## Archimedes of Syracuse

Source: The Works of Archimedes, Sir Thomas L. Heath, ed., Dover, 1953, pp. 233-235, 248-252.

### Quadrature of the Parabola

Archimedes to Dositheus greeting. 1
When I heard that Conon, who was my friend in his lifetime, was dead, but that you were acquainted with Conon and withal versed in geometry, while I grieved for the loss not only of a friend but of an admirable mathematician, I set myself the task of communicating to you, as I had intended to send to Conon, a certain geometrical theorem which had not been investigated before but has now been investigated by me, and which I first discovered by means of mechanics and then exhibited by means of geometry. 2   Now some of the earlier geometers tried to prove it possible to find a rectilineal area equal to a given circle and a given segment of a circle; and after that they endeavoured to square the area bounded by the section of the whole cone 3  and a straight line, assuming lemmas not easily conceded, so that it was recognized by most people that the problem was not solved.  But I am not aware that any one of my predecessors has attempted to square the segment bounded by a straight line and a section of a right-angled cone [a parabola], of which problem I have now discovered the solution.  For it is here shown that every segment bounded by a straight line and a section of a right-angled cone [a parabola] is four-thirds of the triangle which has the same base and equal height with the segment 4, and for the demonstration of this property the following lemma is assumed: that the excess by which the greater of (two) unequal areas exceeds the less, can by being added to itself be made to exceed any given finite area.  The early geometers have also used this lemma; for it is by the use of this same lemma that they have shown that circles are to one another in the duplicate ratio of their diameters, and further that every pyramid is one third part of the prism which has the same base with the pyramid and equal height; also that every cone is one third part of the cylinder having the same base as the cone and equal height they proved by assuming a certain lemma similar to that aforesaid.  And, in the result, each of the aforesaid theorems has been accepted no less than those proved without the lemma.  As therefore my work now published has satisfied the same test as the propositions referred to, I have written out the proof and sent it to you, first as investigated by means of mechanics, and afterwards too as demonstrated by geometry.  Prefixed are, also, the elementary propositions in conics which are of service in the proof.  Farewell.

Proposition 1. 5If from a point on a parabola a straight line be drawn which is either itself the axis or parallel to the axis, as PV, and if QQ' be a chord parallel to the tangent to the parabola at P and meeting PV in V, then

QV  =  VQ'

Conversely, if QV = VQ', the chord QQ' will be parallel to the tangent at P. Proposition 2. 6If in a parabola QQ' be a chord parallel to the tangent at P, and if a straight line be drawn through P which is either itself the axis or parallel to the axis, and which meets QQ' in V and the tangent at Q to the parabola in T, then

PV  =  PT Proposition 3. 7If from a point on a parabola a straight line be drawn which is either itself the axis or parallel to the axis, as PV, and if from two other points Q, Q' on the parabola straight lines be drawn parallel to the tangent at P and meeting PV in V, V' respectively, then And these propositions are proved in the elements of conics.

Proposition 20. 8If Qq be the base, and P the vertex, of a parabolic segment, then the triangle PQq is greater than half the segment PQq. For the chord Qq is parallel to the tangent at P, and the triangle PQq is half the parallelogram formed by Qq, the tangent at P, and the diameters through Q, q.
Therefore the triangle PQq is greater than half the segment.

Corollary.  It follows that it is possible to inscribe in the segment a polygon such that the segments leftover are together less than any assigned area.

Proposition 21. If Qq be the base, and P the vertex, of a parabolic segment, and if R be the vertex of the segment cut off by P, then

DPQq  =  8 DPRQ The diameter through R will bisect the chord PQ, and therefore also QV, where PV is the diameter bisecting Qq 9  Let the diameter through R bisect PQ in Y and QV in M.  Join PM.
By Prop. 19, PV =  (4/3)RM. 10   Also  PV = 2YM. 11   Therefore  YM = 2RY, and DPQM = 2 DPRQ 12  Hence  DPQV = 4 DPRQ13, and  DPQq = 8 DPQR .
Also, if RW, the ordinate14  from R to PV, be produced to meet the curve again in r, RW= rW15, and the same proof shows that

DPQq = 8 DPrq

Proposition 22. 16If there be a series of areas A, B, C, D, ... each of which is four times the next in order, and if the largest, A, be equal to the triangle PQq inscribed in a parabolic segment PQq and having the same base with it and equal height, then

(A + B + C + D + ...)  <  (area of segment PQq)

For since DPQq = 8 DPRQ= 8 DPqr, where R, r are the vertices of the segments cut off by PQ, Pq, as in the last proposition,

DPQq = 4(DPQR + DPqr) Therefore, since DPQq = ADPQR + DPqr = B.
In like manner, we prove that the triangles similarly inscribed in the remaining segments are together equal to the area C, and so on.
Therefore A + B + C + D + ...  is equal to the area of a certain inscribed polygon, and is therefore less than the area of the segment.

Proposition 23. 17Given as series of areas A, B, C, D, ... , Z, of which A is the greatest, and each is equal to four times the next in order, then

A + B + C + D + ... + Z + (1/3)Z = (4/3)A

Take areas b, c, d, ... such that b = (1/3)B, c = (1/3)C, d = (1/3)D, and so on.  Then since b = (1/3)B, and B = (1/4)AB + b = (1/3)A.  Similarly, C + c = (1/3)B...
Therefore

B + C + D + ... + Z + b + c + d + ... + z = (1/3)(A + B + C + D + ... + Y)

But b + c + d + ... + z = (1/3)(B + C + D + ... + Y) Therefore, by subtraction, B + C + D + ... + Z + z = (1/3)A, or

A + B + C + D + ... + Z + (1/3)Z = (4/3)A

[The algebraical equivalent of this result is represented by the following computation: Proposition 24. Every segment bounded by a parabola and a chord Qq is equal to four-thirds of the triangle which has the same base and equal height.

Suppose K = (4/3)DPQq, 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 be greater or less. 18 I. Suppose the area of the segment greater than K.
If we then 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 + D + ... + Z <  (4/3)A

where A = DPQq.  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.
If then DPQq = A, B = (1/4)A, C = (1/4)B, 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 Now, since K exceeds A + B + C + D + ... + 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 + D + ... + 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, Read the commentary on the text
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last modified 9/6/02
Copyright (c) 2000. Daniel E. Otero