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For which $n$ is it possible to divide an equilateral triangle into $n$ equal (i.e., obtainable from each other by a rigid motion) parts?

It is easy to come up with a partition for $n \in \{1, 2, 3, 4, 6\}$. Also very easy to come up with a partition into $k^2$ equal equilateral triangles(you need to build a triangular grid). It follows that if we can build a partition into $n$ equal parts, then we can also build a partition into $nk^2$ for any $k \in \mathbb{N}$.

It is also well known that Mikhail Patrakeev invented the division into $5$ equal parts.

So my question is, what is known about the remaining $n$ that are not represented as $ck^2$, where $c \in \{1, 2, 3, 5, 6\}$ (It is $7, 10, 11, 13, 14, 15, 17...$ parts)? Is there any other known division? Is it proved for some $n$ that there is no corresponding partition into $n$ equal parts?

Vosatorp
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  • I guess there should be some constraint on the "parts", e.g. should they be measurable? But it's already a very interesting question. One might ask the same question for e.g. squares or basically any shape. – WhatsUp Mar 02 '21 at 12:09
  • Oh, yes, this is it, but I don't understand how they were able to divide the triangle into $10n^2$ equal parts. – Vosatorp Mar 02 '21 at 12:52
  • This question has been closed, but I wanted to note that I have added a new answer on the duplicate question from 2016; squarefree parts of $5,6,10,13,14,15,21,30,35,39,55,65,66,70,\ldots$ are all possible using triangular pieces. – RavenclawPrefect Mar 02 '21 at 16:11

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