You have a sheet of paper that is 6 units wide and 25 units long, placed so that the short side is facing you.

geometry/shortest.crease New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

You have a sheet of paper that is 6 units wide and 25 units long,
placed so that the short side is facing you.  Fold the lower right
corner to touch the left side.  Your task is to fold the paper in such
a way that the length of the crease is minimized.  What is the length
of the crease?

Show solution

For the rectangle in question I labeled the lower left corner A, the
lower right B, the upper right C, and the upper left D.  Let B' be the
point on AD onto which corner B is folded.  The crease is determined by
the perpendicular bisector of BB'.

Case I: The crease intersects AD and BC.

        It is easy to show that the length of the crease decreases
        as the length of AB' increases.  Therefore, let B' = D.
        From similar triangles we get the length of the crease to be

               sqrt(23796/625) ~= 6.17038...

Case II: The crease intersects AB and BC.

         Pick B' on AD such that AB' < AB ( otherwise we have Case I ).
         Let the perpendicular bisector intersect AB and BC at points
         F and G respectively.  Let y = FG ( the length of the crease ).
         Let x be the angle BFG.  Then

               FB' = FB = y cos(x)

               angle AFB' = pi - 2x

               AF = AB - FB = 6 - y cos(x)

         Therefore,

               AF = FB' cos(pi - 2x)

                     or

               6 - y cos(x) = y cos(x) cos(pi - 2x)

                     or

               6 = y cos(x) [1 - cos(2x)]

         A little calculus and we get

               y = sqrt(243/4) ~= 7.7942...

         This agrees with a Sketchpad test.

-- Gerald D. Brown <gdbrown@ananov.remnet.rockwell.com>
   (geometry.puzzles, May 1995)

[ Editor: the posting does not say which case answers the question;
of the two values, case I's 6.17 is the smaller. ]