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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.

enter image description here

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:

enter image description here

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:

enter image description here

Image grabbed from here

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  • $\begingroup$ two procedures? cut and fold? and at least one fold? are we gonna do it by folding or cutting? $\endgroup$ – Oray Feb 26 at 10:21
  • $\begingroup$ @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. $\endgroup$ – Przemyslaw Remin Feb 26 at 10:24
  • $\begingroup$ 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? $\endgroup$ – Oray Feb 26 at 10:26
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    $\begingroup$ 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. $\endgroup$ – Przemyslaw Remin Feb 26 at 10:29
  • $\begingroup$ I see, let me look into that then :) $\endgroup$ – Oray Feb 26 at 10:30
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Here's one 5 piece solution:

5 pieces. The blue pieces are the construction: 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.square

To make it into an equilateral triangle:

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


Here's a different 5 piece solution:

Here's an answer with

5 pieces. haberdasher adjusment

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::

notch fix

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  • $\begingroup$ 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. $\endgroup$ – Przemyslaw Remin Feb 27 at 8:53
  • $\begingroup$ There are no isosceles triangles. I'll add some diagrams to make it clearer $\endgroup$ – Dr Xorile Feb 27 at 17:51
  • $\begingroup$ @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. $\endgroup$ – Dr Xorile Feb 27 at 22:08
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    $\begingroup$ 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. $\endgroup$ – Przemyslaw Remin Feb 28 at 9:22

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