| The text is a thesis on geometry in the Heronian tradition. Compared to the impressive
list of mathematical papyri (see D.H. Fowler, `A Catalogue of Tables', ZPE75,1988,
273-) there are few close parallels to this text. Some recently published papyri with
geometrical problems, drawings and explanations, presumably meant for practical purposes
are:;PGen. inv. 259 Jean Rudhart `Trois problemes de geometrie, conserves Sur un papyrus
genevois', MH 35, 1978, 233-40 (=SB XIV 11973). PVidob. G 353 a-c + 59.529 all from
the same roll. See R. Pintaudi and P.J. Sijpesteijn, Neue Texte aus dem antiken Unterricht
(Pap. Flor. 18) n 172-173 and WH.M. Liesker and P.J. Sijpesteijn, 'Bruchstucke antiker
Geometrie', ZPE 113. PVindob. G 26.740 R. Pintaudi - P.J. Sijpesteijn o.l. no 178.
PGent inv. I (ed pr. M. Hombert RBPH4,1925, re-ed. J. Shelton, ZPE 42, 1981, 91-94.)
The purpose of this text is different to the one in question here. It concerns practical
taxation, but the form is the same. Photographs of two further fragments from Vienna
(PVindob G 6o.157-8) are published by Liesker and Sijpesteijn at the end of their
article in ZPE 113 (v. supra). 1he same article also contains an updated bibliography
of mathematical papyri published after Pack2, but not included in Fowler's list (v.
supra). Among the older publications, listed both by Pack, and G. Zalateo, 'Papiri
scolastici', Aeg. 41,1961,160-235, there are some that are remarkably similar to the
one treated here, both in age and in scope, e.g.: PAyer = Pack2 2318 = ed pr. E.J.
Goodspeed,'On the Mensuration of Land', AJP 19, 1898, 25-39 = PChicago Goodspeed 3.
PBerol. inv. 11529 = Pack2 2319 = ed pr. W Schubart, in Amt. Ber. a.d. Kgl. Kunstsamml.
Berlin 1915-16, 161. We have not been able to consult this edition, but there is a
photograph in Karl Weitzmann, Illustrations in Roll and Codex. A Study o f the Origin
and Method of Text-Illustrations. (Studies in Manuscript Illustrations a) Princeton,
1970, Pl. XIII, 35. It is very similar to the present text in lay-out and scope. P
Vindob. inv. 19996 = Pack2 2322 = PRainer 1,1 (= MPER I,1) although all problems treated
there concern solid geometry.;;General Remarks on the Mathematical Problems that Seem
to be Solved in this Papyrus;The figures preserved in Col. II are reminiscent of the
initial propositions in Heron's Metrika, so one might hope to find parallels in his
text, but we have not yet been that lucky. The following dare be said with some confidence,
though:;The number 18 in I.4 could be part of the solution to the following problem,
variants of which are found in the Metrika:;Suppose that in some right-angled triangle
the hypotenuse plus one of the sides about the right angle is 42, while the square
on the third side of the triangle is 756. To find the hypotenuse and the side.;From
Euclid's Elements I.47 (`The Pythagorean Theorem', c2 = a2+b2) we know that the given
square, b2, is the difference between the squares on the other two sides, c2-a2.;From
Elements II.5 or 6 we learn that c2-a2= (c-a)(c+a); from the given numbers we have
that c2-a2 = 756 and that c+a = 42, whence c-a = 756:42 =18. And therefore c = (42+18):2
= 30, a = (42-18):2 =12. These were quite well-known operations to people who knew
the Elements, being among the standard inventory of Pythagorean mathematics. The length
of the side b is irrational, but is not asked for.;All this may be too far-fetched,
considering that we have but 756 and 18 in the text. That some right-angled triangles
are involved, and that some quadratic problems are being solved in the text seems
quite sure.;In Col II, top, the area of a trapezium is being calculated from the given
lengths of the sides, 4 and 8 (some-unit) of the parallel sides, 13 and 15 of the
other two. The method is taught us by Heron in Metrika L13, about a trapezium with
an obtuse angle. The efficient tool is 11.12 of Euclid's Elements (which can be anachronistically
interpreted within trigonometry as the so-called Law of Cosines for Triangles).;Let
the trapezium be ABCD (see above), with AD = 4, AB =13, BC = 8, CD =15, and let CE
be drawn parallel to AB, and CH and AK perpendicular to AD. Now ED = 4, because EA
is equal to BC; Elements ILIZ says that;CD2 - z DE*HE = CE2 + ED2, that is 225 - 8
HE =169 + 16,;whence HE = 5. Now the height of the trapezium, CH, can be calculated
from the right-angled triangle CHD, since;CD2 = CH2 + HD2, that is 225 = CH2 + 81,;or
(similarly, "omoios") from the right-angled triangle AKB, since;AB2 = AK2 + BK2, that
is 169 = AK2 + 25.;In either case CH comes out as 12, so these are nice Pythagorean
triangles, not real areas measured by the "geometres".;Most of the numbers involved
above can be traced in the papyrus (plus a few others, 14 and 18, which escape our
interpretation, although half of 18, 9 that is, does come in as the length of HD,
but we do not see how 18 comes into it in the first place.);The formula of the area
of a trapezium could now be applied:;ABCD = (AD+BC)/2 *CH, that is 72 (square-units).;Obviously,
however, in IL17 another method is applied, namely to calculate the area of the trapezium
as the difference between the area of the rectangle AKCH and of the two triangles
enclosing it (ABK and DCH); which areas can easily be calculated since their bases
and the height are known.;The peculiar numbers (if they are numbers, all of them)
in 11.21 must remain puzzling.;The other diagram, published by Otto Neugebauer in
The Exact Sciences in Antiquity p. 179, illustrates how, from the lengths of sides,
to calculate the area of a quadrangle with one right angle. This is, in fact, not
quite trivial, as Heron demonstrates in Metrikd 1.14, but the example in our papyrus
is so special that it hardly deserves Heron's method, so our interpretation may be
quite wrong.;Let the quadrangle be ABCD, with AB = AD =15, BC = CD = 5 (measured in
some unit), and the angle at B a right angle. To determine the area we must divide
the quadrangle into calculable parts. Let DE be drawn parallel to EC, and CH parallel
to BA. The first step is to realise that,the angle D is also right, since the triangle
ACD is congruent with ABC (three sides respectively equal). Therefore the area of
the quadrangle is twice the area of the triangle, namely 75 (square units). But the
text apparently did not take this route, probably because this is a very special case
with symmetrical parts; so he sets about finding the lengths of CH and DH. The triangle
CHD is obviously similar to DEA, with the linear ratio of CD:DA = 5:15, that is 1:3;
therefore AE is thrice DH, and ED is thrice CH. Anachronistically, we may use equations
by putting CH = x and DH = y:;BE+EA =15 and ED-HD = 5 ;x+3y =15 and 3x-y = 5.;We may
eliminate y by multiplying the latter equation by 3 and rearrange it to 3y = 9x-I5,
which substituted in the former leads to lox = 30, whence x = 3 and y = 4. (We knew
that already, since the results are written on the diagram.);Very little of all this
is recognizable in the papyrus, so we may be barking up the wrong tree. However, it
is beyond doubt that the problem is a nice construction and not a calculation made
from real measurement. |
