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On the Equilibria of Planes, Table of Contents

II Prop. 6
II Prop. 8

7. The centers of the weights of two similar segments enclosed by a straight-line and the section of a right angled cone cut the diameters in the same ratio.

(general diagram)

familiarly inscribed polygon(diagram 1) Let there be two segments in the way stated, ABG, EZH, whose diameters are BD, ZQ, and let the center of the weight of segment ABG be point K, and of EZH point L. One must prove that K, L cut the diameters in the same ratio. (diagram 2) For otherwise, as KB is to KD, so let ZM be to MQ,(*1*) (diagram 3) and let a rectilinear-figure be inscribed in segment EZH, so that the line between the center of the segment and the inscribed rectilinear-figure is less than LM, and let point X be the center of the weight of the inscribed rectilinear-figure, (diagram 4) and let a rectilinear-figure be inscribed in segment ABG similar to that [rectilinear figure inscribed] in EZH[, that is similarly familiarly].(*e1*)(*2*) The center of its weight is nearer to the vertex than that of the segment,(*3*) which is impossible (II Prop. 5). And so it is clear that BK has the same ratio to KD that ZL has to LQ.

Commentary of Eutocius, pp. 290.11-.290.30:

On the 7th

familiarly inscribed polygon“And let a rectilinear-figure be inscribed in segment ABG similar to that in EZH.”

That is similarly familiarly. For when the sections of parabola ABG become equal to those of EZH, so that the sides of the figure inscribed familiarly in segment ABG are equinumerous with those in the rectilinear figure inscribed in EZH. For since points B, Z, in fact, are the vertices of similar segments, the figures inscribed familiarly in this way are similar.(*2*)



familiarly inscribed polygon(*1*) Let the center of the figure familiarly inscribed in segment ABG be N. Clearly, if MZ < ZL and BK : KD = MZ : ZQ, then we can pick a point on BD, say P, such that BP : PD = ZL : LQ, then BP > BK.

Brief proof: componendo BK : BD = MZ : ZQ, where MZ < ZL. So BK : BD < ZL : ZQ. So too BP : BD = ZL : ZQ. Hence, BK : BD < BP : BD, so that BK< BP. This merely shows that we can pick the counter-factual point between the center of weight and the base. If the situation isn't in one segment, it will be in the other.



familiarly inscribed polygon(*2*) If one takes the notion of similar segments from Apollonius, Conics VI, it will follow that the familiarly inscribed figures will be similar. The bracketed text is probably to be explained as the creeping of Eutocius' notes into Archimedes' text. There is no need for the familiarly inscribed figures to be similar. They need only have the same number of sides. And this would seem to be the point of Eutocius' gloss on the text, that 'similar' implies 'familiarly inscribed in the same way'. Is this the sense of similar in the text (eggegrafqô de eis to ABG tmama tôi en tôi EZH omoion euqugrammon)? One is tempted to emend omoion to omoiôs and possibly to transpose it nearer to the verb: let a a rectilinear be inscribed similarly to that in EZH.

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familiarly inscribed polygon(*3*) BN : BD = ZX : ZQ (by II Prop. 3). By the hypothesis, BK : BD = ZM : ZQ. By the construction XL < ML. And we have also supposed (cf. note 1) that ZM > LM.

MZ > XZ > LZ (by II Prop. 5 and hypothesis). BN : BD = ZX : ZQ < ZM : ZQ = BK : BD, so that BN : BD < BK : BD. Hence, BN < BK.