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Is it possible to divide an equilateral triangle into 5 equal (i.e., obtainable from each other by a rigid motion) parts?

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Does "reflection" count as rigid motion? – KennyTM Oct 29 '10 at 17:20
I think it does not leave the arrangement of triangle angles unchanged so no. – Jaska Oct 29 '10 at 20:10

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You might want to look at:

http://www.michaelbeeson.com/research/papers/TriangleTiling.pdf

and the references given there.

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In particular, look at the pinwheel tiling, which is for right triangles. See en.wikipedia.org/wiki/Pinwheel_tiling – lhf Oct 30 '10 at 2:18
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Umm. I was looking a solution for equilateral triangle so how right triangle helps me? – Jaska Oct 30 '10 at 12:02
It seems that, according to Theorem 7 (on page 116) of the reference given by Joseph Malkevitch, the answer to your question is "no." – Joel Reyes Noche Nov 16 '11 at 3:59
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I -think- there's a proof related to sums of angles that says this can't be done.

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Could you elaborate? I don't see how that might work. – Jaska Oct 29 '10 at 20:11

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