About the ratio of the areas of a convex pentagon and the inner pentagon made by the five diagonals I've thought about the following question for a month, but I'm facing difficulty. 
Question : Letting $S{^\prime}$ be the area of the inner pentagon made by the five diagonals of a convex pentagon whose area is $S$, then find the max of $\frac{S^{\prime}}{S}$.
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It seems that a regular pentagon and its affine images would give the max. However, I don't have any good idea without tedious calculations. Can anyone help?
Update : I crossposted to MO.

Here is a partial answer.
Say point C of the large pentagon ABCDE is free to move, so long as the area does not change.  We are curious what happens to the red area.
So C can move parallel to BD.  As it does so, the red area increases on one side and decreases on the other side, due to CA and CE moving.
Consider CE, and the change in red area due only to CE as C moves by $\epsilon$ (parallel to BD).  The point where CE crosses BD moves by $k \epsilon$, where $k$ is the constant ratio $m/n$ of the altitude $m$ from E to BD to the altitude $n$ from E to C's locus line.  Indeed, the change in area is monotonic in the distance $s$ from B$'$ (the intersection of AD and CE) to BD, which itself is monotonic in the position of C.
The area of the red pentagon is affected in this way both by the change in area due to CE and by that due to CA, and therefore it is maximized when C is somewhere in the middle, and it is strictly concave in C's position.
Due to the symmetry of the regular pentagon, we can see the red area is at a local maximum in this case.
The question asks us to prove this is the global maximum, but I would guess an even stronger result, that this is the only local maximum, and even that the red area is strictly concave in the positions of the vertices (subject to constant area of the large pentagon).
If we could show that single-vertex area-preserving moves can transform any pentagon into any "nearby" one, with no vertex ever traveling far from its original position, then that would probably finish the proof of total strict concavity.  It seems to be true, requiring $O(1/\epsilon)$ moves.
