Return to Vignettes of Ancient Mathematics
Introduction to Hippocrates
Introduction to Lunules
Alexander's Method
Comparison of Alexander's and Eudemus' Methods
(diagram 1) The basic idea is to construct an outer arc A and an inner arc B, concave in the same direction. (diagram 2) Suppose (condition 1 of 2) that we have constructed the arcs A and B so that each can be divided into respectively into m arcs A_{i} and n arcs B_{j}, all of which are similar, i.e .,
A_{1} = … = A_{m} and A_{1} + … + A_{m} = A
B _{1} = … = B_{n} and B _{1} + … + B_{n} = B
A_{1} … A_{m} B _{1} … B_{n}
Although this condition is not formally necessary, it greatly simplifies matters. Observe that m and n must be whole numbers, which means that the two whole arcs must be commensurable.
(diagram 3) If A_{i} and B_{j} are similar segments bounded by the similar arcs A_{i} and B_{j} and their respective bases a_{i} and b_{j}, we can stipulate segments A, B,with respective bases a and b:
A_{1} = … = A_{m} = A
a_{1} = … = a_{m} = a
B_{1} = … = B_{n} = B
b_{1} = … = b_{n} = b
Also, by the basic theorem of similar segments, given that A is similar to B, A : B = a^{sqr} : b^{sqr}. Now suppose (condition 2 of 2) that we have constructed the arcs and segments so that the bases have the ratio a^{sqr} : b^{sqr} = n : m.
Hence A : B = a^{sqr} : b^{sqr} = n : m.
Now: A_{1} + … + A_{m} = m*A
and B_{1} + … + B_{n} = n*B
Hence, m*A : n * B = m * n : n * m.
That is, A_{1} + … + A_{m} = B_{1} + … + B_{n}.
(diagram 4) If we can add some area C such that C + m*A = C + n*B, where C + m*A will be a lunule
(diagram 5)
and C + n*B will be rectilinear (diagram 6). This is how Eudemus conceives of the problem.
The different lunules mentioned by Eudemus may have been discovered simply by attempting all simple ratios.
A brief note on Eudemus' method of finding a circle and lunule that can be squared is also in order. (diagram 7) Suppose that one were to look for a lunule to be squared that is the next simplest after the lunule on the semicircle with ratio 2 : 1. This might be another lunule, whose outer arc A has as its base an inscribed equilateral triangle, also with ratio 2 : 1. The outer arc is thus immediately divided into two sides of a regular hexagon, so that the inner arc also needs to be on an inscribed hexagon. (diagram 8) Since the ratio of the square on the side of the equilateral triangle to the square on the side of the hexagon (equal to the radius of the circle) is 3 : 1, one quickly finds that each outer segment is 1/3 the inner segment B. Hence, 3 A = B, (diagram 9) so that the lunule will be 2/3 B + C, i.e. it is less than the rectilinear area (here a triangle) by 1/3 B = A. Hence, the triangle is greater than the lunule by A. (diagram 10) From here it is merely a matter of finding a mathematically interesting figure equal to A. We can add a triangle to the third segment to get a sector of the circle, where the curvilinear figures (diag. 11) = the rectilinear figures (diag. 12):
lunule + segment-on-hexagon-side = triangle-in-segment-on-equilateral-triangle |
lunule + segment-on-hexagon-side + triangle-in-1/6-sector = triangle-in-segment-on-equilateral-triangle + triangle-in-1/6-sector |
lunule + 1/6-sector = triangle-in-segment-on-equilateral-triangle + triangle-in-1/6-sector |
(diagram 13) We could note that these two triangles in the equality are equal as the right triangles in the diagram have equal respective angles while the radii and sides of the hexagon are all equal. Eudemus does something different. It is a matter of making this equality more elegant. (diagram 14) He chooses a circle that is 1/6 the initial circle (equal to the 1/6 sector of the circle) and a hexagon that is 1/6 the hexagon (equal to either of the triangles in the equality). In the account of Eudemus, this will be the segments on an hexagon inscribed in a circle, with each segment equal to 1/6 A. Hence, (diagram 15) the triangle and the hexagon will equal (diagram 16) the lunule and these segments together with the hexagon, i.e., the lunule and the circle. The aesthetics behind this decision lie, one might well infer, in his having a whole circle in the equality and one complete rectilinear figure associated with each of the two curvilinear figures.