Problem: if exists a hexagon ABCDEF such that:
(ABCD) = (BCDE) = (CDEF) = (DEFA) = (EFAB) = (FABC)
For short we name such hexagon as T hexagon.

I construct one such hexagon as following:
Suppose XYZ is any triangle with centroid G
- B, D, F are midpoints of XY, YZ, ZX respectively.
- Fa, Ba are two points on segment FB (by this order) such that:
FaBa/FFa = FaBa/BaB = Phi = (1 + Sqrt(5))/2
- Ly is a line from Fa and parallel with ZX
- Lz is a line from Ba and parallel with XY
- A = intersection of Ly and Lz. Of course A is on median XD
- C = intersection of Lz and YF
- E = intersection of Ly and ZB
ABCDEF is a T hexagon and it holds many interesting properties.
- Lx = line connected C, E
- BD intersect Lz, Lx at Bc, Dc respectively
- DF intersect Lx, Ly at De, Fe respectively
Some interesting propeties:
1.
FaBa/FFa = BaFa/BBa = Phi
BcDc/BBc = DcBc/DDc = Phi
DeFe/DDe = FeDe/FFe = Phi
2.
ABa/BaBc = CBc/BcBa = Phi
CDc/DcDe = EDe/DeDc = Phi
EFe/FeFa = AFa/FaFe = Phi
3.
(BaBcDcFa) = (DcDeFeBc) = (FeFaBaDe)
(FaBaBcFe) = (BcDcDeBa) = (DeFeFaDc)

I hope that we can find other interesting properties of those hexagons.
Best regards,
Bui Quang Tuan

The proof of my construction is easy:
- Because Ly//ZX, Lz//XY therefore GE/GZ = GA/GX = GC/GY and Lx // YZ. So we have following parallel lines:
FB // CE // YZ
BD // EA // ZX
DF // AC // XY
From these we can show facts 1) and 2) then we can show
BcFe // YZ
BaDe // ZX
DcFa // XY
and they devide ACE and BDF into two parts with the same area ratio.
From these facts we can show 3) and other facts. For example we can show that three medians XD, YF, ZB divide ABCDEF into six area equal triangles.

I still not get my constructed hexagon from T hexagon. I also don't know if are there any other T hexagon or not?

Now I suggest another problem which can be generalized for convex polygon.

First we define TP-polygon concept:
A TP-polygon is a convex polygon inside it exists at least one point P such that all areas of triangles bounded by P and each polygon side are equal.
1. Any triangle is TP-polygon and its centroid can be accepted as point P.
2. A quadrilateral is TP if and only if it is parallelogram and its center can be accepted as point P.
Proof:
Suppose ABCD is TP quadrilateral and P is a P-point.
(PAB) = (PAD) => PA passes through midpoint of BD. Similarly PC passes through midpoint of BD. Therefore P is midpoint of BD. Similarly P is also midpoint of AC. It can happen if and only if ABCD id parallelogram.
3. Easy to show that "golden ratio" pentagon in Prasolov's problem is TP and constructed by me hexagon is TP.
4. Two open problems:
- If is golden ratio pentagon only unique TP pentagon? If is my hexagon only unique TP hexagon?
- If can we find all TP polygon for given n-gon (n>4)?

Thank you and best regards,
Bui Quang Tuan