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 ]

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.

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.

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*.

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

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

or

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.

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

*x ^{2} = ay, y^{2}
= bx,* and

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*.

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 = ON ^{2}
= PM^{2}, *(multiply out the last
equality) so that

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)*

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*.*

(1) *OA.ON = PN ^{2}*

(2)

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

*OA:OM = OM:ON = ON:OB*

*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,

*MO ^{2} = AO.ON* and

The original position of parallelograms and triangles are shown here

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.

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.

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

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,*

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

Therefore

and hence

But by construction, *HK = AD. *Hence *MD = FK *and* MD ^{2}
= FK^{2}*

Now *MD ^{2} = BM.MA
+ DA^{2 } *and

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.*

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.

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 = FK ^{2}*

and, if we add *KE ^{2} *to both,

Similarly,

But *BE = CE *and *EF = EG, *hence

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

But since the circle *BDHC* passes through *A,*

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

Now, the rest follows by similar triangles:

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

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

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.

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

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.*

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.

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*.

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

It follows that

*AC:BC* = *PH:BH = AG:PG*

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

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

Criterion. Let *z*_{1},
*z*_{2},…,
*z*_{n}**C**
and put .
Then a complex number *z* is constructible from *z*_{1},
*z*_{2},…,
*z*_{n}
if and only if *z* is contained in a subfield of **C** of the form
*F*(*u*_{1},
*u*_{2},…,
*u*_{n})
where *u*_{1}^{2}*
F* and every *u _{i}*

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 *p*_{2}*(x)*
is a quadratic polynomial with coefficients in *C*, then its roots
(real or not) are also in *C.*

Corollary. Let *F*(*u*_{1},
*u*_{2},…,
*u*_{n}).
Then any complex number *z* which is constructible
from *z*_{1}, *z*_{2},…,
*z*_{n}
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 *x*^{3}
– 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.

- A very complete site which contains almost all mathematical definitions: Eric's Treasure Trove of Mathematics
- Constants
- Constructions
- Geometric Construction
- Neusis Construction
- Matchstick Construction
- Steiner Construction
- Mascheroni Construction
- Poncelet-Steiner Theorem
- Origami
- Curves
- Elements
- Geometric Problems of Antiquity
- Numbers
- Philo Line
- A very well known encyclopaedia which is now on the web: Encycloaedia Britanica
- A historical site devoted to Greec Mathematics
- Another historical site devoted to Greec Mathematics
- A very complete history site devoted to Mathematicians

[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

[ 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 ]