I found this interesting problem from a friend (From Arthur Engel's-Problem Solving Strategies book).
The method to begin the problem is as follows-
Step 1.Take a rectangular strip of paper
Step 2.We all know how to make a knot.Do the same (make a knot) using the strip of paper and fold it along the creases to form a polygon.
(If you are having a problem to make the knot please go to this link to see how to do it. See here )
Step 3.Prove that the polygon $abcde$ formed is regular.
What I could make out from the thing-
When I open the fold again I get this-
Note:-e is mentioned by dotted lines with pencil since e lies in the opposite side of the paper.
Now,my intuition somehow suggest that $a$ and $b$ must be mutually parallel and $d$ and $e$ must also be parallel.(Since then they would become congruent trapeziums).
But unfortunately I have no clue on how to prove this or how to prove that they are equal to $c$.
Any help or response is highly appreciated and thanks a lot in advance!!
Not sure if it's a complete answer, but an attempt to check a converse statement...
Consider the following illustration:
The quadrilateral $ABCD$ is a rhombus (the other $D$ on the right is the length of the diagonal, sorry for the confusion between letters) We have that \begin{equation} d=\frac{1}{\sin\theta},\qquad D=\frac{1}{\cos\theta} \end{equation} and \begin{equation} a = \frac{1}{2}\sqrt{D^2+d^2},\qquad b = a-2d\cos\theta \end{equation} Let's suppose that we want a regular polygon. Then we have to impose that the segments $DE$ and $EB$ are of equal length. Since $\overline{DE}=d$ and $\overline{EB}=b$, that means imposing $d=b$: \begin{equation} d=b\Longrightarrow \frac{1}{\sin\theta}\left(\frac{1}{2\cos\theta}-\frac{2}{\tan\theta}-1\right)=0 \end{equation} i.e. \begin{equation} \left(\frac{1}{2\cos\theta}-\frac{2}{\tan\theta}-1\right)=0 \end{equation} The only solution to this equation in the interval $[0,\pi/2]$ is actually $\theta=2\pi/5$.