Does this mean that one can construct the cube root of two in three dimensions?

Question

The idea here is to extend to three dimensions what ordinary compass-and-straightedge constructions do in two dimensions. The first thing is to define the tools and rules for their use. For instance, in two dimensions, the tools are a compass and straightedge (like a ruler, but with only one edge and no markings), and with them, one may: Draw a line between any two distinct points. Draw a circle with one point as the center, and any other point on its circumference. Draw an arbitrary point on a line or a circle, or off it. Draw the point at the intersection of two lines (if they intersect). Draw the point (or two) at the intersection of two circles (if they intersect). Draw the point (or two) at the intersection of a line and a circle (if they intersect).

In three dimensions, the canvas is not a flat plane, as it is in two dimensions, but all of space. And we introduce a new tool, which I will call a flatiron, which permits you to draw planes. The flatiron rules are as follows; in addition to the above, one may: Draw the unique plane containing any three non-collinear points. Draw a sphere with one point as the center, and any other point on its surface. Draw an arbitrary point on a plane or a sphere, or off it. Draw the line at the intersection of two planes (if they intersect). Draw the circle (or point) at the intersection of two spheres (if they intersect). Draw the circle (or point) at the intersection of a plane and a sphere (if they intersect). Draw the point (or two) at the intersection of a line or circle with a plane or sphere (if they intersect). As an example of what one might do in a three-dimensional construction, consider the following fairly simple task: Given points P and Q, construct a regular tetrahedron with PQ as edge. We proceed as follows: Draw spheres of radius PQ around both P and Q. Draw the circle C at the intersection of spheres P and Q. Draw R, an arbitrary point on circle C. Draw a sphere of radius PR around R. Draw S, one of the two points of intersection between circle C and sphere R. PQRS is then a regular tetrahedron.

Given these new abilities, would it be possible construct a cuberoot?

Answer 1

Probably not with spheres as intersection of a sphere with a plane just gives a circle like conventional construction. Try a conical drawing tool, which gives a greater variety of planar sections such as parabolas, instead: Doubling the cube with the help of a parabola

Answer 2

If you assume that when you intersect two spheres, or a sphere with a plane, or two planes, that $all$ of the intersection points are called constructed points, then any real length is constructible.

But suppose that you only assume a limited amount of information (which I am too tired too specify now. I will get back to this) Then for about the same reason as in $2$ dimensions: Examine the equations for the co-ordinates of lines, planes,circles, and spheres and for the co-ordinates of the intersections of any two of those types of figures.

Choose a $2$-D or $3$-D co-ordinate system where you start by being given the origin and a finite number of other points with rational co-ordinates. The co-ordinates of any constructible point will be members of $\;\mathbb Q[\sqrt {.}\;]\;$.... which is the smallest sub-field of $\mathbb R$ that contains the square roots of all its positive members.

The number $2^{1/3}$ does not belong to this field.

If we could construct two points whose distance apart is $2^{1/3}$ or two segments such that the ratio of their lengths is $2^{1/3}$ then we could construct the point whose first co-ordinate is $2^{1/3}$ and whose other co-ordinate(s) is (are) $0.$

Answer 3

A good way to approach the problem is to define algebraic equations which describe all of your tools and the rules governing the use of each tool. The structure of the algebraic equations will determine what can be constructed.

Tool 1 lets you: Draw a line between any two distinct points. So:

Given $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$. Use the line equation.

Tool 2 the Flatiron lets you: Draw the unique plane containing any three non-collinear points. So:

Given $(x_1, y_1, z_1), (x_2, y_2, z_2)$ and $(x_3, y_3, z_3)$. Use the plane equation.

Tool 3 lets you: Draw a sphere with(I assume 'given' is better than with) one point as the center, and (taking) any other (constructed or known)point (to be)on its surface. So:

Given $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$. Use the sphere equation with center $(x_1, y_1, z_1)$ and radius as the distance between the points.

Tool 4 lets you: Draw an arbitrary point on a plane or a sphere, or off it. So:

I am stumped as to the meaning of tool 4 to be honest.

Tool 5 lets you: Draw the line at the intersection of two planes (if they intersect). This is a given. When you draw the planes you automatically draw that line.

Tool 6 lets you: Draw the circle (or point) at the intersection of two spheres (if they intersect). Another given with the same reasoning as above.

Tool 7 lets you: Draw the circle (or point) at the intersection of a plane and a sphere (if they intersect). Another given with the same reasoning as above.

The following two families of equations govern what values can be constructed under this ruleset:

$$(x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2 - (x_2-x_1)^2 - (y_2-y_1)^2 - (z_2-z_1)^2 = 0$$

$$(x-x_1)[(y_2-y_1)(z_3-z_1)-(z_2-z_1)(y_3-y_1)] - (y-y_1)[(x_2-x_1)(z_3-z_1)-(z_2-z_1)(x_3-x_1)] + (z-z_1)[(x_2-x_1)(y_3-y_1)-(y_2-y_1)(x_3-x_1)] = 0$$

You can show that you don't need Tool 1 since anything constructed by Tool 1 can be constructed by Tool 2. It isn't too much work to show that.

Before continuing the analysis I would like to clarify the mysterious tool 4. I don't understand it. Sounds like it lets you construct anything which isn't very interesting. I must be misunderstanding your wording.