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Given two congruent regular hexagons, we should partition them into a total of $n$ pieces. What is the smallest value of $n$ so that the $n$ pieces together can be formed into an equilateral triangle?

If we start with only one hexagon, it is possible to use five pieces. But we can't combine an equilateral triangle and a hexagon, or two equilateral triangles together. In addition, from a regular hexagon we can make two equilateral triangles by cutting segments $AC,CE,EA$ if the hexagon is $ABCDEF$.

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  • $\begingroup$ Which evidence do you have that this is even possible? $\endgroup$ – Henning Makholm Aug 30 '16 at 12:43
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    $\begingroup$ @HenningMakholm, it's an old theorem that any two simple polygons of equal area are equidecomposable. So start with the equilateral triangle and first cut it in half. Then decompose each half to produce a hexagon. That shows it's possible. (Of course Aretino's answer does better, in terms of fewer pieces.) $\endgroup$ – Barry Cipra Aug 30 '16 at 15:22
  • $\begingroup$ @BarryCipra: When I commented, the question was asking specifically for a decomposition into $6$ parts. $\endgroup$ – Henning Makholm Aug 30 '16 at 16:59
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Six pieces: green hexagon is cut into five pieces and red one is a single piece:

enter image description here

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  • $\begingroup$ Nice construction! But can you prove $6$ is the least number of pieces for which a construction is possible? $\endgroup$ – Barry Cipra Aug 30 '16 at 15:05
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    $\begingroup$ @BarryCipra Of course I cannot: this is just the best I can do at the moment. $\endgroup$ – Aretino Aug 30 '16 at 15:06
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Here is another six piece dissection.

Two equally sized regular hexagons can be cut into two and four pieces as follows:

The six pieces can then be reassembled to form a single equilateral triangle as follows:

Like @Aretino, I cannot prove 6 is the least number of pieces needed for such a construction.

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