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.** 5*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 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.** 6*If
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.** 7*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*

And these propositions are proved in the elements of conics.

**Proposition 20.** 8*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 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*

D*PQq
= *8 D*PRQ*

By Prop. 19,

Also, if

D*PQq*
= 8 D*Prq*

**Proposition
22.** 16*If
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 D*PQq*
= 8 D*PRQ*=
8 D*Pqr*,
where *R*, *r* are the vertices of the segments cut off by *PQ*,
*Pq*,
as in the last proposition,

D*PQq*
= 4(D*PQR*
+ D*Pqr*)

In like manner, we prove that the triangles similarly inscribed in the remaining segments are together equal to the area

Therefore

**Proposition
23.** 17*Given
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)*A*, *B* + *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*)

*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)D*PQq*,
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

If we then inscribe in the segments cut off by

Therefore the polygon so formed must be greater than the area

*A* + *B* + *C* +
*D*
+ ... + *Z* < (4/3)*A*

where *A* = D*PQq*.
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 D*PQq*
= *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

Return to the calendar

last modified 9/6/02

Copyright (c) 2000. Daniel E. Otero