An interesting construction

Let rectangle $ABCD$ be the paper with a width of $x$ and a length of $y$ so that $AB = CD = x$ and $AD = BC = y$ . Let $E$ be the intersection of folds on $BC$ , $F$ be the intersection of folds on $CD$ , and $G$ be the intersection of folds on $AD$ .

From the first fold we have $BE = AB = x$ , and from the second fold we have $CF = CE = BC - BE = y - x$ , and from the third fold we have $DG = DF = CD - CF = x - (y - x) = 2x - y$ . Since triangles $\triangle CEF$ and $\triangle DGF$ are right isosceles triangles, $EF = \sqrt{2}(y - x)$ and $FG = \sqrt{2}(2x - y)$ .

For the ratio of the final product to be the same as the original, either $\frac{y}{x} = \frac{\sqrt{2}(2x - y)}{\sqrt{2}(y - x)}$ or $\frac{y}{x} = \frac{\sqrt{2}(y - x)}{\sqrt{2}(2x - y)}$ . The first solves to $y = \sqrt{2}x$ for a paper ratio of $\frac{y}{x} = \sqrt{2}$ , and the second solves to $y = \frac{1 + \sqrt{5}}{2}x$ for a paper ratio of $\frac{y}{x} = \frac{1 + \sqrt{5}}{2}$ .

Therefore, the sum of all possible ratios is $\sqrt{2} + \frac{1 + \sqrt{5}}{2} \approx \boxed{3.032}$ .