Neusis I

this page uses \(z^*\) as an expression for complex conjugate of \(z\)

Mechanism | Cline

If \(l_1\) and \(l_2\) are (defined as) straight lines across two same points

$$\begin{array}{lllll}\text{line } l_1\ni&a_0 & ,a_1 & ,a_2 \\ \\ \text{line } l_2 \ni&a_0 & ,a_1 & ,a_3\end{array}$$

then

$$l_1 = l_2$$

If \(c_1\) and \(c_2\) are (defined as) two circles across three same points

$$\begin{array} {lllll} \text{circle } c_1 \ni&a_0 & ,a_1 & ,a_2 & ,a_3 \\ \\ \text{circle } c_2 \ni&a_0 & ,a_1 & ,a_2 & ,a_4 \end{array}$$

then

$$c_1 = c_2$$

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addition by \(z\) corresponds to translation preserve both orientation, angle, and distance

multiplication by \( z \neq 0\) corresponds to rotation centered at \( 0\) and scaling centered at \( 0\), preserves angles

if \( z=z^*\) then the multiplication corresponds to pure scaling centered at \( 0\) without rotation, preserving orientation

if \( zz^* = 1\) then the multiplication corresponds to pure rotation centered at \( 0\) without scaling, preserving distances

conjugation \( f(z) = z^*\) corresponds to reflection over horizontal line preserves angle and distance

complex inversion \( f(z) = 1/z\) corresponds to reflection over unit circle (geometric inversion) and reflection over horizontal line, preserves angle \( \forall z\neq 0\)

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Expand definition of point into complex number on a complex plane. If

$$\begin{array}{llll} z_0 & ,z_1 & ,z_2 &\text{ is collinear}\end{array}$$

then

$$\begin{array}{llll} z_0 + z & ,z_1 + z & ,z_2 + z &\text{ is collinear} \\ \\z_0 \cdot z & ,z_1 \cdot z & ,z_2 \cdot z &\text{ is collinear} \\ \\{z_0}^* & ,{z_1}^* & ,{z_2}^* &\text{ is collinear}\end{array}$$

if \(z_0 ,z_1 ,z_2\) are not collinear with \(0\)

then

$$\begin{array}{lllll} \dfrac{1}{z_0} & ,\dfrac{1}{z_1} & ,\dfrac{1}{z_2} & ,0\text{ is concyclic}\end{array}$$

if \(z_0 ,z_1 ,z_2\) are collinear with \(0\)

then

$$\begin{array}{lllll} \dfrac{1}{z_0} & ,\dfrac{1}{z_1} & ,\dfrac{1}{z_2} & ,0\text{ is collinear}\end{array}$$

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if

$$\begin{array}{lllll} z_0 & ,z_1 & ,z_2 & ,z_3\text{ is concyclic}\end{array}$$

(concyclic/cocircular means lying on the same circle) then

$$\begin{array}{lllll}z_0 + z & ,z_1 + z & ,z_2 + z & ,z_3 + z&\text{ is concyclic} \\ \\z_0 \cdot z & ,z_1 \cdot z & ,z_2 \cdot z & ,z_3 \cdot z&\text{ is concyclic with scaling by }\sqrt{zz^*} \\ \\{z_0}^* & ,{z_1}^* & ,{z_2}^* & ,{z_3}^* &\text{ is concyclic}\end{array}$$

if \(z_0 ,z_1 ,z_2 ,z_3\) are not concyclic with \(0\) then

$$\begin{array}{lllll}\dfrac{1}{z_0} & ,\dfrac{1}{z_1} & ,\dfrac{1}{z_2} & ,\dfrac{1}{z_3}&\text{ is concyclic}\end{array}$$

if \(z_0 ,z_1 ,z_2 ,z_3\) are concyclic with \(0\) then

$$\begin{array}{lllll}\dfrac{1}{z_0} & ,\dfrac{1}{z_1} & ,\dfrac{1}{z_2} & ,\dfrac{1}{z_3}&\text{ is collinear}\end{array}$$

let \(c\) be the center of circle across \(z_0, z_1, z_2\) and \(c'\) be the center of circle across \(\dfrac{1}{z_0},\dfrac{1}{z_1},\dfrac{1}{z_2}\) then

$$c'=\dfrac{c^*}{z_0 c^*+z_0 ^*c-z_0 z_0 ^*}$$

unless the circle crosses \(0\), then \(c'\) will be indefinite and instead \(\dfrac{1}{z_0},\dfrac{1}{z_1},\dfrac{1}{z_2}\) lies on a perpendicular bisector (fold line) between \(0\) and \(\dfrac{1}{c}\)

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let \(A\) be a set of complex numbers (that forms a shape on a complex plane)

$$\begin{array}{llll} z_0 & ,z_1 & ,z_2 & ,\dots\in A \end{array}$$ 

and \(ss^*=1\) then 

$$\begin{array}{llll} z_0 + z & ,z_1 + z & ,z_2 + z & ,\dots\in A + z \\ \\ z_0 \cdot s & ,z_1 \cdot s & ,z_2 \cdot s & ,\dots\in A \cdot s \end{array}$$ 

forms the same shape as the shape original set \(A\) form with the same size. If the shape formed by \(A\) has reflectional symmetry, then 

$$\begin{array}{llll}  {z_0}^* & ,{z_1}^* & ,{z_2}^*& ,\dots\in A^* \end{array}$$

forms the same shape as the shape original set \(A\) form

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the set of real numbers \(\mathbb{R}\) and the set of complex number on complex unit circle (also known as the circle group) \(\mathbb{T}\) or \(\mathbb{S}^1\) contains points that, respectively, satisfy 

$$\begin{array}{ll}r&=r^*\\ \\s\cdot s^*&=1\end{array}$$

addition and multiplication between two real numbers result in real numbers. If

$$\begin{array}{lllll}r_1 &,r_2 &\in\mathbb{R}\end{array}$$

then

$$\begin{array}{llllllll}0&, 1&, r_1 &,r_2 &,r_1+r_2&, r_1\cdot r_2 &\in\mathbb{R}\text{ is collinear}\end{array}$$

any complex number collinear to \(0\) and \(1\) is a real number

multiplication between two unit complex number results in unit complex number. If

$$\begin{array}{lllll}s_1 &,s_2 &\in\mathbb{S}^1\end{array}$$

then

$$\begin{array}{lllllll}1&,-1&,s_1 &,s_2 &,s_1\cdot s_2 &,s_1^r &\in\mathbb{S}^1\end{array}\text{ is concyclic}$$

\(1\) and \(-1\) are the only unit complex number that is also a real number

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complex number \(z\) that lies on a line across \(a_0\) and \(a_1\) satisfy

$$(z - a_0)(a_1 - a_0)^* = (z - a_0)^*(a_1 - a_0)$$

$$z(a_1 - a_0)^* -z^*(a_1 - a_0) = a_0 a_1 ^* - a_0 ^* a_1$$

complex number \(z\) that lies on a circle centered at \(a_0\) across \(a_1\) satisfy

$$(z - a_0)(z - a_0)^* = (a_1 - a_0)(a_1 - a_0)^*$$

$$zz^* - a_0 ^* z - a_0 z^* = a_1 a_1 ^* - a_1 a_0^* - a_1 ^* a_0 $$

complex number \(z\) that lies on a perpendicular bisector between \(a_0\) and \(a_1\) satisfy

$$z(a_1 - a_0)^* + z^*(a_1 - a_0) = a_1 a_1 ^* - a_0 a_0 ^*$$

in inversive geometry, lines and circles are indistinguishable. A complex number \(z\) that lies on a line or circle across \(a_0\) ,\(a_1\) , and \(a_2\) satisfy

$$(z - a_1)(z - a_2)^*(a_0 - a_2)(a_0 - a_1)^* = (z - a_1)^*(z - a_2)(a_0 - a_2)^*(a_0 - a_1)$$

or

$$p_0 zz^* + p_1 z + p_2 z^* + p_3 = 0$$

where

$$\begin{array}{llll} p_0 &= a_0^* a_1 - a_0 a_1^* + a_1^* a_2 - a_1 a_2^* + a_2^* a_0 - a_2 a_0^* \\ \\ p_1 &= a_0 a_0^* (a_1 - a_2)^* + a_1 a_1^* (a_2 - a_0)^* + a_2 a_2^* (a_0 - a_1)^*\\ \\ p_2 &= -p_1^* \\ \\ &= a_0 a_0^* (a_2 - a_1) + a_1 a_1^* (a_0 - a_2) + a_2 a_2^* (a_1 - a_0)\\ \\ p_3 &= a_0 a_0^* (a_1 a_2^* - a_1^* a_2) + a_1 a_1^* (a_2 a_0^* - a_2^* a_0) + a_2 a_2^* (a_0 a_1^* - a_0^* a_1) \end{array}$$

the center is

$$z = \dfrac{(\;a_1 a_1^* - a_0 a_0^*\;)(\;a_2 - a_1\;) - (\;a_2 a_2^* - a_1 a_1^*\;)(\;a_1 - a_0\;)}{(\;a_1 - a_0\;)^*(\;a_2 - a_1\;) - (\;a_2 - a_1\;)^*(\;a_1 - a_0\;)}$$

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Mechanism | Element of \(\mathbb{Q}(f)\)

$$\mathbb{Z} \quad \mathbb{Q} \quad \mathbb{A} \quad \mathbb{R} \quad \mathbb{S}^1 \quad \mathbb{R}^2 \quad \mathbb{C} \quad \mathbb{Z}_n \quad \mathbb{Q}\sqrt{\phantom{^|}} \quad \text{...}$$

the set of algebraic numbers with degree of 2 \(\mathbb{Q}\sqrt{\phantom{^|}}\) contains rational numbers as well as algebraic combination between rational numbers with addition, multiplication, and square roots like

$$\sqrt2 \quad ,1+\sqrt3 \quad ,\sqrt{-1} \quad ,\sqrt{1+\sqrt2} \quad, \text{...}$$

addition, subtraction, multiplication, and square roots are closed for this set

the numbers in this set are constructible by means of compass and straightedge

let \(f\) be \(n^{th}\) roots of unity$$f^x = f(x) = e^{2xi\pi/n}=\cos\dfrac{2x\pi}{n} + i \sin\dfrac{2x\pi}{n}$$

let \(x\in\mathbb{Z}\) then following applies

$$\begin{array}{lllll}fn &= 1 \\ \\f(x) &= f(x \mod n) \\ \\ f(x) \cdot f(y) &= f(x + y)\\ \\ \dfrac{f(x)}{f(y)} &= f(x - y)\\ \\f(x)^* & = \dfrac{1}{f(x)} & =f(n-x)\end{array}$$ and $$\begin{array}{lllll} f\cdot f^* &= 1 & &f(x) \text{ lies on unit circle}\\ \\ f+\dfrac{1}{f}&= \left(f+\dfrac{1}{f}\right) & &\in\mathbb{R}\\ \\ f-\dfrac{1}{f}&= -\left(f-\dfrac{1}{f}\right)^* & &\in\left\{ir\mid r\in\mathbb{R}\right\} \end{array}$$

combine roots of unity \(fx\) with integers \(a_1, a_2, \text{...}\) using addition and multiplication to obtain

$$a = a_0 + a_1 f1 + a_2 f2 + \cdots + a_n fn$$

which is an element of \(\mathbb{Z}(f)\)

combine roots of unity \(fx\) with rational \(q_1, q_2, \text{...}\) using addition and multiplication to obtain

$$q = q_0 + q_1 f1 + q_2 f2 + \cdots + q_n fn$$

which is an element of \(\mathbb{Q}(f)\)

addition, subtraction, and multiplication between two element of \(\mathbb{Z}(f)\) yields another element of \(\mathbb{Z}(f)\) but not necessarily division

addition, subtraction, multiplication, and division between between two element of \(\mathbb{Q}(f)\) yields another element of \(\mathbb{Q}(f)\), given the divisor \(\neq 0\)

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Using pentagon (roots of unity order 5) as a case example for working with numbers in \(\mathbb{Q}(f)\). Let \(f\) be \(5^{th}\) roots of unity,

$$1 \quad, f1 \quad, f2 \quad, f3 \quad, f4$$

forms a regular pentagon on a unit circle and a solution of polynomial

$$f^5 - 1=0$$

which factors into 

$$(f-1)(f4 + f3 + f2 + f1 + 1) = 0$$

removing \((f-1)\) gives

$$f4 + f3 + f2 + f1 + 1 = 0$$

where \(f\) no longer \(=1\)

combination between integers and \(5^{th}\) roots of unity will have the form of 

$$a = a_0 + a_1 f1 + a_2 f2 + a_3 f3 + a_4 f4$$

an element of \(\mathbb{Z}(f)\). Addition between two \(\mathbb{Z}(f)\) element is straightforward

$$\begin{array}{lllll}a &= a_0 &+ a_1 f1 &+ a_2 f2 &+ a_3 f3 &+ a_4 f4 \\ \\b &= b_0 &+ b_1 f1 &+ b_2 f2 &+ b_3 f3 &+ b_4 f4 \\ \\ a+ b &= (a_0 + b_0) &+ (a_1 + b_1) f1 &+ (a_2 + b_2) f2 &+ (a_3 + b_3) f3 &+ (a_4 + b_4) f4 \end{array}$$

multiplication between two \(\mathbb{Z}(f)\) element can be demonstrated as following

$$\begin{array}{lllllllllll}a\cdot b &=\phantom{+f0} a_0 \Big( b_0 &+ b_1 f1 &+ b_2 f2 &+ b_3 f3 &+ b_4 f4\Big) &= \phantom{+}a_0 \Big( b_0 &+ b_1 f1 &+ b_2 f2 &+ b_3 f3 &+ b_4 f4\Big) \\ \\&\phantom{=}+ a_1f1 \Big( b_0 &+ b_1 f1 &+ b_2 f2 &+ b_3 f3 &+ b_4 f4\Big)&\phantom{=}+ a_1 \Big( b_0 f1 &+ b_1 f2 &+ b_2 f3 &+ b_3 f4 &+ b_4\phantom{f0}\Big) \\ \\&\phantom{=}+ a_2f2 \Big( b_0 &+ b_1 f1 &+ b_2 f2 &+ b_3 f3 &+ b_4 f4\Big)&\phantom{=}+ a_2 \Big( b_0 f2 &+ b_1 f3 &+ b_2 f4 &+ b_3 &+ b_4 f1\Big) \\ \\&\phantom{=}+ a_3f3 \Big( b_0 &+ b_1 f1 &+ b_2 f2 &+ b_3 f3 &+ b_4 f4\Big)&\phantom{=}+ a_3 \Big( b_0 f3 &+ b_1 f4 &+ b_2 &+ b_3 f1 &+ b_4 f2\Big) \\ \\&\phantom{=}+ a_4f4 \Big( b_0 &+ b_1 f1 &+ b_2 f2 &+ b_3 f3 &+ b_4 f4\Big)&\phantom{=}+ a_4 \Big( b_0 f4 &+ b_1 &+ b_2 f1 &+ b_3 f2 &+ b_4 f3\Big) \\ \\\end{array}$$

$$\begin{array}{lllllllllll}=\phantom{+} a_0 \Big( b_0 &+ b_1 f1 &+ b_2 f2 &+ b_3 f3 &+ b_4 f4 \Big)&=\phantom{+} a_0 b_0 &+ a_0 b_1 f1 &+ a_0 b_2 f2 &+ a_0 b_3 f3 &+ a_0 b_4 f4 \\ \\\phantom{=}+ a_1 \Big( b_4 &+ b_0 f1 &+ b_1 f2 &+ b_2 f3 &+ b_3 f4 \Big)&\phantom{=}+ a_1 b_4 &+ a_1 b_0 f1 &+ a_1 b_1 f2 &+ a_1 b_2 f3 &+ a_1 b_3 f4 \\ \\\phantom{=}+ a_2 \Big( b_3 &+ b_4 f1 &+ b_0 f2 &+ b_1 f3 &+ b_2 f4 \Big)&\phantom{=}+ a_2 b_3 &+ a_2 b_4 f1 &+ a_2 b_0 f2 &+ a_2 b_1 f3 &+ a_2 b_2 f4 \\ \\\phantom{=}+ a_3 \Big( b_2 &+ b_3 f1 &+ b_4 f2 &+ b_0 f3 &+ b_1 f4 \Big)&\phantom{=}+ a_3 b_2 &+ a_3 b_3 f1 &+ a_3 b_4 f2 &+ a_3 b_0 f3 &+ a_3 b_1 f4 \\ \\\phantom{=}+ a_4 \Big( b_1 &+ b_2 f1 &+ b_3 f2 &+ b_4 f3 &+ b_0 f4 \Big)&\phantom{=}+ a_4 b_1 &+ a_4 b_2 f1 &+ a_4 b_3 f2 &+ a_4 b_4 f3 &+ a_4 b_0 f4 \\ \\\end{array}$$

$$\begin{array}{llllll}=\phantom{+}\Big(a_0 b_0 &+ a_1 b_4 &+ a_2 b_3 &+ a_3 b_2 &+ a_4 b_1 \Big)  \\ \\\phantom{=}+\Big(a_0 b_1 &+ a_1 b_0 &+ a_2 b_4 &+ a_3 b_3 &+ a_4 b_2 \Big)f1\\ \\\phantom{=}+\Big(a_0 b_2 &+ a_1 b_1 &+ a_2 b_0 &+ a_3 b_4 &+ a_4 b_3 \Big)f2\\ \\\phantom{=}+\Big(a_0 b_3 &+ a_1 b_2 &+ a_2 b_1 &+ a_3 b_0 &+ a_4 b_4 \Big)f3\\ \\\phantom{=}+\Big(a_0 b_4 &+ a_1 b_3 &+ a_2 b_2 &+ a_3 b_1 &+ a_4 b_0 \Big)f4\\ \\\end{array}$$

addition and multiplication above also applies to \(\mathbb{Q}(f)\) where \(a_0, a_1, \dots, b_0, b_1, \dots\) are rational instead of integers. However, because 

$$f1 + f2 + f4 + f3 + 1 = 0$$

we have

$$\begin{array}{llllll} \phantom{=} a_0 &+ a_1 f1 &+ a_2 f2 &+ a_3 f3 &+ a_4 f4 \\ \\= 0 &+ (a_1 -a_0) f1 &+ (a_2 -a_0) f2 &+ (a_3 -a_0) f3 &+ (a_4 -a_0) f4 \\ \\= (a_0 -a_1) &+ 0&+ (a_2 -a_1) f2 &+ (a_3 -a_1) f3 &+ (a_4 -a_1) f4 \\ \\= (a_0 -a_2) &+ (a_1 -a_2) f1 &+ 0&+ (a_3 -a_2) f3 &+ (a_4 -a_2) f4 \\ \\= (a_0 -a_3) &+ (a_1 -a_3) f1 &+ (a_2 -a_3) f2 &+ 0&+ (a_4 -a_3) f4 \\ \\= (a_0 -a_4) &+ (a_1 -a_4) f1 &+ (a_2 -a_4) f2 &+ (a_3 -a_4) f3 &+ 0\\ \\\end{array}$$

being different expression for the same number. 'Nullifying' one of the \(a_i\) can be used to simplify expression when all of \(a_i\) are congruent to any number modulo some integer. For example

$$4+7f1+10f2+13f3+16f4$$

can be simplified into

$$\begin{array}{lllll}\phantom{=}4 &+7f1 &+10f2 &+13f3 &+16f4 \\ \\=0 &+(7-4)f1 &+(10-4)f2 &+(13-4)f3 &+(16-4)f4 \\ \\=0 &+3f1 &+6f2 &+9f3 &+12f4 \\ \\=3\Big( &\phantom{+}f1 &+2f2 &+3f3 &+4f4 \Big) \\ \\\end{array}$$

One may also choose which \(a_i\) to nullify by aiming for expression with least number of sum of \(n^{th}\) roots of unity. For example,

$$\dfrac12 +\dfrac12 f1 +\dfrac12 f2 -\dfrac12 f3 +\dfrac12 f4$$

is equal to $$\begin{array}{lllll}\dfrac12 &+\dfrac12 f1 &+\dfrac12 f2 &-\dfrac12 f3 &+\dfrac12 f4 \\ \\\dfrac12-\dfrac12 &+\left(\dfrac12-\dfrac12\right) f1 &+\left(\dfrac12-\dfrac12\right) f2&+\left(-\dfrac12-\dfrac12\right) f3 &+\left(\dfrac12-\dfrac12\right) f4 \\ \\0 &+0 f1 &+0 f2 &-1 f3 &+0 f4 \\ \\\end{array}$$

a phenomenon where a number in \(\mathbb{Z}(f)\) expressed with each \(a_i\) being non-integers

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division 

let

$$a = a_1 f1 + a_2 f2 + a_3 f3 + a_4 f4$$

(notice \(a_0\) is nullified) there exists an inverse \(a^{-1}\) in \(\mathbb{Q}(f)\) such that

$$a\cdot a^{-1}=1$$

let 

$$a^{-1} = b_1 f1 + b_2 f2 + b_3 f3 + b_4 f4$$

then

$$\begin{array}{llllll}a\cdot a^{-1}&=\phantom{+}\Big(a_1 b_4 &+ a_2 b_3 &+ a_3 b_2 &+ a_4 b_1 \Big)  \\ \\&\phantom{=}+\Big(a_2 b_4 &+ a_3 b_3 &+ a_4 b_2 &\Big)f1\\ \\&\phantom{=}+\Big(a_1 b_1 &+ a_3 b_4 &+ a_4 b_3 &\Big)f2\\ \\&\phantom{=}+\Big(a_1 b_2 &+ a_2 b_1 &+ a_4 b_4 &\Big)f3\\ \\&\phantom{=}+\Big(a_1 b_3 &+ a_2 b_2 &+ a_3 b_1 &\Big)f4\\ \\\end{array}$$

to obtain a system of linear equations (for \(b_i\) given \(a_i\))

$$\begin{array}{llllll}a_4 b_1 &+ a_3 b_2 &+ a_2 b_3 &+ a_1 b_4 &= 1 \\ \\0   b_1 &+ a_4 b_2 &+ a_3 b_3 &+ a_2 b_4 &= 0 \\ \\a_1 b_1 &+ 0   b_2 &+ a_4 b_3 &+ a_3 b_4 &= 0 \\ \\a_2 b_1 &+ a_1 b_2 &+ 0   b_3 &+ a_4 b_4 &= 0 \\ \\a_3 b_1 &+ a_2 b_2 &+ a_1 b_3 &+ 0   b_4 &= 0\\ \\\end{array}$$

methods such as Gaussian Elimination can be used to solve the system above

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Proof | Pentagon Construction by Neusis

let \(x\in\mathbb{Z}\), then

$$f(5x)=1$$

which turns into polynomial shown earlier 

$$\Big(f(x)\Big)^5- 1= 0$$

factors into 

$$\Big(f(x) -1\Big)\Big(f(x)^4 + f(x)^3 + f(x)^2 + f(x) +1\Big)= 0$$

to obtain 

$$\begin{array}{ll}(x)^4 + f(x)^3 + f(x)^2 + f(x) +1&= 0\\ \\(4x) + f(3x) + f(2x) + f(1x)&= -1\end{array}$$

all possible values of \(f(x)\) 

$$\begin{array}{lllllll}f1 &,f2 &,f3 &,f4 &,1 \quad&\text{ lies on unit circle}\\ \\ f1 +1&,f2  +1&,f3  +1&,f4  +1&, 2&\text{ lies on circle centered at 1 across 2 and 0}\\ \\ \dfrac{1}{f1 +1}&,\dfrac{1}{f2  +1}&,\dfrac{1}{f3  +1}&,\dfrac{1}{f4  +1}&, \dfrac12 &\text{ lies on a line between }0\text{ and }1\\ \\\end{array}$$

the word "between" here means that the line is a perpendicular bisector between \(0\) and \(1\). Because 

$$\Big(f(x) + 1\Big)\Big(f(4x)+f(2x)+1\Big)= 1$$

we can rewrite the equation into

$$\begin{array}{lllllll}f4 + f2 + 1&,f3 + f4 + 1&,f2 + f1 + 1&,f1 + f3 + 1&, \dfrac12 &\text{ lies on a line between }0\text{ and }1\\ \\f4 + f2&,f3 + f4&,f2 + f1&,f1 + f3&, -\dfrac12 &\text{ lies on a line between }-1\text{ and }0\\ \\ \end{array}$$

let \(g(x)=f(x)+f(2x)\) 

$$\begin{array}{lllllll}g2&,g4&,g1&,g3&, -\dfrac12 \quad&\text{ lies on a line between }-1\text{ and }0\\ \\ \end{array}$$

> for all \(x\in\{1,2,4,3\}\), \(g(x)\) lies on a perpendicular bisector between \(-1\) and  \(0\) 

using all possible values of \(f(x)\) again 

$$\begin{array}{lllllll}f1 &,f2 &,f3 &,f4 &,1 \quad&\text{ lies on unit circle, centered at }0\text{ across }1\\ \\-f1&,-f2&,-f3&,-f4 &,-1\quad&\text{ also lies on the same circle }\\ \\-f1 -1&,-f2 -1&,-f3 -1&,-f4 -1&,-2\quad&\text{ lies on circle centered at }-1\text{ across } 0\\ \\\end{array}$$

because

$$\begin{array}{llll}f(4x) + f(3x)+ f(2x)+ f(x)&= -1 \\ \\ f(4x) + f(3x)+ f(2x) &= -1  -f(x) \end{array}$$

the expression can be rewritten as sum of positive sum of \(f(x)\)

$$\begin{array}{lllllll}f4+f3+f2&,f3+f1+f4&,f2+f4+f1&,f1+f2+f3&,-2\quad&\text{ lies on circle centered at }-1\text{ across } 0\\ \\\end{array}$$

let \(h(x)=f(x)+f(2x)+f(4x)\)

$$\begin{array}{lllllll}h2&,h4&,h1&,h3&,-2\quad&\text{ lies on circle centered at }-1\text{ across } 0\\ \\\end{array}$$

> for all \(x\in\{1,2,4,3\}\), \(h(x)\) lies on a circle centered at \(-1\) across \(0\)

on the other hand

$$f(2x)+f(3x) = f(2x)+\Big(f(2x)\Big)^*$$

for all \(x\in\mathbb{Z}\) and

$$\begin{array}{lllllll}f(2x) + f(3x)&\quad&\text{ lies on real axis}\end{array}$$

which also collinear with \(0,1,-1,\dots\)

$$\begin{array}{lllllll}0&,1&,f(2x) + f(3x)&,f(2x) + f(3x)+1\quad&\text{ are collinear}\end{array}$$

multiply the entire line with \(f(4x)\) shows that

$$\begin{array}{lllllll}0&,f(4x)&,f(1x) + f(2x)&,f(1x) + f(2x)+f(4x)\quad&\text{ are collinear}\\ \\0&,f(4x)&,g(x)&,h(x)\quad&\text{ are collinear}\end{array}$$

furthermore, the distance between \(g(x)\) and \(h(x)\) is

$$\begin{array}{lll}\sqrt{\Big(h(x)-g(x)\Big)\Big(h(x)-g(x)\Big)^*}&=\sqrt{f(4x)\Big(f(4x)\Big)^*}\\ \\&=\sqrt{1}=1\end{array}$$

consequently, for all \(x\in\{1,2,4,3\}\)

$$g(x)\text{ lies on a line between }0\text{ and }1$$

$$h(x)\text{ lies on a circle centered at }-1\text{ radius }0$$

$$f(x)\text{ lies on unit circle which also includes }f(4x)$$

$$\begin{array}{llllll} 0&,f(4x)&,g(x)&,(hx) \text{ are collinear }\end{array}$$

the distance between \(g(x)\) and \(h(x)\) is 1

will cover 7-gon, 9-gon, 11-gon, 13-gon, and 25-gon

figure graphs will be provided in the future