A history of Greek mathematics - Wilbourhall.org

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80 ARCHIMEDES
area. Lastly (Prop. 7), if there be two parabolic segments,
their centres of gravity divide their diameters in the same
ratio (Archimedes enunciates this of similar segments only,
but it is true of any two segments and is required of any two
segments in Prop. 8).
Prop. 8 now finds the centre of gravity
of any segment by using the last proposition. It is the
geometrical equivalent of the solution of a simple equation in
the ratio (m, say) of AG to A0, where G is the centre of
gravity of the segment.
Since the segment = § (A ABB'), the sum of the two segments
AQB, AQ'B' = ±(AABB').
Further, if QD, Q'D' are the diameters of these segments,
QD, Q'D' are equal, and, since the centres
of gravity H, H' of the segments divide
QD, Q'D' proportionally, HH' is parallel
to
QQ', and the centre of gravity of the
two segments together is at K, the point
where HH' meets A0.
Now A0 = 4AV (Lemma 3 to Prop.
2), and QD = ±A0-AV=AV. But
H divides QD in the same ratio as G
divides A (Prop. 7)
; therefore
VK = QH = m.QD = m.AV.
Taking moments about A of the segment, the triangle ABB'
and the sum of the small segments, we have (dividing out by
A V and A ABB')
§(1 + m) + §. 4 = -1 .4m,
or 15m = 9,
and m = §
That is, AG = %A0, or AG : GO
= 3 : 2.
The final proposition (10) finds the centre of gravity of the
portion of a parabola cut off between two parallel chords PP',
BB'. If PP' is the shorter of the chords and the diameter
bisecting PP', BB' meets them in N, respectively, Archimedes
proves that, if
NO be divided into five equal parts of
which LM is the middle one (L being nearer to N than M is),