# Algebraic Properties of Euclidean Geometry with Transcendental Curves

Nicole Venner The author thanks John Carter for his wonderful advice on this
project.

(January 1, 2023)

###### Abstract

While geometry with transcendental curves, like the Quadratrix of Hippias and
the Spiral of Archimedes, played a significant role in our modern developments
of geometry and algebra. The investigation has fallen off in the modern era
despite advancements in algebraic tooling. This paper 111Discussion of
original results begins with Definition 28 gives a description of the fields
using modern techniques such as Galois theory while solving an open conjecture
in a 1988 paper [2] to provide an answer to if these curves can solve the
problem of doubling the cube.

## A Quick Review of Geometries

Over 2000 years ago, the Ancient Greeks developed a system of mathematics in
response to the discovery that some numbers such as $\sqrt{2}$ were not
constructible as proportions of integers.

######  Definition 1 (Euclidean Geometry).

A point $\alpha$ is constructible if and only if $\alpha$ can be constructed
as a point using an unmarked ruler and compass. More specifically those tools
allow you to:

  1. 1.

Draw a line through any two points.

  2. 2.

Draw a circle centered at one point while passing through another point.

  3. 3.

For any 2 curves (either a line or a circle), you can draw points wherever
they intersect.

An extremely important step in the understanding of problems in euclidean
geometry is the invention of thinking of points on the plane as pairs of real
numbers and more recently as complex numbers $a+bi$. Thinking of individual
points as numbers allow you to prove some very interesting theorems:

######  Theorem 2.

Given any 2 constructible points $a,b\in\mathcal{C}$ represented as complex
numbers. Then the points with complex representations of $a+b$, $a-b$, $ab$
and $\frac{a}{b}$ are constructible.

###### Proof.

For any constructible numbers $a,b\in C$ that $0=a-a$,and $1=a/a$ and
$a+b=a-((a-a)-b)$ and $a\cdot b=\frac{a}{\frac{b}{b}\div b}$. You can subtract
and divide complex numbers using ruler and compass constructions as shown in
Figure 1 and 2:

Figure 1: Complex Division Figure 2: Complex Subtraction

∎

######  Lemma 3.

For any numbers $a$ exist geometric constructions for:
$\bar{a},\text{Re}(a),\text{Im}(a),|a|$ and $\sqrt{a}$

###### Proof.

Constructing the $|a|$ is relatively easy since the construction involves just
constructing a circle with $a$ as a point on the radius and finding the
intersection with the real line. From there you can find $\bar{a}$ since
$\bar{a}=\frac{|a|^{2}}{a}$. This also lets you say construct
$\text{Re}(a)=\frac{a+\bar{a}}{2}$ and $\text{Im}(a)=a-\text{Re}(a)$. The
construction involving square roots in the complex plane follows from the
general square root construction from Euclid combined with an angle bisector.
∎

As the Ancient Greeks were first developing geometry three classical problems
remained unsolved.

  1. 1.

Construct a square with the same area as a circle.

  2. 2.

Construct a cube with double the volume of another cube.

  3. 3.

Trisect an angle. (In general arbitrary angle partitions)

  4. 4.

Construct a regular-sided heptagon (In general an any-sided regular polygon)

Under the classical rules of euclidean geometry, all three of these problems
are impossible and we shall spend the next few sections giving a brief
overview of why. The key insight comes from the fact that a solution to each
one of these problems must involve constructing certain numbers, namely
$\sqrt{\pi}$, $\sqrt[3]{2}$ and $\sin(20^{\circ})$. (Importantly $\sin(20)$
happens to be the root of $8x^{3}-6x+\sqrt{3}=0$)

So what numbers can be constructed with Ancient Greek geometry? Since you are
dealing with the geometry of a two-dimensional plane it makes sense to involve
numbers that carry the same structure, namely complex numbers. From here you
can define

It turns out that these are all the numbers you can construct with classical
euclidean geometry, via the following theorem:

######  Theorem 4.

Any Intersections that can be created with classic euclidean geometry (line
intersected with line, circle and line, or circle and circle), can be
represented using roots of quadratic polynomials

###### Proof.

Since in the cartesian plane, all circles can be written as
$(x-c)^{2}+(y-d)^{2}=r^{2}$ and lines as $y=ax+b$. Then you can algebraically
represent the x and y coordinates of intersections as solutions to
polynomials, the calculations for a line intersecting a line and a circle
intersecting a circle are relatively straightforward, so all that is left is
to work out the algebra in the case of a circle intersecting with another
circle.

$\displaystyle\sqrt{r^{2}-(x-a)^{2}}+c$
$\displaystyle=\sqrt{p^{2}-(x-b)^{2}}+d$ $\displaystyle\sqrt{r^{2}-(x-a)^{2}}$
$\displaystyle=\sqrt{p^{2}-(x-b)^{2}}+(d-c)$ $\displaystyle r^{2}-(x-a)^{2}$
$\displaystyle=\left(\sqrt{p^{2}-(x-b)^{2}}+(d-c)\right)^{2}$ $\displaystyle
r^{2}-(x-a)^{2}$
$\displaystyle=p^{2}-(x-b)^{2}+2(d-c)\sqrt{p^{2}-(x-b)^{2}}+(d-c)^{2}$
$\displaystyle\frac{(r^{2}-p^{2})+(x-b)^{2}-(x-a)^{2}-(d-c)^{2}}{2(d-c)}$
$\displaystyle=\sqrt{p^{2}-(x-b)^{2}}$
$\displaystyle\frac{(2b-2a)x+(b^{2}+r^{2}-p^{2}-a^{2}-d^{2}+2dc-c^{2})}{2(d-c)}$
$\displaystyle=\sqrt{p^{2}-(x-b)^{2}}$
$\displaystyle\frac{\left((2b-2a)x+(b^{2}+r^{2}-p^{2}-a^{2}-d^{2}+2dc-c^{2})\right)^{2}}{4(d-c)^{2}}$
$\displaystyle=p^{2}-(x-b)^{2}$

Expanding out yields:

$\displaystyle\frac{2x^{2}(b-a)^{2}+4x(b-a)(b^{2}+r^{2}-p^{2}-a^{2}-d^{2}+2dc-c^{2})+(b^{2}+r^{2}-p^{2}-a^{2}-d^{2}+2dc-c^{2})^{2}}{4(d-c)^{2}}=(p^{2}-(x-b)^{2})$
$\displaystyle
2x^{2}(b-a)^{2}+4x(b-a)(b^{2}+r^{2}-p^{2}-a^{2}-d^{2}+2dc-c^{2})+(b^{2}+r^{2}-p^{2}-a^{2}-d^{2}+2dc-c^{2})^{2}=4(d-c)^{2}(-x^{2}+2xb-b^{2}+p^{2})$
$\displaystyle
2x^{2}\left((b-a)^{2}+2(d-c)^{2}\right)+4x\left((b-a)(b^{2}+r^{2}-p^{2}-a^{2}-d^{2}+2dc-c^{2})-2b(d-c)^{2}\right)+(b^{2}+r^{2}-p^{2}-a^{2}-d^{2}+2dc-c^{2})^{2}+4(d-c)^{2}(b^{2}-p^{2})=0$

Since the $x$ is solvable, we can obtain the $y$ value by substituting $x$
into either $ax+b=y$ or $(x-a)^{2}+(y-b)^{2}=r^{2}$ yielding a quadratic as
well. ∎

Since any quadratic can be solved using the quadratic formula, we can
construct every constructible number, using the operations we have defined so
far $(+,-,\times,\div,\sqrt{})$ suffice to construct any constructible number.
Having reduced the problem to algebra we can show it just isn't enough to
solve those classic geometry problems.

### The Foldable Numbers

Knowing that it is only possible to construct solutions of quadratics, it
seems natural to ask if other possible sets of constructions could create more
numbers. A recent modern example is the trigonometric numbers:

######  Definition 5.

A number $\alpha$ is in $\mathcal{O}$ if the point representing the complex
number $\alpha$ is constructible using simple reduced222Every other kind of
simple paper fold can be completed using repeated applications of operation 3
origami folds

  1. 1.

Given two points P1 and P2, we can fold a crease line connecting them.

  2. 2.

Given two lines, we can locate their point of intersection, if it exists.

  3. 3.

Given two points P1 and P2 and two lines L1 and L2, we can, whenever possible,
make a fold that places P1 onto L1 and also places P2 onto L2.

This may seem like an oversimplification of origami, but helpfully a proof
from [3] (originally proved by Lang in 2003) shows us that

######  Theorem 6 (Lang).

If we only allow one fold at a time and assume all our creases are straight
lines. Then any folding operation can be described by our three origami
operations

Using these techniques it becomes possible to create solutions to cubic
equations, consider the following construction for the cube root of 2 in
Figure 3

Figure 3: Construction of the cube root of 2 (Hull) (Maybe replace with own?)

This is done by moving the $x$ and $y$ axes to intersect at (1,2) and then
using our third rule to map $(-1,0)$ and $(0,-2)$, to the new axes. From there
it is possible to use conventional geometry and an argument from similar
triangles to argue that the side length of the square is the cube root of 2.

This exact argument can be easily generalized to solving any cubic as first
proved by Lil.

######  Theorem 7 (Lil).

The roots of any cubic can be expressed using simple paper folds.

The author's first exposure to this was in a video produced by [6].
Furthermore using a similar argument from algebra it turns out to completely
describe the numbers you can make with simple folds in origami. 333There do
exist extensions of origami involving multi-folds that create multiple creases
at once. While these drastically expand the numbers that can be produced. The
author believes there are significant issues in using them for real-life
demonstrations or educational purposes. Not only since the constructions
require a large amount of technical skill to execute, but inaccuracies in the
construction are known to propagate and grow, making developing intuition
quite difficult.

######  Theorem 8.

The intersections created by any origami fold can be described by a cubic
polynomial.

Proof for this can be found in [3]

### Galois Theory and Unconstructible Numbers

From here since we have 2 notions of geometry that can be modeled as
algebraically solving polynomial equations, it remains if it is possible to
construct different numbers like $\sqrt[5]{2}$ or transcendental numbers like
$\pi$ or $e$.

The original idea that these problems are impossible to construct in either
comes from a branch of mathematics called Galois Theory, which we shall
briefly stroll-through for the sake of brevity.

######  Definition 9.

A polynomial is irreducible if it cannot be factored.

######  Definition 10.

A Field $\mathcal{F}$ is a set of numbers that is closed under the operations
required for traditional arithmetic (addition, subtraction, multiplication,
and division). A Field Extension is a process of extending a field
$\mathcal{F}$ with new numbers to form a field $\mathcal{G}$, this extension
is often written as $G/F$

######  Definition 11.

A field $\mathcal{F}$ splits over a polynomial $p(x)$ if every root of $p(x)$
is contained in $\mathcal{F}$. An extension $\mathcal{F}/\mathcal{G}$ is a
splitting extension if for any irreducible polynomial in $\mathcal{G}$ then
either all or none of its roots are in $\mathcal{F}$.

While this is a helpful definition, it might seem hard to come up with
examples where this is true.

######  Theorem 12.

For any irreducible polynomial $p(x$ with coefficients $\mathcal{F}$, then the
extension smallest field containing all roots
$\mathcal{F}(x_{1},x_{2},\dots)/\mathcal{F}$ is splitting.

this can be shown to not be the case.

Some examples and a short lemma might be helpful to understand the concept.

######  Example 13.

Our first example is going to be the smallest field containing both the
rationals and $\sqrt{5}$, written as $\mathbb{Q}(\sqrt{5})$. It is reasonable
to ask what the elements of this field look like, and it turns out it is
completely describable using 2 rational numbers like so:
$\mathbb{Q}(\sqrt{5})=\left\\{a+b\sqrt{5}|a,b\in\mathbb{Q}\right\\}$. These
form a field since you can prove they are closed under subtraction and
division like so:

$\displaystyle(a+b\sqrt{5})-(c+d\sqrt{5})$ $\displaystyle=(a-b)+(b-d)\sqrt{5}$
$\displaystyle\frac{a+b\sqrt{5}}{c+d\sqrt{5}}$
$\displaystyle=\left(\frac{ac-5bd}{c^{2}-5d^{2}}\right)+\left(\frac{bc-
ad}{c^{2}-5d^{2}}\right)\sqrt{5}$

We can also see that this field is the splitting field for the polynomial
$x^{2}-5$ since both of its roots of $\sqrt{5}$ and $-\sqrt{5}$ are contained
in it.

######  Lemma 14.

If a field $\mathcal{F}$ is composed of a tower of fields
$\mathbb{Q}\subset\mathcal{F}_{1}\subset\mathcal{F}_{2}\subset\mathcal{F}_{3}\subset\dots$,
where each extension is splitting. Then $\mathcal{F}/\mathbb{Q}$ is splitting.

As proved in [1].

######  Definition 15.

A field automorphism is an invertible function on a field such that.

$f(a-b)=f(a)-f(b)\qquad f\left(\frac{a}{b}\right)=\frac{f(a)}{f(b)}$

Based on those properties we can conclude that $f(a)\cdot f(b)=f(ab)$ and
$f(a)+f(b)=f(a+b)$. And for any rational $q$ then $f(q)=q$.

Through the power of field automorphisms, we can prove some useful results:

######  Lemma 16.

If $f$ is a field automorphism defined on a splitting field $\mathcal{F}$
defined by a polynomial with rational coefficients $p(x)$, for any root $z$,
then $f(z)$ is also a root.

###### Proof.

If $z$ is a root of $p(x)$, then we know by definition that

$0=a_{n}z^{n}+a_{n-1}x^{n-1}+\dots+a_{1}x^{n}+a_{0}$

We can now apply $f$ to both sides of this equation, and since by the
properties of a field automorphism we know that you can distribute across
integer powers $f(a^{n})=f(a)^{n}$ (proven by using induction with
multiplication) we can finish the proof.

$\displaystyle f(0)$
$\displaystyle=f\left(a_{n}z^{n}+a_{n-1}z^{n-1}+\dots+a_{1}z+a_{0}\right)$
$\displaystyle 0$
$\displaystyle=f(a_{n}z^{n})+f(a_{n-1}z^{n-1})+\dots+f(a_{1}z)+f(a_{0})$
$\displaystyle 0$
$\displaystyle=a_{n}f(z^{n})+a_{n-1}f(z^{n-1})+\dots+a_{1}f(z)+a_{0}$
$\displaystyle 0$
$\displaystyle=a_{n}f(z)^{n}+a_{n-1}f(z)^{n-1}+\dots+a_{1}f(z)+a_{0}$

Thus $f(z)$ is a root of $p(x)$. (This applies generally to polynomials with
irrational coefficients if the automorphism $f$ keeps the coefficients fixed
in place.) ∎

######  Corollary 17.

If $\mathcal{F}$ is a splitting field of a polynomial $p(x)$ with rational
coefficients, for any root $z$ of $p(x)$, then both the conjugate $\bar{z}$ as
well as the real, imaginary parts $\Re(z)$ and $i\cdot\Im(z)$ are in
$\mathcal{F}$,

###### Proof.

Consider that the field automorphism $f(a+bi)=a-bi$. We can prove that this is
a field automorphism with some basic algebra

$\displaystyle f(a+bi)-f(c+di)=a-bi-c+di=(a-c)-(b-d)i=f(a-c+(b-d)i)$

Furthermore

$\displaystyle\frac{f(a+bi)}{f(c+di)}$ $\displaystyle=\frac{a-bi}{c-di}$
$\displaystyle=\frac{(a-bi)(c+di)}{c^{2}+d^{2}}$
$\displaystyle=\frac{ac+bd}{c^{2}+d^{2}}+i\frac{ad-bc}{c^{2}+d^{2}}$
$\displaystyle=f\left(\frac{ac+bd}{c^{2}+d^{2}}+i\frac{bc-
ad}{c^{2}+d^{2}}\right)$ $\displaystyle=f\left(\frac{a+bi}{c+di}\right)$

any root $z=a+bi$ of a polynomial with rational coefficients, then consider
the ∎

Now

######  Theorem 18.

Every field and field extension can be put into correspondence with a group,
often written as $Gal(\mathcal{F})$ or $Gal(\mathcal{F}/\mathcal{G})$. Defined
to be the group formed by all field automorphisms on your field. (The Galois
Group of a field extension $\mathcal{F}/\mathcal{G}$ are the field
automorphisms of $\mathcal{G}$ that keep the values of $\mathcal{F}$
unchanged.)

A true understanding of galois theory is going to require a fair bit of
understanding of the properties of groups. But even then an example might be
helpful:

######  Example 19.

Let $\mathcal{F}$ be the smallest field where $x^{3}-2$ splits over the
rationals. Since $\mathcal{F}$ is the smallest field that contains all the
roots of $x^{3}-2$ we can say
$\mathcal{F}=\mathbb{Q}(\sqrt[3]{2},\omega\sqrt[3]{2},\omega^{2}\sqrt[3]{2})$,
where $\omega^{3}=1$. It also does have a representation using rational
numbers like so:

$\mathcal{F}=\left\\{a+b\cdot 2^{1/3}+c\cdot 2^{2/3}+d\cdot\omega+e\cdot\omega
2^{1/3}+f\cdot\omega 2^{2/3}\quad|\quad a,b,c,d,e,f\in\mathbb{Q}\right\\}$

Since $\mathcal{F}$ is a splitting extension over the rationals it is
relatively easy to calculate its Galois group. Since the rationals (or your
preferred base field) must remain fixed by any field automorphism, by the
properties of the automorphism for any root $z$, then $f(z)$ must also be a
root. Likewise, since every element of your field can be written as an element
of the base field $p_{n}$, times a product of roots to some polynomial
$\alpha$, applying the permutation yields:

$f\left(\sum_{i=1}^{k}a_{i}\cdot\alpha_{1}^{n_{1_{i}}}\cdot\alpha_{2}^{n_{2_{i}}}\cdot\alpha_{3}^{n_{3_{i}}}\cdot\dots\right)=\sum_{i=1}^{k}a_{i}\cdot
f(\alpha_{1})^{n_{1_{i}}}\cdot f(\alpha_{2})^{n_{2_{i}}}\cdot
f(\alpha_{3})^{n_{3_{i}}}\cdot\dots$

Thus the field automorphisms are given by any permutation of the roots of the
polynomial while keeping all of their algebraic properties intact.

For our specific polynomial, the only algebraic properties the roots need to
respect are:

$\displaystyle x_{0}^{3}=x_{1}^{3}=x_{2}^{3}=x_{0}\cdot x_{1}\cdot x_{2}=2$

Thus any permutation leads to a valid automorphism. Showing that the Galois
group of $x^{3}-2$ is the symmetric group $S_{3}$.

######  Definition 20.

The order of any (Galois) group $G$ written as $|G|$ is defined to be the
number of elements (field automorphisms) in the group.

######  Theorem 21.

For any fields $F\subset G$ with a splitting field extension $G/F$. Then

$|Gal(G/F)|\cdot|Gal(F)|=|Gal(G)|$

Furthermore

######  Theorem 22.

For any irreducible polynomial $p$ of degree $n$, the splitting field $F$
containing all the roots $F(p_{1},p_{2},\dots)$ then the order of
$Gal(F(p_{1},p_{2},\dots)/F)$ is divisible by $n$ and divides $n!$.

For interested readers, the proofs for these statements are a combination and
simplification of several proofs in the Galois theory chapter of [1]. Using
these 2 theorems we can show that the three problems are impossible. First
starting by saying that

######  Theorem 23.

Every constructible number $c$ is contained in a field with a Galois group
with an order of $2^{n}$ for some natural number $n$.

###### Proof.

Since every constructible number is created using a finite number of ruler and
compass constructions, you can interpret it as extending the base field of the
rationals using a finite number of solutions to quadratic polynomials.
Combining the previous 2 theorems, we get that the order of the final field
containing our number $c$ must be contained in a field with degree $2^{n}$. ∎

Likewise following the same argument one can see that

But notice that classic problems of doubling the cube and trisecting the angle
involve extensions of irreducible cubic polynomials, $x^{3}-2$ and
$8x^{3}-6x+\sqrt{3}$. Thus any 'nice' field containing them must be divisible
by 3444If you calculate them out it turns out that the first polynomial has a
Galois group of order 6 and the second has order 3. But since every number
constructible with classic euclidean geometry has a degree with no factors of
3.

We can further see that the field of foldable numbers can solve these since:

From here we have one final result

######  Corollary 24.

For any number $o\in\mathcal{O}_{1}$ then the smallest field containing $o$ is
of degree $2^{n}\cdot 3^{k}$ for any integers $n$ and $k$

Since it is possible to describe doubling the cube and trisecting the angle as
the solutions of the cubics $x^{3}-a$ and $3x^{3}-2x+c$ for $c<1$ and any $a$.
However, some numbers are still out of reach, arbitrary angle division and
arbitrary regular polygons are still unconstructible as well as the elusive
square of the circle.

## The Quadratrix of Hippias

In Ancient Greece, the first partial solution to the angle trisection problem
comes from Hippias where he imagines a new curve in the plane drawn with a
compass-like construction. The definition of the curve can be given as so.

Figure 4: Definition of the Quadratrix

######  Definition 25.

Given a circle with center $A$ and point on the radius $B$. The quadratrix is
defined so that for any point on the circle $X$. Then the point on the
quadratrix $Z$ on the line $\overline{AX}$ is defined so that the arc-length
of  A BX is equal to the distance from the line $\overline{AB}$ As seen in
Figure 4

######  Example 26.

(Construction of the number $\pi/7$) Using the definition of the quadratrix,
we can try to find the points where it intersects the imaginary axis, since
the imaginary line is angled at $\pi/2$ radians from the real line, then by 4,
it must intersect at the points $\pm\frac{i\pi}{2}$. Since the number $i\pi/2$
is constructible, we can multiply by the constructible number $\frac{-2i}{7}$
to construct the point $\frac{\pi}{7}$.

######  Example 27.

(Partition of an angle into 5 pieces)

Figure 5: Partition of an angle into 5 parts.

Hippias also proved that this curve can partition an arbitrary angle, and can
therefore create any regular-sided polygon. 130 years later the mathematician
Dinostratus proved that it is also possible to square the circle as well.
(Both techniques will be proved later by saying that the field containing
quadratrix is also inside) But from here to the best of the author's knowledge
no one has done a further dive into the quadratrix aside from reiterating the
results from over 2 millennia ago.

## Algebraic Definition of $\mathcal{T}_{1}$

As we have just seen, the majority of the power of the quadratrix in classic
constructions comes from its ability to convert angles into segment lengths
and segment lengths into angles, thus let us consider a field generated by
those 2 constructions:

######  Definition 28.

The field of ''Perpendicular Trigonometric'' Numbers $\mathcal{T}_{1}$ is
defined to be the smallest field satisfying

  1. 1.

The set of rational numbers $\mathbb{Q}$ is contained in $\mathcal{T}_{1}$
(Lemma 49 lets us show that every constructible number is in
$\mathcal{T}_{1}$)

  2. 2.

The field is closed under the operation defined in Definition 30

  3. 3.

The field is closed under the operation defined in Definition 29

######  Definition 29 (Arc $\to$ Segment: T1).

For any arc  A AB it is possible to construct a segment $\overline{CD}$ with
length equal to the arc-length of  A AB

######  Definition 30 (Segment $\to$ Arc: T2).

For any segment $\overline{CD}$ it is possible to construct a arc  A AB with
arc-length equal to the length of $\overline{CD}$

It is quite apparent that this field is already solving ancient problems from
antiquity as since it is easy to construct a segment of length $\pi$, you can
easily square the circle (Also showing that
$\mathcal{C}(\pi)\subset\mathcal{T}_{1}$). It goes significantly deeper than
that as the following theorem should show

######  Theorem 31.

For any real trigonometric $\alpha$, then $\alpha$ is trigonometric if and
only if $e^{i\alpha}$ is trigonometric. (Alternatively stated, for any
$|\beta|=1$ is trigonometric if and only if $\ln(\beta)$ is trigonometric)

Figure 6: A Geometric View of Complex Exponentiation

###### Proof.

If a real number $\alpha$ is constructible then via T2 we can wrap the segment
from $0$ to $\alpha$ around the unit circle with an endpoint of
$\cos(\alpha)+i\sin(\alpha)=e^{i\alpha}$. Likewise for any trigonometric point
on the unit circle is expressible as
$\cos(\alpha)+i\sin(\alpha)=e^{i\alpha}$then we can use T1 to unwind that to a
segment length, thereby constructing $\alpha$. (This can also be expressed by
taking the natural log of any point on the unit circle in the complex plane.)
∎

######  Theorem 32 (Trigonometric Closure).

Any real number $\alpha$ is trigonometric if and only if $\sin(\alpha)$,
$\cos(\alpha)$, $\tan(\alpha)$, are trigonometric.

###### Proof $\Rightarrow$.

Its notable that if $\alpha$ is real and trigonometric then by theorem 31,
then $e^{i\alpha}$ is trigonometric and since:

$\displaystyle\sin(x)$ $\displaystyle=\frac{e^{ix}-e^{-ix}}{2i}$
$\displaystyle\cos(x)$ $\displaystyle=\frac{e^{ix}+e^{-ix}}{2}$
$\displaystyle\tan(x)$ $\displaystyle=-i\frac{e^{ix}-e^{-ix}}{e^{ix}+e^{-ix}}$

Every number of this type is therefore constructible. ∎

###### Proof $\Leftarrow$.

Using the functions above, it is possible to derive the following formulas
using the socks and shoes rule ie. $(f\circ g)^{-1}=g^{-1}\circ f^{-1}$.

$\displaystyle\sin^{-1}(x)$
$\displaystyle=-i\ln\left(\sqrt{1-x^{2}}+ix\right)$
$\displaystyle\cos^{-1}(x)$
$\displaystyle=\frac{\pi}{2}+i\ln\left(\sqrt{1-x^{2}}+ix\right)$
$\displaystyle\tan^{-1}(x)$
$\displaystyle=\frac{i}{2}\ln\left(\frac{1-ix}{1+ix}\right)$

Where the inverse trig functions are defined for real values ($[0,1]$ for
$\sin^{-1}$ and $\cos^{-1}$, and the entire real line for $\tan^{-1}$), then
the values that are given to $\ln(x)$ are all of magnitude 1. ∎

## Geometric Definitions of $\mathcal{T}_{1}$

So how does this curve relate to the quadratrix and other transcendental
curves throughout history that have been used to solve similar problems? While
it might be tempting to look at the fields generated by creating a
"quadratrix" compass, or "trigonometric" compass that can draw these shapes
and think about what intersections you might get, however trying this method
can easily lead to intractable problems since by definition of your field is
going to include solutions to equations like this:

$\displaystyle\alpha x=\sin(x)$

Since it must include the intersections of a sloped line and the standard
trigonometric curve. Based on that it makes sense to examine subfields with
fewer allowed constructions.

### Trigonometric Curves

######  Definition 33.

Consider the field of numbers generated by allowing intersections with the
curves $x\mapsto(x,\sin(x))$ or $x\mapsto(x,\cos(x))$, then consider the
subfield:

  1. 1.

All the standard constructions in classic euclidean geometry, (ie, creating
lines and circles, and finding intersections of any two lines, two circles, or
a line and a circle)

  2. 2.

Given the curve $x\mapsto(x,\sin(x))$ or $x\mapsto(x,\cos(x))$ when graphed on
$\mathbb{R}^{2}$, you can construct the intersection of the curves, and any
vertical line.

  3. 3.

Given the curve $x\mapsto(x,\sin(x))$ or $x\mapsto(x,\cos(x))$ when graphed on
$\mathbb{R}^{2}$, you can construct the intersection of the curves, and any
horizontal line.

######  Theorem 34.

The field described in Definition 33 is equal to $\mathcal{T}_{1}$.

###### Proof.

The first direction of proving that $\mathcal{T}_{1}$ contains the
construction amounts to showing that all the intersections with vertical lines
are constructible since this happens to be the same as evaluating the
function, and horizontal lines are the same as finding function inverses. Thus
applying Theorem 32 deals with both halves of our equivalence. Showing every
number constructible using classic Euclidean constructions will wait until
lemma 49 ∎

### Quadratrix of Hippias

######  Definition 35.

Consider a subfield of numbers constructible with the quadratrix, generated
using these constructions:

  1. 1.

All the standard constructions in classic euclidean geometry, (ie, creating
lines and circles, and finding intersections of any two lines, two circles, or
a line and a circle)

  2. 2.

Given the standard quadratrix as constructed in Figure 4, you can construct
the intersection of the quadratrix and any line passing through the origin.

  3. 3.

Given the standard quadratrix as constructed in Figure 4, you can construct
the intersection of the quadratrix and any horizontal line.

######  Theorem 36.

The field described in Definition 35 is equal to $\mathcal{T}_{1}$.

###### Proof.

To proceed with this proof we shall show that every number in each field is
present in the other $(Quad\Rightarrow\mathcal{T}_{1})$. Notice that given any
angle measure $\alpha$, you can construct a ray at the origin where the angle
formed with the real axis is $\alpha$. Then by definition, the distance
between the intersection and the real axis is $\alpha$. Likewise, creating a
horizontal line given by $\\{x+i\cdot\alpha\quad x\in\mathbb{R}\\}$. Then find
the intersection with the quadratrix and draw the line to the origin creating
an angle with the real axis of value $\alpha$. Showing every number
constructible using classic Euclidean constructions will wait until lemma 49

$(\mathcal{T}_{1}\Rightarrow Quad)$. Consider a line drawn through the origin
with angle $\alpha$, by definition, the point on the quadratrix can be written
as $|\alpha|+i\cdot|\alpha|\cdot\cot(\alpha)$. Since the point is expressible
using trigonometric numbers, it does not extend the field further. Likewise,
taking a horizontal line, by the first property we can turn its distance from
the real axis into an angle value and see where that line intersects

∎

### Archimedean Spiral

######  Theorem 37.

The field $\mathcal{T}_{1}$ is equal to the field of constructible numbers
with the additional ability to construct intersections with the Archimedean
Spiral with lines passing through the origin and circles centered at the
origin.

Figure 7: The Spiral of Archemedies

######  Definition 38.

Consider a subfield of numbers constructible with the Spiral of Archimedes,
generated using these constructions:

  1. 1.

All the standard constructions in classic euclidean geometry, (ie, creating
lines and circles, and finding intersections of any two lines, two circles, or
a line and a circle)

  2. 2.

Given the standard Spiral of Archimedes as constructed in Figure 7, you can
construct the intersection of the spiral and any line passing through the
origin.

  3. 3.

Given the standard Spiral of Archimedes as constructed in Figure 7, you can
construct the intersection of the spiral and any circle centered at the
origin.

######  Theorem 39.

The field described in Definition 38 is equal to $\mathcal{T}_{1}$.

###### Proof.

To proceed with this proof we shall show that every number in each field is
present in the other

$(Archm\Rightarrow\mathcal{T}_{1})$. Notice that if you have any angle, you
can create a ray from the origin with an angle $\alpha$ the distance from the
point of intersection on the spiral and the origin will be $|\alpha|$, thus
showing it is constructible. Likewise for any constructible real number $r$,
then taking that number we can create a circle of radius $r$ centered at the
origin. Drawing a ray from the origin to this point lets us create a ray of
$z$ radians from the real axis. Showing every number constructible using
classic Euclidean constructions will wait until lemma 49

$(\mathcal{T}_{1}\Rightarrow Archem)$. Notice the parameterization of the
spiral of Archimedes as $(t\cdot\cos(t),t\cdot\sin(t))$., this lets us
algebraically solve for the intersection points of any circle centered at the
origin with radius $r$, namely $r\cos(t)+i\cdot r\sin(t)$. Likewise for any
ray from the origin with an angle $\theta$ its intersection point will be
defined by $\theta\cos(\theta)+i\cdot\theta+\sin(\theta)$ ∎

## Transcendence of $\mathcal{T}_{1}$

Now having established some various formulas for the algebra of
$\mathcal{T}_{1}$ we can continue deriving results first citing an important
theorem in transcendental number theory.

######  Definition 40.

The transcendence degree of a field $\mathcal{F}$ is the smallest number of
transcendental elements $a_{1},\dots,a_{n}$ such that every element of
$\mathcal{F}$ can be expressed as the roots of polynomials with coefficients
in $\mathbb{Q}(a_{1},\dots,a_{n})$

######  Theorem 41 (Lindemann–Weierstrass).

For any linearly independent set of algebraic numbers
$a_{1},a_{2},\dots,a_{n}$, then the transcendence degree of
$\mathbb{Q}(e^{a_{1}},e^{a_{2}}\dots,e^{a_{n}})$ is $n$

A proof for this is available in [4]

######  Example 42.

$\pi$ is transcendental (not the root of any polynomial with rational
coefficients)

###### Proof.

To show that $\pi$ is not algebraic, we must show that $\pi$ being algebraic
leads to a contradiction. If $\pi$ is algebraic, then $i\pi$ is algebraic, and
therefore the field $\mathbb{Q}(e^{i\pi})=\mathbb{Q}(-1)=\mathbb{Q}$ has
transcendence degree 1. But $\mathbb{Q}$ trivially has a transcendence degree
of $0$, so $\pi$ is transcendental. ∎

######  Theorem 43.

The transcendence degree of $\mathcal{T}_{1}$ is infinite.

###### Proof.

The key to this will be the Lindemann–Weierstrass theorem. Since the set

$\\{i\sqrt{p}\quad|\quad p\text{ prime}\\}$

gives you an infinite set of algebraic numbers that are trigonometric and
linearly independent. Via theorem 31 the number $e^{i\sqrt{p}}$ is
constructible. And since $\mathcal{T}_{1}$ is contained in the extension

$\mathbb{Q}(e^{i},e^{i\sqrt{2}},e^{i\sqrt{3}},e^{i\sqrt{5}}\dots)\subset\mathcal{T}$

Thus the transcendence degree of $\mathcal{T}_{1}$ is infinite. ∎

######  Theorem 44.

For $a,b\in\mathcal{T}_{1}$ and $\beta\in\mathbb{R}$, then $\alpha^{\beta}$ is
constructible if and only if $|\alpha|^{\beta}$ is constructible.

###### Proof.

One direction is simple since for real $\beta$ then we can take the absolute
value of $\alpha^{\beta}$ to give us $|\alpha^{\beta}|=|\alpha|^{\beta}$. For
the other direction notice that for any trigonometric $\alpha$ we can break it
down into $\alpha=|\alpha|e^{i\delta}$. Like so

$\displaystyle\alpha$ $\displaystyle=|\alpha|e^{i\delta}$
$\displaystyle\alpha^{\beta}$
$\displaystyle=\left(|\alpha|e^{i\delta}\right)^{\beta}$
$\displaystyle\alpha^{\beta}$ $\displaystyle=|\alpha|^{\beta}e^{i\delta\beta}$

Since $e^{i\delta\beta}$ is trigonometric then if $|\alpha|^{\beta}$ is
trigonometric then $\alpha^{\beta}$ is trigonometric. ∎

######  Lemma 45.

For $\alpha\in S_{1}$ and trigonometric then $\sqrt[n]{\alpha}$ is
trigonometric. (Setting $\alpha=1$ gives you the roots of unity.)

###### Proof.

The special case of Theorem 44 where $|\alpha|=1$ since $1^{\beta}=1$. Any nth
root of unity is constructible since $e^{\frac{2\pi i}{n}}=\zeta_{n}$. ∎

For the next result, we need a quick definition:

######  Definition 46 (Abelian Group).

A group is abelian, if for any 2 elements $a,b$ then $a\star b=b\star a$.

######  Theorem 47 (Kronecker-Weber).

If $\mathcal{F}$ is a field with a finite abelian Galois extension then
$\mathcal{F}$ is contained in the field generated by some $n$th root of unity.

Due to Kronecker and Weber the author originally found the theorem in [1] on
page 600.

######  Corollary 48.

Every number inside a finite abelian extension of $\mathbb{Q}$ is contained in
$\mathcal{T}_{1}$.

This also with a quick lemma lets us take care of showing that our algebraic
definition of $\mathcal{T}_{1}$ also contains all constructible numbers

######  Lemma 49.

Every constructible number is contained in $\mathcal{T}_{1}$, alternatively
written as $\mathcal{C}\subseteq\mathcal{T}_{1}$

###### Proof.

Every quadratic has degree $2$ and thus has galois group of at most degree
$2!=2$, since every group of order $2$ is abelian it is contained in
$\mathcal{T}_{1}$ ∎

This shows that there is a vast array of polynomials that can be created
through this process, including the roots of irreducible polynomials to any
degree.

## Roots of polynomials with nonabelian galois groups in $\mathcal{T}_{1}$ and
Doubling the Cube is not possible with Angle Partitons

The main goal of this section will be to show that it is possible to construct
algebraic numbers outside of the abelian numbers generated by the Kronecker-
Weber theorem. And build the foundations for why certain numbers such as
$\sqrt[3]{2}$ are probably not contained in $\mathcal{T}_{1}$

######  Lemma 50.

For any nontrivial Pythagorean triple $a^{2}+b^{2}=c^{2}$ Then

$\frac{a+bi}{c}$

is not a root of unity.

###### Proof.

Via the previous theorem, all we need to do is check that $a+bi$ is not a root
of unity of degree $n$ where $\varphi(n)=2$, namely we have to check $3$, $4$
and $6$. And since the 3rd and 6th roots of unity involve Gaussian integers
divided by square roots. And the fact that $a$ and $b$ have to be not equal to
zero takes care of the 4th roots of unity. ∎

######  Lemma 51.

Minimal polynomial of the trigonometric number
$\sqrt[n]{\frac{a+bi}{\sqrt{a^{2}+b^{2}}}}$ for $a,b\in\mathbb{R}$ is
$x^{2n}-(2a(a^{2}+b^{2})^{-3/2})x^{n}+1$

###### Proof.

$\displaystyle\sqrt[n]{\frac{a+bi}{\sqrt{a^{2}+b^{2}}}}$ $\displaystyle=x$
$\displaystyle a+b\sqrt{-1}$ $\displaystyle=x^{n}\sqrt{a^{2}+b^{2}}$
$\displaystyle b\sqrt{-1}$ $\displaystyle=x^{n}\sqrt{a^{2}+b^{2}}-a$
$\displaystyle-b^{2}$ $\displaystyle=(x^{n}\sqrt{a^{2}+b^{2}}-a)^{2}$
$\displaystyle 0$
$\displaystyle=(a^{2}+b^{2})x^{2n}-2ax^{n}\sqrt{a^{2}+b^{2}}+a^{2}+b^{2}$
$\displaystyle 0$ $\displaystyle=x^{2n}-\frac{2a}{\sqrt{a^{2}+b^{2}}}x^{n}+1$

∎

######  Theorem 52.

$\mathcal{T}_{1}$ contains algebraic numbers with nonabelian field extensions.

###### Proof.

Consider the number $\frac{3+4i}{5}$, via Theorem 44, the number
$\sqrt[3]{\frac{3+4i}{5}}$ is trigonometric. Since it's algebraic it must have
a minimal polynomial and using lemma 51 it has a minimal polynomial of
$5x^{6}-6x^{3}+5$. The Galois group of this subgroup of the permutation group
$S_{6}$ is generated by the following elements.

$\left\\{(24),(35),(12),(34),(56)\right\\}$

This group is not abelian since $(24)(12)\neq(12)(24)$.555Calculated using the
Magma CAS ∎

We can solve many more polynomials with nonabelian galois extensions.

######  Theorem 53.

The solutions for a reduced cubic polynomial $x^{3}+px+q=0$ are expressible
using trigonometric functions:

$x_{k}=2\sqrt{-\frac{p}{3}}\cos\left[\frac{1}{3}\arccos\left(\frac{3q}{2p}\sqrt{\frac{-3}{p}}\right)-\frac{2\pi
k}{3}\right]\quad\text{ for }k=0,1,2$

This quite helpful result is given by [5] paper on Descartes and the cubic
equation. Furthermore doing some results on the cubic discriminant quantity
$\frac{3q}{2p}\sqrt{\frac{-3}{p}}$. Lets us conclude that the polynomial has 3
real roots if the discriminant has a magnitude less than or equal to 1. And
has 2 complex roots and 1 real root otherwise.

######  Corollary 54.

If a cubic has three real roots then it splits over $\mathcal{T}_{1}$

###### Proof.

The proof for this hinges on the proof that when the input value to $\arccos$
namely $\left(\frac{3q}{2p}\sqrt{\frac{-3}{p}}\right)$ is between $-1$ and
$1$, then the polynomial will have three real solutions. The alternative case
where a cubic has 1 real and 2 complex roots, (ie $x^{3}-2$), will involve
putting values with a magnitude greater than or equal to 1. ∎

Does this describe all the polynomials we can create with our algebra? This
and other questions have a nonzero probability of potentially being answered
in this paper!

######  Lemma 55.

The polynomial $x^{2n}+2cx^{n}+1$ has solutions in $\mathcal{T}_{1}$ for
trigonometric and real $|c|\leq 1$

###### Proof.

Consider the minimal polynomial of

$x=\sqrt[n]{\frac{-1+\sqrt{c^{2}-1}}{\sqrt{1+c^{2}-1}}}$

Since it is in our special form we can use Lemma 51 to say that the minimal
polynomial is:

$\displaystyle 0$ $\displaystyle=x^{2n}-\frac{2(-1)}{\sqrt{c^{2}}}x^{n}+1$
$\displaystyle 0$ $\displaystyle=x^{2n}-(-2c)x^{n}+1$ $\displaystyle 0$
$\displaystyle=x^{2n}+2cx^{n}+1$

Unfortunately, the limitation is that $\sqrt{c^{2}-1}$ must be real forces $c$
to be in $0\leq c\leq 1$. Luckily doing some variable substitution and
flipping the sign of $a$ in our initial number gives us the minimal polynomial

$\displaystyle x=\sqrt[n]{\frac{1+\sqrt{c^{2}-1}}{\sqrt{1+c^{2}-1}}}$
$\displaystyle 0=x^{2n}-2cx^{n}+1$

Thus letting us conclude that $|c|\leq 1$. ∎

######  Lemma 56.

Given a root $z$ of $x^{2n}+2cx^{n}+1$, the polynomial splits over
$\\{z,\bar{z},z\zeta_{n}^{1},\bar{z}\zeta_{n}^{1},\dots,z\zeta_{n}^{n},\bar{z}\zeta_{n}^{n}\\}$

###### Proof.

The step is to verify that if $z$ is a root of $x^{2n}+2cx^{n}+1$ then so is
$z\zeta^{k}_{n}$ we can do this with substitution and the definition that
$(\zeta_{n}^{k})^{n}=1$ we can simplify down

$\displaystyle 0$
$\displaystyle=(z\zeta_{n}^{k})^{2n}+c(z\zeta_{n}^{k})^{n}+1$ $\displaystyle
0$ $\displaystyle=z^{n}(\zeta_{n}^{k})^{2n}+cz^{n}(\zeta_{n}^{k})^{n}+1$
$\displaystyle 0$
$\displaystyle=z^{n}(\zeta_{n}^{k})^{2n}+cz^{n}(\zeta_{n}^{k})^{n}+1$
$\displaystyle 0$ $\displaystyle=z^{n}+cz^{n}+1$ $\displaystyle 0$
$\displaystyle=0$

Thus verifying that $z\zeta^{k}_{n}$ is a root. Since the polynomial has all
real coefficients the conjugate of every root $\overline{z\zeta^{k}_{n}}$ is
also a root. Since the conjugate of any root of unity is also a root of unity,
you can write the set of $2n$ roots as
$\\{z\zeta^{k}_{n},\bar{z}\zeta^{k}_{n}\\}$ ∎

######  Theorem 57.

If $p(x)$ is an irreducible polynomial over the rationals with real and
complex roots then it has no roots in $\mathcal{P}$

###### Proof.

Consider what would happen if a real root $k_{R}$ of $p(x)$ was included,
since $\mathcal{P}(x)\cap\mathcal{R}$ is splitting then every root of $p(x)$
must be in $\mathcal{P}(x)\cap\mathcal{R}$, this is impossible since some
roots of $p(x)$ are complex. Likewise any complex root $k_{I}$ cannot be
included since $\mathcal{P}$ is splitting, but $k_{R}$ cannot be included so
neither can $k_{I}$. ∎

The first form I have seen of a proof for this in the literature is [2], where
he proves a weaker version for cubics and angle trisection only using
Cardano's closed-form solution. (There is an error in this earlier formulation
where he states: "A real cubic equation can be solved geometrically using
ruler, compass, and angle-trisector if and only if its roots are all real",
this is true of irreducible cubics, but simple examples like $x^{3}-1$
disprove the general case.)

######  Corollary 58.

Every irreducible polynomial with an abelian galois extension has all real or
all complex roots.

###### Proof.

Via Theorem 47 our polynomial must have all its roots constructible using
angle partitions. Thus our polynomial must have all real or all complex roots
via Theorem 57. ∎

######  Corollary 59.

Any odd $k$th root of a prime number $\sqrt[k]{p}$, is not constructible using
angle partitions, for any odd $k$ greater than 3.

###### Proof.

Since the minimal polynomial $x^{n}-p$ is irreducible via Eisenstein's
criterion since $p$ divides every coefficient except for the first and $p^{2}$
does not divide $p$. For $n\geq 3$ the polynomial has both real and complex
roots then $\sqrt[n]{p}$ is not constructible using partitions. (Also serves
as an argument that doubling the cube is impossible with conventional
euclidean geometry.) ∎

This leads us to the following result that will be further expanded upon in
the conclusion.

######  Conjecture 60.

Since doubling the cube is impossible with angle partitions it seems unlikely
that it is possible in $\mathcal{T}_{1}$.

## General Intersections of Transcendental Curves

Limiting ourselves to certain intersections let us use the similar geometric
interactions of all curves to produce a field with an interesting algebraic
structure. Letting in other kinds of intersections introduces a great amount
of complexity to the problem. Consider first the field created by allowing any
sloped line intersections with simple trigonometric functions like $\sin$ and
$\cos$

######  Definition 61.

The field generated by allowing sloped line intersections with $\sin$ or
$\cos$ is the field $\mathcal{T}_{1}$ adjoined with the roots of the
expression $\sin{x}+ax+b$

######  Theorem 62.

If both $a$ and $b$ are algebraic then any solution $x$, q$\sin^{-1}(x)$ are
transcendental.

###### Proof.

First, assume that $x$ is algebraic, then since $a$ and $b$ are algebraic then
$ax+b$ is algebraic, and therefore $\sin{x}$ is algebraic, but if $x$ is
algebraic then $\sin{x}$ is algebraic, but via [Lindemann–Weierstrass]
$\sin{x}$ and $x$ cant both be algebraic so by contradiction $x$ is
transcendental. Since $ax+b$ is transcendental then $\sin{x}$ is also
transcendental. ∎

######  Corollary 63.

Under the constraints of Theorem 62 then for any solution $x$ then
$\sin^{-1}(x)$ must be transcendental

###### Proof.

Because $\sin{\pi\cdot\frac{p}{q}}$ is algebraic, it contradicts the
aforementioned theorem. Thus $\pi\cdot\frac{p}{q}$ is not a valid solution. ∎

######  Corollary 64.

For any numbers $a\in\mathbb{R}$ and $b\in\mathcal{T}_{1}$ the equation
$\sin(x)=ax+(\sin(b)-ab)$ has a trigonometric solution

###### Proof.

A solution of $x=b$ satisfies the equation. ∎

######  Corollary 65.

The problem statement is equivalent to finding all roots of
$bx-\frac{x^{3}}{6}+\frac{x^{5}}{25}+\dots$

###### Proof.

Using the Taylor series expansion for $\sin(x)$ we have that we are trying to
find the roots of

$-\left(b_{1}x+a_{1}\right)+\left(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}\right)$

∎

######  Theorem 66.

Any sloped line intersecting the quadratrix involves finding the solution to
$a+\frac{b}{x}=\cot(x)$

###### Proof.

$\displaystyle ax+b$ $\displaystyle=x\cot(\pi x/2)$ $\displaystyle
a+\frac{b}{x}$ $\displaystyle=\cot(\pi x/2)$ $\displaystyle a+\frac{b}{x}$
$\displaystyle=\cot(x)$

∎

Furthermore, some basic plots seem to indicate that there is a trove of
complex solutions to even the simple base case of $\sin(x)-ax$

Figure 8: $\sin(x)-x/2$ in the complex plane

Furthermore, the where $\sin(x)$ and $ax$ intersect and are tangent has a
fascinating expression in the complex plane, where the complex solutions
combine on the real line into a solution with a multiplicity of 2, before
separating on the real line as seen in Figure 9.

Figure 9: $\sin(x)-0.1283x$ in the complex plane

What's interesting is that the characteristic equations for the quadratrix
seem to buck the trend and only have solutions appearing on the real line.

Figure 10: $\cot(x)-.2-3/x$ in the complex plane

## Conclusion

Despite the best efforts of the author some important questions on this matter
remain unresolved,

######  Conjecture 67.

Does the field created by allowing arbitrary partition of angles $\mathcal{P}$
completely encapsulate the algebraic numbers in $\mathcal{T}_{1}$

The strategy used for this involved turning our geometric constructions into
algebra and then using our base algebraic objects to construct the following
class of functions that take in algebraic numbers and output algebraic
numbers.

$\displaystyle f_{q}:\mathbb{A}\rightarrow\mathbb{A}$ $\displaystyle
f_{q}(z)=e^{q\cdot\ln(z)}$

However, saying that this class of functions generates all algebraic numbers
would involve proving that there are no other "ways" of making algebraic
numbers. For example, consider the function

$\displaystyle
g:\mathbb{A}\times\mathbb{A}\times\mathbb{A}\rightarrow\mathbb{C}$
$\displaystyle g(a,b,c)=e^{\frac{\ln(a)\ln(b)}{\ln(c)}}$

We know that this function does produce algebraic output when $c$ equals 0. Or
when either $a$ or $b$ are expressible as $c$ raised to a rational power.
(Actually plugging in values gives you results that look like
$2^{\log_{3}(5)}$.) There might very well be a theorem in transcendental
number theory that could easily resolve this.

Likewise, there are a bunch of additional questions that could be answered
about allowing general intersections, namely.

######  Conjecture 68.

Are all the fields allowing general intersections with the transcendental
curves (trig functions, Quadratrix of Hippias, Archemedian Spiral) identical?
(Do they form a larger $\mathcal{T}_{2}$)

I suspect that if it is true it would be very hard to prove, but on the other
hand, it would be very pretty if it was true.

######  Conjecture 69.

Does allowing general intersections in any of the transcendental curves allow
the construction of algebraic numbers that are impossible in $\mathcal{T}_{1}$

There might very well be a clever trick in the algebra to allow this and if
any reader would be interested in continuing this work further I would
recommend they consider starting here.

## Acknowledgements

I would like to sincerely thank Dr. John Carter for advising me on this
project and helping me through the publication process, as well as teaching
the History of Mathematics class where the quadratrix was first introduced to
me.

## References

  * [1] David Steven Dummit and Richard M. Foote  ``Abstract algebra''  New York: Wiley, 2009
  * [2] Andrew M. Gleason  ``Angle Trisection, the Heptagon, and the Triskaidecagon''  In _The American Mathematical Monthly_ 95.3, 1988, pp. 185  DOI: 10.2307/2323624
  * [3] Thomas C. Hull  ``Origametry: Mathematical Methods in Paper Folding''  Cambridge University Press, 2020  DOI: 10.1017/9781108778633
  * [4] Saradha Natarajan and Ravindranathan Thangadurai  ``Pillars of Transcendental Number Theory''  Singapore: Springer Singapore, 2020  DOI: 10.1007/978-981-15-4155-1
  * [5] R… Nickalls  ``Viète, Descartes and the cubic equation''  In _The Mathematical Gazette_ 90.518, 2006, pp. 203–208  DOI: 10.1017/S0025557200179598
  * [6] Burkard Polster  ``Why don't they teach this simple visual solution? (Lill's method)'', 2019  URL: https://www.youtube.com/watch?v=IUC-8P0zXe8