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Pappus of Alexandria, Mathematical Collection vi §§1-11, pp. 474.1-488.25

Contents:
Lemmata:

These concern Menelaus trilaterals (spherical triangles whose sides are circular-arcs of great circles less than a semicircle).

Prop. 1: Any two sides of a trilateral > the third side.

Prop. 2: Let a trilateral be erected on the same base as another trilateral with it vertex inside the vertex of the other. Then the two sides from the vertex of the inside trilateral on the same base < the two sides of the outer.

Prop. 3: Let a circular-arc be drawn from the vertex to the bisection of the base of a trilateral. The two sides from the vertex > 2 the this circular-arc.

Prop. 4: Let two arcs be drawn from the vertex of a trilateral to points on the base equidistant (in arcs) from the end points of the base. These form a trilateral inside a larger trilateral. The two sides from the vertex of the larger trilateral are larger than the two sides of the smaller.

Theorems:

The following theorems use this figure: A great-circle through the poles of two circles, one of which is an equator parallel to some latitudes and the other is at an angle to the equator. The initial great circle is at right angles to the others. We can call this (from its astronomical use), a colure-equator-ecliptic configuration. We will only be concerned with the quadrant between the intersection of the ecliptic/equator and the colure as marked.

Prop. 5: (= Theodosius, Sphaerica, iii 5) Let two adjacent equal arcs be marked off on the ecliptic and let latitudes be drawn to the colure. These mark off two unequal arcs on the colure, where the one nearer the equator is larger than the one further away.

Prop. 6: (not in Theodosius, Sphaerica) Let two non-adjacent equal arcs be marked off on the ecliptic and let latitudes be drawn to the colure. These mark off two unequal arcs on the colure, where the one nearer the equator is larger than the one further away.

Prop. 7: Second proof of Prop. 6 (props. 7-9). Step in the proof of the case of Prop. 6 where the arcs marked off on the ecliptic are commensurable with the arc between them.

Prop. 8: Step in the proof of case of Prop. 6 where the arcs marked off on the ecliptic are incommensurable with the arc between them, to show that the arc on the colure nearer the equator is not smaller than the arc further away.

Prop. 9: Step in the proof of case of Prop. 6 where the arcs marked off on the ecliptic are incommensurable with the arc between them, to show that the arc on the colure nearer the equator is not equal to the arc further away.

Prop. 10: If the adjacent arcs marked off on the colure are equal, then the arc marked off on the ecliptic nearer to the equator is smaller than the one further away.

Prop. 11: If the arcs on the ecliptic are unequal with the arc nearer the equator being larger, then the arc on the colure nearer the equator will be much larger than the one further away.


 

 

Pappus of Alexandria, Collection 6

This contains solutions of difficulties in the Small Collection of Astronomy.

 

Preamble (p. 474.3-14)

Many of those who teach the subject of astronomy, when they hear the propositions less mindfully add some things as necessarily true and omit others as unnecessary. For they say of Theodosius, Sphaerics book six, theorem six, that it is required that each of two great circles that intersect one another be cut by the circle through the poles at right angles. But this is not always the case. Similarly, they omit in Euclid, Phaenomena, theorem 2, how many times the zodiacal circle will be at right angles to the horizon. Although they give a fallacious account of Theodosius for On Nights and Days, theorem 4, they even omit several other things which follow as not being necessarily the case, for each of which we will present a proof.


 

 

 

(general diagram)

Prop. 1 (pp. 474.15-476.17). If three circular-arcs of a great circle on a sphere intersect one another, where each is smaller than a semicircle, two of them are larger than the remaining one, however they are taken.

(diagram 1) For let the circular-arcs of great circles intersect one another at points A, B, G. I say that two of them are larger than the remaining one, however they are taken.

(diagram 2 = general diagram) For let the center of the sphere be taken, and the same point is also the center of circular-arcs AB, BG, GA, and let it be A. And let DA, DB, DG be joined. (diagram 3) And so since a solid angle, at D, is enclosed by 3 planes, ADB, BDG, GDE, two of them are larger than the remaining one, however they are taken (Elements xi 20). (diagram 4) And angles ADB, BDG, GDA stand on circular-arcs ADB, BDG, GDA. Therefore, two of them are larger than the remaining one, however they are taken.

In the Sphaerics, Menelaus calls this sort of figure a trilateral.

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(general diagram)

Prop. 2 (p. 476.18-31).. If two circular-arcs of a great circles are erected on one side of a trilateral and inside it, the erected circular-arcs will be smaller than the two remaining sides of the trilateral.

(diagram 1) For let two circular-arcs of a great circles are erected on one side, BG, of a trilateral, ABG, and inside it, the erected circular-arcs will be smaller than the two remaining sides of the trilateral. (diagram 2 = general diagram)I say that BDG is smaller than circular-arcs BAG.

(diagram 3) Since two sides of every trilateral are larger than the remaining side, therefore GE, ED are larger than GD. (diagram 4) Let DB be added in common. Therefore, circular-arcs GEB are larger than GDB. (diagram 5) Again since two sides of every trilateral are larger than the remaining side, therefore circular-arcs BAE are larger than EB. (diagram 6) Let EG be added in common. Therefore, circular-arcs BAG are larger than BEG. But BEG are larger than BDG. Therefore, BAG are much larger than BDG.

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(general diagram)

Prop. 3 (p. 478.1-21)

(diagram 1) Let circular-arcs of three great circles, AB, AG, AD, intersect circular-arc BD of a great circle, and let each of AB, AG, AD be smaller than a fourth-part, but let BG be equal to GD. To prove that BAD together is larger than twice AG.

(diagram 2) Let GE be placed equal to AG. Since AG is smaller than a fourth-part, GE is also smaller than a fourth-part. Therefore, AE is smaller than a semicircle. (diagram 3) Therefore, circle AD, when filled out, will not come through E. (diagram 4 = general diagram) And so let great circle EDZ be inscribed through points E, A, (diagram 5) and since DG is equal to GB, while AG is equal to GE, therefore the line from D to E is equal to the line from A to B. Therefore circular-arc BA is equal to circular-arc DE. (diagram 6) But since two sides of every trilateral are larger than the remaining side, but DE is equal to AB while EG is equal to GA, therefore BAD together is larger than twice AG.

AD + DE > AGE
AD + AB > 2AG

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(general diagram)

Prop. 4 (pp. 476.18-480.6).

(diagram 1) Let circular-arcs of four great circles, AB, AG, AD, AE intersect circular-arc BE of a great circle, and let BG be equal to DE, and let each of AB, AG, AD, AE be smaller than a fourth-part. To prove that BAE is larger than GAD.

(diagram 2) Let GD be bisected at D, and let great-circle AZH be inscribed through A, Z, and let ZH be placed equal to AZ, (diagram 3) and let great circle HEK be inscribed through H, E, with great circle HDQ inscribed through H, D. (diagram 4) And since HZ is equal to ZA, with DZ equal to ZG, therefore, DH is also equal to GA. (diagram 5) For the same reasons, EH is also equal to EA. (diagram 6) Since two circular-arcs, AD, DH, were erected on one side, HA, of trilateral HEA and inside it, therefore circular-arcs ADH are larger than circular-arcs AZH, so that AEH are larger than ADH (theorem 2). (diagram 7) But EH is equal to AB, with HD equal to AG. Therefore, BAE together is larger than GAD, which it was required to prove.

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(general diagram)

Prop. 5. (pp. 480.7-482.8) Given these as previously proved, let it be required to prove theorem 5 of book 3 of the Sphaerica of Theodosius in a different way.

(diagram 1) For let point A on the circular-arc of great circle ABG be the pole of latitudes, and let two great circles cut this circle at right angles, where BG is one of the latitudes and EZ is inclined to the latitudes, (diagram 2) and let equal circular-arcs in succession on the same side, HQK, be taken from EZ (diagram 3) and through points H, Q, K let there be inscribed circles parallel to BG, namely MN, XO, PR; (diagram 4) to show that PX is larger than MX.

(diagram 5 = general diagram) For let there be inscribed through A and each of K, H, Q great circles AK, AQ, AH. It is in fact obvious that each of AK, AQ, AH circular-arcs is smaller than a fourth-part (since the circular-arc from A to great circle BG is a fourth-part). (diagram 6) And so since when the circular-arcs of three great circles, AK, AQ, AH intersect the circular-arc of great circle EZ, and KQ is equal to QH, each of AK, AQ, AH is smaller than a fourth part, therefore because of what was previously proved (theorem 3) KAH together is more than twice as large as AQ (diagram 7) where KAT is twice AS (for the three, AS, AK, AT, which go through the pole, are equal to one another). Therefore, the remainder TH is more than twice as large as SQ. But SQ is equal to TU. Therefore HU is larger than TU. But HU is equal to PX, and UT to XM. Therefore PX is larger than XM, which it was required to prove.

AK + ATUH > 2 ASQ
AK + AT = 2 AS
Hence, TUH > 2 SQ
TU = SQ
Hence, HU > SQ
XN = SQ and PS = HU
Hence, MX > PX

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(general diagram)

Prop. 6 (p. 482.9-22)

(diagram 1) Let it be required to prove this when the equal arcs are not continuous (for Theodosius did not prove this), and let the figure be the same, but let the equal circular-arcs be HQ, KL, and let the parallel circles be MN, XO, PR, ST, (diagram 2 = general diagram) and let there be inscribed through A and each of H, Q, K, L, great circles, AH, AQ, AK, AL. They will be smaller than a fourth-part. (diagram 3) And because of theorem 4 above, LAH together will be larger than KAQ together, (diagram 4) and LAC together is equal to UAF together (for they are lines from the pole of circle MN). Therefore the remainder CH is larger than FQ, UK together. (diagram 5) But FQ is equal to CY. Therefore the remainder YH is larger than UK. But YH is equal to SP, while UK is equal to MX. Therefore SP is larger than MX, which it was required to prove.

AL + ACYH > AUK + AFQ
AL + AC = AU + AF
Hence, CYH > UK + FQ
CY = FQ
Hence, YH > UK
MX = UK and SP = YH
Hence, SP > MX

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(general diagram)

Prop. 7 (pp. 482.23-484.20)

(diagram 1) Let it be now required to prove the same thing in a different way. For let the pole of latitude lines be on the ciruclar-arc of great circle ABG, and let two great circles DE, BG intersect this at right angles, where BG is one of the latitudes and DE is inclined to the latitudes, and from DE let there be equal circular-arcs ZH, QK, and let there be inscribed latitude circles LM, NX, OP, RS. I say that OR is larger than NL.

(diagram 2) For ZH is either commensurable with HQ or it isn't. Let it first be commensurable. But HZ is equal to QK. And QK is therefore commensurable with QH. Therefore, the three, ZH, HQ, QK, are commensurable with one another. (diagram 3 = general diagram) And so, let them be divided into the measures at T, U, F, C, Y, and let there be inscribed through T, U, F, C, Y latitude circles, WT, A1U, FB1, CG1, YD1. (diagram 4) And since ZT, TU, UH, HF, FQ, QC, CY, YK are circular-arcs equal to one another, therefore RW, WA1, 1O, OB1, B1N, NG1, G1D1, DL are unequal and intitially starting from the largest RW (by prop. 5). And the number of RW, WA1, A1O, is equal to the number of NG1, G1D1, D1L. Therefore RO is larger than NL.

ZT = TT = UH = HF = FQ = QC = CY = YK
Hence, RW > WA1 > A1O > OB1 > B1N > NG1 > G1D1 > D1L
Hence, RW + WA1 + A1O > NG1 + G1D1 + D1L
or 3 RW > 3 NG1
or RO > NL

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(general diagram)

Prop. 8 (pp. 482.23-484.20)

(diagram 1) But with the same things supposed, let ZH not be commensurable with HQ. I say that here too RO is larger than LN.

(diagram 2) If not, it is either equal or smaller. Let it first be smaller, and let N1G be placed equal to RO, (diagram 3) and with three lines of the same kind, LN, NG1, NO, let ND1 be taken a line commensurable with NO, larger than NG1, and smaller than NL, and let it be ND1. (diagram 4 = general diagram) or (close-up diagram) And let there be latitude circles, XG1, YD1, and let HT be placed equal to YQ, and let TW be a latitude. (close-up diagram 5) And so, since each of YQ, HT is commensurable with HQ, WO is larger than ND1 (by prop. 7). Therefore, RO is much larger than ND1. But RO is equal to NG1. Therefore NG is larger than ND1, the smaller than the larger, which is impossible. Therefore, RO is not smaller than NL.

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(general diagram)

Prop. 9 (pp. 482.23-484.20): the argument would seem to be a little garbled. In the Greek text, it doesn't follow that RO > 2 ND1, only that RO > 2 D1L, while, in his commentary, Commandino seems to assume Prop. 6 (as does ver Eecke explicitly), which is the theorem to be proved. In any case we expect the theorem to use Prop. 8. A simple emendation will solve the problem.

(diagram 1) With the same things supposed, I say that it is not equal either. For if it is possible, let it be, (diagram 2 = general diagram) and let HZ, QK be bisected at T, Y, and let there be latitude circles TW, YD1. (diagram 3: close-up) And so, since TZ, TH are equal, therefore RW, WO are unequal starting from the largest, RW (Prop. 5).Again, since QYK are equal, therefore ND1, D1L are unequal starting from the largest, ND1 (Prop. 5)

(a textual problem begins here; see below) And so, since RW is larger than WO, but ND1 is larger than D1L, therefore RO is more than double ND1 (why?), which is impossible (for it was previously proved that ND1 > D1, so that NL < 2 ND1). Therefore, RO is not equal to NL. But it was proved that it isn't smaller either. Therefore RO is larger than NL.

Commandino's comment on the last part, after quoting the codex:
Some things seem to be lacking in this place. Therefore, I should think it is to be proved in this way (adjusting for this lettering). And so since ND1 is larger than D1L, therefore NL is less than twice as large as ND1. Again since RW is larger than WO, but ND1 is larger than D1L, but WO is also larger than ND1, as was proved above (does he mean vi 6 or 8, which doesn't prove this?), therefore RO is larger than double ND1. But NL is equal to RO. Therefore NL is larger than double that ND1, which is impossible, since is was proved to be smaller.

Suggest text(diagram 4: close-up): And so, since RW is larger than WO but ND1 is not larger than OW (by theorem 8), therefore RO is more than double ND1, which is impossible (for it was previously proved that ND1 > D1, so that NL < 2 ND1). Therefore, RO is not equal to NL. But it was proved that it isn't smaller either. Therefore RO is larger than NL.
RO > 2 WO
WO ND1
Hence, RO > 2 ND1

Codex (p. 486.18-21):
Suggested emendation:

Another alternative would be to fill in the lacuna after "it was previously proved" as
"for it was previously proved that ND1 is not larger than WO" and to understand the rest of the argument along the same lines.

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(general diagram)

Prop. 10 (pp. 482.23-484.20)

(diagram 1) Again, let the pole of latitudes be on a great circle, and let BG, DE cut it at right angles, and let there be latitude circles KL, MN, XO and let XM be equal to MK. I say that ZH is smaller than HQ.

For if not, it is either equal or larger. And so, ZH is not equal to HQ. For XM would then be larger than MK. But it isn't. Therefore ZH is not equal to HQ. I say that it is not larger either. (diagram 2 = general diagram) For if it is possible, let it be and let HP be placed equal to QH. And so since HP is equal to HQ, therefore RM is larger than MK. Therefore XM is much larger than MK, which is impossible. For it was supposed equal. Therefore ZH is not larger than HQQ. But is was proved that it is not equal either. Therefore ZH is smaller than HQ, which it was required to prove.

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(general diagram)

Prop. 11 (pp. 482.23-484.20)

(diagram 1) And so it was proved that if there is a circle, ABG, and two great circles, BG, DE, cut it at right angles, and there equal circular arcs are taken, AH, HQ, and latitude circles are taken, KL, MN, XO, then XM becomes larger than MK. Let ZH be larger than HQ. I say that XM is much larger than MK.

(diagram 2 = general diagram) For since ZH is larger than HQ, let HP be placed equal to HQ, and let a latitude circle, PR, be inscribed. And so, since HP is equal to HQ, RM is larger than MK. Therefore XM is much larger than MK, so that if ZH is larger than QH, then XM also becomes larger than MK, which it was required to prove.

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