I posted that already and nobody answered. I think the answer should be yes.

Any solution?

Bump .

We use $\frac{n(n+1)}{2}$ such hexagons to stack them in triangular pattern ($n$ to be determine later), each of three "edges" of the figure obtained consist of $n-1$ triangle area needed to be fill in order to make the whole figure convex. We hope to fill all such triangle area with some hexagons. Note that we do not need to (and will not) cover endpoints of "edges" of our almost-triangle figure. It's not hard to see that we can cover each of such area using two more hexagons so that no area appear exterior of the figure we hoped. So we need, in total, $\frac{n(n+1)}{2}+3\times (2\times (n-1))$ hexagons. Also, the area of obtained figure is at least of those non-overlapped $\frac{n(n+1)}{2}$ hexagons, which is equal to $\frac{n(n+1)}{2}$. Finally, use $n=1949$ and Wolfram guarantee it's satisfy both constraints.

Is there any other approach, different from the official solution?

The problem is that the problem is so easy that it becomes hard just because of that fact. But another thing it is to keep in mind that yes/no questions on EMC are 95% yes

JANMATH111 wrote: The problem is that the problem is so easy that it becomes hard just because of that fact. But another thing it is to keep in mind that yes/no questions on EMC are 95% yes Yes, since problems 1 and 4 were easy, I supposed that P2 must be hard, so i didn't try to solve it more than five minutes. *facepalm*