Duplication of the cube



 

by Francois Rivest and Stephane Zafirov
 

[ History | Hippocrate | Archytas | Eudoxus | Menaechmus1st, 2nd | Plato | Erathostenes | Nicomedes ]
[ Apollonius, Heron and Philon | Diocles | Sporus and Pappus | Philo line | Impossibility of construction ]
[ Applet | Links | References | Home | CS 507 HomePage | Stephane | Francois | Email us | End ]



 

History

Although Euclid solves more than 100 geometric construction problems in the Elements, many more were posed whose solutions required more than just compass and straightedge. Three such problems stimulated so much interest among later geometers that they have come to be known as the "classical problems": doubling the cube, i.e., constructing a cube whose volume is twice that of a given cube; trisecting the angle, i.e., constructing the third of an angle; and squaring the circle, i.e., constructing a square of area equal to that of a given circle. In what follows, we will be mainly concerned with the first of these problems: how to double a cube?

One possible issue for the origin of the cube duplication problem is the following. An ancient tragic poet had represented Minos as dissatisfied with a tomb which he had put up to Glaucus, and which was only 100 feet each way. He therefore ordered it to be made double the size, the poet making him add that each dimension should be doubled for this purpose(!). The poet was, as showed von Wilamowitz, not Aeschylus or Sophocles or Euripides, but some obscure person who owes the notoriety of his lines to his ignorance of mathematics. Geometers took up the question and made no progress for a long time, until Hippocrates of Chios showed that the problem was reducible to that of finding two mean proportionals in continued proportion between two given straight lines. Again, after a time, the Delians were told by  the oracle that, if they would get rid of a certain plague, they should construct an altar of double the size of the existing one. They consulted therefore Plato who replied that the oracle meant, not that god wanted an altar of double the size, but that he intended, in setting them the task, to shame the Greecs for their neglect of mathematics and their contempt for geometry. According to Plutarch, Plato would have referred the Delians to Eudoxus and Helicon of Cyzicus for a solution. The problem was thereafter studied in the Academy, and constructions of the problem of the two means were proposed in the 4th century B.C. by Archytas, Eudoxus, Menaechmus and even (though erroneously) to Plato himself. After Hippocrates' reduction of the problem, it was always attacked in the form of finding two mean proportional between two given straight line segments.

In seeking the solutions of problems, geometers developed a special technique, which they called "analysis." They assumed the problem to have been solved and then, by investigating the properties of this solution, worked back to find an equivalent problem that could be solved on the basis of the givens. To obtain the formally correct solution of the original problem, then, geometers reversed the procedure: first the data were used to solve the equivalent problem derived in the analysis, and from the solution obtained the original problem was then solved. In contrast to analysis, this reversed procedure is called "synthesis." Menaechmus' cube duplication is an example of analysis: he assumed the mean proportionals x and y and then discovered them to be equivalent to the result of intersecting the three curves whose construction he could take as known. (The synthesis consists of introducing the curves, finding their intersection, and showing that this solves the problem.) It is clear that geometers of the 4th century BC were well acquainted with this method, but Euclid provides only syntheses, never analyses, of the problems solved in the Elements. Certainly in the cases of the more complicated constructions, however, there can be little doubt that some form of analysis preceded the theses presented in the Elements.
 



 

Reduction to the two mean proportionals

Hippocrate showed that the problem could be reduced to that of finding two mean proportionals: if for a given line segment of length a it is necessary to find x such that x3 = 2a3, line segments of lengths x and y respectively may be sought such that a:x = x:y = y:2a; for then a3/x3 = (a/x)3 = (a/x)(x/y)(y/2a) = a/2a = 1/2.



 

Archytas

Of all the solution that of Archytas is the most remarkable of all, especially when his date is considered (first half of fourth century B.C.), because it is not a construction in a plane but a bold construction in three dimensions, determining a certain point as the intersection of three surfaces of revolution, (1) a right cone, (2) a cylinder, (3) a tore or anchor-ring with inner diameter nil. The intersection of the two later surfaces gives (says Archytas) a certain curve (which is in fact a curve of double curvature), and the point required is found as the point in which the cone meets this curve.

Suppose that AC, AB are the two straight lines between which two mean proportionals are to be found, and let AC be made
the diameter of a circle and AB a chord in it.

Intersection projected to the plane

Draw a semicircle with AC as diameter, but in a plane at right angles to the plane of the circle ABC, and imagine this
semicircle to revolve about a straight line through A perpendicular to the plane of ABC (thus describing half a tore with inner
diameter nil).

Next draw a right half-cylinder on the semicircle ABC as base; this will cut the surface of the half-tore in a certain curve.

Lastly let CD, the tangent to the circle ABC at the point C, meet AB produced in D; and suppose the triangle ADC to revolve
about AC as axis. This will generate the surface of a right circular cone; the point B will describe a semicircle BQE at right
angles to the plane of ABC and having its diameter BE at right angles to AC; and the surface of the cone will meet in some
point P the curve which is the intersection of the half-cylinder and the half-tore.

Intersection in space
Let APC' be the corresponding position of the revolving semicircle (the white triangle is APC'), and let AC' meet the circumference ABC in M.

Drawing PM perpendicular to the plane of ABC, we see that it must meet the circumference of the circle ABC because P is on the cylinder which stands on ABC as base.

Let AP meet the circumference of the semicircle BQE in Q, and let AC' meet its diameter BE in N. Join PC', QM, QN.

Then, since both semicircles are perpendicular to the plane ABC, so is their line of intersection QN (Eucl.XI. 19).

Therefore QN is perpendicular to both BE and AM.

Therefore QN.QN = BN.NE = AN.NM, (Eucl. III. 35) so that the angle AQM is a right angle.

But the angle APC' is also right; therefore MQ is parallel to C'P.

It follows, by similar triangles, that

C'A:AP = AP:AM = AM:AQ
that is
AC:AP = AP:AM = AM:AB,

and AB, AM, AP, AC are in continued proportion, so that AM, AP are the two mean proportionals required.

In the language of analytical geometry, if AC is the axis of x, a line through A perpendicular to AC in the plane of ABC the axis of y, and a line through A parallel to PM the axis of z, then P is determined as the intersection of the surfaces.

where AC = a, AB = b.

From the first two equations we obtain

and from this and (3) we have

or

 AC:AP = AP:AM = AM:AB.

Compounding the ratios, we have

therefore the cube of side AM is to the cube of side AB as AC is to AB.

In the particular case where AC = 2 AB,

and the cube is doubled.



 

Eudoxus

Eudoxus' solution is unfortunately lost. The epigram of Erathostenes says that Eudoxus used a species of curves which he called "curved lines" (look at Kampyle of Eudoxus). Eutocius had evidently seen some version of the solution which was not correct. For, he says in his preface that Eudoxus claimed to have discovered a solution by means of "curved lines" but not only did he make no use of such lines in his proof, but he actually used a certain discrete proportion as if it were continuous. The latter part of this statement compels us to assume that Eutocius' source was in some way defective, for it is inconceivable that a mathematician of the calibre of Eudoxus could have made such  a mistake.


Menaechmus

  • Menaechmus' solution is probably the first one using the conic sections and their properties. The two properties which he used were the ordinate property of a parabola and the assymtote-property of a rectangular hyperbola.

  •  

     
     
     

    If x, y are the required two mean propotionals between two straight line segments a, b, that is, if a:x = x:y = y:b, then clearly

    x2 = ay, y2 = bx, and xy = ab

  • The properties of the parabola and hyperbola used by Menaechmus are precisely those expressed by these relations when x, y are the Cartesian coordinates referred to rectangular axes. Meneachmus' first solution used the second and third of the conics so represented, the second solution the first and second.

  •  

     
     
     

    First solution.

    Let AO, OB,  be placed at right angles, represent the two given straight line segments, AO being the greater.

    Suppose the problem is solved, and let the two mean proportionals be OM measured along BO and ON measured along AO. Complete the rectangle OMPN.

    Figure 1 - One parabola and one hyperbola

    Then, since AO:OM = OM:ON = ON:OB, (i.e. a:x = (x:y) =  y:b is equialent to a:y = (y:x) = x:b inversed),
    we have (1)     OB.OM = ON2 = PM2, (multiply  out the last equality) so that P lies on a parabola with vertex O, with axis OM, and with latus rectum OB.

    Again (2)    AO.OB  = OM.ON = PN.PM, (i.e. ab = xy), so that P lies on a hyperbola with O as a centre and OM, ON as assymptotes, and such that the rectangle contained by  the straight lines PM, PN drawn from any point P of the curve parallel  to one assymptote and meeting the other respectively is equal to the given rectangle AO.OB.

    If, then, we draw the two curves in accordance with the data, we determine the point P by their intersection, and

    AO:PN = PN:PM = PM:OB
    (a:x = x:y = y:b)

    Second solution.

    In this case we draw two parabolas, namely

    (1) the parabola with O as vertex, ON as axis, and OA as latus rectum,
    (2) the parabola with O as vertex, OM as axis, and OB as latus rectum.

    Figure 2 - Two parabolas
    These parabola determine by their intersection a point P such that

    (1) OA.ON = PN2
    (2) OB.OM = PM2

    Since PN = OM, PM = ON, it follows that

    OA:OM = OM:ON = ON:OB

    and OM, ON are the required mean proportionals between OA, OB.



     

    Plato

    The solution, which Eutocius alone gives or mentions, can hardly be Plato's, only because we know Plato objected on principle to solutions by mechanical means. He considered them as destroying the good geometry.
    Plato's Instrument
    It may have been evolved in the Academy by some one contemporary with or junior to Menaechmus, for the arrangement in the figure of the two given straight lin segments and the two means between them is exactly the same as in Menaeucmus' figure. They are arranged in straight lines mutually at right angles and in descending order order of magnitude when taken in clockwise order. The difference is that what Menaechmus does by means of conics is here done by means of mechanical contrivance.

    FGH is a rigid right angle made of wood say. KL is strut which can move along GF, while always remaining parallel to GH.

    We have to place the machine so that the inner side of GH always passes through B, and the inner side of of the strut always passes through A, and then to move the machine, and the strut along it, until (1) the inner angle at G lies on AO produced and (2) the inner angle (towards G) at K lies on BO produced. (This seems challenging but doable).

    Then the four line segments OA, OM, ON, OB take up the same position as in Menaechmus' figures and, since the angles at M, N are right,

    MO2 = AO.ON and NO2 = MO.OB,

    Whence   AO:MO = MO:ON = ON:OB.


    Erathostenes

    This, too, is a mechanical construction. The device consists of a rectangular frame along which slide three parallelograms (or the triangle which are the halves of them) of height equal to the width of the frame. The parallelograms or triangles move always so that their bases describe one straight line (one edge, say the upper of the frame), and they can slide onver one another.

    The original position of parallelograms and triangles are shown here

    Figure 1 - Original Position
    AX, EY are the sides of the frame; AMF, MNG, NQH (the halves of the parallelograms ME, NF, QG) are the triangles which slide along the frame.

    The second figure shows the result of sliding all triangles except the first (which remains stationary) along their original positions to positions in which trey overlap one another, as AMF, M'NG, N'QH.

    Let AE and DH (perpendicular to EY) in the second figure be the two given straight line segments.

    Figure 2 - Ovelaping triangles
    Let N'QH be the position of the triangle NQH in which QH passes through D, and let the triangle M'NG be such a position of the triangle MNG that the points B, C in which MF, M'G, and NG, N'H respectively intersect are in a straight line with A, D.

    Produce AD to meet EY in K.

    Then AE:BF = EK:FK = AK:KB = FK:KG = BF:CG and similarly, BF:CG = CG:DH.

    Therefore AE, BF, CG, DH are in continued proportion and BF, CG are the required mean proportionals.

    It should be observed that Eratostenes' epigram is a much simpler tool then the above ones and also that it can be used to construct any number of means in continued proportion to be interpolated.



     

    Nicomedes

    Nicomedes' solution uses a construcion by means of a curve called the conchoid.

    Let AB, BC, placed at right angles, be the two given straight line segments. Complete the parallelogram ABCL.

    Nicomedes' construction
    Bisect AB, BC at D, E respectively. Join LD and let LD, CB be produced to meet at G. Draw EF at right angles to BC and of length such that CF = AD. Join GF, and draw CH parallel to it.

    Now from the point F draw FHK cutting CH in H and BC produced in K in such a way that  the intercept HK = CF = AD.
    This is done by means of a conchoid in which F is the pole, CH the 'ruler', and the 'distance' is equal to AD or CF. Then by the property of the conchoid, HK = the 'distance'.

    Join KL, and produce it to meet BA produced in M. Then CK, MA are the claimed mean proportionals.
    For, since BC is bisected at E and produced to K,

    BK.KC + CE2 = EK2
    Adding EF2 to each, we have
    BK.KC + CE2 = KF2

    Now, by parallels,  MA:AB = ML:LK = BC:CK. But AB = 2AD and 2BC = GC.

    Therefore

    MA:AD = GC:CK = FH:HK

    and hence

    MD:DA = FK:HK.

    But by construction, HK = AD. Hence MD = FK and MD2 = FK2

    Now    MD2 = BM.MA + DA2     and  FK2 = BK.KC + CF2 (from above)

    and therefore CK:MA = BM:BK = LC:CK while at the same time BM:BK = MA:AL.

    Therefore    LC:CK = CK:MA = MA:AL   or   AB:CK = CK:MA = MA:BC.



     

    Apollonius, Heron and Philon

    It is convinient to group those originately different solutions together for the reason that they are basically equivalent.

    Let AB, AC, placed at right angles, be the two given straight line segments. Complete the rectangle ACDB, and let E be the point at which the diagonals bisect one another.

    Apolloius'  construction
    Then a circle with centre E and radius EB will circumcribe the rectangle ABCD.

    Now (Apollonius) suppose a circle drawn with centre E and cutting AB, AC produced in points F, G such that F, D, G are in one straight line.

    Or (Heron) place a ruler so that its edge passes through D and turn it about D until the edge intersects AB, AC produced in points F, G which are equidistant from E.

    Or (Philon) turn the ruler about D until it cuts AB, AC produced and the circle about ABCD respectively F, G, H such that the intercepts FD, HG are equal.

    All three constructions give the same points, F, G.

    First we have to prove that AF.FB = AG.GC

    (a) With Apollonious' and Heron's constructions we have, if K is the middle point of AB,

    AF.FB + BK = FK2

    and, if we add KE2 to both,

    AF.FE +BE2 = EF2

    Similarly,

    AG.GC + CE2 = EG2

    But BE = CE and EF = EG, hence

    AF.FB = AG.GC

    (b) With Philon's construction, since GH = FD

    HF.FD = DG.GH

    But since the circle BDHC passes through A,

    HF.FD = AF.FB and DG.GH = AG.GC

    therefore    AF.FB = AG.GC as before.

    Now, the rest follows by similar triangles:

    FA:AG = DC:CG = FB:BD
    therefore
    DC:CG = CG:FB = FB:BD
    i.e.,
    AB:CG = CG:FB = FB:AC



     

    Diocles and the cissoid

    The cissoid is evolved as follows. AC, DC are perpendicular diameters of a circle. Let E, F be points on the quadrants BD, BC respectively such that the arcs BE, BF are equal.

    Draw EG, FH perpendicular to DC. Join CE, and let P be the point in which CE, FH intersect.

    Figure 1 - Cissoid
    The cissoid is the locus of all the points P corresponding to different positions of E and F, E on the quadrant BD, F on the quadrant BC, such that EB = BF (Figure above).

    If P is any point on the cissoid, we need to show that FH, HC are two mean proportionals between DH, HP, or that

    DH:HF = HF:HC = HC:HP

    It is clear from the construction that EG = FH and DG = HC so that CG:GE = DH:HF.

    Now FH is a mean proportional between DH, HC. Therefore  DH:HF = HF:HC. And, by similar triangles,

    CG:GE = CH:HP

    Hence

    DH:HF = HF:CH = CH:HP

    If we look at the cissoid from an analytic geomtry point of view, we get that since DH.HP = HF.CH, we have (where a is the radius of the circle, OH = x, HP = y and we use OC, OB as axes of coordinates),

    which is the Cartesian equation of the cissoid. It has cusp at C, and the tangent to the circle at D is an assymptote to it.

    Figure 2 - Cissoid

    Suppose that the cissoid is represented in the above figure by the magenta curve. Diocles shows how to find two mean proportionals between two straight lines a, b as follows.

    Take K on OB such that DO:OK = a:b. (Now we refer no longer to the last but the first figure of this section).

    Join DK, and produce it to meet the cissoid in Q. Through Q draw the ordinate LM perpendicular to DC. Then, by the property of the cissoid, LM, MC are two mena proportionals in continued proportion between DM, MQ.
    And

    DM:MQ = DO:OK = a:b

    In order to find the two mean proportionals between a, b, take straight line segments x, y bearing to the same ratio to LM, MC respectively that a bears to DM and b to MQ. Then x, y are the required mean proportionals between a, b.



     

    Sporus and Pappus


    These solutions, given separately by Eutocius, are really identical, and also in a way, equivalent to the Diocles' one. The difference being that instead of using the cissoid, Sporus and Pappus use a ruler that they can turn about C (look at figure 1 of last section) until, intersecting DK produced in Q,OB in T, and the circle in R, it makes the intercepts QT, TR equal.

    Sporus was known to Pappus, and may even have been either his master or fellow student. From Pappus' own account, we gather that Pappus took the credit for the solution.



     

    Reduction to the Philon line

    The transversals drawn throught a given point within a given angle so that the sides of the angle intercepts on the transversal a segment of minimal length has become known as the Philo Line of the point for the given angle. Here is one characterization of the Philo Line. Given a point P and an angle RCS, let AB, A on the side CR and B on the side CS, be the Philo Line under consideration, and let Q be the foot of the perpendicular from C on AB. Then P and Q must isotomically divide the segment AB; that is, we must have AP = QB.

    Now consider the rectangle CGPH shown below and suppose that the line segment AP has been drawn so that, as in the characterization, AP = QB where Q is the intersection of AB and its perpendicular through C.

    Philon's Construction
    Observe that C, G, P, Q and H are concyclic (i.e. lie on the same circle) and

    (AP)(AQ) = (BQ)(BP)

     But,
    (AP)(AQ) = (AG)(AC),    (BH)(BC) = (BQ)(BP)

    It follows that

    AC:BC = BH:AG
    But, since the triangles BCA, PGA and BHP are similar,

    AC:BC = PH:BH = AG:PG

    or
    GC:BH = BH:AG = AG:HC

    and BH and AG are two mean proportionals between GC and HC. It follows, in particular that if GC = 2(HC), then (AG)3 = 2(HC)3, and the duplication problem is solved. The whole solution, it is seen, depends upon drawing the segment APB so that AP = QB.

    A parallel web project on the Philo Line can give you more inside on this subject.



     

    Impossibility of construction

    by means of straight edge and compass

    Eventhough many solutions have been found to the classical problems since the early stages of Greec geometry, no progress at all has been made towards a purely geometric construction. The reason for that being that such a construction is impossible and we have to wait until the late 19th century when sudden developement of algebra solves the problem. Proving that such constructions are impossible in general involves some notions of field theory as well as some basic Galois theory which we will assume here. In order for us to state the results (without proof), we will need to define what is called constructible numbers (also called Euclidian numbers). Since geometric constructions are done in the plane, a usefull structure in algebra to represent it is the field of complex numbers C. Here is a criterion for constructible number:

    Criterion. Let z1, z2,, znC and put . Then a complex number z is constructible from z1, z2,, zn if and only if z is contained in a subfield of C of the form F(u1, u2,, un) where u12 F and every ui2F(u1, u2,, ui - 1).

    To get a better feeling of what a constructible number is we can say that if one starts with the set S = {0,1} as constructed points, then the set of constructible numbers from S which we call C, can be defined recursively as:

    C contains Q and if  p2(x) is a quadratic polynomial with coefficients in C, then its roots (real or not) are also in C.

    Corollary. Let F(u1, u2,, un). Then any complex number z which is constructible from z1, z2,, zn is algebraic of degree a power of two over F.

    Impossibility proof. Here we have to show that the Delian constant, is not a constructible (complex) number. This
    follows from [Q():Q] = 3 (not a power of two!), since x3 2 is irreducible over Q[x].

    For more on this subject, you can look either in [3] or in any algebra book of your choice which covers some basic Galois theory.


    Applet


     


    Links
    on related subjects


    References

    [1] Journal Scripta Mathematica, vol.26, 1959, pp. 141-148

    [2] Sir Thomas L. Heath, A Manual of Greek Mathematics, Oxford University Press, 1963, pp.154-170

    [3] Jacobson, Basic Algebra I, Second Ed., 1985, p. 221



    [ History | Hippocrate | Archytas | Eudoxus | Menaechmus1st, 2nd | Plato | Erathostenes | Nicomedes ]
    [ Apollonius, Heron and Philon | Diocles | Sporus and Pappus | Philo line | Impossibility of construction ]
    [ Applet | Links | References | Home | CS 507 HomePage | Stephane | Francois | Email us | End ]