Cutting a hexagon to make an equilateral triangle

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?

-
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

Solution discovered by Harry Lindgren (1961)

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)
-
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