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Archimedes
“The Method of Archimedes Treating of Mechanical Problems- To Eratosthenes”
Thomas L. Heath (Ed., Trans.)
Proposition 1
[The following is quotation, except for double bracketed material]
Let ABC be a segment of a parabola bounded by the straight line AC and the parabola ABC, and let D be the middle point of AC.
Draw the straight line DBE parallel to the axis of the parabola and join AB, BC.
Then shall the segment ABC be 4/3 of the triangle ABC.
From A draw AKF parallel to BE, and let the tangent to the parabola at C meet BBE in E and AKF in F.
Produce CB to meet AF in K, and again produce CK to H, making KH equal to CK.
Consider GH as the bar of a balance, K being its middle point.
Let MO be any straight line parallel to ED, and let it meet CF, CK, AC in M, N, O and the curve in P.
Now, since CE is a tangent to the parabola and CD the semi-ordinate,
EB = BD;
" for this is proved in the Elements [of Conics]*." [* i.e. the works on conies by Aristaeus and Euclid. Cf. the similar expression in On Conoids and Spheroids, Prop. 3, and Quadrature of Parabola, Prop. 3.] [[My note: Elements of Conics is lost, and the other two references Heath makes are to propositions that as well refer to the proofs in the Conics]].
Since FA, MO are parallel to ED, it follows that
FK = KA,
MN = NO.
Now, by the property of the parabola, " proved in a lemma,"
MO : OP = CA : AO [Cf. Quadrature of Parabola, Prop. 5]
= CK : KN [Eucl. VI. 2]
= HK : KN.
Take a straight line TG equal to OP, and place it with its centre of gravity at H, so that TH = HG;
then, since N is the centre of gravity of the straight line MO,
and
MO : TG = HK : KN,
it follows that TG at H and MO at N will be in equilibrium
about K. [On the Equilibrium of Planes, I. 6, 7] [[P6, 7. Two magnitudes, whether commensurable or incommensurable, balance at distances reciprocally proportional to the magnitudes.]]
Similarly, for all other straight lines parallel to DE and meeting the arc of the parabola, (1) the portion intercepted between FC, AC with its middle point on KC and (2) a length equal to the intercept between the curve and AC placed with its centre of gravity at H will be in equilibrium about K.
Therefore K is the centre of gravity of the whole system consisting (1) of all the straight lines as MO intercepted between FC, AC and placed as they actually are in the figure and (2) of all the straight lines placed at H equal to the straight lines as PO intercepted between the curve and AC.
And, since the triangle CFA is made up of all the parallel lines like MO, and the segment CBA is made up of all the straight lines like PO within the curve, it follows that the triangle, placed where it is in the figure, is in equilibrium about K with the segment GBA placed with its centre of gravity at H. [[Again see above figure]]
Divide KC at W so that CK = 3KW;
then W is the centre of gravity of the triangle ACF ; " for this is proved in the books on equilibrium ". [Cf. On the Equilibrium of Planes I. 15]
Therefore ΔACF : (segment ABC) = HK : KW
= 3 : 1.
Therefore segment ABC = 1/3ΔACF.
But ΔACF = 4ΔABC.
Therefore segment ABC = 4/3ΔABC.
"Now the fact here stated is not actually demonstrated by the argument used; but that argument has given a sort of indication that the conclusion is true. Seeing then that the theorem is not demonstrated, but at the same time | suspecting that the conclusion is true, we shall have recourse to the geometrical demonstration which I myself discovered and have already published*." (Heath 1|2)
Pages from the text:
Archimedes. “The Method of Archimedes Treating of Mechanical Problems- To Eratosthenes”. In The Method of Archimedes, Recently Discovered by Heiberg: A Supplement to the Works of Archimedes 1897. Ed. Thomas L. Heath. Cambridge: Cambridge UP, 1912.
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