QUADRATURE OF THE PARABOLA. 


“ ARCHIMEDES to Dositheus greeting. 


“When I heard that Conon, who was my friend in his life- 
time, 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. 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* and a straight line, assuming lemmas 
not easily conceded, so that it was recognised 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 dis- 
covered 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, and for the demonstration 

* There appears to be some corruption here: the expression in the text is 
τᾶς ὅλου τοῦ κώνου τομᾶς, and it is not easy to give a natural and intelligible 
meaning to it. The section of ‘the whole cone’ might perhaps mean a section 
cutting right through it, i.e. an ellipse, and the ‘ straight line’ might be an axis 
or a diameter. But Heiberg objects to the suggestion to read τᾶς ὀξυγωνίου 
κώνου τομᾶς, in view of the addition of καὶ εὐθείας, on the ground that the former 


expression always signifies the whole of an ellipse, never a segment of it 
(Quaestiones Archimedeae, p. 149). 


234, ARCHIMEDES 


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 earlier 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 that spheres are to one another in the 
triplicate 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 afore- 
said 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 send 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 (στουχεῖα κωνικὰ χρεῖαν 
ἔχοντα és τὰν ἀπόδειξιν). Farewell.” 


Proposition 1. 


If from a point on a para- 
bola a straight line be drawn 3 
which is either itself the axis or 
parallel to the axis, as PV, and 
uf QQ’ be a chord parallel to 
the tangent to the parabola at P P ν 
and meeting PV in V, then 


OV =VQ. 
Conversely, if QV = VQ’, the 
chord QQ’ will be parallel to the 
tangent at P. 


* The Greek of this passage is: συμβαίνει δὲ τῶν προειρημένων θεωρημάτων 
ἕκαστον μηδὲν ἧσσον τῶν ἄνευ τούτου Tov λήμματος ἀποδεδειγμένων πεπιστευκέναι. 
Here it would seem that πεπιστευκέναι must be wrong and that the passive 
should have been used. 


QUADRATURE OF THE PARABOLA. 235 


Proposition 2. 


Tf in a parabola QQ’ be a chord parallel to the tangent at P, 
and if a straight line be drawn through P which 1s either rtself 
the axis or parallel to the aais, and which meets QQ’ in V and 
the tangent at Q to the parabola in T, then 


a eae) Pa 


Proposition S. 


If 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 

PV PV SOV OV 2 


“ And these propositions are proved in the elements of conics.*” 


Proposition 4. 


If Qq be the base of any segment of a parabola, and P the 
vertex of the segment, and vf the diameter through any other point 
R meet Qq in O and QP (produced if necessary) in F, then 


OV V0 OF FR; 
Draw the ordinate RW to PV, meeting QP in Kk. 


* i.e. in the treatises on conics by Euclid and Aristaeus. 


236 ARCHIMEDES 


Then PVG Wi OV eR Ws 
whence, by parallels, 
POI PO*: PE. 


In other words, PQ, PF, PK are in continued proportion ; 
therefore 


POH = PP: Pe 
=PQ+PF:PF+PK 


= QF: KF. 
Hence, by parallels, 


QV: VO=OF: FR. 
[It is easily seen that this equation is equivalent to a change of 
axes of coordinates from the tangent and diameter to new axes 
consisting of the chord Qq (as axis of «, say) and the diameter 
through Q (as axis of y). 
2 

For, if QV=a, PV= = where p is the parameter of the 
ordinates to PV. 

Thus, if QO = a, and RO = y, the above result gives 


ως OF 
v—-a OF-y’ 
4: 
whence acy, meee ee 


or py =x (2a—2).] 


QUADRATURE OF THE PARABOLA. 237 


Proposition 5. 


If Qq be the base of any segment of a parabola, P the vertex 
of the segment, and PV its diameter, and if the diameter of the 
parabola through any other point R meet Qq in O and the 
tangent at ᾧ in E, then 

Q0 209 = ER: RO. 


Let the diameter through R meet QP in F. 


Then, by Prop. 4, 
OViViG= OF OER: 
Since QV = γᾷ, it follows that 
VG ξεν εν (1). 
Also, if VP meet the tangent in 1, 
PT=PYV, and therefore HF= OF. 
Accordingly, doubling the antecedents in (1), we have 
Cg 00 — OL OR: 
whence QO: 0Og=ER: RO. 


238 ARCHIMEDES 


Propositions 6, 7*. 


Suppose a lever AOB placed horizontally and supported at 
its middle point O. Let a triangle BCD in which the angle C is 
right or obtuse be suspended from B and Ὁ, so that C 1s attached 
to O and CD is in the same vertical line with O. Then, if P be 
such an area as, when suspended from A, will keep the system i 
equilibrium, 

P={ABCD. ) 

Take a point # on OB such that BE = 20", and draw EFH 
parallel to OCD meeting BC, BD in F, H respectively. Let G 
be the middle point of FH. 


A ie) E 


Then G is the centre of gravity of the triangle BCD. 


Hence, if the angular points B, C be set free and the 
triangle be suspended by attaching F to £, the triangle will 
hang in the same position as before, because EFG is a vertical 
straight line. “For this is proved +.” 


Therefore, as before, there will be equilibrium. 


Thus P:ABCD=OE: AO 
=e 
or P=1ABCD. 


* In Prop. 6 Archimedes takes the separate case in which the angle BCD of 
the triangle is a right angle so that C coincides with O in the figure and F with 
E, He then proves, in Prop. 7, the same property for the triangle in which 
BCD is an obtuse angle, by treating the triangle as the difference between two 
right-angled triangles BOD, BOC and using the result of Prop. 6. I have com- 
bined the two propositions in one proof, for the sake of brevity. The same 
remark applies to the propositions following Props. 6, 7. 

+ Doubtless in the lost book περὶ ζυγῶν. Cf. the Introduction, Chapter IL., 
ad jin. 


QUADRATURE OF THE PARABOLA. 239 


Propositions 8, 9. 


Suppose a lever AOB placed horizontally and supported at 
its middle point O. Let a triangle BCD, right-angled or obtuse- 
angled at C, be suspended from the points B, EK on OB, the 
angular point C being so attached to EK that the side CD is in the 
same vertical line with E. Let Q be an area such that 

AO: 0H=ABCD: ἢ: 
Then, if an area P suspended from A keep the system in 
equilibrium, 
P< ABCD but > Q. 

Take G the centre of gravity of the triangle BCD, and draw 

GH parallel to DC, i.e. vertically, meeting BO in H. 


A OME H B 


We may now suppose the triangle BCD suspended from H, 
and, since there is equilibrium, 


ESB CDE MANOR OUT 5, τ vate sons (i); 
whence P<eABCD. 
Also ABCD :Q=A0: OZ. 
Therefore, by (1), ABCD:Q>ABCD: P, 
and PQ. 


Propositions 10, 11. 


Suppose a lever AOB placed horizontally and supported at O, 
its middle point. Let CDEF be a trapezium which can be so 
placed that its parallel sides CD, FE are vertical, while C ds 
vertically below O, and the other sides CF, DE meet in B. Let 
EF meet BO in H, and let the trapezium be suspended by attaching 
F to H and Cto O. Further, suppose Q to be an area such that 


AO : OH = (trapezium CDEF) : Q. 


240 ARCHIMEDES 


Then, if P be the area which, when suspended from A, keeps the 


system in equilibrium, 


Pee: 
The same is true in the particular case where the angles at 
C, F are right, and consequently C, F coincide with O, H 
respectively. 
Divide OH in K so that 
(20D + FE) :(2FE+CD)=HK : KO. 


A 


Draw KG parallel to OD, and let G be the middle point of 
the portion of AG intercepted within the trapezium. Then G 
is the centre of gravity of the trapezium [On the equilibrium of 
planes, I. 15]. 

Thus we may suppose the trapezium suspended from K, and 
the equilibrium will remain undisturbed. 

Therefore 

AO: OK = (trapezium CDEF): P, 
and, by hypothesis, 
AO: OH = (trapezium CDEF) : Q. 
Since OK < OH, it follows that 
PQ): 


Propositions 12, 13. 

If the trapezium CDEF be placed as in the last propositions, 
except that CD is vertically below a point L on OB instead of 
being below O, and the trapezium is suspended from L, H, 
suppose that Q, R are areas such that 

AO : OH = (trapezium CDEF) : Q, 
and AO: OL = (trapezium CDEF) : R. 


QUADRATURE OF THE PARABOLA. 241 


If then an area P suspended from A keep the system in 
equilibrium, 
PS he bute < Q: 
Take the centre of gravity G of the trapezium, as in the 
last propositions, and let the line through @ parallel to DC 
meet OB in K. 


A oO LK H B 


Then we may suppose the trapezium suspended from K, 
and there will still be equilibrium. 


Therefore (trapezium CDEF): P= AO: OK. 
Hence 
(trapezium CDEP) : P > (trapezium CDEF) : Q, 
but < (trapezium CDEF) : R. 
It follows that P=@Q but > &. 


Propositions 14, 15. 


Let Qq be the base of any segment of a parabola. Then, if 
two lines be drawn from Q, q, each parallel to the axis of the 
parabola and on the same side of Qq as the segment is, either 
(1) the angles so formed at Q, q are both right angles, or 
(2) one is acute and the other obtuse. In the latter case let 
the angle at q be the obtuse angle. 


Divide Qq into any number of equal parts at the points 
O,, O2,... On. Draw through gq, 0,, O,,... On diameters of the 
parabola meeting the tangent at Qin 2, Z,, #,,... Ε΄, and the 
parabola itself m g, R,, R,,...R,. Join OR,, QR,; ... QRu 
meeting G2, Onis, Oia One fs el, Fy, PB 00 Prat 

HA 16 


242 ARCHIMEDES 


Let the diameters Hg, ,0,,... H,O, meet a straight line 
QOA drawn through Q perpendicular to the diameters in the 
points O, H,, H,, ... H, respectively. (In the particular case 
where Qq is itself perpendicular to the diameters q will coincide 
with O, O, with H,, and so on.) 


It is required to prove that 
(1) A£qQ<3(sum of trapezia FO,, F,02,...Pn1~0,andA £,,0,Q), 
(2) ALqQ >3(sumof trapezia R,0,, R,O;,...R,,~O,and A R,0,Q). 


Suppose AO made equal to OQ, and conceive QOA as a 
lever placed horizontally and supported at O. Suppose the 
triangle HqQ suspended from OQ in the position drawn, and 
suppose that the trapezium ΜΚ, in the position drawn is 
balanced by an area P, suspended from A, the trapezium £0, 
in the position drawn is balanced by the area P, suspended 


QUADRATURE OF THE PARABOLA. 243 
from A, and so on, the triangle #,,0,Q being in like manner 
balanced by Pri. 


Then P,+P,+...+Pni will balance the whole triangle 
EqQ as drawn, and therefore 


PP we a tA BQ: [Props 6; 7] 
Again AO OH. = O00 > OH, 
= Qq : 90; 
= E,0, : OR, [by means of Prop. 5] 


= (trapezium HO,): (trapezium FO,); 
whence [Props. 10, 11] 


(Oi) >eP%. 
Next AO; OF ΞΞ 0, ΟΣ 
Ξ τ: (Les a). waco conc ses bes (a), 
while AQ ΟΞ ἡ ΣΝ 
=i Oban HO) iret atts, cabs (B); 


and, since (a) and (8) are simultaneously true, we have, by 
Props. 12, 13, 
(Op Ps E0,). 


Similarly it may be proved that 
(F,0;) > 2. => (8,0;), 
and so on. 
Lastly [Props. 8, 9] 
AE OO ΡΞ 8020. 
By addition, we obtain 
(1) (FO,)+(410,)+...+(Pn+On)+ ΔῊ, ΟΞ ΡΊΞΕ. Ἐς Εν 
»έδδᾳρ, 

or A EqQ <3 (FO, + F,0, +... + Fr On + A E,0nQ). 
(2) (R,0,)+(R,0;)+...+(Rn+10n)+ARn0O,Q<P.+P,+...+Prais 

<P,\+ Pot ...4+ Pai, a fortiori, 

«ἐδῆᾳρῳ, 
or A EqQ > 3 (RO, + R,0;+... + Rp On + A RnOnQ). 

16—2 


24.4. ARCHIMEDES 


Proposition 16. 


Suppose Qq to be the base of a parabolic segment, q being 
not more distant than Q from the vertex of the parabola. Draw 
through q the straight line qE parallel to the axis of the parabola 
to meet the tangent at Qin EF. It is required to prove that 


(area of segment)=1 AEqQ. 


For, if not, the area of the segment must be either greater 
or less than + AXqQ. 


I. Suppose the area of the 
segment greater than 1 A EqQ. 
Then the excess can, if con- 
tinually added to itself, be 
made to exceed AXgQ. And 
it is possible to find a submul- 
tiple of the triangle HgQ less 
than the said excess of the 
segment over + AXgQ. 


Let the triangle FgQ be such 
a submultiple of the triangle 
EqQ. Divide Eq into equal 
parts each equal to qf, and let 
all the points of division in- 
cluding F' be joined to Q meet- 
ing the parabola in f,, Ro, ... 
R, respectively. Through R,, R,, ... R, draw diameters of the 
parabola meeting σῷ in O,, Ος, ... O, respectively. 


Let O,R, meet QR, in F,. 
Let O,R, meet QR, in D, and QR, in F,. 
Let O,R, meet YR, in D, and QR, in F,, and so on. 


We have, by hypothesis, 
A FqQ < (area of segment) — 1A £qQ, 
or (area of segment)— AFgQ >1 AE@Q ......... (a). 


QUADRATURE OF THE PARABOLA. 245 


Now, since all the parts of gH, as gF and the rest, are equal, 
O,R, = R,F,, 0,D, = DR, = R.F,, and so on; therefore 
A FqQ =(FO, + R,0,+ D0; +...) 


i" =(FO,+ FD, + FD, +... +FraDnit+A EnRnQ)...(8). 
ut 


(area of segment) < (FO, + F,0, +... ἘΠ, +A L,0,Q). 
Subtracting, we have 
(area of segment) —A FgQ < (2,0, + B,0;+4+... 


213 Tan +A R,,0,Q), 
whence, ὦ forteorr, by (a), 


LA EqQ < (2,0, + 2,0; +... + Rp On + A R,OnQ). 
But this is impossible, since [Props. 14, 15] 
1A EqQ > (#0, + BL0;+...+ Rp On+ AR,0,Q). 
Therefore 
(area of segment) 1A HqQ. 
II. If possible, suppose the area of the segment less than 
5A EQ. 
Take a submultiple of the triangle HqQ, as the triangle 
FqQ, less than the excess of A HqQ over the area of the 
segment, and make the same construction as before. 


Since AFqQ<1ALqQ—(area of segment), 
it follows that 
A FqQ + (area of segment) < $A HqQ 
< (FO, + FO, +... + Fa1On + AEn0nQ). 
[Props. 14, 15] 
Subtracting from each side the area of the segment, we have 
A FqQ < (sum of spaces gFR,, RF R2, ... EnRnQ) 
< (PO, + FD, +... + FriDnit ΔΑ, ΠΟ), α fortiori ; 
which is impossible, because, by (8) above, 
A FqQ = FO, + FD, +... + FPaDniat AEnRnQ. 
Hence (area of segment) ¢ LA HqQ. 
Since then the area of the segment is neither less nor 
greater than +A EqQ, it is equal-to it. 


246 ARCHIMEDES 


Proposition 17. 


It is now manifest that the area of any segment of a 
parabola is four-thirds of the triangle which has the same base 
as the segment and equal height. 


Let Qq be the base of the segment, P its vertex. Then 
PQg is the inscribed triangle with the 


same base as the segment and equal 2 


ν 
height. 
q 

Since P is the vertex* of the seg- 
ment, the diameter through P bisects 
Qq. Let V be the point of bisection. 

Let VP, and g# drawn parallel to 
it, meet the tangent at Qin 7, £ re- a 
spectively. 

Then, by parallels, 

gh =2VT, 

and JE led [Prop. 2] 
so that Vi 2 Vv. E 


Hence AXqQ =4APQq. 
But, by Prop. 16, the area of the segment is equal to tA HqQ. 
Therefore (area of segment) = 4A PQq. 


Der. “In segments bounded by a straight line and any 
curve I call the straight line the base, and the height the 
greatest perpendicular drawn from the curve to the base of the 
segment, and the vertex the point from which the greatest 
perpendicular is drawn.” 


* Tt is curious that Archimedes uses the terms base and vertex of a segment 
here, but gives the definition of them later (at the end of the proposition), 
Moreover he assumes the converse of the property proved in Prop. 18. 


QUADRATURE OF THE PARABOLA. 247 


Proposition 18. 
Tf Qq be the base of a segment of a parabola, and V the 
middle point of Qq, and tf the diameter through V meet the 
curve in P, then P ws the vertex of the segment. 


Q 


τὴ 


For Qq is parallel to the tangent at P [Prop. 1]. Therefore, 
of all the perpendiculars which can be drawn from points on the 
segment to the base ἕῳ, that from P is the greatest. Hence, 
by the definition, P is the vertex of the segment. 


Proposition 19. 

If Qq be a chord of a parabola bisected in V by the diameter 
PV, and if RM be a diameter bisecting QV in M, and RW 
be the ordinate from R to PV, then 

PV =4RM. 
Q 


q 
For, by the property of the parabola, 
Vere Ors ΤῊ: 
a es 8: 
so that PV =P W,. 
whence PV =4kRM. 


248 ARCHIMEDES 


Proposition 20. 


If 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 Qa 
formed by Qq, the tangent at P, and the 
diameters through Ὁ, q. 


Therefore the triangle PQq is greater 
than half the segment. P ν 


Cor. It follows that τέ a possible 
to inscribe in the segment a polygon such 
that the segments left over are together 
less than any assigned area. 


Proposition 21. 


If Qq be the base, and P the vertex, of any parabolic 
segment, and if R be the vertex of the segment cut off by PQ, 
then 

ἌΡ ΞΕ ΑΕ ΡΣ 

The diameter through R will bisect the chord PQ, and 
therefore also QV, where PV is the 
diameter bisecting Qg. Let the dia- 
meter through & bisect PQ in Y and 
QVin M. Join PM. 


By Prop. 19, 
PV =<RM. 
Also PVi=2Y M. 
Therefore YM —=2RY, 
and AOPOM IAP RO. 
Hence APOV =4ANPRQ, 


and A PQq =S8APRYQ. 


QUADRATURE OF THE PARABOLA. 249 
Also, if RW, the ordinate from R to PV, be produced to 
meet the curve again in 7, 
RW=rW, 
and the same proof shows that 
APQg=S8A-Pr¢. 


Proposition 22. 


If there be a series of areas A, B, Οὐ, 1), ... each of which is 
Sour times the next in order, and if the largest, A, be equal to the 
triangle PQ 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 APQq =8APRQ =8APgqr, where R, 7 are the 
vertices of the segments cut off by PQ, 
Pq, as in the last proposition, 

A PQ =4(APQR + APqr). 

Therefore, since A PQq = A, 

APQR+ APqr= B. 

In lke manner we prove that the 
triangles similarly inscribed in the re- 
maining 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. 


Given a 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+...4274+34Z7=4A. 


250 ARCHIMEDES 


Take areas 0, c, αἰ, ... such that 


= 4B, 
c=10C, 

d =1D, and so on. 
Then, since b= 4B, 
and B=}A, 
Bab = +A 
Similarly C+c=4B. 

ἜΝ meena 


B+C+D+...4+Z4+b+ce+d4+...4¢2=1(A+B4+C+.. 


But = b+ce+d+...4¢y=1(B+C0+D+4...4+ Y). 


13 
“κ᾽. Ὁ 


Therefore, by subtraction, 
B+04+D+4+...4Z+2=1A4 
or A+B+C+...444+4Z=4$A4. 


ΤῊ 


QUADRATURE OF THE PARABOLA. 251 


| The algebraical equivalent of this result is of course 
Ie EG) =e 3 
e ie ( 1 Ng | 


I=] 


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 as 
the segment and equal height. 


Suppose K=$APQq, 


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 

Κ΄, 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, Le. 
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+4...42Z< 4A, 
where A=/< Pio, 


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


II. Suppose, if possible, that the area of the segment is 
less than ΚΑ. 


252 ARCHIMEDES, 


If then APQqg=A, B=14A, C=1B, 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+4+...4X+1X =4A [Prop. 23] 
we 


Now, since K exceeds 4+B+C+...+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 [΄. 
Thus, since the segment is neither greater nor less than K, 


(area of segment PQq) = ἡ =4APQq.