Return to Vignettes of Ancient Mathematics
Introduction to Hippocrates
The Methods of Eudemus and Alexander Compared
(diagram 1) Consider two sectors A and B with different radii RA and RB, with angles of the sectors being and , and where the endpoints of the arcs coincide. The lunule is the figure formed by the arcs of the two sectors. (diagram 2) Consider the segments A and B formed by the arcs of sectors A and B and the line connecting their endpoints. The area of the lunule = the area of segment A - the area of segment B.
(diagram 3) The area of segment A = the area of the sector A - the area of the triangle of A (the part of sector A not part of the segment). Let = 1/2 . Then the area of triangle of A is:
2* (1/2 RA cos RA sin ) = 1/2 RA2 sin 2 = 1/2 RA2 sin
The area of the sector is (for in radians): 1/2 RA2
Therefore, the area of the segment is: 1/2 RA2 - 1/2 RA2 sin
(diagram 4) The area of segment B = the area of the sector B - the area of the triangle of B (the part of sector B not part of the segment). Let = 1/2 . Then the area of triangle of B is: 2* (1/2 RB2 cos RB2 sin ) = 1/2 RB2 sin 2 = 1/2 RB2 sin
The area of the sector is (for in radians): 1/2 RB2
Therefore, the area of the segment B: 1/2 RB2 - 1/2 RB2 sin
(diagram 5) The area of the lunule = segment A - segment B
= (1/2 RA2 - 1/2 RA2 sin ) - (1/2 b RB2 - 1/2 RB2 sin )
= 1/2 ( RA2 - RB2) + 1/2 (RB2 sin - RA2 sin )
(diagram 6) Note that RB2 sin > RA2 sin . If the lunule is to be constructible without transcendental curves (in particular, without a solution to squaring a circle), then the first term should be 0, in which case:
RA2 = RB2
RA2 : RB2 = :
In other words, there is a constructible solution only if sectors A and B are equal. Sectors A and B are equal iff the squares of their radii are in inverse proportion to their angles.
(diagram 7) The area of the lunule = segment A - segment B. Since we suppose:
1/2 ( RA2 - RB2) = 0
The area of the lunule = the difference between the triangle of A and the triangle of B
= 1/2 (RB2 sin - RA2 sin )
(diagram 8) Note that if < p/2 and > p/2, sin < 0 and the area will be the sum of the two areas.
(diagram 9) We now consider the question of whether such a lunule can be found. The chord connecting the endpoints of the lunule is the common base of segment A and segment B. This is:
2 RA sin = 2 RB sin y
Let p : 1 = RA2 : RB2 = : = : , where p is a real number.
Then RA2 = p RB2 and = p . Hence,
2 RB√p sin = 2 RB sin p
Since p is a real number, for convenience, let's think of
RA = √n
RB = √m
√n sin m = √m sin n
If n and m are whole numbers (i.e., m/n = p is rational), there will be a simple expansion of each sine in terms of sin .
In other words, where a solution can be constructed (in whatever sense of 'can be constructed' we care to choose), there will be (in the same sense) a constructible lunule.
There are 5 lunules where the equation √m sin n = √n sin m, can be reduced to either a linear or quadratic equation. Hence, there are five solutions with straight line and circle. Their ratios are:
(diagram 10) 2 : 1. This lunule is discussed by Alexander and Eudemus
(diagram 11) 3 : 1. This lunule is discussed by Eudemus
(diagram 12) 3: 2. This lunule is discussed by Eudemus
(diagram 13) 5 : 1
(diagram 14) 5 : 3
The following reduce to cubic equations and so are solvable using conic sections.
(diagram 15) 4 : 1
(diagram 16) 4 : 3
(diagram 17) 7 : 1
(diagram 18) 7 : 3
(diagram 19) 7 : 5