Take the weight of the piece to be 1 Then if the partition is possible, then you get sum(a^i/(1+a)^j)=1/2 for some i,j with the property in any two terms the values of both i,j of one term are greater than the other Adding the terms you get in the denominator some power of (1+i) and some terms relating to a and 1+a in the numerator Take the second least power of 1+a in the numerator For that condition you will not get two or more such terms Now that power of 1+a divides the whole numerator But this is only possible if 1+a divides 2 So I think no solution for a not equal to 1

I am probably missing something but what if we take $a=\sqrt{2}-1$. If the starting piece has mass $1$ and if we cut it into four pieces then we will have two pieces with masses $\frac{\sqrt{2}-1}{2}$, one with mass $\frac{(\sqrt{2}-1)^2}{2}$ and one with mass $\frac{1}{2}$. But $\frac{(\sqrt{2}-1)^2}{2}+2\cdot\frac{\sqrt{2}-1}{2}=\frac{1}{2}$.