S=> Small Side of the rectangle
L=> Large side of the rectangle
X=> the point at which you need to rotate the paper so that the opposite sides touch
Such an X must form a right angled triangle
Therefore from Pythogaras Theorem we get
S^2 + X^2=(L-X)^2
Simplifying this we get
X=(L^2-S^2)/2L -(i)
Now, the crease will form a right angled triangle if we draw an imaginary line
from X parallel to the side S(which will be offcourse of length S),
The small RAT on the top left & the one at the bottom are similar (it has something to do with a side & and two angle being same) which makes th base of the crease triangle of length L-2X, for our problem the crease(hypotenus of the imaginary triangle) has to be L,
So, again using Pythogaras Theorem we get
S^2+(L-2X)^2=L^2
simplifying this we get
S^2-4LX+4X^2=0 - (ii)
Substitute the Value of X from (i) in (ii)
S^4+S^2L^2-L^4=0
This equation will have 4 roots,
2 Imaginary,1 negative (which we will discard as not applicable in our case) & One
will lead
l/s=Sqrt(2)
1, 2, 5 + 3, 4, 2+4, 2, 7