## 12 Apr 2014

### Archimedes’ [P24] ‘Quadrature of the Parabola’, Proposition 24

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Archimedes

Proposition 24 [quoting]

Proposition 24.

P24. Every segment bounded by a parabola and a chord Qq is equal to four-thirds of the triangle which has the same base as the segment and equal height.

Suppose K = 4/3ΔPQq,

where P is the vertex of the segment ; and we have then to prove that the area of the segment is equal to K.

For, if the segment be not equal to K, it must either be greater or less.

I. Suppose the area of the segment greater than K.

If then we inscribe in the segments cut off by PQ, Pq triangles which have the same base and equal height, i.e. triangles with the same vertices R, r as those of the segments, and if in the remaining segments we inscribe triangles in the same manner, and so on, we shall finally have segments remaining whose sum is less than the area by which the segment PQq exceeds K.

Therefore the polygon so formed must be greater than the area K; which is impossible, since [Prop. 23]

A + B + C+ ... + Z < 4/3A,

where A = ΔPQq.

Thus the area of the segment cannot be greater than K.

II. Suppose, if possible, that the area of the segment is less than K. (Heath 252)

If then ΔPQq = A, B = 1/4A, C = 1/4B, and so on, until we arrive at an area X such that X is less than the difference between K and the segment, we have

A + B + C + ... + X + 1/3X = 1/4A [Prop. 23]

= K.

Now, since K exceeds A + B + G + ... + X by an area less than X, and the area of the segment by an area greater than X, it follows that

A + B + C + ... + X > (the segment);

which is impossible, by Prop. 22 above.

Hence the segment is not less than K.

Thus, since the segment is neither greater nor less than K,

(area of segment PQq) = K = 4/3ΔPQq.

Archimedes. “Quadrature of the Parabola.” In The Works of Archimedes. Ed. T.L. Heath. Cambridge UP, 1897. Obtained at

https://archive.org/details/worksofarchimede00arch