Update: The answer to the title question is no, as pointed out by Tapio and Willie. I would be more interested in lower bounds.


[Monsky's famous theorem](https://en.wikipedia.org/wiki/Monsky%27s_theorem) with amazingly tricky proof says that if we dissect a square into an odd number of triangles, they cannot have the same area. However, this proof does not give any quantitative bounds, while from the compactness it follows that there should be some bound, e.g. on the size of the largest triangle that is well separated from the reciprocal of the number of triangles. I don't think the original proof gives any bounds, and I am not aware of any (substantially) different proofs.


So if a unitsquare is partitioned into n (odd) triangles, then what is the smallest $f$ such that the largest will have area at least $f(n)$? Monsky says that $f(n)>1/n$, but how much bigger? If we partition to $n-1$ (even) equal area triangles and then cut one of them, then this gives $f(n)\le 1/(n-1)$, which is not sharp as shown by Tapio and Willie. But can anyone give a better lower bound than $1/n$?



If a unitsquare is partitioned into 101 triangles, is the area of one at least 1%?
As suggested in a comment above, I had asked a version of this question years ago. It makes sense to look at upper and lower bounds for the quantity $f(n)-\frac1n$. It is easily seen that 
$$ f(n) - \frac1n \le \frac1{n^2-n} $$
and indeed has shown that 
$$ f(n) - \frac1n = O\big(\frac1{n^3}\big), $$
see Bernd Schulze's paper "[On the area discrepancy of triangulations of squares and trapezoids](http://www.combinatorics.org/ojs/index.php/eljc/article/view/v18i1p137)", The Electronic Journal of Combinatorics 18(1) (2011), #P137, 16 pp.


Computational experiments from an unpublished Diploma thesis (Katja Mansow, Ungerade Dreieckszerlegungen, TU Berlin 2003, 49 pages, in German) suggest that the true order of this difference is, however, singly-exponentially small. From gap-theorems in semi-algebraic geometry there is a doubly-exponential lower bound (details to be worked out).


