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On the Equilibria of Planes, Table of Contents

I Prop. 11-12
I Prop. 15

Book I Prop. 13 version 1: The center of weight of a triangle lies on the median of the triangle.
Book I Prop. 13 version 2: The center of weight of a triangle lies on the median of the triangle.
Book I Prop. 14: The center of weight of a triangle is the intersection of the medians of the triangle.
Book I Prop. 14 corollary:The center of weight cuts the median in a ratio of 2 : 1 (not in Equilibria of Planes).

Note: Footnote markers to Eutocius will have ‘e‘ as (*e1*), while notes of the translator will be plain, (*1*)

13. The center of weight of every triangle is on the straight-line which is drawn from the angle to the middle of the base.

(diagram 1 = general diagram)

(diagram 2) Let there be a triangle, ABG, and let AD in it be against the middle of base BG. One must prove that the center of weight of ABG is on AD.

(diagram 3) For otherwise, still, if it is possible, let it be Q, and (diagram 4) let QI be drawn through it parallel to BG. (diagram 5) And, in fact, what remains of the repeatedly bisected DG will eventually be less than QI. Let each of BD, DG be divided into equals, and though the cuts parallel to AD let lines be drawn and (diagram 6) let EZ, HK, LM be joined.(*e1*) They will, in fact, be parallel to BG. (diagram 7) The center of weight of parallelogram MN, in fact, is on US, the center of weight of KC on TU, and that of ZO on TD. (prop. I 9) (diagram 8) Therefore, the center of weight of the magnitude composed from all of them is on straight-line SD.(*2*) Let it be, in fact, R, and (diagram 9) let RQ be joined and extended, and let GF be drawn parallel to AD. (diagram 10) Triangle ADG, in fact, has this ratio to all the triangles inscribed up from AM, MK, KZ, ZG similar to ADG, that which GA has to AM, since AM, MK, ZG, KZ are equal.(*e3*) (diagram 11) And since triangle ADB also has the same ratio to all the similar triangles inscribed up from AL, LH, HE, EB, which BA has to AL, (diagram 12) therefore triangle ABG has this ratio to all the mentioned triangles, that which GA has to AM. (diagram 13) But GA to AM has a larger ratio than FR to RQ. For the ratio of GA to AM is the same as that of [a whole] FR to RP[since the triangles are similar](*e4*) (*5*) (diagram 14) Therefore, triangle ABG also has a larger ratio to the mentioned triangles than FR to RQ. (diagram 15) Thus, separando parallelograms MN, KC, ZO also have a larger ratio to the triangles left over than FQ to RQ. (diagram 16) And so let XQ come to be to QR in the ratio of the parallelograms to the triangles. And so since there is a certain magnitude, ABG, whose center of weight is Q, and there has been taken away from it a magnitude composed of parallelograms MN, KC, AO, and point R is the center of weight of the magnitude taken away, (diagram 17) therefore the center of weight of the remaining magnitude composed from the left over triangles is on straight-line RQ when extended and cut off as having this ratio to QR, which the taken away magnitude has to the remainder. Therefore, point X is the center of weight of the magnitude composed from those left over (I prop. 8), which is impossible. For all of them are on the same sides of the straight-line drawn through X parallel to AD [that is, on either side (of AD?)].(*e6*) (*7*) And so, what was proposed is clear.

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(*1*)Eutocius, Commentary on Plane Equilibria, 272.2-10: On 13. “And let EZ, HK, LM be joined. They will, in fact, be parallel to BG.” For since BO is equal to YG and DB to DG, as DB is to OB, DG will be to YG. But as DO is to OB, AE is to EB. For EO is parallel to AD. But as DY is to YG so is AZ to ZG. Therefore, as AE is to EB, AZ is also to ZG. Thereore, EZ is parallel to BG. The others will be also proved similarly.
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(*2*) Heiberg cites I prop. 4 to justify this step, which cannot be right, since the parallelograms are unequal. Dijksterhuis (p. 386) cites I prop. 6. One may presume that his reasoning is that all the parallelograms are commensurable as they are always measured by the first parallelogram. Nonetheless, the justification is presumably a much weaker proposition, “If the center of weights of the whole of two magnitudes is on the straight line joining the centers of the weights of the two magnitudes.” The closest proposition to this is I prop. 8, or rather a converse that is trivially provable. Suppose A and B are two weights with centers a and b, and that the whole is not on the straight line ab. Let c be the center of the whole AB and join ac. Then the center of B is on the extension of line ac on the opposite side from a, which contradicts the assumption. Of course, there may be a more basic way of proving the required proposition, e.g. directly from I prop. 6.

For absence of this lemma as casting doubt on Archimedes' authorship of at least important parts of Book 1, cf. L. Berggren, “Spurious Theorems in Archimedes' Equilibrium of Planes: Book I,” Archive for History of Exact Sciences, 16 (1976): 87-103.

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back to note 2 to the alternative version

(*3*)Eutocius, Commentary on Plane Equilibria, 272.11-25 (note the difference in Eutocius' version of the text): “ADG, in fact, has this ratio to all the triangles inscribed up from AM, MK, KZ, ZG similar to ADG, that which GA has to AM, since the straight-lines are equal.” For since triangles ADG, ASM are similar, they have double ratio to one another that AG has to AM. But since AG is now supposed quadruple AM, triangle ADG to ASM has a ratio which 16 has to one and has to all the triangles from AM, MK, KZ, ZG the ratio that 16 has to four. Therefore, there is a proportion, as ADG triangle is to the triangles from AM, MK, KZ, ZG, similar to ASM, so are the triangles themselves to ASM, that is GA to AM. For they are similar and on equal bases and for this reason equal, and they are to one another as the bases.
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Diagram 13(diagram 13)

(*4*)(diagram note 3) The bracketed text makes little sense. Basically, MW was constructed so that aI < IQ. If RF is parallel to DG, then it is obvious that GA : AM = GD : DW = FR : PR, since RFGD and RPWD are parallelograms. Otherwise, QI : aI = QR : PR, so that QR > PR. Hence, FR : PR > FR : QR. If we draw Rb parallel to DG, it is easy to see that FR : FP = Rb : cR = GD : WD = GA : AM.
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(*5*) (diagram note 4) Eutocius, Commentary on Plane Equilibria, 272.26-274.5: “But GA has a larger ratio to AM than FR to RQ. For the ratio of GA to AM is the same as that of FR to RP.” For it you conceive RF, GD as extended and coming together, because of the parallels as FR is to RP so is GD to DW. But as GD to DW, GA is to AM. Therefore, as GA is to AM, so too is FR to RP. But FR has a larger ratio to RP than FR to RQ. Therefore, GA also has a larger ratio to AM than FR to RQ.
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Diagram 13(diagram 17) (*6*)Eutocius, Commentary on Plane Equilibria, 274.6-274.12 (note the difference in Eutocius' version of the text): “which is impossible. For all the centers will be in the plane on the same sides of the straight-line drawn through X parallel to AD.” That is on the other side. And all the magnitudes will incline clearly to that side and will not equibalance, which is not supposed. For R is supposed as the center of the parallelograms, and X of the triangles.
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(*7* ) Note that the line parallel to AD through X is not part of Heiberg's diagram. This claim may not be obvious. It is enough to show that XR > FR, since FG is on one side of all the little triangles (FG is parallel to AD). Clearly, since X is the center of weight of the little triangles (i.e., ASM +...), R of the big parallelograms, and Q of the sum of the two, it follows that the large parallelograms : small triangles = XQ : QR. Componendo, the large parallelograms + small triangles : small triangles = triangle ABG : small triangles = XR : QR = FR : PR. But FR : PR > FR : QR, so that XR : QR > FR : QR. Hence, XR > FR.

(diagram 20) Since each small triangle is concave-on-the-same-sides, the center of weight of each must be inside the figure (I Assumption 7). Since the number of triangles is even, doubling from two (i.e., a power of two), just keep applying I prop. 4. Since they are all similar and equal, the pairs will be equal and the points halfway between will be inside or on triangle ABG. Similarly, the centers of each quadruple will be equal and the points halfway between them will be inside the triangle. So the center of weight of all of them must be somewhere inside triangle ABG. If there were more triangles one would just continue the process.

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The same in a different way.
(diagram 1 = general diagram)

(diagram 2) Let there be a triangle, ABG, and let AD be drawn to the middle of BAG. I say that the center of weight of triangle ABG lies on AD.

(diagram 3) For otherwise, still, if it is possible, let it be Q, and (diagram 4) let AQ, QB, QG be joined, and (diagram 5) ED, ZE, ZD(*1*) be drawn to the middles of BA, AG, and (diagram 6) EK, ZL parallel to AQ, and (diagram 7) let KL, LD, DK, DQ, MN be joined. (diagram 8) Triangle ABG is similar to triangle DZG since BA is parallel to ZD, and point Q is the center of weight of triangle ABG, and, therefore, point L is the center of weight of triangle ZDG. For point Q, L are placed similarly in each of the triangles [since they make equal angles and proportional sides. For this is obvious.] (I prop. 11) (diagram 9) For the same reasons, in fact, point K is also the center or weight of EBD. (I prop. 11) (diagram 10) Thus, the center of weight of the magnitude composed from both, triangles EBD, ZDG, is on the middle of straight-line KL [since triangles EBD, ZDG are equal] (I prop. 4). And N is the middle of KL, since as BE is to EA so is BK to QK, but as GZ is to ZA so is GL to LQ. But if this is the case, BG is parallel to KL. And DQ has been joined. Therefore, as BD is to DG, so is KN to NL. Thus, N is the center of the magnitude composed from both of the mentioned triangles. (diagram 11) But point M is also the center of weight of parallelogram AEDZ. (I prop. 10) (diagram 12) Thus the center of weight of the magnitude composed from all of them is on straight-line MN.(*2*) But, point Q is also the center of weight of ABG. Therefore, when extended MN goes through point Q, which is impossible.(*e3*) (*4*) Therefore, it is not the case that the center of weight of triangle ABG is not on straight-line AD. Therefore, it is on it.
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(*1*) ZD is not constructed, but is essential to the proof.
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(*2*) This is the same issue that arise in the first version of the proof (cf. note 2 there). Heiberg refers here to Eutocius' comment on I prop. 4, which is hardly more helpful.

Diagram of triangle ABG with center of weight(*3*) Eutocius, Commentary on Plane Equilibria, 274.14-18 (note the difference in Eutocius' version of the text): On what's in a different way of 13. “For the Q, K, L are placed similarly in the triangles.” For AQ, EK, ZL, being parallel, divide the angles similarly, and QLG, QKB are the same in all the triangles, and KD, DL are left. (Eutocius presumably means: angle KBD, KBE are the same in triangles ABG, EBD so that triangle EBK ≈ triangle AQB. Hence, BK : BQ = EK : AQ. It remains to show that BK : BQ = EK : AQ = DK : GQ. It will also be the case that BE : BA = BD : BG, since triangle EBD ≈ ABG, so that triangle KBD ≈ triangle BQ, so that BK : BQ = KD GQ. THe proof that L is similarly situated in ZDG as Q in ABG is similar). back

Diagram of triangle ABG with center of weight(*4*)Since EZLK is a parallelogram, and EM = MZ and NK = NL, MN is parallel to ZL which is constructed as parallel to AQ. So if MN goes through Q, AQ will bisect EZ and will be on the median, AD, contrary to the hypothesis.

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14. In every triangle, the point at which the straight-lines drawn from the angles to the middles of the sides of the triangle meet is the center of weight.

Diagram of triangle ABG with center of weightLet there be a triangle, ABG, and let AD be drawn to the middle of BG, and BE to the middle of AG. In fact, the center of weight of triangle ABG is one each of AD, BE. For, this has been proved. Thus, point Q is the center of weight.


 

Corollary: A corollary that is used in II Prop. 5, and in Quadrature of the Parabola 6. It is proved in Heron, Mechanica II 35, where it is attributed to Archimedes. Elsewhere in the Equilibria of Planes, I shall refer to this proposition as the corollary to II Prop. 14.

Diagram of triangle ABG with center of weightThe center of weight of a triangle cuts the median in a ratio of the segment towards the vertex to the segment towards the base as 2 to 1 (i.e. AQ : QD = 2 : 1)

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