**Archimedes,**__On the Equilibria of Planes__Prop. II 3,**with the notes of Eutocius (pp. 284.24 - 286.9)**

- translated by Henry Mendell (Cal. State U., L.A.)

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of Ancient Mathematics

On the Equilibria of Planes, Table of Contents

3. If a rectilinear-figure is familiarly inscribed in each of two similar segments enclosed by a straight-line and the section of a right angled cone, but the inscribed rectilinear-figures have the their sides equal in multitude to one another, the centers of the weights of the rectilinear-figures cut the diameters of the segments similarly.(*Eutocius*)(*note 1, on Eutocius*)

(diagram 1 = general diagram) Let there be two segments, ABG, XOP, and let rectilinear-figures be inscribed familiarly in them, and having the number of all the sides equal to one another, and let BD, OR be the diameters of the segments, and let EK, ZI, HQ be joined as well as ST, UG, CY. (diagram 2) And so, since BD is divided by parallel [lines] into ratios of successive odd numbers and RO as well (II Prop. 2 intro.), and the segments of them are equal in multitude, it is clear that the segments of the diameters will be in the same ratios, and the parallels will have the same ratios. (diagram 3) And the centers of the weights of trapezoids AEKT and XSTP will be positioned similarly on straight-lines LD, WR, since AG, EK have the same ratio as XP, ST (II Prop. 15). Again the centers of the weights of trapezoids EZIK, SUFT will similarly be dividing LM, WVand the centers of the weights of trapezoids ZHQI, UCYF will be similarly dividing MN, VJ. But the centers of the weights of triangles HBQ, COY will be positioned similarly on BN, OJ.(*2*) (diagram 4) The trapezoids and triangles in fact have the same ratio. And so, it is clear that the center of the weight of the whole rectilinear-figure in segment ABG similarly divides BD as the center of weight of the figure inscribed in segment XOP divides OR (I prop 8 corollary and some work with proportions), which it was required to prove.

Commentary of Eutocius, pp. 284.24-286.9:

On the 3rd (the diagram is based on Toomer's translation of Apollonius, but rotated, *Conics* VI 11, p. 754, where AK : AP = GO : GC)

Apollonius defined similar segments of the section of the cone in the sixth book of the *Conics*, in which, when parallels to the base, equal in multitude, are drawn in each, the parallels and the bases are in the same ratio to the segments of the diameters towards the vertices, i.e., the segments to the segments (i.e.,where AZ : AQ : AK= MG : XG : OG, then IE : ZA = tL : MG and RH : QA = TN : XG, and xB : KA = ED : OG); and that all parabolas are similar (VI 11). But the familiarly inscribed figure has been discussed in the previous lemma. The reason for the diameters being divided similarly is so that their segments have the same ratio. The rest of the theorem is clear from the previously described figure.(*1*)

There are two relevant definitions in Apollonius. The first is: VI 2a (trans. from the Arabic translation by G. Toomer, p. 264): And **similar** [conic sections) are such that, when ordinates are in them to fall on the axes, the ratios of the ordinates to the lengths they cut off from the axes from the vertex of the section are equal to one another, while the ratios to each other of the portions which the ordinates cut off from the axes are equal ratios.

The second is VI 7 (trans. from the Arabic translation by G. Toomer, p. 264-6): And segments that are called **similar** are those in which the angles formed between their bases and their diameters are equal, and for which, an equal number of lines having been drawn in each of them parallel to their base, the ratios of these lines (and also the ratio of [each] base) to the lengths which they cut off from the diameter from the vertex of the section are equal for every segment; similarly the ratio of part cut off from the diameter of one to the [corresponding] part cut off from the diameter of the other [must be equal in all cases]s.

Here for comparison is my translation of the definition in Eutocius: when parallels to the base, equal in multitude, are drawn in each, the parallels and the bases are in the same ratio to the segments of the diameters towards the vertices, i.e., the segments to the segments...

Theorem VI 11 states that all parabolas are similar, not that all parabolic segments are, which certainly is not true, since the angle between the diameter and the base may vary. So it is possible that Eutocius is thinking of a notion of similar segments that omits the first condition. The third condition will hold of parabolas. In fact, as Eutocius implies, there is no need, except for emphasis, for Archimedes to state the condition, here or later in prop. II 7. It is difficult to understand the import of Dijksterhuis' comment, “... it is important that for the theorem to be true the condition of similarity of the segments (III: 2.81) is not necessary,” given that at III: 2.81, he cites the definition of Apollonius. It is not necessary that Archimedes mean by ‘similar segment’ the same as Apollonius, and it is possible that he intends the weaker sense suggested by Eutocius.

Cf. G. Toomer. *Apollonius, Conics Books V to VII: The Arabic Translation of the Lost Greek Original in the Version of Banu Musa* (New York: Springer-Verlag, 1985).

E.J. Dijksterhuis. *Archimedes*, trans. by C. Dikshoorn with a forward by Wilbur Knorr (Princeton: Princeton University Press, 1987).

(*2*) There are two ways of interpreting this step. If the proposition that justifies the step is I Prop. 12, then Archimedes must be using the Apollonius notion of similar segment (see previous note). If so, the proposition essentially depends on the segments being similar in this way, and so will the subsequent II Prop. 7. This will eviscerate the logical structure of the book, as II Prop. 8 depends on II Prop. 7 and on the weaker notion of similar segments. If, on the other hand, the step depends on I Prop. 14, the similarity of the triangles is unimportant, only a weaker sense of similarity, that the medians and the base are proportional. The fact that the earlier propositions depended on I Prop. 15 might be thought to support this view. Both propositions are stronger than what the respective steps require.

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