Correct solutions were submitted by Bradley University community members Limin Liang, Aleksandar Nedyalkov, and one anonymous solver.  Other correct solutions were submitted by Francisco del León-Sotelo y Esteban, Iñigo Picaza, Bill Webb, Juan Carlos Marivela, Ahmed AboAdham, Paul Botham, Lou Cairoli, Francesc Suñol i Esquirol, Thiry Yves, Paul Lee.

As many of you pointed out, the sketch I gave was woefully deficient.  The sketch on the left is, I hope, more accurate. 

 From triangle ABE we have AB² + AE² = BE² and s² + AE² = (t - AE)² , so (1) AE = (t² - s²)/2t.  Note that it makes no difference whether we fold the corner C down to B or the corner B up to C; by symmetry we conclude that AE = DF.  Draw a line from E perpendicular to AC intersecting BD at G, we have FG = t - 2AE which, after substituting from (1) and simplifying, gives (2) FG = s²/t.  In the triangle EFG we have EG² + FG² = EF² and using (2) we can conclude that

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©2005 Alberto L. Delgado