Solution

The minimal green fraction is ½.

First it is seen that the problem satisfies translational symmetry of the small square centre, since the sum of the triangle heights is constant. We translate it such that the two squares are concentric and the figure has fourfold rotational symmetry.

We scale the large square to 1. Naming the vertex in the first quadrant (x,y), we can calculate the small square area as half the square of the diagonal, which evaluates to 2(x²+y²).

In the concentric case, all four triangles are congruent and have area (1/2-x)/2. Therefore the green area sums to A(x,y)=2(x²+y²-x+1/2). To find the extremes, we set dA/dx and dA/dy to 0. This gives x=1/2 and y=0. This is indeed a minimum and it evaluates to 1/2.