Prove that every convex polygon with area $1$ is contained in a parallelogram of area $\frac{4}{3}$

Question

Prove that every convex polygon with area $1$ is contained in a parallelogram of area $\frac{4}{3}$

I can only show that polygon is contained in a rectangle of area $2$.

Answer 1

It seems the following.

This is the best result, because the smallest area of a polygon which has a center of symmetry and contains a given triangle of area 1 is 2, see D.O. Shklyarskiy, N.N. Chentsov, I.M.Yaglom “Geomerical estimations and combinatorial geometry problems”, Moskow, Nauka, 19741, Problem 52.b (in Russian)). In particural, no triangle of area 1 is contained in a parallelogram of area less than 2.