Any solution?

Since the specific area of the hectogon drawn is arbitrary (we can always just scale it), we'll instead consider all areas relative to the total area for convenience. The answer is $0$, that is, the Devil will always be able to generate a sequence of hectogons such that the 'covered' area tends to $0$. Consider the regular hectogon. Accept this to be true, we'll show it later: There is an 'uncovered' triangle formed among $3$ pairs of vertices, some of which may be shared. Say $A_1$,$B_1$,$C_1$,$A_2$,$B_2$,$C_2$ in clockwise order. We can then move $A_1$ and $B_1$, and all vertices in between, within a small $\epsilon$ of each other, and do the same with $C_1$,$A_2$ and $B_2$,$C_2$. The net result is the 'uncovered' triangle taking up the bulk of the area, and all other areas being of negligible size. As $\epsilon$ tends to $0$, the uncovered area tends to $1$. Now to show the lemma, that no $n$ gon can be fully covered by $n-3$ triangles. We'll do it by induction. Note that any triangle comprises at most $2$ edges on the perimeter. If a triangle that comprises exactly $2$ exists, we can easily finish with the induction hypothesis. Otherwise, there exists at least one edge that is not part of the perimeter of a triangle. That edge's midpoint is not contained in any triangle's interior or perimeter, so the $n$-gon is not fully covered.

joyce_tan wrote: ...There is an 'uncovered' triangle formed among $3$ pairs of vertices, some of which may be shared. Say $A_1$,$B_1$,$C_1$,$A_2$,$B_2$,$C_2$ in clockwise order. We can then move $A_1$ and $B_1$, and all vertices in between, within a small $\epsilon$ of each other, and do the same with $C_1$,$A_2$ and $B_2$,$C_2$. The net result is the 'uncovered' triangle taking up the bulk of the area... Could you elaborate it, because I don't quite get all of it. I think to carry out that idea you need entire triangle, inside the $100$-gon, any point of it not covered by any other triangle among the given $97$. I think, it's false.

Yep, that's what I needed. I showed it in the next 2 paragraphs.

joyce_tan wrote: Yep, that's what I needed. I showed it in the next 2 paragraphs. What you had shown is that the $n$-gon couldn't be fully covered, that's, there exists a point inside it not covered by any triangle. What you "need" the way you've proceeded is that there is an entire triangle any point of which not covered by any triangle. The latter is not true. Do you see the difference? In fact your approach can be carried out if you prove that vertices of $100\, (n)$-gon can be partitioned into $3$ sets $A,B,C$, each set consisting of consecutive vertices, such that there is no listed triangle $A_iA_jA_k$ with $A_i\in A, A_j\in B, A_k\in C$. It's true. Btw, the author of the problem is Nikolai Beluhov (Bulgaria).