Archimedes to Dositheus, greetings.
After I heard that Conon, who fell no way short in our friendship, had died and that you had
become an acquaintance of Conon and were familiar with geometry, we were saddened on
behalf of someone both dear as a man and admirable at mathematics, and we resolved to
write and send to you, just as we had meant to write to Conon, one of the geometrical
theorems that had not been observed earlier, but which now has been observed by us, it
being earlier discovered through mechanical <means>, but then also proved through
geometrical <means>. Some of those who worked earlier in geometry attempted to prove
(draw) how it would be possible to find a rectilinear area equal to a circle that’s given or a
segment of a circle that’s given, and afterwards they attempted to square the area enclosed
by the section of the cone as a whole (ellipse? or Beere, cone generally, as opposed to any of
the three classes) by assuming some lemmas that were not readily admissible. For this
reason, these were condemned by most people as not being discovered by them. But we do
not know of anyone previously who attempted to square the segment enclosed by the
straight-line and right-angled section of a cone, which has not been found by us. For it is
proved that every segment enclosed by a straight-line and right-angled section of a cone is
a third-again the triangle having its base as the same and height equal to the segment, i.e.,
when this lemma is assumed for its demonstration: that is possible for the excess of
unequal areas by which the larger exceeds the smaller, by being added to itself, to exceed
any proposed finite area. Earlier geometers have also used this lemma. For they use this
lemma itself to demonstrate that circles have to one another double ratio of the diameters,
and that spheres have triple ratio to one another of the diameters, and further that every
pyramid is a third part of the prism having the same base as the pyramid and equal height.
And because every cone is a third part of the cylinder having the same base as the cone and
equal height, they would assume some lemma similar to the mentioned one to prove
(draw) <it>. It follows that we have no less conviction in each case (? we expect dative or
cognate accusative, each theorema) of the mentioned theorems than those demonstrated
without this lemma. It is adequate given that those presented by us have been raised to a
conviction similar to these. And so, having written up the demonstrations of it we are
sending first, how it was observed through mechanical <means> and afterwards how it is
demonstrated through geometrical. But first there are also proved (drawn) conic elements
that are useful for the demonstrations. Farewell.
Preliminary theorem
If a straight line is cut at random, then the sum of the square on the whole and that on one
of the segments equals twice the rectangle contained by the whole and the said segment
plus the square on the remaining segment.
If a perpendicular to the diameter is drawn and a line is drawn from the intersection with
the orthotome to a distance twice the abcissa, then the line is tangent at the intersection.
Suppose not. Let p be the latus rectus
x^2 = 2 y p
take a, b as ordinate and abcissa.
a^2 = 2 b p
then
a : 2 b is the ratio of lines from intersection of ordinate and abcissa to intersection of
abcissa and hypothesized tangent.
suppose we have c^2 = 2 d p and c : d+b = a : 2b
2bc=ad+ba
a^2 = 2 b p
c^2 = 2 d p
suppose b > d
2 b : d+b = a : c
4 b^2 : (d+b)^2 = a^2 : c^2
4 b^2 : (d+b)^2 = 2 p b : 2 d p
4 b^2 : (d+b)^2 = b : d
4 b^2 : (d+b)^2 = 4 b^2 : 4 d b
(d+b)^2 = 4 d b
d^2 + b^2 + 2 b d = 4 d b
d^2 + b^2 = 2 db
d^2 + b^2 = 2 d b + (b-d)^2 (ii 7)
2 d b + (b-d)^2 = 2 db, which is impossible
If b<d, we get 2 d b + (d-b)^2 = 2 db, which is also impossible
1.
If there is a section of a right-angled cone, on which ABG, and <straight-line> BD is
parallel to the diameter or is the diameter itself, while AG is parallel to the tangent of the
section of the cone at B, AD will be equal to DG. And if AD is equal to DG, AG and the
tangent to the section of the cone at B will be parallel.
Let parbola ABG be drawn with latus rectus X. Let the diameter be BD, and let there be a
parallel to the diameter EZ. Then let there a tangent at E and extend it to H. Take a point G
on the parabola and let there be drawn a parallel to EH, QKG, with Q on the parabola, with
K the intersection of EZ and QG. I say that QK = KG.
Let DB be extended to G, and let perpendiculars, QL to AG, EM to BD, and KL to AG be
drawn.
Since E is a tangent and BD the diameter, HB = BM and T(EM) = 2 O(BM,X). Hence, T(EM) =
O(MH,X). Hence, EM : X = MH : EM.
Clearly triangles QLG, GDG, KZG, and HME are all similar. Hence, HM : ME = QL : LG = KZ :
ZG = EM : X.
However, T(QN) = 2 O(BN,X).
Hence, T(LD) = T(QN) = 2 O(BN,X) = 2 O(BD-DN,X) = 2 O(BD,X) - 2 O(DN,X) = 2 O(BD,X) - 2
O(QL,X)
Or, T(LD) = 2 O(BD,X) - 2 O(QL,X)
And, T(DG) = 2 O(BD,X)
Hence, T(LD) = T(DG) – 2 O(QL,X)
Hence, 2 O(QL,X) = T(DG) - T(LD) = O(DG+LD,DG-LD) = O(GL,DG-LD)
Or QL : GL = DG-LD : X
However, QL : LG = EM : X = DZ : X
Hence, DZ = DG – LD
Or DG = DZ + LD = LZ.
Hence QK = KG.
Suppose that QK is equal to EG. Then do everything in reverse.
2.
If there is a section of a right-angled cone, ABG, and BD is parallel to the diameter or
is the diameter itself, while ADG is parallel to the tangent of the section of the cone at B, and
EG is tangent to the section of the cone at G, BD and BE will be equal.
3.
If there is a section of a right-angled cone, ABG, and BD is parallel to the diameter or
is the diameter itself, and certain <straight-lines> AD, EZ are led out parallel to the tangent
to the right-angled cone at B, then AD will be in power to EZ as BD to BZ.
But these were demonstrated in the conic elements.