There is no hypothesis about the area of the polygon?

Nope, this is the problem. Only one student succesfully solved it. Another one was pretty close. Both can re-take the Junior TSTs next year.

ok, let's go : Let $A,B$ be two given vertices of the $n$-gon. If $M$ is another vertex such that $AMB$ has area 1. Then $M$ belongs to one of the two lines parallel to the line $AB$ at distance $ \frac {2} {AB}$ from it. Using the convexity of the $n$-gon, we deduce that there are at most 4 such points $M$, two on each side of the line $AB$. - There are exactly $n$ choices of $A,B$ such that $AB$ is a side of the $n$-gon. For such a choice, there are at most two points $M$ as above, so there are at most 2 triangles of area 1. - There are exactly $n$ choices of $A,B$ such that there is exactly one another vertex between $A$ and $B$, in the clockwise sense. For such a choice, there are at most three triangles of area 1. - There are exactly $ \frac {n(n-1)} {2} - 2n$ choices of $A,B$ such that there are at least 2 vertices of the n-gon in both sides of the line $AB$. For such a choic, there are at most 4 triangles of area 1. Summing, it leads to a number $T$ of triangles of area 1 such that : $T \leq 2n + 3n + 4( \frac {n(n-1)} {2} - 2n) = n(2n-5)$. Moreover, since a triangle has exactly three sides each triangle of area 1 is counted three times in the above reasoning according to the choice of the height. Thus the number of triangles of area 1 is at most $ \frac T 3$, and we are done. Pierre.

I think you have missed a condition concerning the area

No, he didn't. Basically you only need to prove that there exist at most $n(2n - 5)/3$ triangles with vertices among the vertices of the $n$-gon, the areas of which are equal. I agree that it's a weird way to state the problem, but it is correct.

Arne could you post your solution? I am really curious about this. thanks

I didn't say I solved the problem [I don't have time right now anyway]. I just looked at the problem statement and I figured out that it is actually correct. I might think of a solution later.