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Introduction to Hippocrates
Eudemus' Method
Alexander's Method
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

or

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 

= m

= n

RA = √n

RB = √m

Hence, 

√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

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