Neusis II

 



Proof | Heptagon Construction by Neusis


let \(f(x)\) be \(7^{th}\) 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

$$f(x)^7=1$$

which factors into

$$\Big(f(x) - 1\Big)\Big(f(x)^6 + f(x)^5 + f(x)^4 + f(x)^3 + f(x)^2 + f(x)^1 + 1\Big)=0$$

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

$$f(6x) + f(5x) + f(4x) + f(3x) + f(2x) + f(x) = -1$$

let

$$p(x)=f(x) + f(2x) + f(4x)$$

we can find that

$$\begin{array}{llll}p(x)&=p(2x)&=p(4x)\\ \\p(3x)&=p(6x)&=p(5x)\\ \\p(7x)&=3\end{array}$$

$$\begin{array}{ll}\Big(p(x)\Big)^*&=f(x)^*+f(2x)^*+f(4x)^*\\ \\&=f(6x)+f(5x)+f(3x)\\ \\&=p(3x)\end{array}$$

$$p(x) + p(3x) = -1$$

which shows that \(p(x), p(3x)\) lies on line \(z+z^*=-1\), and

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

which shows that \(p(x),p(3x)\) lies on a circle at \(0\) with radius \(\sqrt{2}\). For completeness, the following shows the algebraic expression for both \(p(x)\)

$$\begin{array}{llllllll}p(x)\cdot p(x) &=& &f(x) &f(2x) &(4x)&=p(x)+p(3x)+p(3x)&=p(x)+2p(3x)\\ \\&&f(x)&f(2x)&f(3x)&f(5x)\\ \\&&f(2x)&f(3x)&f(4x)&f(6x)\\ \\&&f(4x)&f(5x)&f(6x)&f(x)\\ \\ \end{array}$$

$$\begin{array}{ll}\Big(p(x) - p(3x)\Big)^2 &= p(x)^2 + p(3x)^2 - 2 p(x) p(3x)\\ \\ &=p(x)+2p(3x)+p(3x)+2p(x)-4\\ \\ &=-3-4\\ \\ &=-7\\ \\ p(x)-p(3x)&=\pm\sqrt{\overset{\phantom{0}}{-7}}\\ \\ \end{array}$$

$$\begin{array}{lll}2p(x)&=\Big(p(x)+p(3x)\Big)+\Big(p(x)-p(3x)\Big)&=-1+\sqrt{\overset{\phantom{0}}{-7}}\\2p(3x)&=\Big(p(x)+p(3x)\Big)-\Big(p(x)-p(3x)\Big)&=-1-\sqrt{\overset{\phantom{0}}{-7}}\end{array}$$

meanwhile

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

the distance between \(\Big(f(x)+f(3x)\Big)\) and \(f(x)\) is 

$$\begin{array}{lllll}\sqrt{\Big(f(x)+f(3x)-f(x)\Big)\Big(f(x)+f(3x)-f(x)\Big)^*}&=\sqrt{f(3x)\Big(f(3x)\Big)^*}\\ \\&=\sqrt{1}\\ \\ \end{array}$$

and

$$\begin{array}{lllll}0&,1&,f(x)+f(6x)&\text{ are collinear}\\ \\0&,f(3x)&,f(2x)+f(4x)&\text{ are collinear}\\ \\f(x)&,f(3x)+f(x)&,f(2x)+f(4x)+f(x)&\text{ are collinear}\\ \\ \end{array}$$

consequently, for \(x\in\{1,3,2,6,4,5\}\)

$$f(x)\text{ lies on unit circle}$$

$$f(x)+f(3x)\text{ lies on a circle at }-1\text{ radius }\sqrt{2}$$

$$f(x)+f(2x)+f(4x)=\dfrac12\left(-1\pm\sqrt{\overset{\phantom{0}}{-7}}\right)$$

and the distance between \(\Big(f(x)+f(3x)\Big)\) and \(f(x)\) is \(1\)

this yields 3 distinct possible neusis construction excluding mirrored image for regular heptagon











additionally,

$$\begin{array}{lll}f(4x)&\text{ lies on unit circle}\\ \\ f(4x)+1&\text{ lies on circle at }1\text{ through }0,2\\ \\ \dfrac{1}{f(4x)+1}&\text{ lies on a line between }0\text{ and }1\end{array}$$
because (a circle passing across 0 maps to a line not passing through 0 under inversion)
$$\Big(f(4x)+1\Big)\Big(1+f(x)+f(2x)+f(3x)\Big)=1$$
we can rewrite into

$$\begin{array}{lll}1+f(x)+f(2x)+f(3x)&\text{ lies on a line between }0\text{ and }1\\ \\ f(x)+f(2x)+f(3x)&\text{ lies on a line between }-1\text{ and }0\end{array}$$
the distance between \(f(x)+f(3x)\) and \(f(x)+f(2x)+f(3x)\) is

$$\begin{array}{lllll}\sqrt{\Big(f(x)+f(2x)+f(3x)-f(x)-f(3x)\Big)\Big(f(x)+f(2x)+f(3x)-f(x)-f(3x)\Big)^*}&=\sqrt{f(2x)\Big(f(2x)\Big)^*}\\ \\&=\sqrt{1}\\ \\ \end{array}$$

and

$$\begin{array}{lllll}0&,1&,f(x)+f(6x)&&\in\mathbb{R}\\ \\0&,1&,f(x)+f(6x)&,f(x)+f(6x)+1&\in\mathbb{R}\\ \\0&,f(2x)&,f(x)+f(3x)&,f(x)+f(3x)+f(2x)&\text{ are collinear}\\ \\ \end{array}$$

gives another distinct 3 neusis constructions (excluding mirrored image) for regular heptagon. the difference with the former is that the former use two circles as directrices and the latter use a line and a circle as directrices (remember that lines and circles are indistinguishable under inversive geometry)


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