# Four ways of looking at a Pentagon, part 1

Suppose there is a pentagon. How do we know the relationship between the edges?

Well, we can imagine the Pentagon is made of Triangles, and look at the Spreads between the lines that make up those triangles.

Now if we imagine the Pentagon triangle’s lines form spokes, as though the Pentagon were inside a wheel, we can imagine there is a relationship between those lines orientation. We could think of the Angle between the lines… but what about Rational Trigonometry, and it’s Spread? I’m going to use my amateur reading of Wildberger’s Divine Proportions for inspiration here.

Let’s imagine the lines are all related by the Spread between them. For example we could imagine splitting the pentagon in half, then splitting that into 5 triangles, and each triangle has one point that forms the axle of the “wheel”. Then, each axle point has two lines coming out, and they each have the same spread, s.

But how does this help us calculate this relationship as a number? It can help us if we imagine that we can group the triangles together and look for patterns in the Spread equations.

The tricky bit is that Spreads do not add simply like Angles do. For example a 30 degree angle has a Spread of $\dfrac{1}{4}$. Doubling that angle gives us 60 degrees, but the Spread corresponding to a 60 degree angle is $\dfrac{3}{4}$. Clearly $\dfrac{3}{4}$ is not the same as two times $\dfrac{1}{4}$.

What you do instead is look at the Spread Polynomials. Two lines separated by $s$ each form a new spread $s_2$, which is calculated as follows:

$s_2=4s_1(1-s_1)$

This is called a Spread Polynomial. In fact, it is the Spread Polynomial called $S_2$ in Wildberger’s book on page 104, it is basically used for doubling a spread and shows up in all kinds of patterns involving spreads.

So if we had a spread of $\dfrac{1}{4}$ for 30 degrees, and we double it, we get a spread of

$s_2=4\dfrac{1}{4}(1-\dfrac{1}{4})$

$s_2=(1-\dfrac{1}{4})$

$s_2=\dfrac{3}{4}$

That is the Spread corresponding to 60 degrees, which makes sense as 60 is double the angle of 30.

Spread polynomials are not just analogues for doubling an angle… there is a spread polynomial $S_3$ for tripling an angle, $S_4$ for quadrupling, and on and on. But they are not quite as simple as that.

The tricky bit, the interesting bit, is that each Spread actually represents two different angles. The spread of $\frac{1}{4}$ actually represents both the 30 degree angle and the 150 degree angle. In the angle system, these angles are distinct, but in the Spread system, they are the same number. $\frac{1}{4}$. We are just thinking about the lines themselves and their separation, and there is only one number that describes that separation.

So when we are adding Spreads, maybe we can think about it as if we are adding multiple different possibilities of angles, at the same time. For example, if I say I’m “adding” Spread $\frac{1}{4}$ to Spread $\frac{1}{4}$, does that mean I’m adding a 30 degrees angle to another 30 degrees angle to get 60? Or, does it mean I’m adding a 150 degree angle to a 30 degree angle to get 180 degrees?

Let’s take a closer look at the “Equal Spreads” theorem of Divine Proportions on page 94. It shows that when calculating the combination of two equal spreads, $r$, you basically get two answers:

$r=0, r=4s(1-s)$

For our little example above of 150 degrees + 30 degrees, that makes sense, because 180 degrees is represented by a Spread of $0$. So adding spreads of $\frac{1}{4}$ is actually like adding either 30+30 to get 60 (spread $\frac{1}{4}$) or 30+150 to get 180 (spread $0$).

What of 150+150? Yes, that makes 300, which is basically 60, which corresponds to $\frac{3}{4}$, which is the $r=4s(1-s)$ solution as we calculated above.

But what if we go for the tripling of a spread?

As shown on page 101 of Divine Proportions, the calculation of a Spread equation for combining three equal spreads, when derived using the Triple Spread Formula, actually has two solutions, $r$.

$r=s, r=s*(3-4s)^2$

That means that when you are presented with three equal spreads, and try to find the spread of their combo, you can get multiple answers. Imagine again our example of 30 degree angle, or two lines separated by a spread of $\frac{1}{4}$. What would ‘triple’ that angle bring? In angle based trigonometry, it brings you 90 degrees. If you use the Three Equal Spread formula … you get this:

$r=\frac{1}{4}, r=\frac{1}{4}(3-4\frac{1}{4})^2$

$r=\frac{1}{4}, r=\frac{1}{4}(2)^2$

$r=\dfrac{1}{4}, r=1$

Well, $r=1$ makes sense… that is a right angle in Spread land. Perpendicular. 90 degrees. But what about this $r=\frac{1}{4}$ business?

This is where my amateur understanding is going to have to speculate a bit. From what I can gather, $\frac{1}{4}$ is kind of like a special spread solution, a bit like 0 was for doubling the spread in the example above. I get this impression from Spread polynomials, rotations and the butterfly effect, by S. Goh and N.J. Wildberger where they describe The Special Spreads.

$0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1$

If you think about adding 150 degrees to a 30 degree angle to another 150 degree angle… you wind up with 180 degrees plus 150, which is basically 330 degrees. But 330 degrees is basically describing two lines that are at a 30 degree angle from each other. Again, this spread would be $\frac{1}{4}$.

In that sense, maybe you can think of combining three spreads of $\frac{1}{4}$ and getting a result of $\frac{1}{4}$. An addition of two 150 degree angles and a single 30 degree angle gives you 330, which is a spread of $\frac{1}{4}$. An addition of two 30 degree angles and a single 150 degree angle would give you an angle of 210 degrees, which is again a 30 degree separation, which gives us a spread of $\frac{1}{4}$ again.

I could be wrong about all this, but it might be worth thinking about as a geometrical interpretation of “why” trying to use Spread polynomials to “Add angles” doesn’t work exactly the same way as you might think, if you grew up thinking about adding angles together.

It’s almost like spreads are some sort of simple version of Quantum Computing where you calculate multiple solutions to multiple possibilities at the same time.

But what about Pentagons again? There we have 5 spreads! How to combine those?

Well there are multiple ways. Let’s first consider Wildberger’s Divine Proportions, page 166, Exercise 14.3, we can imagine that all five spreads together form a spread of 0. This means solving this equation for $S_5$:

$s(5-20s+16s^2)^2=0$

I’m sure there is a way to do this by hand, but I was never good at factoring polynomials by hand. Fortunately in the modern day there is something called Symbolic Mathematics and computers that can manipulate it. Here is a simple program using the Python language and an add-on for it called SymPy .

>>> from sympy import *
>>> s=symbols(‘s’)
>>> print solve( s*(5-20*s+16*s**2)**2, s )

[0, -sqrt(5)/8 + 5/8, sqrt(5)/8 + 5/8]

Cool! That is the same answer that Wildberger gives in the book, if we assume that 0 here is a “trivial zero” of the equation. Not bad for a three line program.