firstname.lastname@example.org (Yoav Ilan Haim Parish) poses the problem
How can I fold an approximate (regular) pentagon from a quadratic sheet of paper easily ?
In addition to the nice constructions of G.J. McCaughan and Marvin Minsky (the man with the finite automata?) I have the following contribution:
One of my students once showed me this beautiful paper folding procedure, wondering whether or not it was exact:
Start with a square ABCD. Fold along the diagonal BD, folding C to A. Find the midpoint E on AB (folding B to A and opening it again). Then fold D to E. By this last folding you get two new points: F on BD and G on AD. The angle EFG is about twice the angle BFE. Bisect angle EFG by folding such that G will be lying on FE, and fold angle BFE to the backside. Now you may use a scissor to cut out a regular pentagon (cut at right angle) or a nice star (cut more arbitrarily)!
The folding procedure above is not exact; the angle BFE being slightly less than 37 degrees, instead of the exact value of 36 degrees. But when you are folding, you are not able to see the inaccuracy. (The McCaughan folding from an A4-sheet wasn't exact either, but also very close.) But F divides BD in an exact rational ratio: DF:FB = 5:7 !
o o Havard Johnsbraten, Telemark College of Education, Notodden, Norway E-mail email@example.com or G=haavard;S=johnsbraaten;O=tmlh;P=uninett;A= ;C=no