A rectangular sheet of paper is folded so that two diagonally opposite corners come together. The crease thus formed is as long as the longer side of the rectangle. What is the ratio of the longer side of the rectangle to the shorter?

• 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
• 11 years agoHelpfull: Yes(3) No(0)