# Where is the pentagon in the Fibonacci sequence?

It is common wisdom that "When you see $$\pi$$, there is a circle close at hand". For example:
The periods of sine and cosine equal $$2\pi$$? Properly constructed, the right triangles that define them trace out a circle.
$$e^{i\pi} = -1$$? An artifact of polar complex coordinates.
The factor of $$4\pi$$ in Coulomb's Law? The electric force is spherically symmetric.
This makes sense, as $$\pi$$ is the ratio between a circle's circumference and its diameter.

Similarly, the golden ratio $$\phi = \frac{1+\sqrt{5}}{2}$$ is the ratio between a regular pentagon's diagonal and its side, and the two do seem to go hand in hand:
The Penrose tiling, composed of golden ratio kites and darts, is pentagonally symmetric.
The icosahedron and dodecahedron have pentagons in their structure, either vertex figures or faces, and are riddled with $$\phi$$s.
To construct a golden rectangle, you need to construct an angle that is $$\frac{1}{5}$$ of a turn, which is also how you start the construction of a regular pentagon.

However, the golden ratio is also found in the Fibonacci sequence as the limit of the ratio between adjacent terms. And there are plenty of cases where $$\phi$$ pops up because of this: Whythoff's nim, Lucas sequences, coverings with mono- and dominoes.

So now the question is Where's the pentagon in the Fibonacci sequence? Or is it that there's a Fibonacci sequence in the pentagon?

It's been long enough, time to accept an answer. mr_e_man's 'proof without words' is exactly what I'm looking for, so he gets the tick. That said, everyone's answers here were useful, especially as they reminded me that the pentagon in the Fibonacci sequence would only be approximate. Pointing this out is why I'm adding this rather than just clicking the tickmark.

• @JyrkiLahtonen It's nice, yes, and it looks like the beginnings of a proper answer, but I'm not sure how to integrate this into the question Commented Jul 3 at 5:29
• @JyrkiLahtonen Maybe someone else can find the connection. The edit will stay until I can find a better place for it Commented Jul 3 at 5:36
• May be the following: Using "obvioiusly" similar triangles we can easily see that the lengths of the sequence zigzagging line segments $\ell_1,\ell_2,\ldots$, follow the Fibonacci recurrence: $$\ell_{n+2}=\ell_{n+1}+\ell_n?$$ Commented Jul 3 at 5:39
• Fibonacci sequence is an approximation of &varphi;: this means that a pentagon cannot contain Fibonacci, or if you prefer Fibonacci creates just an approximation of a regular pentagon.
– mau
Commented Jul 3 at 17:25
• The Fibonacci sequence is related to, but not equal to the golden ratio. There is no reason to expect that the sequence mimics the geometric series $\varphi^n$ than there is to expect that the Fibonacci spiral is the same as the golden spiral. Commented Jul 3 at 20:19

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).

• Will delete, if a suitable duplicate is located. This has to have appeared in dozens of textbooks. Commented Jul 3 at 8:56
• Don't delete, cite! Link to the "duplicates" you find as prior art, that way I and any one curious can find them again Commented Jul 3 at 16:29
• @mr_e_man Then you don't get exact regular pentagons. The images in your answer are ok, but don't quite explain that small discrepancy. Commented Jul 5 at 5:44
• Animated version of the missing square puzzle: raw.githubusercontent.com/gist/PM2Ring/… Commented Jul 5 at 7:24
• In the missing square puzzle, the rectangle has a gap only if the side of the square is an even-argument Fibonacci number like $F_6=8$. If an odd-argument Fibonacci number like $F_7=13$ is used for the square, the rectangle will give an overlap instead. Commented Jul 5 at 14:51

$$\phantom{xxxxxxxxxxxxxxxx}$$

• +1. This was the visualization I'd considered posting. Now I don't have to! :)
– Blue
Commented Jul 5 at 6:01
• +1. If only we could also show how/why the pentagons don't come out quite right :-) Commented Jul 5 at 6:06
• @JyrkiLahtonen: "If only we could also show how/why the pentagons don't come out quite right" One way to do that would be to show angle measures (at least for the later iterations), indicating that they aren't $72^\circ$ but that they keep getting closer to it.
– Blue
Commented Jul 5 at 7:15
• A lovely proof without words. Well, proof of concept at any rate Commented Jul 5 at 8:36

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.

• I like this because it brings an integer sequence into a pentagon. +1 has been there for quite some time. Commented Jul 7 at 17:42

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 $$P_3$$:

• Is there a nice way to see geometrically that the side length of the green pentagon is the sum of the side lengths of the black and purple pentagons? I can kinda see it if I squint but it'd be nice if there was a way to draw the diagram that made it really obvious. Commented Jul 3 at 4:17
• @QiaochuYuan You can somewhat deduce it by considering the three equivalent isoceles triangles in the bottom right.
– Zuy
Commented Jul 3 at 6:30
• "This construction should naturally follow for the other metallic ratios as well." But not in general with polygons. You can use octagons for the silver ratio $1+\sqrt2$, but that's it for polygonal constructions unless you inelegantly use superpositions of different diaginal lengths. Commented Jul 20 at 10:00

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 OP is already aware of this. The question is about the relationship to the Fibonacci numbers, not the golden ratio. Commented Jul 3 at 4:53
• I know what you mean ,@QiaochuYuan , though my thinking is something like this : [[1]] Say we make a triangle with two Consecutive Sides , then calculate the largest internal angle ($108$) & then calculate the Diagonal (Side of the triangle) to get $2\sin(54)$ which will involve $\sqrt{5}$ , then we might ask : Where is the FIB Series ? We have used trigonometry to hide that relation which is then not easy to see. [[2]] What I have shown here is that we actually get the FIB Series relation via 2 "Similar" triangles. That will more easily highlight where / why the FIB Series is involved here.
– Prem
Commented Jul 3 at 5:20
• Some of the side-lengths of the blue pentagon in the bottom image might be illuminating. Namely, "base" has side length x-1, as do a few others. And, (x-1)+1 is x
– Yakk
Commented Jul 4 at 14:34
• Here's what you may want to do. Let the first pentagon have side $s$ and the second have side $d$ (from the diagonal of the first pentagon). Show that subsequent pentagons have sides $s+d$, then $s+2d$, $2s+3d$, etc, generating the Fibonacci numbers as coefficients of the initial side and diagonal. Commented Jul 9 at 23:43
• A similar construction using regular octagons, and choosing the unique set of diagonals parallel to the sides, generates Pell number coefficients. Commented Jul 9 at 23:45

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.

• Observation: Look at the circles that approach the lower right vertex of each pentagon. They are nearly concurrent there, whereas with regular pentagons they would be exactly concurrent. The concurrency is closer with the larger pentagon. Commented Jul 10 at 20:54

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.

• "there is neither a pentagon in the Fibonacci sequence nor a Fibonacci sequence in the pentagon". We have an answer that identifies the pentagon with the Lucas sequence and triangular pieces of it with the Fibonacci sequence. Keep up! Commented Jul 4 at 9:40
• @OscarLanzi -- I'll keep up looking for your triangle counting idea in textbooks or a paper related to Fibonacci sequence. Just to overcome my qualms. Commented Jul 4 at 13:30