Construct a rectangle ABCD. Note that we are looking for the ratio of shorter side to longer side, so we can set AB>BC and further set BC=1 and AB=x for convenience (resulting in the answer to the problem being length AB, or x).
Label the endpoints of the resultant crease E and G (E on AB and G on CD) and label the center of the rectangle F.
Note that F is the midpoint of AC, the midpoint of EG and that AC and EG are perpendicular.
Note that triangles AEF and ABC are similar, thereby giving us that EF/BC=AF/AB=AE/AC.
From geometry and pythagorus, we can see that:
EF=x/2
AF=sqrt(x²+1)/2
AE=sqrt(2•x²+1)/2
AC=sqrt(x²+1)
This can now be solved using any two of the three parts of the equation above. The easiest is probably the following:
EF/BC=AF/AB
x/2=sqrt(x²+1)/2x
x^4-x^2-1=0
This is easiest solved by substituting u=x^2, and gives
u=(1+/-sqrt(1+4))/2
and thereby
x=sqrt((1+sqrt(5)/2), or approximately 1.26 |