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You are given a square piece of paper. You can cut it into pieces and rearrange them to form a new shape. You are allowed to rotate and flip pieces, provided that they are all used. Can you cut the square into pieces to form these shapes?

  1. An equilateral triangle.
  2. A regular hexagon.
  3. A regular octagon.

Bonus: What is the minimum number of pieces needed for each shape?

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    $\begingroup$ Must the pieces have Jordan curve boundary and are you assuming the Axiom of Choice? $\endgroup$ – hexomino May 14 at 12:17
  • $\begingroup$ @hexomino I am not really sure what these mean? The piece shape can be arbitrary - whatever you can cut from the original paper. $\endgroup$ – Dmitry Kamenetsky May 14 at 12:30
  • $\begingroup$ It's just that I know the answer to 4 already but it depends a lot on your definition of pieces. $\endgroup$ – hexomino May 14 at 12:52
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    $\begingroup$ final shapes have the same area as original. $\endgroup$ – Dmitry Kamenetsky May 14 at 23:46
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    $\begingroup$ I removed the circle as it's too speculative. Also added bonus question about minimising pieces. $\endgroup$ – Dmitry Kamenetsky May 14 at 23:51
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Assuming my last comment on the question, then the answer is

Yes.

This is due to Wallace-Bolyai-Gerwien Theorem: "Two polygons are congruent by dissection iff they have the same area. In particular, any polygon is congruent by dissection to a square of the same area".

For the bonus, the best known dissections are

due to Theobald, G. "Geometric Dissections."

enter image description here

For the original question (before the circle was removed),

"Laczkovich (1988) also proved that a circle is congruent by dissection to a square (furthermore, the dissection can be accomplished using translations only)."

~ "Weisstein, Eric W. "Wallace-Bolyai-Gerwien Theorem." From MathWorld--A Wolfram Web Resource."

~ "Laczkovich, M. "Von Neumann's Paradox with Translation." Fund. Math. 131, 1-12, 1988."

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  • $\begingroup$ Correct! This is the answer I was looking for. I was surprised that a circle can be done too. $\endgroup$ – Dmitry Kamenetsky May 15 at 23:50
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    $\begingroup$ @DmitryKamenetsky Note that the circle proof relies on axiom of choice and on non-measurable subsets (This is one of the reasons why hexomino asked for clarifications). $\endgroup$ – Vepir May 15 at 23:59
  • $\begingroup$ Vepir thanks for the clarification. I also wonder if curved cuts can be used? I posted a new puzzle about this topic: puzzling.stackexchange.com/questions/110080/… $\endgroup$ – Dmitry Kamenetsky May 16 at 1:58

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