Duplication of the Cube : Darrell Mattingly, Cateryn Kiernan

The ancient Greeks originated numerous mathematical questions, most of which they learned to solve using simple mathematical tools, such as the straight edge and the collapsable compass. Three of these problems persist today, challenging students in contemporary classrooms. This triology of problems, the trisection of a given angle, the squaring of a circle, and the duplication of the cube, have since been proved impossible using exclusively the straight edge and the compass. In the quest to solve these problems using those specific tools, however, mathematicians developed numerous alternate solutions using other mathematical tools. The last problem of the trilogy is the focus of this discussion, and it challenged mathematicians for centuries, due to the restriction of using only the aforementioned tools.

Origin of the Problem

The origin of the Duplication of the Cube Problem proves to be an interesting tale. In the early 4th Century BC, a plague existed in Athens, that eventually eliminated about a quarter of the population. At the height of the plague, a delegation traveled to the Oracle of Apollo at Delos to inquire about how to stop the continuation of the plague. The oracle replied that if the delegation was able to sucessfully duplicate Apollo's cubicle altar, the plague would end. Plato is said to have believed that the oracle did not give the Athenians this problem in order to actually double the size of Apollo's altar. Rather, the oracle "intended, in seting them the task, to shame the Greeks for their neglect of mathematics and their contempt for Geometry" (Heath, 155). This theory seems to be well founded, for the Greeks were unable to correctly duplicate the volume of the cube. In their attempt, they doubled the length of each side, thus increasing the volume by a scale of eight, and thus prolonging the length of the plague. This task set forth by the oracle originated the problem of the duplication of the cube and is often referred to as the "Delian Problem" (Boyer, 64). Although the Athenians were unable to solve the problem on this attempt, they were able to originate other solutions, but not exclusively with a straight edge and a compass.

Proof that NO Platoic Solution Exists for the "Delian" Problem

After centuries of mathematicians had worked on this problem, a proof developed that it could not be done using exclusively the straight edge and compass. This proof is based on theorems about the powers of degrees of subfields generated by the x and y coordinates of the side of the cube to be duplicated. Although the desired point can be approximated, it cannot in fact be found based on these theorems.

"Non-Platoic" Solutions

Few Ancient Greek mathematicians are credited with developing working proofs of the solutions of this problem. Eudoxus has been credited with a solution, of which his pupil, Eutocius, reported the development. He believed that Eudoxus had developed a solution by means of "curved lines," but in his extension of the proof, neglected to use them (Heath, 156). Historians believe that Eutocius' sources may have been wrong, due to this discrepancy. In addition to Eudoxus' solution, Hippocrates of Chios also developed a working solution of the "Delian" problem. Hippocrates claimed that this problem was "reducible to finding two mean proportionals in continued proportion between two given straight lines" (155). Other mathematicians, such as Manaechmus, fed off of this supposition, developing some proofs of their own of this problem, although none of them could solve it with just a straight edge and a compass. However, as soon as two marks could be added to the straight edge to form a ruler, a working solution was possible.

Rule and Collapsable Compass Solution

Description

The aim of this section is to show that, in fact, a " ruler " and collapsable campass is enough to solve the "Delian" problem if one strange pre-step is allowed, namely the following:

* Given: a straight edge, compass, and a segment of unit length

* Action: Mark two points on the straight edge one unit apart using the compass

Solution Algorithm:

* Given a distance "1 unit", construct an equalsided triangle ABC, with side length of 1 unit.

* Prolong the line through A,B and construct the point D such that |DB| = 1.

* Draw lines through B,C and through D,C making both of them "long enough".

* Finally, and this is the only step requiring the marked straight edge , adjust the ruler by hand so that it passes through the vertex A, and the distance between the points where it intercepts the lines the points where it intercepts the lines BC and DC is just 1 unit.

* Align the straight edge with point A and the elongated segment BC so that the line BC intersects the straight edge at the distant marked point. E is the point where the line containing the segment DC intercepts the straight edge at the closer marked points on the straight edge.

It is not difficult to show that the distance between point A and E is just .

Geometric Representation

> A:=[1/2,sqrt(3)/2]: B := [0,0]: C:= [1,0]:

> unit := plot([B,C],color=red,thickness=3):

> triangle:=plots[polygonplot]([A,B,C],color=yellow,thickness=3):

> R1:=[0,1.07]: R2:=[3,-.19]: R3:=[3.15,-.03]: R4:=[.15,1.25]:

> ruler:=plots[polygonplot]([R1,R2,R3,R4],color=blue,thickness=3):

> lne1 := [[-1/2,-(1/2)*sqrt(3)],[1,1*sqrt(3)]]:

> lne2:=[[-1/2,(-sqrt(3)/2)],[((sqrt(6)/2)+1),(sqrt(2)/2)]]:

> lne3:=[[1/2,(sqrt(3)/2)],[2.56,0]]:

> unit2:=plot([[0,0],[-1/2,-sqrt(3)/2]],color=red,thickness=3):

> unit3:=plot([[2.56,0],[1.66,.375]],color=red,thickness=3):

> mark1:=plot([[2.56,0],[2.66,.1]],color=green,thickness=3):

> mark2:=plot([[1.66,.375],[1.76,.475]],color=green,thickness=3):

> line1:=plots[polygonplot](lne1,color=red,thickness=3):

> line2:=plots[polygonplot](lne2,color=red,thickness=3):

> line3:=plots[polygonplot](lne3,color=red,thickness=3):

> plots[display]([unit,triangle,unit2,unit3,ruler,mark1,mark2,line1,line2,line3],scaling=constrained);

Archytas of Tarentum's Solution
The most interesting and complex of the solutions to the "Delian" Problem is the solution developed by Archytas of Tarentum. His solution, however, lies in a geometric proof, contrary to what Plato felt Athenians should avoid. Archytas developed a solution using three geometric figures and their intersects. Archytas used a right cone, a cylinder, and a torus, all placed on the same coordinate system, and claimed that the intersection of these figures gave the length of the desired duplicated cube, , where is the length of the side of the orginal cube. To illustrate his method to solve the "Delian" problem, we will first decide on the . For this illustration, is 1. Next, we need the three formulas, Archytas used . They are:

Cone:

Cylinder:

Torus:

We need to form this geometric shapes. So, we will ask maple to help us.

> f:= x^2=y^2+z^2:

> g:=2*(1)*x=x^2+y^2:

> h:=(x^2+y^2+z^2)^2=4*(1)^2*(x^2+y^2):

> cone:=plots[implicitplot3d](f,x=2^(1/3)..2^(1/3)+2,y=-3..3,z=-3..3,grid=[20,20,20], color=green):

> cyl:=plots[implicitplot3d](g,x=2^(1/3)..2^(1/3)+2,y=-3..3,z=-3..3,grid=[20,20,20], color=red):

> torus:=plots[implicitplot3d](h, x=-1..3,y=-3..3,z=-3..3,grid=[20,20,20], color=blue):

> plots[display]({cone,cyl,torus},scaling=constrained,style=patchnogrid,orientation=[153,90],lightmodel= light3, title="Archytas Method for The Duplication of the Cube");

The human eye has a difficult time reading a 3-D figure, so we will take a slice of this figure to make it easy to see.

Right. You need to slice it at the appropriate z = constant or y = constant plane in order to "see" that the x-coordinate of the intersection is Of course you can only approximately see that, but I think you ought to use Maple to calculate the coordinates of the intersection point of the cone, torus, and cylinder. Then you can use the exact value of z rather than .81 By playing playing with the solutions to the equations, you can work out that is the exact value for z.

> fs:=subs(z=sqrt(2*2^(2/3)-2*2^(1/3)),f):

> gs:=subs(z=sqrt(2*2^(2/3)-2*2^(1/3)),g):

> hs:=subs(z=sqrt(2*2^(2/3)-2*2^(1/3)),h):

> ts:=plots[implicitplot](hs,x=1.22..1.28,y=-1.1..-.9,grid=[30,30],color=blue,thickness=3):

> cys:=plots[implicitplot](gs,x=1.22..1.28,y=-1.1..-.9,grid=[30,30],color=red,thickness=3):

> cs:=plots[implicitplot](fs,x=1.22..1.28,y=-1.1..-.9,grid=[30,30],color=green,thickness=3):

> plots[display]({ts,cys,cs},title="Slice of Archytas Figure for The Duplication of the Cube");

>

>

>

>

As you can see, the intersect of these three geometric figures gives the duplicated cube a side of length is . A cube of this size does have volume 2, which duplicates the original volume 1. Let's use maple to verify our answer.

well, I can sort of see it, approximately, but it is still only approximate.

> x,y,z;

>

> f;g;h;

> assume(x>0,y>0,z>0);

> solve({f,g,h},{x,y,z});

> psols :=allvalues(%[2]);

> sol := psols[1];

> evalf(sol);

> evalf([%]);

> evalf(2^(1/3));

To check this out for values other than = 1, create your own figures using the same prompts as above, and find the intercept that these figures create. Cubing it, you will see that the volume is in fact doubled every time.

Although Archytas develop a convincing solution to this complex problem, he does so using more than the simple straight edge and compass. The fact is that this problem cannot be solved with these simple tools, proved after many great thinkers had spent the majority of their lives striving to solve this problem.

Exercises

Problem #1:

Using the same prompts as we used above, check for the value a = 2. Simply substitute in a = 2 where we substituted in a = 1 in our formulas of the cylinder and torus. Then copy the prompts, and you will discover that the intersection proves to be the desired length of the duplicated cube.

Problem #2:

In similar fashion to the precious problem, choose any value of a of any magnitude and check for the intersection. Cubing this x value will give you a cube of duplicated volume.

An intersecting problem: Use maple to draw two cubes, one of whose volume is double that of the other.