For young students it is an interesting surprise to discover that a knot tied in a strip of paper is a regular pentagon. I'm interested to find a simple, but rigorous, geometrical proof of this "experimental" fact.
The classical book Mathematical Models by H. Martyn Cundy and A.P. Rollett (cited in Wikipedia) extend a similar construction to other regular poligons, but don't gives a proof.
It is easy to prove that in a regular pentagon a diagonal is parallel to the opposite side, but here we have to prove the converse.
Let the pentagon be ABCDE labelled clockwise with C at the top.
The simplest way is to first prove that whenever you fold the strip the overlapping region is an isosceles triangle. (I shall leave the easy proof of this to you.) Hence AD = AC = EC = EB. That implies ∠ABC = ∠BCD = ∠CDE by an easy angle chasing (or by going back to the fold property), which implies AB = CD and BC = DE. Thus △ABC ≅ △CDE, and hence ∠ACE = ∠DEC = ∠BCA. Symmetrically, ∠ACE = ∠ECD. Thus △ABC and △CDE are isosceles, so AB = BC = CD = DE. It remains to observe that ∠ACE = ∠BCA = ∠DAC, and so △ACE ≅ DAC, which gives the fifth side equality AE = DC. Thus △BCD ≅ △DEA ≅ △EAB, yielding the final two needed angle equalities ∠BCD = ∠DEA = ∠EAB.
In case you want to see a diagram: