By Bob Robinson
“If I were at the outside, say at the heavens of the fixed stars, could I stretch my hand or my stick outward or not? To suppose that I could not is absurd; and if I can stretch it out, that which is outside must be either body or space (it makes no difference which it is, as we shall see). We may then get to the outside of that again, and so on, asking at our arrival at each new limit the same question; and if there is always a new place to which the stick may be held out, this clearly involves extension without limit. If now what so extends is body, the proposition is proved; but even if it is space, then, since space is that in which body is or can be, and in the case of eternal things we must treat that which potentially is as being, it follows equally that there must be body and space without limit.” Archytus, circa 400-365 B.C.
I have invented a pedagogical -that is, teachable- model of the ancient Greek Archytus’ geometric solution to the classical problem of finding two mean proportionals between two extreme magnitudes, often also called the problem of “the duplication or doubling of the cube”. (If unfamiliar with the term “mean proportionals” think, for first approximation, in terms of numbers. What are the two mean proportionals between 1 and 8, 1 and 27, and 1 and 125?) Archytus’ solution requires no numbers; instead the smaller of the two extreme magnitudes is any chord of a circle, while the larger of the two extremes is the same circle’s diameter. His solution is three dimensional, involving the intersection in a point of three “solid” surfaces: a cylinder, a torus (doughnut shape), and a cone. But it is not just three dimensional, and to call it “three dimensional” is, as we shall see, somewhat misleading for a truthful understanding of the problem, its solution, and Plato’s friend Archytus himself. To situate Archytus’ solution to the problem of the doubling of the cube, consider the following quote of Eratosthenes (circa 200 B.C.), who developed his own solution to that problem, by Theon of Smyrna. “Eratosthenes in his work entitled Platonicus relates that, when the god proclaimed to the Delians by the oracle that, if they would get rid of a plague, they should construct an altar double of the existing one, their craftsman fell into great perplexity in their efforts to discover how a solid could be made double of a similar solid; they therefore went to ask Plato about it, and he replied that the oracle meant , not that the god wanted an altar of double the size, but that he wished, in setting them the task, to shame the Greeks for their neglect of mathematics and their contempt for geometry.” Archytus’ intention was not merely to move from the two dimensional realm (doubling the square) to the three dimensional realm (doubling the cube), but was to move outside the realm of three abstract dimensions into the realm of physical geometry. Three abstract dimensions appear in Archytus’ construction as a sphere. The determination of two mean proportionals lies outside that sphere, on the surface of a torus and a cylinder which surround the sphere in Archytus’ construction. One must, in effect, poke a stick through the three dimensional sphere of Archytus’ construction, to see where the stick intersects the cylinder and torus. Compare this with Archytus’ astronomical notion of “stretching my hand or my stick outwards” at “the heavens of the fixed stars”. Consider the analogous case, of how it is necessary to move into three dimensions, to double the square. The diagonal of a square, which forms the side of a square with double the area of the original one, is formed by folding the original square in half, that is, by rotation outside the plane of the original square!
Situate a transparent sphere (such as a Lenart Sphere) in a cardboard box, such that the sphere sits snugly in a hole in the top of the box and the equator of the sphere is level with the top of the box. The hole in the top of the box should be put tangent to one side (edge) of the box. Trace with a spherical compass (also included with the Lenart Sphere) a number of concentric circles of latitude on the sphere which, unlike the circles of latitude on the earth, are all made perpendicular to the equator of the sphere. That is, the compass forming the circles is pivoted on a point on the equator of the transparent sphere. Be sure to use erasable ink of a certain color, say red. Next, with the point of tangency of the hole in the top of the box and the side of the box as center, trace with an ordinary plane compass a circle on the top of the box, such that the circle has twice the diameter of the hole in the top of the box. The diameter of the sphere and hole form the radius of the new circle, so that the new circle will be just tangent to the hole at a point directly opposite where the hole is tangent to the side of the box. This new circle, which we can only represent a portion of in our model, forms the outer perimeter of the torus in Archytus’ construction. The “hole” in the middle of the torus is of null diameter, and is represented by the center around which we pivoted the new circle, that is, the point of tangency of the hole in the top of the box and and the side (edge) of the box. Obtain from an arts and crafts store some wooden hoops with about the same diameter as the sphere, and some clear acetate. Cut some of the wooden hoops in a continuously growing array of arc lengths, up to and including a couple that are semicircles. Wrap the acetate around one of the semicircular hoops, such that it forms a cylinder, and with the sphere put into its hole in the top of the box, snuggle the acetate half cylinder in between the sphere and the hole. Next, position the smaller hoop lengths in order of ascending length, starting from near the point of tangency of the larger circle traced on the top of the box and the hole in the top of the box. The base of each hoop length should be on the larger circle, and its top should rest on the acetate wall of the cylinder. The effect should be that, were the hoops to continue past the barrier of the acetate cylinder, they would all converge on the point of tangency of the hole and the side of the box, or, what is the same thing, the center of the circle on the circumference of which their bases rest. Silicone should be used to secure their positions on the top of the box and the acetate wall of the cylinder. When the construction is completed, the wooden hoops should approximate a partial (quarter) torus, which if completed would wrap around the sphere and cylinder. Trace with a dry erase marker the line of intersection, so formed, between the cylinder and the torus. When the sphere is placed in the hole in the box, maneuver it so that the center of the concentric latitude circles traced on its surface coincides with point of tangency of the hole in the top of the box and the side of the box. Obtain a laser pointer. Shine it through the sphere, from the point where the hole in the top of the box is tangent to the side of the box, so that it hits the side of the acetate cylinder. Play around with it a bit. Trace along the curved line of intersection of the torus and cylinder, so that the laser crosses in succession all the lines of latitude traced on the sphere. Next, perform the inverse operation, by tracing along each circle of latitude to see where it intersects the line of intersection of the torus and cylinder traced on the clear acetate cylinder. As you do so, note how the laser’s motion along each circle of latitude forms a distinct cone. The integral of the bases of all such possible cones is nothing but the sphere itself! That is, the sphere is the only truthful representation of all possible cones formed by rotating all possible chords of a circle emanating from a single point on that circle’s circumference.
Archytus’ Creation of Two Mean Proportionals
Archytus wishes to find the two mean proportionals between any chord of the circle, contained in the great circle of the equator of the sphere in our construction, and the diameter of that same circle, which is also the diameter of the sphere, the cylinder, and the “tube” of the torus in Archytus construction. Place the chord with its origin at the point of tangency of the hole and the side of the box. (From now on we will simply call this point, which by our construction is on the equator of the sphere and also is the center “hole of null dimension” of the torus, and is the point from which we will direct the laser pointer, the origin.) For any such chord, there is implicitly a circle of latitude on the sphere, which is everywhere equidistant from the origin, forming, as we have indicated, a distinct cone. Next, trace with the laser along one such circle of latitude on the sphere, until the laser beam crosses the line of intersection of the torus and the cylinder. There will now be three points of light associated with the laser beam, one at the origin, one on the surface of the sphere, and one on the curved line of intersection of the torus and the cylinder. Looking down from directly on top of the model, imagine a plane including the three aforesaid points of light, and perpendicular to the top of the box slicing through the sphere, the cylinder and the torus. That plane, being perpendicular to both the top of the box and to the equator of the sphere, will form a circle cut exactly in half by the equator of the sphere, just as the circles of latitude were cut in half by the plane of the equator. This cut is most efficiently represented by a semicircle drawn on a clear overlay (provided with the Lenart Sphere), and placed on the sphere so that it coincides with the origin and the point on the sphere through which the laser shines. It should be of different color than the (red) concentric circles of latitiude on the sphere, say green. Now draw a diagram of a cross section of the portion of Archytus’ model cut by the perpendicular plane. The diagram will not be precise, but simply representative of certain geometric relations that exist in the three dimensional model. Construct two tangent circles, one inside the other. At the point of tangency, mark O. The smaller circle represents the green circle in our three dimensional model, while the larger one represents a cross section of the torus. Through O, draw a straight line that forms a diameter OB of the smaller circle, and OD of the larger circle. This equatorial line represents a cross section of the plane forming the top of the cardboard box in our three dimensional model. In the diagram, through B, draw a line perpendicular to OB, that forms a tangent to the smaller (green) circle, and intersects the larger circle at C. Draw OC and CD. Where OC intersects the smaller circle, mark A. Triangle OAB, being inscribed in the smaller semicircle of the diagram, will have a right angle at A. In triangle OBC, diameter OB and tangent BC will be perpendicular to each other, so there will be a right angle at B. Triangle OCD, being inscribed in the larger semicircle of the diagram, will have a right angle at C. All three triangles share angle DOC (identical to angle BOA). So, by similar right triangles OAB, OBC, and OCD, the continued proportions OA/OB=OB/OC=OC/OD are produced. OA represents the distance from O to the surface of the sphere at A, and is equivalent in length to the original chord forming the lesser extremity of Archytus’ demonstration. OD of the diagram represents the cross sectional diameter of the torus “tube” in the three dimensional model. By the way the model was constructed, the diameter of the tube of the torus is the same as the diameter of the sphere. And diameter of the sphere is the same as the diameter of the equatorial great circle of the sphere, which formed the larger extreme in Archytus’ demonstration. Straight line BC, perpendicular to OB in the diagram, is a cross section of the cylinder, which is perpendicular to the equatorial plane in Archytus’ three dimensional model. Thus, by reference to the schematic diagram, it is easily seen that Archytus’ model really does create two mean proprtionals OB and OC between two extremes OA and OD. Well, who needs the three dimensional model if a two dimensional drawing gives us the required solution? But the diagram is only schematic, and does not give us true values for OB and OC. Only the three dimensional model does. (Indeed, even Eratosthenes’ later mechanical method for finding two mean proportionals, which is two dimensional, only minimizes the problems of approximation inherent in any two dimensional model that uses only straight lines.) The more interesting question is, is Archytus’ model simply three dimensional, or of a higher power? Consider the fact, that in Archytus demonstration, we must go outside, not only the circle, but the three dimensional sphere, to find the two mean proportionals OB and OC. On the surface of the sphere itself, this is reflected in the different orientations of the red and the green circles.