# How to prove a regular pentagon is formed by knotting a rectangular strip of paper?

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!!

• A link showing how to make a knot would be very useful! (PS: I don't know) – N.S.JOHN Apr 24 '16 at 16:17
• @N.S.JOHN Follow this link to learn how to make it-youtu.be/q1zAwHIXoYI – tatan Apr 25 '16 at 13:35
• Thanks. It is better to tell it in the question. – N.S.JOHN Apr 25 '16 at 13:35
• @N.S.JOHN Its very difficult to describe the process in words.You may please do it if you are able to...Thanks!! – tatan Apr 25 '16 at 13:38
• Here's a purely geometrical formulation of the problem: Let the vertices of the pentagon be labeled $PQRST$ (with $P$ being the one where edges $a$ and $e$ meet). Because $QT$ and $RS$ are opposite sides of a rectangular paper strip, they are parallel and a unit distance apart. The same is true for the pairs of line segments $PQ$ and $RT$, $QS$ and $TP$, and $QR$ and $PS$. Given this information we have to prove that the pentagon $PQRST$ is regular. – Rahul Apr 25 '16 at 14:26

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 $$d=\frac{1}{\sin\theta},\qquad D=\frac{1}{\cos\theta}$$ and $$a = \frac{1}{2}\sqrt{D^2+d^2},\qquad b = a-2d\cos\theta$$ 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 $d=b$: $$d=b\Longrightarrow \frac{1}{\sin\theta}\left(\frac{1}{2\cos\theta}-\frac{2}{\tan\theta}-1\right)=0$$ i.e. $$\left(\frac{1}{2\cos\theta}-\frac{2}{\tan\theta}-1\right)=0$$ The only solution to this equation in the interval $[0,\pi/2]$ is actually $\theta=2\pi/5$.
• You have drawn two $D$ s...What does the $D$ between $AC(=a)$ actually signify?What is that point actually? – tatan Apr 26 '16 at 15:49
• that $D$ is the length of the longest diagonal of the rhombus. – marco trevi Apr 26 '16 at 17:19