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On the Equilibria of Planes, Table of Contents

II Prop. 1
II Prop. 3

Definition and lemmata

Proposition 2

(diagram 1 = general diagram)

familiarly inscribed polygon(diagram 2) 2. If a triangle is inscribed in a segment enclosed by a straight-line and the section of a right angled cone, that has the same base as the segment and an equal height, and again in the remaining segments triangles are inscribed, and repeatedly triangles are inscribed in the remaining segments in the same way, let the figure that comes about in the segment be said to be familiarly inscribed. (diagram 3) It is obvious that the straight-lines joining the angles of the inscribed figure, those nearest the vertex and those successively, will be parallel to the base and will be bisected by the diameter of the segment, (diagram 4) and cut the diameter into ratios of successive odd numbers with one stated for the vertex of the segment. One must prove these things in order.(*Eutocius*)(*note on theorem*)

(diagram 1 = general diagram)

If a rectilinear-figure is inscribed familiarly in the segment enclosed by a straight-line and the section of a right angled cone, the center of weight of the inscribed [figure] will be on the diameter of the segment.

(diagram 2) Let there a segment ABG as stated, and let there be inscribed familiarly in it a rectilinear-figure AEZHBQIKG. One must prove that the center of the weight of the rectilinear-figure is on BD. (diagram 3) For since the center of the weight of trapezoid AEKG is on LD, the center of trapezoid EZIK on ML, the center of trapezoid ZHQI on MN (all I prop. 9), and furthermore the center of the weight of triangle HBQ on BN (I prop. 13), (diagram 4) it is clear the center of weight of the whole rectilinear-figure is on BD.

Commentary of Eutocius, pp. 278.284.23:

On the 2nd.

Eutocius diagram Plane Equilibria II 2He prefaces the second theorem with certain clarifications, how it is possible in the section of a right-angled cone to inscribe a figure familiarly and says, “One must prove these things in order. And so, since what is said is unclear, it is necessary to say some brief things about it found in the Conics of Apollonius.

Let there be a figure enclosed by a parabola, ABG and a straight-line, AG; let BD be is diameter. It is obvious, in fact, that point B is the vertex of the segment. For Apollonius called the end-points of diameters at the lines vertices of the lines. If we join AB, BG, there will be the triangle [from] ABG having the same base as the segment and equal height, the perpendicular drawn from B to AG. For BD is not always an axis. If in fact by taking the vertices of segments AB, BG we draw EZ through them parallel to BD, as EHQ, ZJ1K, these will be diameters of segments AB, BG. For it has been proved in the case of the parabola that all [lines] drawn parallel to the diameter are diameters of the section. E, Z will, in fact, be vertices of the segments and the tangents through E, Z will be parallel to AB, BG, respectively. ELZ will also, in fact, be parallel to ADG, since EQ, ZK are parallel and are equal diameters of equal sections and fit one another, as is proved in the 6th of the Conics (*1*). And since EHQ is parallel to BD, as BH is to HA so is DQ to QA. [similar triangles ABD, AHQ] But HB is equal to AH. For diameter EH bisects it as it is parallel to the tangent. Therefore, QA is also equal to DQ. For the same reasons, in fact, DK is also equal to KG. But a whole, AD, is equal to DG. Therefore, DQ is also equal DK and for this reason EL to LZ as well. Thus he says truthfully that the [lines] joining the vertices of the segments will be parallel to the base of the segment and will be divided in two by the diameter of the segments.(*2*)

Let, in fact, AE, EB, BZ, ZG also be joined and let them be bisected at points M, N, X, O, and PMRS, TNUF, CXYW, JOQ1S1, and TA1C and P1B1G1D1E, J1, J. It is obvious, in fact, from what was just proved that TC and EZ and PJ are parallel to AG, and that TA1 is equal to A1C, and EL to LZ, and PD1 to D1J. I say that they cut BD into successive odd numbers, that is, where BA1 is of one sort of thing, A1L of three such things, LD1 of five, and D1D of seven.

Eutocius diagram Plane Equilibria II 2For since AH is equal to HB and EQ is parallel to BD, therefore AQ is also equal to QD. Therefore, AD is double DQ. Thus it is also double EL. Therefore the square on AD is also four-times the square on EL. But as the square on AD is to the square on EL, so, it has been proved, is BD to BL. Therefore, DB is also four-times BL. Therefore, DL is three-times LB. Therefore, where LB is of one sort of thing, LD is of three such things. For the same reaasons, in fact, where LB is of four sorts of things, LD is also of twelve. And since EN is equal to NB and EZ1 to Z1L and QF to FD, EL is double LZ1, that is double TA1. Therefore, the square on EL is four-times the square on TA1. Therefore, LB is also four-times BA1. Thus LA1 is three-times A1B. Therefore where LB is of four sorts of things, BA1 is of one of these sorts, but A1L is of three, and LD of twelve. Again, since AM is equal to ME and AR to RH and AS to SQ, so too are AS, SQ, QF, FD equal. Therefore, where AD is of four sorts of thing, SD, that is, PD1, is of three. Therefore, where the square on AD is of sixteen sorts, the square on PD1 is of 9 of these sorts. Therefore, where DB is of sixteen sorts, BD1 is also of nine. Therefore, a remainder, D1D, is also of seven. And so, since it has been proved that where BD is of sixteen of some sorts, BA1 is of one of these sorts, and A1L of three, and DD1 of seven, a remainder, LD1, is of five. Therefore, BD is cut by the parallels into the ratios of the successive odd numbers where one is said of the segment at the vertex. And so it is clear from the diagram that the lines drawn are cut by the diameters into numbers placed successively from a unit. For, where TA1 is of one sort of thing, EL is of two of these sorts, and P1D1 of three, and AD of four. For, as they are all parallel they divide one another into equals by making parallelograms. Figure APETBCZJG has been named by Archimedes, inscribed familiarly.


(*1*) In his edition, Heiberg cites Conics VI 19, probably from Halley's translation:

Ductis ad Axem Parabolae vel Hyperbolae normalibus, erunt segmenta à duabus quibusvis normalibus, ad utroque Axis latere abscissa, similia et aequalia; segmentum autem quodvis quius ejusdem sectionis non erit iisdem simile. (when perpendiculars are drawn to an axis of a parabola or hyperbola, the segments cut off on either side of the axis by any two such perpendiculars will be similar and equal; however, any other segment of the same section will not be similar to them.

In his edition of Apollonius, Conics V-VII (p. 584, n. 97), Toomer suggests Halley has rewritten the proposition, which he translates from the Arabic (p. 318), “VI 7 When lines are drawn in parabola or hyperbola as perpendiculars to the axis, then the two segments cut off by each pair of perpendiculars on either side [of the axis] are similar and similar in position; but as for the other segments [in that section], they are dissimilar to them [the other pairs].”

In any case, Eutocius requires a much more general theorem where the lines are ordinates and are merely parallel to the tangent at B, where BD is a diameter.


(*2*) This may be proved simply and generally from propositions in Quadrature of the Parabola (QP)

Eutocius diagram Plane Equilibria II 2Lemma: If E is the vertex of AB, and EL is parallel to AG, then DL is three-times LB.

Join EQ parallel to diameter (or parallel to the diameter BD. By QP prop. 2, AH = HB. Since QELD is a parallelogram, QD = EL. Since ADB is similar to AQH, AH : HB = AQ : QD. Hence, AQ = QD.
However, by QP prop. 3, T(AD) : T(EL) = DB : LB.

Since BAD is similar to BZ1L, AD : Z1L = DB : LB.
Hence, T(AD) : T(EL) = AD : Z1L = T(AD) : O(AD,Z1L).
Hence, T(EL) = O(AD,Z1L) or AD : EL = EL : Z1L
But EL = GD = 1/2 AD.
Hence AD : 1/2 AD = 1/2 AD : Z1L
Or AD : 1/2 AD = AD : 2 Z1L
so that 2 Z1L = 1/2 AD or Z1L = 1/4 AD
But AD : Z1L = DB : LB = 4 : 1.
Or DL : LB = 3 : 1


Eutocius diagram Plane Equilibria II 2Theorem A: If B is the vertex of segment ABG with BD its diameter and Z, E are vertices of the adjacent segments AEB, BZG, then ZE is parallel to AG.

Repeat the lemma for point Z. Draw the parallels to AG, EL with L on the diameter and ZL1 with L1 on the diameter. Clearly, DL : LB = 3 : 1 and DL1 : L1 = 3 : 1. Hence, L coincides with L1 and DLZ is a straight line.

Theorem B: The segments on the diameter cut off by corresponding vertices in a familiarly inscribed polygon are in the ratio of increasing odd numbers from 1.

The proof is similar to that of Eutocius, but uses the lemma. Basically, we show that AS = SQ = QF = FD, etc. for the parallels to the diameter drawn from each vertex to AG. In each case, base of the segment for each vertex by the parallel to the diameter, so that there will be two similar triangles, e.g., AMS and AEQ, where AM = ME, so that AS = SQ. Similarly, BEL ≈ NEU with EN = NB, so that EU = UL and QF = FD. Hence, the ratios FD : DQ : DS : DA = 1 : 2 : 3 : 4, and again by QP prop 3, BA1 : BL : BD1 : BD = T(FD) : T(DQ) : T(DS) : T(DA) = 1 : 4 : 9 : 16. Separando (see note), BA1 : A1L : LD1 : D1D = 1 : 3 : 5 : 7.


Note: one might want to prove this version of separando. If a1 : a2 : ... : an = m1 : m2 : ... : mn, then a1 : a2 - a1 : ... an - an-1 = m1 : m2 -m1 : ... : mn - mn-1.

Since ai : ai+1 = mi : mi+1, separando,

ai+1- ai : ai+1 = mi+1 - mi : mi+1

and since ai+1 : ai+2 = mi+1 : mi+2, separando,

ai+1 : ai+2 - ai+1 = mi+1 : mi+2 - mi+1

Ex aequali, ai+1- ai : ai+2 - ai+1 = mi+1 - mi : mi+2 - mi+1