golden ratiotrigonometry
I just learned that the diagonal of a pentagon (size 1) is the golden ratio (https://twitter.com/fermatslibrary/status/1210561047154872320)
I tried to verify that, but ended up having to show that: $$\cos{\frac{2\pi}{5}}=\frac{1}{\phi}$$
Question: is there a way to calculate the diagonal of a pentagon without having to do any relatively complex trigonometric calculations? For example only by drawing some smart supportive lines and using the Pythagorean Theorem?
Here's a much better argument than my previous attempt. I'm leaving out a few details due to time constraints, but will update later.
Let pentagon $ABCDE$ have side-length $s$, and let $F$ be the apex of an equilateral triangle erected internally upon $\overline{CD}$. The orthocenter of $\triangle ABF$ is the intersection of $\overline{BE}$ with the perpendicular $\overline{FF'}$ from $F$ to $\overline{AB}$. Let $X$ be the point that separates $\overline{DE}$ in the golden ratio: $|DX|/|XE|=\phi$. Here's the key insight (proof left to the reader for now):
The Second Isodynamic Point of $\triangle ABF$ is the apex ($Q$) of an equilateral triangle erected externally upon $\overline{DE}$.
We establish the result by showing the collinearity of $P$, $Q$, $X$ via Menelaus' Theorem applied to $\triangle BEM$, where $M$ is the midpoint of $\overline{DE}$. A little angle chasing gives the various angle measures; we also use the facts $|BE|=s\phi$ and $\phi-1=\phi^{-1}$. Calculating the appropriate signed ratios ...
$$\begin{align} \frac{|EX|}{|XM|}&=\phantom{-}\frac{s/(1+\phi)}{s/2-s/(1+\phi)}=\frac{2}{\phi-1} = 2\phi \\[4pt] \frac{|MQ|}{|QB|}&= -\frac{s\sin 60^\circ}{|EB|\sin\angle BEQ/\sin\angle BQE}= -\frac{s\sin 60^\circ}{s\phi\sin 132^\circ/\sin 30^\circ}= -\frac{s\sin 60^\circ}{2\phi\sin 48^\circ} \\[8pt] \frac{|BP|}{|PE|} &= \phantom{-}\frac{|BF|\sin 48^\circ/\sin\angle FPB}{|EF|\sin 120^\circ/\sin\angle FPE} = \frac{\sin 48^\circ}{\sin 60^\circ} \end{align}$$ ... we see that their product is $-1$. Thus, Menelaus holds, and collinearity is verified. $\square$
Note: When $F$ is the apex of the triangle erected externally on $\overline{CD}$, then the First Isodynamic Point of $\triangle ABF$ is the apex of the triangle erected internally on $\overline{DE}$, and a similar argument holds to give the companion construction I described in my previous answer. The pentagram constructions follow similarly.
Promoting my comment and image explanation into an answer.
The Fibonacci numbers themselves don't readily appear in a pentagon/pentagram, but the golden ratio and the same recurrence relation do show up. As always, the starting point is the golden triangle — an isosceles triangle with respective angles $2\pi/5$, $2\pi/5$ and $\pi/5$:
We have two similar triangles here leading to the familiar equation. If the length of the blue line is $1$ unit and the length of the red line is $x$, then the big isosceles triangle shows that the black line segment has length $1+x$. But, the similarity of the two triangles yields $x^2$ for that length. Hence $$x^2=1+x,\tag{1}$$ from which we can solve $x=\phi=(1+\sqrt5)/2.$
We can always draw a pentagram inside a regular pentagon by including the diagonals. But, by extending the sides of the same pentagon until they meet, we get another pentagram. Joining its star points gives us a bigger pentagon, and we can keep going like in the following image:
Observe that we could equally well look at the smaller pentagon bordered by the diagonals of a bigger one, draw its diagonals to form a smaller pentagram et cetera.
Anyway, it is easy to calculate the angles that appear in these recursive drawings, and see that they form a sequence of similar triangles. It follows that each pentagon (resp. pentagram) is a scaled up version of the preceding one by a factor of $\phi^2$. We also see that the lengths of consecutive segments forming the red zigzag are sides of isosceles triangles similar to the earlier one. Hence their lengths $\ell_0,\ell_1,\ldots$ all satisfy $\ell_{n+1}/\ell_n=\phi$, and thus satisfy the Fibonacci recurrence: $$\ell_n=\ell_{n-1}+\ell_{n-2}.$$
This is somewhat unsatisfactory because the Fibonacci numbers $F_n$ themselves don't show up — only the recurrence relation. We all know that, by Binet's formula, the numbers $F_n$ need a (small) correction term involving powers of the other root $-1/\phi$ of the equation $(1)$.
This discrepancy has been a source of recreational mathematics: Missing square puzzle, locally here and here. The idea is explained in the following image
On the left we have a square of side length $\phi^2$ subdivided into four polygons in such a way that the vertical as well as the horizontal sides have lengths $1,\phi$ or $\phi^2$. Those polygons can then be perfectly rearranged to form a rectangle of dimensions $\phi\times\phi^3$. All due to the identity $\phi^2=1+\phi$.
The missing square puzzles then emerge when we use consecutive Fibonacci numbers instead of consecutive powers of $\phi$ as lengths. Below see the version with respective side lengths $3,5,8$.
The four polygons in it would, again, form a nice $8\times8$ square, but narrowly fail to fill up a $13\times5$ rectangle. Just as well, given that $13\cdot5=8\cdot8+1$. Drawning it coarsely enough the puzzle designers can make that thin gap disappear! Using larger consecutive Fibonacci numbers makes the gap look thinner, but it always exists. All because we have the identity $$F_nF_{n+2}=F_{n+1}^2\pm1$$ with the sign alternating according to the parity of $n$ (and the indexing of the Fibonaccis).
Best Answer
Consider a regular pentagon $ABCDE$ and a vertex inscribed pentagram $ACEBD$ with the additional intersection points $a,...,e$ somewhat closer to the center but on the oposite ray. Then the pentagram will have sides $AdeC, CabE, EcdB, BeaD, DbcA$.
By assumption you have $AB=BC=CD=DE=EA=1$. Let further be $Ad=Bd=Be=Ce=Ca=Da=Db=Eb=Ec=Ac=:x$ and $ab=bc=cd=de=ea=:y$.
Now consider the isoscele triangle $AEd$. Thus you get $1=x+y$. Its base is $x$. Then consider the scaled down isoscele triangle $Ebc$ with sides $x, x, y$ (its tip angle clearly is the same). From those you get the scaling ratio $$\varphi:=\frac1x=\frac xy$$ Together with the above ($y=1-x$) the equation for the golden ratio follows: $$1-x=x^2$$ or, when dividing by $x^2$ and re-inserting $\varphi$: $$\varphi^2=\varphi+1$$ --- rk