Doubling the cube neusis

Question

Can anyone explain me in simple maths the neusis construction at http://en.wikipedia.org/wiki/Doubling_the_cube#Using_a_marked_ruler?

Why and how does it produce the $\root 3 \of 2$?

Answer 1

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.

Answer 2

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.

Answer 3

The foundation is this.

https://baragar.faculty.unlv.edu/papers/TwiceNotch.pdf

Essentially, neusis is equivalent to using a conchoid to assist in a compass-and-straightedge construction. A conchoid is effectively a finite tower of field extensions over the constructible numbers of degrees 2, 3, 5, and 6. As such, the conchoid/neusis enables the solution of cubic and quartic equations with constructible coefficients, and since arbitrary cube roots can be constructed with neusis, and the (very unwieldy) quartic formula only has square root and cube root extractions, all cubics and quartics are solvable by conchoids/neusis.

Answer 4

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 $AG = \sqrt[3]{2}$