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:
- At least one element is flipped to the other side.
- At least one element remains on its original side.
- 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:
- for four elements with 1 element flipped
- 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