Return to Vignettes
of Ancient Mathematicse
II Prop. 5
II Prop. 7
6. Given a segment enclosed by a straight-line and the section of a right angled cone it is possible to inscribe familiarly into the segment a rectilinear-figure so that the straight-line between the centers of weight of the segment and the inscribed rectilinear-figure is less than any proposed straight-line.
(diagram 1) Let there be given a segment, ABG, as stated, the center of whose weight is Q, and let a triangle, ABG, be inscribed familiarly in it, and let the proposed line be Z (diagram 2) and let triangle ABG have to area C this ratio, that BQ has to Z. (diagram 3) Let a rectilinear-figure AKBLG be, in fact, inscribed familiarly in segment AKBLG, so that the remaining segments are less than C, and let E be the center of the weight of the inscribed rectilinear-figure. (I prop. 13 or 14, I prop. 15, II prop. 2 (with introduction), I prop. 8 corollary) I say, in fact, that QE is less than Z. (diagram 4) For, otherwise, either it is equal or larger. Since rectilinear-figure AKBLGG has a larger ratio to the remaining segments than triangle ABG to C, that is QB to Z, BQ also has to Z a ratio no smaller than the one that it has to QE, since QE is not smaller than Z, therefore, rectilinear-figure AKBLG has a much larger ratio to the remaining segments than BQ to QE. (diagram 5) Thus, if we make some other magnitude to QE as rectilinear-figure AKBLG to the remaining segments [since Q is the center of the weight of segment ABG, with EQ extending and some straight-line taken out having to EQ a ratio that rectilinear-figure AKBLG has to the remaining segments], it will be larger than QB. And so, let HQ have it to QE. Therefore H is the center of the weight of the segments composed from the remaining segments (I prop. 8), which is impossible. For since it is drawn through H parallel to AG on the same sides as the segment. And so, it is clear the QE is smaller than Z. It was required to prove this.
On the 6th
The center of the segment is altogether one and nearer to the vertex of the segment than those of the inscribed rectilinear figure. For E, as it happens, is the center of weight of triangle ABG is, where BD is cut in this way, so that EB is double ED. It is obvous that all the centers of the inscribed rectilinear figures fall between points Q, E, and by as much as the familiarly inscribed figure is more sided, by so much does it get nearer to Q. And so, it is obvious that it is impossible for the line between the centers of the familiarly inscribed rectilinear figureand the segment be larger than EQ, but it is possible for it to be not merely smaller than QE, but also than every given line. In fact as one can see from the diagram, the convergence is fast.