- EGYPT 235
a steward, to oversee planting and harvest. As the painting shows, the instrument
used to measure distance was a rope that could be pulled taut. That measuring
instrument has given rise to another name often used to refer to these surveyors:
harpedonaptai, from the words harpedone, meaning rope, and haptein, meaning at-
tach. The philosopher Democritus (d. 357 BCE) boasted, "In demonstration no
one ever surpassed me, not even those of the Egyptians called harpedonaptai."^2
The geometric problems considered in the Egyptian papyri all involve mea-
surement. These problems show considerable insight into the properties of simple
geometric figures such as the circle, the triangle, the rectangle, and the pyramid;
and they rise to a rather high level of sophistication in computing the area of a hemi-
sphere. Those involving flat boundaries (polygons and pyramids) are correct from
the point of view of Euclidean geometry, while those involving curved boundaries
(disks and spheres) are correct up to the constant of proportionality chosen.
1.1. Areas. Since the areas of rectangles and triangles are easy to compute, it
is understandable that very little attention is given to these problems. Only four
problems in the Ahmose Papyrus touch on these questions: Problems 6, 49, 51,
and 52 (see Plate 1).
Rectangles, triangles, and trapezoids. Problem 49 involves computing the area of a
rectangle that has dimensions 1 khet by 10 khets. This in itself would be a trivial
problem, except that areas are to be expressed in square cubits rather than square
khets. Since a khet is 100 cubits, the answer is given correctly as 100,000 square
cubits. Problem 51 is a matter of finding the area of a triangle, and it is illustrated
by a figure (see Plate 1) showing the triangle. The area is found by multiplying
half of the base by the height. In Problem 52, this technique is generalized to a
trapezoid, and half of the sum of the upper and lower bases is multiplied by the
height.
Of all these problems, the most interesting is Problem 6, which involves a twist
that makes it equivalent to a quadratic equation. A rectangle is given having area
12 cubit strips; that is, it is equal to an area 1 cubit by 12 cubits, though not of the
same shape. The problem is to find its dimensions given that the width is three-
fourths of the length (2 4 in the notation of the papyrus). The first problem is to
"calculate with 2 4, until 1 is reached," that is, in our language, dividing 1 by 2 4.
The result is 1 3. Then 12 is multiplied by 1 3, yielding 16, after which the scribe
takes the corner (square root) of 16—unfortunately, without saying how—getting
4 as the length. This is a very nice example of thinking in terms of expressions.
The scribe seems to have in mind a picture of the length being multiplied by three-
fourths of the length, and the result being 12. Then the length squared has to be
found by multiplying by what we would call the reciprocal of its coefficient, after
which the length is found by taking the square root.
Slopes. The beginnings of trigonometry can be seen in Problems 56-60 of the pa-
pyrus, which involve the slope of the sides of pyramids and other figures. There
is a unit of slope analogous to the pesu that we saw in Chapter 5 in the problems
involving strength of bread and beer. The unit of slope is the seked, defined as
the number of palms of horizontal displacement associated with a vertical displace-
ment of 1 royal cubit. One royal cubit was 7 palms. Because of the relative sizes of
(^2) Quoted by the second-century theologian Clement of Alexandria, in his Miscellanies, Book 1,
Chapter 15.