Neusis III




Proof | Hendecagon construction by Neusis

for any complex number \(z\) :

$$z + z^*\in\mathbb{R}$$
$$z z^*\in\mathbb{R}^{+}$$
$$\dfrac{\sqrt{\overset{\phantom{0}}{zz^*}}}{z^*}\in\mathbb{S}^{1}$$
$$z - z^*\in\Big\{ix\mid x\in\mathbb{R}\Big\}$$







let \(f(x)\) be \(11^{th}\) roots of unity 

$$f^x=f(x)$$

for all \(x\in\mathbb{Z}\), \(f(x)\) forms a regular heptagon on unit circle in complex plane and satisfies

$$f(x)^{11}=1$$

which factors into

$$\Big(f(x) - 1\Big)\Big(f(x)^{10} + f(x)^9 + \cdots + f(x)^2 + f(x)^1 + 1\Big)=0$$

which we can obtain from (where \(f(x)\neq1\) 

$$f(10x) + f(9x) + f(8x) + f(7x) + f(6x) + f(5x) + f(4x) + f(3x) + f(2x) + f(x) = -1$$

let

$$p(x)=f(x) + f(4x) + f(5x) + f(9x)+f(3x)$$

we have 

$$\begin{array}{lllll}p(x)&=p(4x)&=p(5x)&=p(9x)&=p(3x)\\ \\p(2x)&=p(8x)&=p(10x)&=p(7x)&=p(6x)\\ \\p(11x)&=5\end{array}$$

$$\Big(p(x)\Big)^*=p(2x)$$

$$p(x) + p(2x) = -1$$

which shows that \(p(x),p(2x)\) lies on line \(z+z^*=-1\) 

$$\begin{array}{llllllll}p(x)\cdot p(2x) &=& &f(2x)&f(8x)&f(10x)&f(7x)&f(6x)&=p(x)+p(2x)+p(2x)+5+p(x)&=3\\ \\&&f(x)&f(3x)&\ddots&\ddots&\ddots&\ddots\\ \\&&f(4x)&f(6x)&\ddots&\ddots&\ddots&\ddots\\ \\&&f(5x)&f(7x)&\ddots&\ddots&\ddots&\ddots\\ \\&&f(9x)&1&\ddots&\ddots&\ddots&\ddots\\ \\&&f(3x)&f(5x)&\ddots&\ddots&\ddots&\ddots\\ \\ \end{array}$$

which shows that \(p(x),p(2x)\) also lies on circle with radius \(\sqrt{3}\). The algebraic expression for \(p(x),p(2x)\)

$$\begin{array}{lll}2p(x)&=\Big(p(x)+p(2x)\Big)+\Big(p(x)-p(2x)\Big)&=-1+\sqrt{\overset{\phantom{0}}{-11}}\\2p(2x)&=\Big(p(x)+p(2x)\Big)-\Big(p(x)-p(2x)\Big)&=-1-\sqrt{\overset{\phantom{0}}{-11}}\end{array}$$

the neusis construction for hendecagon is quite different than neusis construction for pentagon, heptagon, that has been covered previously. Involves two verging shape instead of one line segment on a line

$$\begin{array}{llll}\phantom{=}p(x)&\text{ lies on circle at }&0\text{ radius }\sqrt{3}\\\\ \phantom{=}p(x)f(-x)&\text{ lies on circle at }&0\text{ radius }\sqrt{3}\\\\ =\Big(f(x)+f(4x)+f(9x)+f(5x)+f(3x)\Big)f(-x)\\\\ =1+f(3x)+f(8x)+f(4x)+f(2x)\\\\ \phantom{=}f(3x)+f(8x)+f(4x)+f(2x)&\text{ lies on circle at }&-1\text{ radius }\sqrt{3}\\\\\end{array}$$

$$\begin{array}{llll}0&,1&,f(3x)+f(8x)&&\in\mathbb{R}\\\\0&,1&,f(3x)+f(8x)&,f(3x)+f(8x)-1&\in\mathbb{R}\text{ are collinear}\\\\\end{array}$$

$$\begin{array}{llll}0&,1&,f(2x)+f(9x)&&\in\mathbb{R}\\\\0&,1&,f(2x)+f(9x)&,f(2x)+f(9x)+1&\in\mathbb{R}\\\\0&,f(2x)&,f(4x)+1&,f(4x)+1+f(2x)&\text{ are collinear}\\\\-1&,f(2x)-1&,f(4x)&,f(4x)+f(2x)&\text{ are collinear}\\\\-1+f(3x)+f(8x)&&,f(4x)+f(3x)+f(8x)&,f(4x)+f(2x)+f(3x)+f(8x)&\text{ are collinear}\\\\ \end{array}$$

$$\begin{array}{lllll}0&,1&f(x)+f(10x)&&&\in\mathbb{R}\\\\ 0&,1&&-1&f(x)+f(10x)+1&\in\mathbb{R}\\\\0&,f(3x)&&-f(3x)&f(4x)+f(2x)+f(3x)&\text{ are collinear}\\\\f(8x)&,f(3x)+f(8x)&&f(8x)-f(3x)&f(4x)+f(2x)+f(3x)+f(8x)&\text{ are collinear}\\\\\end{array}$$

$$\begin{array}{llllll}1&,f(3x)&,f(8x)&,f(4x)&\in\mathbb{S}^1&\text{ center }0\\\\ -1&,-f(3x)&,-f(8x)&,f(4x)&\in\mathbb{S}^1&\text{ center }0\\\\ f(3x)+f(8x)-1&,f(3x)+f(8x)-f(3x)&,f(3x)+f(8x)-f(8x)&,f(4x)+f(3x)+f(8x)&\text{ is concyclic radius 1}&\text{ with center }f(3x)+f(8x)\\\\ f(3x)+f(8x)-1&,f(8x)&,f(3x)&,f(4x)+f(3x)+f(8x)&\text{ is concyclic radius 1}&\text{ with center }f(3x)+f(8x)\\\\ \end{array}$$

in addition, the distance between \(f(8x)\) and \(f(8x)-f(3x)\) is 1

and the distance between \(f(3x)+f(8x)+f(4x)+f(2x)\) and \(f(3x)+f(8x)+f(4x)\) is 1

[in-depth details and graph visualization and figures will be provided in future. If reader need visualization in the meantime, this 5 collection of constructions , this animation (x) , or this animation (youtube) is what this post proving for]

Hendecagon | 

reader with a background in group theory might recognize that heptagon, nonagon, tridecagon constructible by angle trisection. Similarly, regular hendecagon is constructible by angle quintisection.


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