Where is the pentagon in the Fibonacci sequence?

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).

The dissection of a regular pentagon into triangles leads to a correlation with the Lucas numbers.

The division process shown below is described in detail at this MO post (Image by this author).

We begin by dividing the pentagon into three triangles by drawing a pair of diagonals. Then the largest triangle is subdivided by setting off 36° from one of its 72° angles. In the next stage the largest triangles which now are the ones having 108° apex angles are subdivided by setting off 36° from that angle. This process is repeated, dividing the largest triangles each time. At each stage we have two groups of congruent isosceles triangles. The first group has 108° apex angles and their count is always an even-argument Lucas number ($L_0=2, L_2=3$, etc). The second group has apex angles of 36° and their count is an odd-argument Lucas number ($L_1=1, L_3=4$, etc). With infinitely many iterations the generated nodes form the tenfold quasilattice.

To get Fibonacci numbers count pieces within just one of the three triangles into which the pentagon is first divided.

One (somewhat artifical) relationship (albeit not limited to a single pentagon) can be seen as coming from the key property of the golden ratio that $$ \phi^2 = \phi+1 $$ Consequently, $$ \phi^n = \phi^{n-1} + \phi^{n-2} $$ (Notice the Fibonacci-like pattern here!) It is easy to show that, by reducing this, we get $$ \phi^n = F_n \phi + F_{n-1} $$ for $F_n$ the $n$th Fibonacci number.

Draw a regular pentagon $P_1$ of unit side length. It has a diagonal of length $D_1 = \phi$.

Draw another regular pentagon $P_2$, with said diagonal as one of its sides. Then $$D_2 = \phi^2 = 1 + \phi$$

Repeat: draw a third regular pentagon, $P_3$, from that diagonal, using it as one of its sides. Then $P_3$ has diagonal $$ D_3 = (\phi+1)\phi = \phi^2 + \phi = 1 + 2\phi $$

The pattern continues on and on: we can easily establish that, however you wish to frame it, $$ D_n = D_{n-1} + D_{n-2} = \phi D_{n-1} = F_{n-1} + F_n \phi $$

This construction should naturally follow for the other metallic ratios as well.

A clunky visual of what the stacked pentagons would look like, green being $P_3$:

I don't know if this is going to be totally satisfying in geometric terms, but one way to think about this is as a special case of the Kronecker-Weber theorem. Namely, let

$$\zeta_5 = e^{ \frac{2 \pi i}{5} } = \cos \frac{2 \pi}{5} + i \sin \frac{2 \pi}{5}$$

be a primitive $5^{th}$ root of unity. The number field $\mathbb{Q}(\zeta_5)$ is Galois with Galois group $(\mathbb{Z}/5)^{\times} \cong \mathbb{Z}/4$, which implies that it has a unique quadratic subfield. That subfield is also the unique real subfield, which is generated by

$$\zeta_5 + \zeta_5^{-1} = 2 \cos \frac{2 \pi}{5} = \frac{\sqrt{5} - 1}{2} = \varphi - 1.$$

So in fact this is the quadratic field $\mathbb{Q}(\varphi)$ generated by $\varphi$. The Kronecker-Weber theorem asserts more generally that every Galois extension $K/\mathbb{Q}$ with abelian Galois group necessarily arises in this way, as a subfield of a cyclotomic field. For quadratic fields this can be proven explicitly using Gauss sums, and $\zeta_5 + \zeta_5^{-1}$ above is one of the simplest examples of a Gauss sum.

From this number-theoretic point of view the relationship between $\varphi$ and the Fibonacci numbers is best explained by the fact, which can be stated in several different equivalent ways, that the Fibonacci numbers occur as the entries of the power of a $2 \times 2$ matrix, namely we have

$$\left[ \begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array} \right]^n = \left[ \begin{array}{cc} F_{n+1} & F_n \\ F_n & F_{n-1} \end{array} \right].$$

This immediately implies that $F_n$ has a closed form in terms of the eigenvalues of this matrix, which we might call the Fibonacci matrix. And these eigenvalues are $\varphi$ and $1 - \varphi$! This matrix is in turn just the matrix of left multiplication by $\varphi$ acting as a linear transformation on $\mathbb{Q}(\varphi)$ in the basis $\{ 1, \varphi \}$. So, in other words, the above identity is equivalent to the identity

$$\varphi^n = F_n \varphi + F_{n-1}.$$

This special relationship between $\varphi$ and the number $5$ has the following concrete number-theoretic consequence: for any prime $p$ except $5$, if you compute the period of $F_n \bmod p$ (the Pisano period) you can show that it divides either $p - 1$ or $2(p + 1)$, and which case is which turns out to depend only on the value of $p \bmod 5$. The only exception is $p = 5$ itself, where the period is $20$. This turns out to be a special case of quadratic reciprocity, which is closely related to the Kronecker-Weber theorem and Gauss sums.

Whether anything more directly geometric than this can be said I don't know. The issue is I don't see a nice way to write down a geometric construction which corresponds to taking powers of $\varphi$. I suppose you could iteratively construct larger and larger golden rectangles but this seems a little contrived to me.

Consider this Diagram :

Sides are $1$ , Diagonals are $x$.
Diagonals are Parallel to Sides.

The Shaded triangles are "Similar" , hence we have :
$1/(x-1)=x/1$

Hence $1=x(x-1)$

Eventually we get :
$$x^2-x-1=0$$ $$x^2=x+1$$

Hence we have the Golden Ratio Relation here.

We can thus justify renaming $x$ to $\phi$ here.

UPDATE :

We might want a "Series of Pentagons" where the ratio is either a Constant or will converge to the Constant.

Consider this Diagram :

Smaller the Pentagon , Closer it gets to $O$.
Extending in the other Direction , we make larger Pentagons , going towards $\infty$.

Deriving the Constant here might give Alternate Ways to look at the Issue.

The figure shows two not regular pentagons constructed from the Fibonacci sequence $A,B,C,D,E,F$, taking consecutive Fibonacci numbers as side and diagonal. The two pentagons have a side with different length from the others. The sequence of pentagons shows how they tend to become a regular pentagon.

There is neither a pentagon in the Fibonacci sequence nor a Fibonacci sequence in the pentagon. You may compute the $n^{th}$ Fibonacci number by $\displaystyle F_n=\left\lfloor\frac{1}{2}+\frac{\varphi^n}{\sqrt{5}} \right\rfloor$ what approximates for larger $n$ a converging ratio $F_n/F_{n+1}$. Taking this as a reason to surmise a relation with a pentagon is as if you try to square the circle.

Note: with the supplement of OP on 2024-07-09 21:00:04Z I see my point of view as affirmed.