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On the Equilibria of Planes, Table of Contents

II Prop. 7
II Prop. 9
II Prop. 9 (paraphrase of Eutocius)

8. The center of the weight of every segment enclosed by a straight-line and the section of a right angled cone divides the diameter of the segment to that the part of it at the vertex of the segment is half-again that at the base.

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familiarly inscribed polygon(diagram 1) Let section BG be as stated, and let its diameter be BD, and the center of weight point Q. One must prove that BQ is half-again QD. (diagram 2) Let triangle ABG, whose center of weight is E, be familiarly inscribed in segment ABG, and let each of segments AKB, BLG be bisected, and let lines KZ, HL be drawn (as in II Prop. 5, parallel to BD). (diagram 3) Therefore, they are diameters of segments AKB, BLG. And so, let the center of weight of segment AKB be M, and N that of BLG, and let ZH, MN, KL be joined. Therefore the center of the weight of the magnitude composed from both segments is C.(I Prop. 4) (diagram 4) Since as BQ is to QD, so too is KM to MZ (II Prop. 7), and componendo and alternando as BD is to KZ so is DQ to MZ, but BD is four-times KZ. For this is proved at the end where the symbol sunsign is.(*1*)(*e1*) Therefore, DQ is also four-times MZ. Thus too, a remainder, BQ, is four-times a remainder, KM, that is SC. (diagram 5) Therefore, a remainder together, BS, CQ, is three-times SC. (diagram 6) Let BS be three-times SX. Therefore, CQ is also three-times XC (since QC+SB is three-times SC). (diagram 7) Also since BD is four-times BS (for this too is proved(*e2*)) and BS is three-times SX, therefore XB is a third part of BD.(*e3*) (diagram 8) But ED is also a third part of DB, since E is the center of weight of triangle ABG.(I Prop. 14 corollary) Therefore, a remainder, XE, also is a third part of BD. (diagram 9) And since point Q is the center of the weight of the whole segment, C is the center of the weight of the magnitude composed from both segments, AKB, BLG, and E of triangle ABG, as triangle ABG is to the remaining segments, so will CQ be to QE. But triangle ABG is three-times the segments(*e4*) [since the whole segment is a third-again triangle ABG]. Therefore CQ is also three-times QE. But CQ was also proved to be three-times CX. Therefore, XE is five times EQ, that is DE of EQ. For it is equal to it. Thus DQ is six-times QE. BD is also three-times DE. Therefore, BQ is half-again QD, which it was required to prove.(*e5*)

(*1*) The symbol is standard for the sun. The reference is obscure, and ‘at the end’ suggests the end of a book. Is this one of the theorems that will be proved in order (cf. II prop. 2)? “Where the symbol (sêmeion) ‘sun’ is...” is peculiar and suggest that the lemma is marked by this symbol, itself eccentric. Heiberg reports in the notes to the text, but not in his edition of Eutocius, that in the Greek mss. Eutocius writes out 'the sun'. Peyrard (Oeuvres d'Archimède, 1807) suggests, “ainsi qu'on démontrera à la fin, à l'endroit où est la lettre theta.”

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(*e1*) Commentary of Eutocius, pp. 278.284.23:

pp. 290.21-292.15: On the 8th

familiarly inscribed polygon[note that Eutocius' quotation of Archimedes is more expansive than the text we have] “and componendo as BD is to DQ, so is KZ to ZM, and alternando, as BD is to KZ so is DQ to MZ, but BD is four-times KZ, but BD is four-times KZ, For this is proved at the end where the sign sunsignis.” We will prove it in order.

Let there be a parabola, ABG, whose diameter is BD, and let AD be drawn ordinatewise, and let AB be joined, and let AB be bisected at Z, and let EZ be drawn through Z, parallel to BD. is a diameter of segment AB. And from E, Z let EH, ZQ be drawn in parallel to the ordinate. And so, since AZ is equal to BZ, AB is double ZB, and DB double BQ, and AD double ZQ, that is EH. Thus the square on AD is four-times the square on EH, and for this reason DB is four-times BH in length. And so, since [on the one hand (Gr. men)] BD is double BQ, BQ is double BH. and QH is equal to HB, that is to EZ because EHZQ is a parallelogram. Therefore, ZE is four-times BD.

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familiarly inscribed polygon(*e2*) pp. 292.16-21: “also since BD is four-times BS. For this too is proved.” For BD has been proved in the lemma as four-times each of BH, EZ. Thus BH is equal to EZ, and for this [reason], BS is here equal to KZ and BD is four-times each of them.

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familiarly inscribed polygon(*e3*) pp. 292.21-26: “Therefore XB is a third part of BD.” (trivially different text) For since BD is four-times BS, therefore where BD is four sorts, BS is of one. Therefore, where BD is of twelve sorts, BS is also of three sorts. But BS is three-times SX. Therefore, where BS is of three sorts XS is of one, and a whole, BX is of four. But BD was of twelve of these. Therefore, BX is a third part of BD.

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familiarly inscribed polygon(*e4*) pp. 292.27-294.6: “But triangle ABG is triple the segments” (trivial variation in the text--triploun does not appear in Heiberg’s edition of Archimedes) For it has been proved by him in On the Right-angled Section of the Cone, that every figure enclosed by a straight-line and section of a right-angled cone is three-again a triangle having the same base as it and equal height. Thus segment ABG is a third-again triangle ABG, and separando triangle ABG is three-times segments AKB, BLG.

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familiarly inscribed polygon(*e5*) pp. 294.7-14: “BD is also three-times DE. Therefore, BQ is half-again QD, which it was required to prove.” For, since BD is triple DE, therefore, where BD is of fifteen sorts, ED is of five such sorts. But where DE is of five sorts, QE is of one such, and a whole, QED, is of six. Therefore, DQ is six-times QE. Therefore, where BD is of fifteen sorts, DQ is of six such and a remainder, QB, is of nine. Thus, BQ is half-again QD.

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