Return to Vignettes of Ancient Mathematics

(preliminary diagram: moving or still) If the pole of parallel lines [latitudes] is on the circular-arc of a great circle, and two great circles cut this [circle] at right angles, where one is one of the latitudes, while the other is inclined to the latitudes, and equal circular-arcs are taken from the inclined circle without being in succession, but on the same side of the great circle among the latitudes, and great circles are inscribed through the resulting points and the pole, then they will take out between them unequal circular-arcs from the great circle among the latitudes, and the one nearer to beginning of the great circle is always larger than one farther.

For let the pole of the latitudes on a great circular-arc of the circle ABG be point A, and let two great circles DEG, BE cut circle ABG at right angles, where BE is one of the latitudes, while DEG is inclined to the latitudes. And from inclined circle DEG let equal circular-arcs ZH, QK be taken, which are not in succession, but on the same side of great circle BE, one of the latitudes, and through points Z, H, Q, K and the pole A let great circles be inscribed, AZL, AHM, AQN, AKX. I say that circular-arc LM is larger than circular-arc NX.

top

Case 1: (general diagram 1)

(diagram 1.1) For HQ either is commensurable with ZH, QK or is not.

(diagram 1.2) Let HQ first be commensurable with ZH, QK, and let ZH, HQ, QK be divided into parts at points O, P, R, S. (diagram 1.3 = general diagram 1) And through points O, P, R, S and pole A, let there be inscribed great circles OT, PU, RF, SC.

And so, since circular-arcs ZO, OH, HP, PR, RQ, QS, SK are in succession and equal to one another, LT, TM, MU, UF, FN, NC, CX in succession are larger than one another beginning from the largest, LT. (cf. Theodosius, Sphaerica iii 6) And so, since LT is larger than NC, and TM than CX, therefore the whole LM is larger than the whole NX.

top

Case 2a: (general diagram 2)

(diagram 2.0) Let HQ not be commensurable with ZH, QK. I say that circular-arc LM is likewise larger than circular-arc NX.

For if LM is not larger than circular-arc NX, either it is less than it or it is equal.

(diagram 2.1) First, if it is possible, let LM be smaller than NX, as it holds in the second diagram.
(diagram 2.2) Let NO lie equal to LM, and through pole A and O let great circle PO be inscribed, (diagram 2.3) and since there are three unequal circular-arcs of the same kind, KQ, QP, HQ, let a circular-arc QR be taken larger than QP, but smaller than QK, and commensurable with HQ (see scholion with lemma to iii 9), (diagram 2.4 = general diagram 2) and let SH lie equal to QR (since ZH = QK and QR < QK, SH < ZH), and through points S, R and pole A let great circles ST, RU be inscribed.

And so since SH is equal to QR, and HQ is commensurable with each of SH, QR, therefore TM is larger than NU (By the commensurable case). But LM is larger than TM. Therefore, ML is much larger than NO. But it is also equal, which is impossible. Therefore, LM is not smaller than NX.

top

Case 2b: (general diagram 3)

(diagram 3.1) I say that LM is not equal to NX either.

For if it is possible, let it hold as in the case of diagram 3, (diagram 3.2) and let ZH, QK be bisected at points O, P, (general diagram 3.3 = general diagram 3) and through points O, P and through pole A, let great circles OR, PS be inscribed.

And so since ZO, OH are equal and in succession to one another, therefore LR, RM in succession with one another are larger than one another, starting with the largest LR. (cf. Theodosius, Sphaerica iii 6) Therefore LR is larger than RM. Therefore LM is more than twice MR. Again, since QP, PK are in succession with one another, NS, SX in succession with one another are larger than one another, starting with the largest NS. Therefore NS is larger than SX, so that XN is less than twice NS. And so since LM is equal to NX, where LM is more than double MR. XN is less than twice NS, therefore RM is less than NS when OH, QP are supposed equal, which is impossible (by case 2a). Therefore LM is not equal to NX.

2 RM < LM
LM = NX
NX < 2NS
Hence, RM < NS. Note that the impossibility comes from the fact that OH = 1/2 ZH = 1/2 QK = QP and case 2a.

It was proved that it is not smaller either. Therefore circular-arc LM is larger than circular-arc NX.

top