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Given $m \times n$ rectangle with area $A$, and $m,n \in \mathbb{N}$. Let $S_k(m,n)$ be the number of way to dissect this rectangle into $k$ non-overlapping triangles whose area is $\frac{A}{k}$.

It is known that when $k$ is odd $S_k(n,n)=0$ (For example see Paul Monsky, "On Dividing A Square Into Triangles", American Mathematical Monthly, Vol 77 no 2, 1970)

Is it true that $S_k(m,n) < \infty$?
Is there any known bound for $S_k(m,n)$ (even when $m=n$ and $k$ even)?

Edit: There are several possible interpretations of the problem, one may assume that the vertices of the triangles are lattice points. Or perhaps (harder?), when we are given an arbitrary rectangle with a fixed dimension (not necessarily has an integer dimension).

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Just to check, are there any conditions on the triangles? Do the vertices need to be lattice points? Then again, judging from your $S_k(m,n) < \infty$, I guess that the answer is no? – Calvin Lin Jan 3 '13 at 7:49
it's entirely an open question, so it can be lattice point or not. – Ajat Adriansyah Jan 3 '13 at 8:14
@AjatAdriansyah, open question for you means unknown answer or no more conditions needed on the triangles? – Sigur Jan 3 '13 at 10:40
To my knowledge it is open, and I would like to present it in a more flexible open-ended fashion. For example one can consider the variant of the problem where the dimension of rectangle is not specified. – Ajat Adriansyah Jan 3 '13 at 12:41
It seems when the triangle’s vertices are nor necessarily the lattice points then $S_k(n,m)$ does not depends on $n$ and $m$, because an affine transformation can transform the dissected rectangle into a square, preserving all triangles and their relative areas. – Alex Ravsky May 1 '13 at 0:22

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