Those strangely intermediate folks who are at once unfamiliar with compass-and-straightedge constructions and yet not intimidated by their spectre may find the Wikipedia page a reasonable starting place. But what I'm proposing today is something different, which I'm a bit startled hasn't been discussed more prominently: three-dimensional constructions. I've had this in my virtual back pocket for a while; this seems as good an opportunity as any to pull it out for a looksee, and figure out if anyone else has encountered anything like this.

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).

*squaring the circle*); they were unable to trisect arbitrary angles (that is, construct an angle with one-third the extent of a given angle); and they were unable to construct the cube root of two (also known as

*doubling the cube*).

It turns out that these are impossible, and can be proved to be so, using some notions from field theory. That has not, of course, stopped people from submitting reams upon reams of alleged constructions of one of these three objects, all of which (you may be assured) are somewhere bogus.

But enough of that for now. 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).

*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.

So, a couple of questions, one easy, and one not so easy:

- Suppose that indeed, the cube root of two is not constructible in three dimensions. What about the
*fourth*root of two? In which dimensions might that be constructible? - Is it possible to construct all five of the regular polyhedra? In addition to the tetrahedron and cube, these include the regular octahedron (eight faces), dodecahedron (twelve faces), and icosahedron (twenty faces).

In two dimensions, given a segment divided in the ratio 1:p, you can construct a segment divided in the ratio 1:sqrt(p). Apply the same procedure to the second segment, and you now have a segment divided in the ratio 1:(fourth root of p).

ReplyDeleteTo shorten the following constructions, I’ll assume a procedure to “draw the one or two points at which three spheres intersect, if they intersect.” Your rules permit this procedure by first finding the circle in which two of the spheres intersect (if they do) and then the one or two points at which that circle intersects the third sphere. I’ll also assume you can construct a segment of length r = (p + sqrt(q))*AB given any segment AB, for any rational numbers p and q; and that you can then draw a sphere of radius r around any point. (These follow from the ability to construct the segment of length r and the circle of radius r around any point using compass and straightedge in a plane.)

ReplyDeleteFor the regular octahedron: construct an equilateral triangle, ABC. Draw point D, one of the intersections of the spheres of radius AB around A, of radius AB around B, and of radius AB*sqrt(2) around C. Then ABD and ABC are two adjacent faces of a regular octahedron. In a similar fashion, draw two more faces of the same octahedron adjacent to ABC. You just have to be careful to choose the correct one of the two possible points of intersection each time. You can ensure this by requiring the new points also to be on the sphere of radius AB around D. You will then have drawn all 6 vertices of the desired octahedron. (Or you can draw all 6 possible points of intersection, and then you will have two regular octahedra joined at a common face.)

Alternative construction: construct a square ABCD. Draw spheres of radius AB around centers A, B, and C. Draw the two points E and F at which these three spheres intersect. Then A, B, C, D, E, and F are the vertices of a regular octahedron.

For the regular icosahedron: construct an equilateral triangle, ABC. Draw point D, one of the intersections of the spheres of radius AB around A, of radius AB around B, and of radius τ·AB around C, where τ = (1 + sqrt(5))/2. Then ABD and ABC are two adjacent faces of a regular icosahedron. Continue adding more adjacent faces until you have drawn all 12 vertices of the icosahedron.

Alternative construction: construct a regular pentagon ABCDE. Draw spheres of radius AB around centers A, B, and C. Draw F, one of the two points at which these three spheres intersect. Then F together with any two adjacent vertices of ABCDE form one face of the regular icosahedron to be constructed. Furthermore, F together with any two non-adjacent vertices of ABCDE form three consecutive vertices of a regular pentagon. Find all the vertices of the five pentagons with vertices at F. Each pentagon produces two vertices, but each vertex is produced by two of the pentagons, so this gives you just five new points. But those five new points also form a regular pentagon, GHIJK. Draw spheres of radius GH around centers G, H, and I; of the two points of intersection of these three spheres, draw the one farther from F, and call it L. Then ABCDEFGHIJKL are the 12 vertices of a regular icosahedron.

For the regular dodecahedron: construct a regular pentagon ABCDE. Draw a sphere of radius AB around B and spheres of radius AC around A and C. Draw one of the two intersections of these three spheres and call it F. Then A, B, and F are three consecutive vertices of another regular pentagon; construct that pentagon, ABFGH. Then ABCDE and ABFGH are adjacent faces of a regular dodecahedron. Continue adding adjacent faces until you have found all 20 vertices.

That's right, well done!

ReplyDeleteIncidentally, since the dodecahedron and icosahedron are duals of each other, one can also construct the dodecahedron by simply finding the centers of the faces of the icosahedron.

The cube and octahedron are also duals of each other, so one can also employ that tactic for one of them, but they are simple enough that it's worthwhile finding separate constructions for both.

Any thoughts on whether the cube root of 2 is constructible in three dimensions?

Yes, you can use the duality of the figures, either to build a dodecahedron from an icosahedron or the other way around.

ReplyDeleteHaving constructed a cube, one can also set the origin of a Cartesian system at one vertex of the cube, aligning the three axes with the three edges that meet at that vertex, and using the length of each edge as the unit. One can then construct any point (x,y,z) where each of x, y, and z is constructible. Then one can construct a regular dodecahedron by drawing points at (±τ,±τ,±τ) and the points at all permutations of (±1,±τ²,0), where τ = (1+√5)/2; and one can construct a regular icosahedron by drawing points at all permutations of (±1,±τ,0).

For the question of what is

ReplyDeletenotconstructible, it has been too long since algebra, so I had to pull a couple of textbooks off my shelf. Jacobson treats the straightedge-and-compass problem as one of constructing a complex number, which is not the most obvious thing to extend to three dimensions, but Herstein uses (x,y) coordinates, and it seems to me that one could simply introduce a z coordinate and the appropriate descriptions of intersecting spheres, circles, planes, and lines in terms of equations in x, y, and z. Thinking of the constructions this way is also a motivation for the last pair of constructions (listing the (x,y,z) coordinates of the figures).I have not worked this through far enough to guarantee that Herstein’s proof completely carries over into three dimensions, but the gist of it is that all constructible points are found by solving systems of linear and/or quadratic equations in the coordinates. Starting with a unit length, the linear systems allow you to set each of x, y, and z to any value in the field of rational numbers, and the quadratic equations allow you to adjoin square roots of field values to the set of possible values. By repeated application you can adjoin fourth roots, then eighth roots, and so forth, but never cube roots.

So, this is not a proof, but enough for me to form a very strong hunch that the cube cannot be duplicated by flatiron-and-compass constructions in three dimensions; or for that matter, in any finite number of dimensions.

Another question is, what is a good set of rules for constructions in three dimensions? Sources available on the Web seem generally to give only five rules, not six, for straightedge and compass; and Jacobson notes that if you assume two points are initially known, there is no need for a rule for introducing “arbitrary” points. The idea is that even without that rule, from two initial points you can easily construct a dense set of points in the plane, from which you can select whatever “arbitrary” point a construction might require.

ReplyDeleteIt is also well known (if a fact that many of us probably heard once in high school geometry and then promptly forgot can be considered “well known”) that the straightedge is redundant; one can perform all straightedge-and-compass constructions with compasses alone (using just two rules). But people like to use straightedges for constructions in the plane, so the straightedge rules persist. Present-day teachers also seem to prefer non-collapsible compasses rather than the classical model.

In three dimensions, I think we need to start out with at least three points in order to avoid the need for arbitrary points in our constructions; but if we have any three known non-collinear points, we can dispense with the rules involving arbitrary points, leaving 11 rules (the 5 from plane geometry plus 6 involving planes and/or spheres), though I would argue that these are really 14 rules: the last rule, for intersecting a line

orcircle with a planeorsphere is really four rules (2 × 2). On the other hand, if you can draw a sphere centered at a point in a plane and can then intersect the sphere and plane, do you still need a separate rule for drawing a circle in a plane? I think not.Consider the following rules, applied an initial set of at least three non-collinear points:

1. Draw the unique plane containing any three non-collinear points.

2. Draw a sphere with one point as the center, and any other point on its surface.

3. Draw the point at the intersection of three planes (if they intersect at a unique point).

4. Draw the point(s) at the intersection of two distinct planes and a sphere (if they intersect at exactly one or two points).

5. Draw the point(s) at the intersection of two spheres and a plane (if they intersect at exactly one or two points).

6. Draw the point(s) at the intersection of three spheres (if they intersect at exactly one or two points).

That’s just six rules, compared to five for plane constructions. Of course these rules do not make it possible literally to “draw the circumcircle of a square” (for example), unless you consider a circle to be “drawn” when you have constructed its center and one point on its circumference (in effect, its center and radius). But then again, how does one draw the edges of a square that is constructed with compasses alone? If you like, you could include rules for drawing a line through two points and/or for drawing the circle (if there is one) at the intersection of a plane and a sphere in order to make certain constructions “pretty.” Or you could remove the restrictions (when drawing intersections of planes and/or spheres) that the intersections must consist of exactly one or two points; one could then draw lines and circles under these rules.

With those considerations, would these rules be sufficient for all “flatiron” constructions?

By the way, if you consider the action of drawing lines, planes, circles, and spheres to be essentially the ability to construct the points of intersection given in the rules, then I think one can “draw” a sphere with a compass: the compass is set to the given center and radius, then swung about the center of the sphere to meet the point of another compass or the surface of a flatiron; it then is constrained to move in a circle, along which it may meet another compass point or flatiron surface. If one considers “drawing a sphere” to require some way of marking every point on the sphere’s surface, the way a compass might mark every point along the circumference of a circle, then one needs some more sophisticated type of tool that can sweep out a curved surface. This seems rather cumbersome to me.

ReplyDeleteAnd oh, yes, thanks to you, my living room now is littered with Zome structures.

ReplyDelete