**Pappus on the Quadratrix**©- trans. by Henry Mendell, Cal. State U., L.A.

Return to Vignettes of Ancient Mathematics

Pappus of Alexandria, __Mathematical Collection__ IV §§30-34,
pp. 252.26-264.2

Also pp. 54.16-22 (classification of curves), 270.26-28 (general discussion of curves), pp. 286.1-18 (cutting a circle in a given ratio), and 292.1-298.2, (three problems of finding a circular-arc and angles)

Other sources for the curve:

Proclus,

Commentary on the First Book of Euclid's Elements, p. 272.7-10 and p. 356.8-12

Iamblichus,Commentary on Aristotle's Categories, as quoted by Simplicius,Commentary on Aristotle's Categories, p. 192.18-24 andCommentary on Aristotle's Physics, p. 60.10-16

Prop. 30: Construction of the Quadratrix

Prop. 31 (first part): Objections to the curve by Sporus.

Prop. 31 (second part)-32: Basic argument for the rectification of circle with a quadratrix. Let AH be the base of the quadratrix, AB the side, and arc BED a quarter of a circumference: AH : AB = AB : arc BED.

Prop. 33: Construction of the Quadratrix from a cylindrical spiral, as the intersection of a plektoid and a plane. An analysis.

Prop. 34: Construction of the Quadratrix from an Archimedes spiral, as the intersection of a plektoid and a plane. An analysis.

NOTE: You may pause the animation
in Netscape® by clicking on it and selecting "Stop Animation"
and restart it by selecting "Reload".

Notes and additions to the text are in Blue.

prop. 30. A certain line was taken up* by Deinostratus and by Nicomedes for the squaring of the circle, as well as by some other more recent people. This gets its name from its symptom. For it is called by them quadratrix (i.e., squarer), and it has this generation.

diagram 1: still figure | generation of the curve |

Let there be displayed a
square ABGD and with center A let there be inscribed circular-arc
BED, and let AB be moved in such a way that point A remains but
B moves along the BED circular-arc, and let BG accompany it while
remaining always parallel to AD and with point B moving down BA.
And in an equal time, both let AB moving evenly over angle BAD,
i.e., point B over the BED circular-arc, converge, and let BG
travel along straight-line BA, i.e., let B move down line BA.
It clearly follows that each line, AB and BG, will coincide with
AD at the same time. When this motion occurs, straight-lines
BG, BA will intersect during the motion at some point which is
always changing together with them. A certain line which
is concave on the same side [cf. Archimedes,
On the Sphere and Cylinder, Axiom 2] is described by this
point in the location between straight lines BAD and the circular-arc
BED, namely BZH. This line is also thought to be needed
for finding a square equal to the given circle. Its principal
symptom is of this sort. For any line drawn to the circumference,
such as AZE, it will be that as the whole circular-arc is to ED,
so is straight-line BA to ZQ. For this is obvious from the
generation of the line.

Prop. 31. Sporus was, reasonably, ill pleased with it for the following reasons. First with regard to what it is thought to be a necessary matter, it assumes this in a hypothesis. For how is it possible if the two points begin to move from B, the one in a straight line to A, the other about the circumference to D to have them arrive together in an equal time without first knowing the ratio of straight-line AB to circular-arc BED? For the speeds of the motions must be in this ratio. How will it be possible to have them arrive together if they use undetermined speeds, unless it happens by sheer luck? How wouldn't this be absurd?

Next, its limit which they use for the squaring of the circle, i.e., the point at which it cuts straight-line AD is not found. Let the stated things be conceived on the proposed diagram. For whenever GB, BA move and are made to arrive together, they coincide with AD and no longer intersect one another. For the section stops before the coincidence with AD. This section then becomes a limit of the line where it falls on straight-line AD. That is, unless someone should say that he conceives the line extended to AD, in the way that we suppose straight-lines are. But this does not follow on the supposed principles, except as point H is taken with the ratio of the circular-arc to the straight line already assumed. Without this ratio being given, there is no need to trust in the opinion of the men who discovered it and to accept the line which is, anyway, very mechanical [and useful for many problems in mechanics].

But one must relate first the problem proved through it.

For historians of 17th century mathematics, this passage is very important. Descartes read it as a general Greek rejection of curves more complex than conic sections. He too was ready to reject non-algebraic curves as a part of geometry and so could both reject Greek parochialism concerning geometrical lines, while at the same time use this argument to reject curves which do not fit with his own methods, such as logarithmic curves as being a part of the science of Mechanics. This way, his geometry could be complete. An interesting question arises. Descartes says that he is quoting the Commendino's Latin translation instead of the Greek original as Latin is more familiar. However, this is surely a conceit. In this regard, a small misprint in the Latin translation is interesting. "Sporo" was misread by the printer as "spero" (I hope), and thus Sporus objecting to the curve, became "Haec autem linea spreo iure ac merito non satisfacit propter haec" (But this line, I hope, rightly and with merit was not satisfactory for these reasons). Combined with the statement that the construction was too mechnical, Descartes could spread the myth that such curves were expelled by Greek mathematicians from geometry and relegated to mechanics. Did Descartes read Book iv §33 (see below)?

On the basis of remarks by Proclus that the inventor of the curve as Hippias, Tannery proposed that the author of the curve was the fifth cent. B.C.E. sophist, Hippias of Ellis. This went almost unchallenged until the work of Wilbur Knorr. One piece of the puzzle for Tannery was the verb (paralambanô) used here by Pappus that seemed to suggest that Deinostratus (4th cent. B.C.E.) inherited the curve from earlier mathematicians, a common use of the verb. A cursory survey of Pappus, however, will satisfy the reader that Pappus uses the verb to mean 'employs' or 'take up' with no nuance that it was taken up from someone (e.g., book iii, p. 54.12-14: "Those problems which are solved when one or more sections of a cone are 'employed' for the discovery are called solid"--other examples are even less suggestive of transmission).

The basic strategy is a compression argument in the style of Archimedes. By this one proves that arc BED (1/4 circumference) : BG = BG : GQ. Hence, all one needs to do is to choose a straight-line X such that arc BED : BG = BG : X and show that X can be neither less than or greater GQ. Although the proof is straight-forward, it is much less evident how the property was discovered. Several routes are possible.

For if there is square ABGD and the circular-arc BED about
center G, and BHQ becoming a quadratrix, as was described, it
is proved that as circular-arc DEB is to straight-line BG, so
is BG to straight-line GQ. For if it is not, either it will
be to a larger than GQ or to a smaller.

Case
I: arc DEB : BG = BG : X and X > GQ

(diagram 1: standard
form)

Let it first, if possible, be to a greater GK, and let circular-arc
ZHK be described about center G and cutting the line at H, and
let there be perpendicular HL, and let GH be connected and extended
to E. (diagram
2) (1) And so since as circular-arc
DEB is to straight-line BG, so is BG, (2-3)
i.e. GD, to GK, (4) but as
GD is to GK, circular-arc BED is to circular arc ZHK (for as the
diameter of the circle is to the diameter, so is the circular-arc
of the circle the circular-arc), (5) it
is obvious that circular-arc ZHK is equal to straight-line BG.
(diagram 3) (6) And since, because of the symptom
of the line, it is the case that as BED is to ED, so is BG to
HL, (7-8) and therefore as ZHK is
to circular-arc HK, so is straight-line BG to HL. (9-10) And circular-arc ZHK was proved
equal to straight-line BG. Therefore, circular-arc HK is
equal to straight-line HL, which is absurd. Therefore, it
is not the case that as circular-arc BED is to straight-line BG,
so is BG to a larger than GQ.

Summary:

1. By
the hypothesis, arc DEB : BG = BG : GK

2. Since
BG = GD,

3. it
follows (1
and 2)
DEB : GD = GD : GK

4. Since
similar arcs are in the same ratio as the the radii, arc DEB :
arc ZHK = GD : GK

5. Hence
(3 and
4) ,
GD = arc ZHK

6. By
the symptom of the quadratrix, arc BED : arc ED = BG : HL

7. Since
(unstated) arc ZHK : arc HK = arc BED : arc ED

8. By
6 and
7, arc
ZHK : arc HK= BG : HL

9. Hence
(by 5
and 8)
GD : arc HK = BG : HL

10. By
2 and
9, arc
HK = HL

11. But,
in fact, arc HK > HL, so that if arc
DEB : BG = BG : X, it is not the case that X > GQ

(diagram 4: lemma
form)

Arc HK > straight-line HL. For
by Archimedes, *On the Sphere and Cylinder* Assumption 1 (Of lines having the same end-points the least is a straight-line), HK > straight-line
HK, while HK is the hypotenuse of right triangle HLK. Hence
HK> HL.

Case II: arc DEB : BG = BG : X and X < GQ

(figure 4)

(diagram 1: standard form)

Prop. 32. I say that it is not to a lesser either. For if it is possible, let it be to KG, and let circular-arc ZMK be inscribed about center G. And let KH be at right angles to GD cut the quadratrix at H, and let GH be joined and extended to E. (diagram 2) (1-3) As in the previously inscribed figures, (diagram 3) we will prove that circular-arc ZMK is equal to straight line BG, (4-6) and that as circular-arc BED is to ED, i.e., as ZMK is to MK, so is straight-line BG to HK. (7-8) From all of this it is obvious that circular-arc MK will be equal to straight-line KH, which is absurd. Therefore, it will not be the case that as circular-arc BED is to straight-line BG, so is BG to a smaller than GQ. It was proved that it wasn't to a larger either. Therefore it is to GQ.

Summary:

Summary:

1. By
the hypothesis, arc BED : BG = BG : GK

Since similar arcs are in the same ratio as the the radii,

2. arc
BED : arc ZMK = BG : GK

3. Hence (by 1 and 2), GD = arc ZMK

4. By
the symptom of the quadratrix, arc BED : arc ED = BG : HK

5. Hence
(by similar arcs),
arc ZMK : arc MK = arc BED : arc ED

6. Hence
(by 4
and 5)
arc ZMK : arc MK = BG : HK

7. By
3 and
6, BG
: arc MK = BG : HK,

8. Hence
(by 7)
arc MK = HK

9. But,
in fact, HK > MK, so that if arc
BED
: BG = BG : X, it is not the case that X < GQ

(diagram 3: lemma
form)

Circular-arc MK < straight-line KH.
Draw a tangent to ZMK at M and let it meet HK at N. By Archimedes,
*On the Sphere and Cylinder *Assumption 2 (if two concave lines in the same plane have the same endpoints and one of them is enclosed by the other and a straight line having the same endpoints or parts are enclosed and other parts are common, the enclosed line is shorter than the containing concave line), MNK > arc MK. But
since MN is tangent to a circle, MN is at right angles to GM,
i.e. to MH. Hence triangle NMH is a right triangle with
hypotenuse HN. Hence, HN > MN. Hence, HN + NK >
MN + NK, and HNK > MNK > arc MK.

This is also obvious, that the third straight-line taken in the proportion of straight-lines QG, GB will be equal to circular-arc BED (if GQ : GB = GB : X, then X = arc BED), and four-times it will be equal to the circular-arc of the whole circle. (4 X = the circumference of circle BED) Given that a straight-line was found equal to the circular-arc of the circle, it is clear first that it is very easy to construct a square equal to the circle itself. For area formed by the circular-arc of the circle and the line from the center of the circle [radius] is double the circle, as Archimedes demonstrated. (GB*X = 2 circle BED)

top

The basic idea of the derivation is to
start off with the definition of a general curve, of which the
quadratrix is one member. Point E is on this curve since the ratio
EZ : DG is given. One then shows that E is the projection of a
point I onto plane ABG, and that I is on a line which is the intersection
of a plane which intersects line BG and a surface (whose name
is unclear in the text, either 'cylindroid' or 'plektoid') which
is defined by rotating a line LQ parallel to plane ABG such that
L is on LB perpendicular to the plane at B and Q is on a cylindrical
spiral QHG.

(general diagram)

Prop. 33. And so, this generation of the line is, as was said, very mechanical, but it is possible to analyze it geometrically through loci on surfaces in this way:

(diagram 1) Let a fourth-part of a circle ABG be in position and let any line BD be drawn through it, and perpendicular EZ to BG with a given ratio (here 3:5) to the circular-arc DG. I say that E is on a line.

(diagram 2)
For let the surface of a right cylinder be conceived on the circular-arc
ADG, and on it a spiral inscribed and given in position GHQ, and
a side of the cylinder QD, and let EI, BL be drawn at right angles
to the plane of the circle, but QL through Q and parallel to BD.
(diagram 3)
Since the ratio of straight-line EI to circular-arc DG is given
because of the spiral, also given is the ratio of EZ to DG, hence
the ratio of EZ to EI will be given. And ZE , EI are in
position. Therefore, ZI being joined is in position.
And it is perpendicular to BG.

(diagram 4)The point here is that no matter where triangle
IEZ is positioned, it keeps the same shape and the ratios IE :
IZ : EZ : arc DG are constant, so that angle IZE remains constant.

(diagram 5) Therefore, ZI is on the intersecting plane, so that so is I. (diagram 6 and diagram 7: generation of surface, perhaps plektoid) It is also on the *** surface, [cylinderlike or, more probably, plektoid (twisted or coil-like figure) in either case we can assume that the motion of QL generates the surface] since QL moves through spiral QHG and straight line LB and since QL is given in position it is always parallel to the hypothesized plane. (diagram 8) Therefore I is on a line (i.e., I is on the intersection of the plane and the plektoid(?)), (diagram 9 = general diagram) so that E is too. (because E is a projection of I onto the plane of ABG) And so this was analyzed generally. (diagram 10) If the ratio of straight-line EZ to circular-arc DG is the same as that of BA to ADG, the previously described quadratrix line comes about.

The text has a lacuna where the curved
surface which intersects the plane is introduced "k... faneiai",
which Hultsch fills in with "kulindroeidei epifaneiai"
(cylindroid surface). Ver Ecke suggested 'plektoid' (twisted-like),
which only occurs twice in Book IV of the
*Collectio* and nowhere else in Greek, but seems more appropriate, i.e.
assuming we can read 'pi' for 'kappa'. At §34 (see below)
the cylindroid is treated as a surface of a figure on an Archimedes
spiral, while the plektoid is generated in the same way as here.

The strategy here is to construct a cylinder
and a cone which determines a line going through K. One then constructs
I from K and E and then shows that I lies on the intersection
of a plektoid generated by rotating line LQ and a plane inclined
at BG, as in the previous construction (§33).

(general diagram),

Prop. 34. It is also possible to analyze it through a spiral inscribed on a plane in a similar manner. (diagram 1) We start with two lines, AB, BG, here at angles 110 degrees; the quadratrix-like curve is drawn. For let the ratio of EZ to circular-arc DG be the same as the ratio of AB to circular-arc ADG, and (moving diagram 2 and still diagram 3) where straight-line AB moves about B and travels along circular-arc ADG, let a point on it starting from A come to be at B with AB getting position GB, and let it produce spiral BHA. (diagram 4) Therefore, it is the case that as AB is to BH, circular-arc ADG is to GD, and in alternation. But also EZ is to DG. Therefore, BH is equal to ZE. (diagram 5) Let KH be drawn at right angles to the plane and equal to BH. (diagram 6) Therefore, K is on a cylindroid surface on the spiral. (diagram 7) But also it is on a conic line (actually a conic spiral), since BK, if joined, comes to be (diagram 8) on a conic surface inclined at half a right angle to the hypothesized plane and drawn through given point B. (a 90 degree cone with axis at right angles to the plane, but side tilted at 45 deg. (diagram 9) This is because the triangle BK is always an isosceles right triangle, so that we can imagine joining BK to a perpendicular to AB at B and rotating the resulting right triangle around) (diagram 10) Therefore, K is on a line. (the intersection of the cone and the cylindroid) (diagram 11) Let LKI be drawn through K and parallel to EB, and let BL, EI be at right angles to the plane. (still diagram 12) and (moving diagram 13: generation of plektoid surface) Therefore LKI is on a plektoid surface [twisted or coil-like figure], since it moves through straight-line BL which is in position and through a line in position at where K is. I, therefore, is also on the surface. (diagram 14) But also it is on the plane, (for ZE is equal to EI, since it is also equal to BH, and ZI comes to be in position, as it is perpendicular to BG. (diagram 15) The point here is that no matter where triangle IEZ is positioned, it keeps the same shape and the ratios IE : IZ : EZ : arc DG are constant, so that angle IZE remains constant. (diagram 16) Therefore, I is on the line as the intersection of the plektoid surface and the plane, (diagram 17=general diagram) so that so is E. (diagram 18) It is clear that if angle ABG is a right angle, the previously discussed quadratrix line comes about.

Note: for those interested in the role of the diagram, note that Q (theta) is implicit in the construction, but is not mentioned in the text.