**Pappus on Nicomedes' cochloid (conchoid)**©2002- trans. by Henry Mendell, Cal. State U., L.A.

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Pappus of Alexandria, __Mathematical Collection__ IV §§26-29,
pp. 242.13-250.32:

Prop. 26: Construction of the first conchoid and its symptom.

Prop. 27, part 1: Claim that the curve decreases monotonically and approaches the canon asymptotically.

Prop. 27, part 2: Given an angle, a point outside the angle, and a length, to construct (with a cochloid) a line from one leg of the angle to the other with the given length and whose extension intersects the given point.

Prop. 28.: Pappus' construction with a 1st cochloid and demonstration of a double mean proportion, i.e. given a : b, to find a : c = c : d = d : b.

Prop. 29 Claim that the discovery of the double mean proportion is a solution to the problem of constructing a cube in a given ratio to a given cube.

Prop. 26 (pp. 242.1-244.20). For the doubling of the cube, a certain line was provided by Nicomedes and has this sort of generation.

(diagram 1: moving
or still) Let
there be displayed a straight-line AB and on it at right angles
GDZ, and let there be taken a certain point E which is given on
GDZ, and with point E remaining in in the place where it is, let
GDEZ be moved spiraling down straight-line ADB through point E
in such a way that D moves through every point
on straight-line AB and never falls off of it while GDEZ is spiraling
(twisting) through E. When this sort
of motion comes about on both sides, it is clear that point G
describes a line such as LGM and its symptom is like this. (diagram 2) When
some straight-line falls from point E to the line, the line taken
out between straight-line AB and line LGM is equal to straight-line
GD. For, since AB remains in place and point E remains, whenever
D arrives at H, straight-line GD will be congruent with HQ and
point G will fall on Q. Therefore, GD is equal to HQ. Similarly,
if some other straight-line falls from point E to the line, it
will make the line cut off by the line and straight-line AB equal
to GD [since the lines falling on it are equal to this line].
(diagram 3) Let
straight line AB, he says, be called the canon (ruler)
and the point a pole, and GD an interval, since the lines falling
on line LGM are equal to this line, and line LGM itself the first
cochloid (conchoid), since the second
and the third and the fourth are displayed as useful for other
theorems.

Prop. 27, part 1 (pp. 244.21-246.3) (diagram 4)

(diagram 1)Nicomedes himself demonstrates that the line can be described with instruments and that it constantly and decreasingly converges on the canon , i.e., that perpendicular GD is the largest of all the perpendiculars from any points on LGQ line to straight-line AB, while a perpendicular nearer to GA is always larger than one further away, (diagram 2)and if there is a straight-line in the place between the canon and the cochloid and it is extended then it is cut by the cochloid. And in the commentary on the sundial(?) of Diodorus, where we wanted to trisect an angle, we made use of the line under discussion.

(diagram 3) This is not difficult to show. We will prove
that straight-line PT parallel to the canon AB, when extended
intersects the cochloid LGM, no matter how near it is to the canon.
(diagram 4) Draw perpendicular TS to AB. (diagram
5) Since TS < GD, we can draw right
triangle TSR. (diagram
6) Now draw a line at E parallel to
the canon AB and copy angle TRS. One now extends this to the canon
and to the cochloid, i.e. one draws straight-line EHQ. (diagram 7) Draw perpendicular QB to AB. (diagram
8) Clearly, by parallel lines, angle
QHB = the angle at E = angle TRS. Since QH = GD = TR, in triangles
TSR, QBH, TS = QB. (diagram
9) Hence, if we may extend PT (parallel
to AB) to Q, and similarly on the other side. Hence, PT may be
extended to intersect to the cochloid. From this it is also clear
that the curve monotonically decreases on each side, since QB
decreases as the angle QHB decreases as angle DEH increases towards
a right angle.

Prop. 27, part 2 (p. 246.3-14), lemma to Prop. 28:

(diagram 1)From what was said, it is obvious that given an angle as HAB and a point outside it G it is possible to draw GH and to make KH between the line and AB equal to the given.

(diagram 2) Let a perpendicular, GQ, from point G to AB be drawn and extended, and let DQ be equal to the given , and with pole G and the interval as given, i.e., DQ, and with canon AB let a first cochloid line be drawn, EDH. Therefore, it meets with AH for the reason given. (diagram 3) Let it meet at H, and let GH be joined. Therefore KH is also equal to the given.

Prop. 28, start (pp. 246.15-18), an alternative
construction:

(diagram 4, continuation
of the lemma of Prop. 27) There are some who, for a convenience
of use, place the canon at G and move it until the line from the
limit between the straight line AB and line EDH becomes equal
to the given. The alternative construction
does not use a cochloid, but the curve is left in the diagram
to show equivalence.

Prop. 28, contintued (pp. 246.18-250.25):

(general diagram)

For when this is the case (previous lemma), what was initially proposed is proved (I mean, a cube which is double a cube is found). Earlier given two straight-lines two means in continuous proportion are taken whose construction Nicomedes merely displayed, but we also adapted the demonstration to the construction in this way.

(diagram 1) For let there be given two straight-lines GL and LA at right angles to one another whose two means in continuous proportion it is necessary to find, (diagram 2) and let parallelogram ABGL be completed, (diagram 3) and let each of AB, BG be bisected at points D, E, and let DL be joined and extended so as to meet the extension of GB at H, (diagram 4) and let EZ be at right angles to BG. (diagram 5) And let GZ be procured equal to AD, (diagram 6) and let ZH be joined and (diagram 7) GQ be parallel to it. (diagram 8) Since angle KGQ is outside given point Z, (diagram 9) let ZQK be drawn producing QK equal to AD or to GZ, since this was proved possible through the cochloid line). (diagram 10 = general diagram) And KL being joined, let it be extended so as to meet the extension of AB at M. I say that as LG is to KG, KG is to MA, and MA is to AL.

1) Since BG is bisected by E and
KG is added to it, therefore BK*KG
with the square of GE is equal to the square of EK (Euclid,
Elements ii 6). Let the square of EZ be added in common.
Therefore, BK*KG with the squares
of GE, EZ, i.e., the square of GZ,
is equal to the squares of KE, EZ,
i.e., the square of KZ. That is, BK*KG +
GZ^{2} = ZK^{2}.

2) (diagram 11) And since as MA is to AB, ML is to LK (MALMBK), but as ML is to LK, so is BG to GK (MALMLGK and AL = BG), so therefore as MA is to AB so is BG to GK. And AD is half AB, and GH is double BG. (MA : AB/2 = 2BG : GK) Therefore as MA is to AD, so is HG to KG. But as HG is to GK, so is ZQ to QK, because of parallel lines HZ, GQ (Euclid, Elements vi 2: parallels cut the sides of a triangle KHZ proportionally). Hence, MA : AD = ZQ : QK. And by ratio composition, therefore, as MD is to DA, ZK is to KQ. But AD is supposed equal to QK [since AD is also equal to GZ]. Therefore MD is equal to ZK. (note: ZQ = MA) Therefore the square of MD is also equal to the square of ZK.

3) And BM*MA with the square of DA is equal to the square of MD (Euclid, Elements ii 6).

4) But BK*KG
with the square of ZG was proved equal to the square of ZK (1), so that BM*MA
+ AD^{2} = BK*KG + GZ^{2} (by
1, 2, 3), where the square
of AD is equal to to the square of GZ (for AD is supposed equal
to GZ). Therefore BM*MA is also equal
to BK*KG. Therefore, as MB is to
BK, GK is to MA.

5) But as BM is to BK, LG is to GK (MBKLGK). Therefore as LG is to GK, GK is to AM. And also as MB is to BK, MA is to AL (MBKMAL). Therefore, as LG is to GK, GK is to AM, and AM to AL.

The basic idea behind Pappus' argument lies in the relations
(1) and (3). (2) is there merely to show that ZK = MD (given by 2), we can see
that (3) really is

(ZK+QK)*ZQ + QK^{2} = ZK^{2}

BK*GK + GZ^{2} =ZK^{2}

whence, BK*GK = (ZK+QK)*ZQ

whence, BK : (ZK+QK) = ZQ : GK

Prop. 29 (p. 250.26-32): When this is proved it is immediately clear how it is required, given a cube, to find another cube in the given ratio.

For let the given ratio be of straight-line A to B, and let two mean in continuous proportion be taken, G, D. Therefore, as A is to B so is the cube of A to the cube of G. For this is clear from the elements.