Ratio of Lengths of Sides of Rectangle When Bisector of a Diagonal Equals Longer Side

Question: A rectangular sheet of paper is folded so that two diagonally opposite corners come together. If the crease formed is the same length as the longer side of the sheet, what is the ratio of the longer side of the sheet to the shorter side?
Solution: Draw a rectangle with longer side  
  and shorter side  
. Draw a diagonal  
  then the crease  
  must bisect this crease at right angles.

The length of the diagonal  
  by Pythagoras Theorem and the distance  
  is half of this:  
\[AO= \frac{\sqrt{l^2+s^2}}{2}\]

Also since the crease  
  is of length  
  and the diagonal bisects the crease,  

  is a right angled triangle since  
  at right angles. We can find two expressions for  
  and equate them.
Drop a perpendicular from P to the base of the rectangle then  
  is a right angled triangle and  
By Pythagoras Theorem  
\[QM= \sqrt{l^2 - s^2}\]
  is half of this:  

Now apply Pythagoras Theorem to the triangle  
\[ ( \frac{l}{2} + \frac{\sqrt{l^2-s^2}}{2})^2 = (\frac{l}{2})^2 + (\frac{\sqrt{l^2+s^2}}{2})^2\]

\[\frac{l^2}{4} + \frac{l \sqrt{l^2-s^2}}{2} +\frac{l^2-s^2}{4}=\frac{l^2}{4} + \frac{l^2+s^2}{4} \]

\[\frac{l \sqrt{l^2-s^2}}{2}= \frac{s^2}{2}\]

\[l^2(l^2-s^2) =s^4\]

\[ l^4-l^2s^2-s^4=0\]

We can treat this as a quadratic in  
  with solutions  
\[l^2 = \frac{s^2 \pm \sqrt{s^2+4s^2}}{2} = s^2 (\frac{1 \pm \sqrt{5}}{2}) \]

We only need the positive option since the negative one mean  
  is negative.
\[l= s \sqrt{\frac{1+ \sqrt{5}}{2}}\]

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