Doubling the cube neusis
Matthew Harrington 
Can anyone explain me in simple maths the neusis construction at ?
Why and how does it produce the $\root 3 \of 2$?
$\endgroup$3 Answers
$\begingroup$By chasing angles, you find that $ACG$ is a right triangle, and if you drop a perpendicular from $G$ to $CH$ at $I$, you find that $CGI$ is half of an equilateral triangle. Let $M$ be the midpoint of $BC$.
Then if $AG=x$, if follows that $CG=\sqrt{x^2-1}$ and $GI=\frac{1}{2}\sqrt{x^2-1}$. From the similar triangles $HGI$ and $HAM$ we have $\frac{1}{2}\sqrt{x^2-1} = (\sqrt{3}/2)/(x+1)$ or $(1+x)\sqrt{x^2-1}=\sqrt{3}$. Squaring both sides produces the equation $x^4+2x^3-2x-4=(x+2)(x^3-2)=0$, as above.
$\endgroup$ $\begingroup$For the purpose of demonstrating the construction's validity, I shall locate the points of the construction as follows:
$A\left(-{1\over2}\mid{{\sqrt3}\over2}\right)$
$B\left(-1\mid0\right)$
$C\left(0\mid0\right)$
$D\left(-{3\over2}\mid-{{\sqrt3}\over2}\right)$
The construction cited by the OP indicates that the line AGH has length $\sqrt[3]2+1$, that the line segment AG has length $\sqrt[3]2$, and that the line segment GH has length $1$. Let us assume that it is so and see the result of the assumption.
The point H on the positive branch of the x-axis at a distance of $\sqrt[3]2+1$ from point A turns out to be $\left(\sqrt[3]4\mid0\right)$.
We determine the point G which divides line AH into two segments, AG and GH, proportional, respectively, to $\sqrt[3]2$ and $1$. Point G is $\left({3\over{2+2\sqrt[3]2}}\mid{\sqrt3\over{2+2\sqrt[3]2}}\right)$.
Finally, The equation of line DC (extended beyond C) is $x=\sqrt3y$. Every $x$ on the line is $\sqrt3$ times as large as its corresponding $y$. From a cursory look at the coördinates of point G, it is obvious that it too is on the extended line CD, which proves the validity of the construction.
$\endgroup$ 1 $\begingroup$Use the law of cosines on triangle $HCA$ to get $(AG + GH)^2 = AC^2 + CH^2 - 2AC \times CH \cos C$. By construction $GH = AC = 1$ and $\cos C = -\frac{1}{2}$.
Therefore, $(AG + 1)^2 = 1 + CH^2 + CH$ or, equivalently, $AG^2 + 2AG = CH^2 + CH$.
Now use Menelaus' theorem on triangle $BAH$ with transversal $DCG$ to get that $BD \times AG \times CH = AD \times GH \times BC$, which becomes $AG \times CH = 2$.
Plugging one into the other gives the equations $AG^2 + 2AG = (\frac{2}{AG})^2 + \frac{2}{AG}$, which becomes $AG^4 + 2AG^3 - 2AG - 4 = 0$, which has positive real solution $AG = \sqrt[3]{2}$
$\endgroup$
