.

I would like to propose the variation of haberdasher's problem. Imagine the square to be made of two colored paper. Like this - one side red and the other yellow:

Here is my variation of the riddle, refined after comments
Cut the square with the least number of cuts to form equilateral triangle, provided that:

1. At least one element is flipped to the other side.
2. At least one element remains on its original side.
3. The flipped element is asymmetric.

(I would most welcome propositions to reword the riddle in less words.)

So say we have square painted with red on one side and yellow on the other. Starting with completely red square we build red-yellow triangle. Third condition prohibits flipping isosceles triangles or squares. In other words, the flipped element may not remain not flipped.

I suspect that there exists a solution:

1. for four elements with 1 element flipped
2. for four elements with 2 elements flipped

These solutions I would like to find. I would like to exclude all trivial solutions based on asymmetric mirror shaped figures like the letters db.

Update. Please be observant. The shapes in original Henry Dudeney solution are not symmetric. Exact measures are presented here:

Image grabbed from here

• two procedures? cut and fold? and at least one fold? are we gonna do it by folding or cutting?
– Oray
Feb 26 '19 at 10:21
• @Oray not fold. Cut and flip on the other side like with pancakes. Say we have a <b>wooden square</b> and you cut it. It's been painted one side red and the other yellow. No folding of wooden boards possible. Feb 26 '19 at 10:24
• then why not just cut mirror image for the given example above for some pieces? and turn it around to make it the same shape?
– Oray
Feb 26 '19 at 10:26
• You mean the green triangle? Any of the four shapes is not symmetric. Each of the sides of the green triangle has different length. It is not that simple. Feb 26 '19 at 10:29
• One neat thing about the haberdasher solution is that you could connect the pieces with hinges so that if you move the hinges to one extreme you get the square and then you can open them the other way to make a triangle. It would make a fun table. Feb 26 '19 at 14:38

Here's one 5 piece solution:

5 pieces. The blue pieces are the construction: In this case, the yellow shows a rectangle that is $$\sqrt{3}\times 1$$, and the red shows a square that is $$\sqrt{\sqrt{3}}\times \sqrt{\sqrt{3}}$$.

To make this into a square:

Simply slide a and e down to the right and move b into the remaining space.

To make it into an equilateral triangle:

Flip a, b, and c vertically as a unit and move them to the right.

Here's a different 5 piece solution:

Here's an answer with

5 pieces.

You will see that

Both b and c need to be flipped (obviously would be better if only one needed to be flipped). The construction is to take the original green piece and flip it upside down (to make b). Then c is equal to the piece that's missing out of the original yellow piece (but also flipped - unless that's isosceles, which I doubt))

However, which is still a problem, it's possible to just solve it the original way without any flips. To solve this, you can simply take a symmetrical sliver out of the blue piece and have it so that this must be filled by a matching extra bit on piece b. This then enforces the need to flip the piece::

• Are both your solutions based on flipping the isosceles triangle to the other side? Probably the square stretched to a rectangle in your second attempt? Please review it. Feb 27 '19 at 8:53
• There are no isosceles triangles. I'll add some diagrams to make it clearer Feb 27 '19 at 17:51
• @PrzemyslawRemin, hopefully now the diagrams are clearer. I think this does what you want by requiring some pieces must be turned over. Both solutions are 5 pieces now and require 2 pieces to be flipped over. Feb 27 '19 at 22:08
• Yes! Very clever idea to use intermediate rectangle! Bravo! Your second approach is now clear to me. It is a correct solution for 5 elements. I will leave the question open for some time inviting ideas for less cuts. Feb 28 '19 at 9:22