Return to Vignettes of Ancient Mathematics

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 two arbitrary points are taken on the same side of the great circle among the latitudes, and great circles are inscribed through the resulting points and the pole, it will be the case that as the circular-arc of the great circle among the latitudes which is between the initial great circle and the next one through the poles is to the circular-arc of the inclined circle between the same circles, so the next circular-arc of the great circle among the latitudes between the great circles through the pole and the points taken is to some circular-arc smaller than the circular-arc of the inclined circle between the two taken points.

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 on the inclined circle DEG let there be taken two points Z, H on the same side of the great circle among the latitudes BE. And through points Z, H and pole A, let great circles AZQ, AHK be inscribed. I say that it is the case that as circular-arc BQ is to circular-arc DZ, so is circular-arc QK to some circular-arc smaller than circular-arc ZH.
For either ZH is commensurable with DZ or it is not.

top

Case 1: Let it first be commensurable, and let DZ, HZ be divided into parts at points L, M, N, and through L, M, N, and pole A let there be inscribed great circles LX, MO, NP.

And so, since DL, LM, MZ, ZN, NH are in succession and are equal to one another, therefore BX, XO, OQ, QP, PK are successively larger than one another starting from the largest BX. And so since BX, XO, OQ, QP, PK are successively larger than one another, but DL, LM, MZ, ZN, NH are in succession and equal to one another, and the number of BX, XO, OQ is equal to the number of DL, LM, MZ, while the number of QP, PK is equal to the number of ZN, NH, therefore BQ to DZ has a larger ratio than QK to ZH. Therefore if we make it that as BQ is to DZ, so is QK to some other circular-arc, it will be to one smaller than ZH. Therefore, as circular-arc BQ is to circular-arc DZ, so is QK to some circular-arc smaller than ZH.

top

Case 2: Let ZH not be commensurable with DZ. I say that it is likewise the case that as circular-arc BQ is to circular-arc DZ, so is circular-arc QK to some circular-arc smaller than circular-arc ZH.

For if it isn't, either it is to one larger than ZH or to it.

Case 2a: First, if it is possible, let it be to one larger than ZH, LZ, as it holds in the case of diagram 2. And since there are three unequal arcs, LZ, ZH, ZD, let there be taken some circular-arc ZM smaller than ZL, larger than ZH, and commensurable with ZD, and through M and pole A let great circle MN be inscribed.

And so, since ZM is commensurable with ZD, it is therefore the case that as BQ is to DZ, so is QN to some circular-arc smaller than ZM. But as BQ is to DZ so is QK to ZL. Therefore as QK is ZL so is QN to a circular-arc smaller than ZM. And by alternate proportion. Therefore as QK is to QN, so is ZL to a circular-arc smaller than ZM. But QK is smaller than QN. Therefore LZ is also smaller than ZM. But it is also larger, which is impossible. Therefore it is not the case that as BQ is to DZ so is QK to some circular-arc larger than circular-arc ZH.

top

I say that it is not to it either.

Case 2b: For if it is possible, let it be that as BQ is to DZ, so is QK to ZH, as it holds in the third diagram, and let each of DZ, ZH be bisected at points L, M, and through each of points L, M and pole A, let great circles LN, MX be inscribed.

And so, since DL, LZ are in succession and equal to one another, therefore BN, NQ are successively larger than one another starting from the largest BN. Therefore, BQ is more than double QN. In fact, we will similarly prove that KQ is also less than double QX. And so, since BQ is more than double QN, and KQ is less than double QX, therefore BQ has a larger ratio to QN than KQ to QX. And by alternate proportion. Therefore BQ to QK has a larger ratio than NQ to QX. But as BQ is to QK, so is DZ to ZH. Therefore NQ to QX has a smaller ratio than DZ to ZH. But as DZ is toZH, so is LZ to ZM. Therefore NQ to XQ has a smaller ratio than LZ to ZM. And by alternate proportion. Therefore NQ to LZ has a smaller ratio than QX to ZM. Therefore, if we make it that as NQ is to LZ so is QX to some other, it will be to one larger than ZM, which was proved impossible. Therefore, it is not the case that as BQ is to DZ, so is QK to ZH.

BQ > 2QN
KQ < 2QX
BQ : QN > KQ : QX
BQ : KQ > QN : QX
BQ : DZ = KQ : ZH (hypothesis)
BQ : KQ = DZ : ZH
DZ : ZH > QN : QX
DZ : ZH = 2*LZ : 2*ZM = LZ : ZM > QN : QX
QX : ZM > QN : LZ
QX : x = QN : LZ, where x > ZM, which is impossible by case 2a.

But it was proved that it is not to a larger either. Therefore, it is to a smaller. Therefore, it is the case that as BQ is to DZ, so is QK to a circular-arc smaller than circular-arc ZH.

top