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Monsky's theorem states that it is impossible to dissect a square into an odd number of triangles of equal area. If $n$ is an even integer, I am interested in the number of ways of dissecting a square into $n$ such triangles. Call $s_n$ this number. My question is:

What are the first terms of the sequence $s_n$?

with an option on this other question:

How can we compute $s_n$ for a given $n$?

Judging from the lack of answers to this question, it does not seem trivial that these numbers are even finite, let alone known. Obviously, $s_2=2$, and I believe that $s_4=25$. Other values are unknown to me.

Added 15/05/2015: I believe that $s_6$ is at least (and perhaps exactly) equal to $818$. This, however, does not lead to any sequence of the OEIS, so I'd be happy to be proved wrong.

Added 03/06/2015: For $n=2,4$ (but not for $n=6$, see mjqxxxx's answer below) the equidissections of the square that I could find all involve triangles whose vertices are rational numbers. For larger $n$, it is not sufficient to look only at equidissections satisfying this property.

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Pierre-Guy Plamondon
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  • Forgive me if this question is stupid, but if two dissections can be made identical by using the symmetry of the square, would you still consider them distinct? (I think that by your note that $s_{2}=2$ the answer to this is "yes" but you may have thought of one I've overlooked.) Also, is the order in which the dissecting edges are added important to you? – Shai May 08 '15 at 18:28
  • Can you enlighten us on how you got $s_4$? I must be overlooking massive amounts of dissections since I can only come up with $9$ ($\times$, $\backslash/$ and its four rotations, $//$ and its four rotations / reflections) – AlexR May 08 '15 at 18:34
  • @Shai Indeed, I am looking for the number of dissections in an "absolute sense", not up to symmetries of the square (although this number would be interesting as well). I am not sure I understand what you mean by the "order" in which the edges are added, but I would guess that the answer is no: I am only interested in the dissections themselves, not their construction (unless this leads to an answer to my second question). – Pierre-Guy Plamondon May 08 '15 at 18:34
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    @AlexR There are two ways of dissecting the square into two isoceles triangles, and each of these can be dissected in 3 ways. This gives $2\times 3\times 3$ dissections, one of these (the $\times$ one) appearing twice. So $17$ dissections. Now, you can split the square into two rectangles in $2$ ways, and each rectangle can be split into two triangles in $2$ ways. This adds $8$ new dissections, for a total of $25$ (I had overlooked the doubly counted dissection, editing my post right now!). – Pierre-Guy Plamondon May 08 '15 at 18:40
  • @Pierre-GuyPlamondon Ah so I forgot about the $16$ with a "ray" pattern. Thanks :) – AlexR May 08 '15 at 18:42

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Even for $n=6$ it doesn't seem necessary for the coordinates to be rational. For instance, consider the following dissection into six right triangles of equal area: enter image description here

The four interior points do not, I think, have rational coordinates. Calling the lower left corner $(0,0)$ and the lower right corner $(1,0)$, I found the interior point at lower right to be at $\left(\frac{1}{2} + \frac{1}{6}\sqrt{5}, \frac{1}{3}\right)$.

mjqxxxx
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  • Nice example! It seems this pattern can be used to produce equidissections into $n$ triangles for any even $n\geq 6$, and that most of them will have irrational coordinates. – Pierre-Guy Plamondon Jun 10 '15 at 09:49
  • I'll award you the bounty for this example that I'm very glad to see, but I won't accept your answer since the question is still not settled. – Pierre-Guy Plamondon Jun 10 '15 at 15:29