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The problem is to cut a regular hexagon into parts that can be put together (without overlaps or wasting any parts) to make an equilateral triangle. The cuts should all be straight.

What is the smallest number of parts that will still let you achieve this?

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I count an upper limit of three cuts, six pieces... What have you tried? –  abiessu Nov 24 '13 at 20:12
    
I can do 6 parts but no lower although I would be interested to see your solution as I use 5 cuts. –  marshall Nov 24 '13 at 20:13
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2 Answers

up vote 5 down vote accepted

Solution discovered by Harry Lindgren (1961)

enter image description here

My explanation on how to compute all points :

  1. $A,B,C$ are the vertices of an equilateral triangle
  2. $M$ is the middle of $AC$ and $N$ is the middle of $BC$
  3. $i$ is the projection of $M$ on $AB$
  4. $P$ is the point between $i$ and $B$ such that $|iP|=|iM|$
  5. $Q$ is the point between $A$ and $P$ such that $|QP|=|AM|$
  6. Consider the triangle $ABN$. Let $\delta$ the bisector line of the angle $A$ ($NAB$ to be precise). $R$ is intersection of $PM$ and the parallel to $\delta$ through $Q$.
  7. $S$ is the point between $R$ and $M$ such that $|RS|=|QR|$
  8. $T$ is the point between $R$ and $S$ such that $|RT|=2|SM|$
  9. $V$ is the point such that $TV=QS$ (QSVT is a parallelogram)
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That's really great, thanks. Can points 3. and 6. be done using ruler and compass? Also, is this solution unique in some sense? –  marshall Nov 27 '13 at 11:59
    
yes, they can because you can build a perpendicular (or parallel) line or bisector line with compass and ruler. –  Xoff Nov 27 '13 at 12:02
    
For the unicity, I don't know any other solution with 5 or less pieces. But I think it's still an open question. Not sure. –  Xoff Nov 27 '13 at 12:09
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If you want a peak at an answer (and more), try http://mathworld.wolfram.com/Dissection.html

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For those avoiding looking so as not to spoil their own fun, I'll point out that the hexagon-to-triangle dissection shown on that page doesn't use straight cuts. (One cut has a kink in it.) –  Blue Nov 24 '13 at 21:15
    
@Blue It looks like there may be one of the apparent line segments which is really two segmenmts making an angle near 180 degrees. Is that what a "kink" is? If so it only increases the count of cuts by 1, and still keeps the number of regions as 5. –  coffeemath Nov 25 '13 at 0:19
    
@coffeemath: The kink I see in the "6-3" figure is a dramatic bend, making a clearly concave piece that wraps around one corner of the (apparently-equilateral) triangular piece in the hexagon arrangement. –  Blue Nov 25 '13 at 0:28
    
The lines look straight to me although some start and stop at unspecified positions. It's not at all clear how you actually make the cuts. I feel a really good answer would come with a set of instructions for using a ruler and compass. Can this be done? –  marshall Nov 25 '13 at 17:14
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