Prove that in an arbitrary convex hexagon there is a diagonal that cuts off from it a triangle whose area does not exceed $\frac16$ of the area of the hexagon. What are the properties of a convex hexagon, each diagonal of which is cut off from it is a triangle whose area is not less than $\frac16$ the area of the hexagon?
Problem
Source: IV All-Ukrainian Tournament of Young Mathematicians, Qualifying p8
Tags: geometry, geometric inequality, hexagon, convex, diagonal, area, Ukrainian TYM