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(diagram 2 = general diagram)
(diagram 1) Again, let segment BQG be enclosed by a straight-line and section of a right-angled cone, but let BG not be at right angles to the diameter. It is necessary, in fact, that either the line drawn from point B parallel to the diameter be on the same sides as the segment or that the line drawn from G make an obtuse angle with BG.(*) (diagram 2) Let it be the line making an obtuse angle at B, and let BD be drawn parallel to the diameter from B, and from G the line GD touching the section of the cone at G, and let BG be divided into equal segments, however many, BE, EZ, ZH, HI, IG, and let ES, ZT, HY, IX be drawn from E, Z, H, I parallel to the diameter, and let them be joined from the points at which these cut the section of the cone to G, and be extended. I say, in fact, that (diagram 3) triangle BDG is also smaller than (diagram 4) three-times trapezoids BF, LZ, MH, NI, and triangle GIX, (diagram 5) but larger than three-times ZF, HQ, IP and triangle GOI.
(diagram 6) Let DB be extended on the other side. And so, having drawn a perpendicular, GK, I demarcated AK as equal to GK. In fact, let there be conceived again a balance AG, and its middle K, and let it be suspended from K, and let triangle GKD be also suspended from half the balance at G, K, holding as it is now positioned, and from the other part of the balance, let areas R, C, Y, W, D, J be suspended at A, and let R incline equally to trapezoid DE, holding as it is now positioned, and C to trapezoid ZS, and Y to TH, and W to YI, and DJ to triangle GIX. In fact, the whole will incline equally with the whole.Thus triangle DBG would also be three-times area RCYWDJ. (Prop. 7) (diagram 7) In fact, trapezoid BF will be proved, similarly to that previously, to be larger than area R (Prop. 11), (diagram 8) and trapezoid WI will be proved to be larger than area C and less than ZF, and trapezoid MH to be larger than area Y and smaller than HQ (Prop. 13), and furthermore trapezoid NI to be larger than area W and smaller than PI, and triangle XIG to be larger than area DJ and smaller than GIO (Prop. 9). (diagram 9) And so it is clear. Namely, triangle BDG is more than three times the inscribed figure and less than three times the circumscribed figure. For details see the proof of Prop. 14 and substitute the odd numbered propositions in the justifications for even numbered ones.
(*) Archimedes point is that whenever one side has an acute angle the other side has an obtuse angle. So it makes no difference how we orient the parabola on the balance. We could just flip it around. The alternative would be to put the parabola on the left side of the balance, a trivial variation.
The basic idea will be to compress the inner and outer trapezoids and triangle into the parabola (requires html5 browser support):