**A basic introduction to the mathematics of the lunules of Hippocrates of Chios**©

by Henry Mendell (Cal. State U., L.A.)

Return to Vignettes of Ancient Mathematics

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 R_{A} and R_{B}, 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 R_{A} cos R_{A} sin ) = 1/2 R_{A}^{2} sin 2 = 1/2 R_{A}^{2} sin

The area of the sector is (for in radians): 1/2 R_{A}^{2}

Therefore, the area of the segment is: 1/2 R_{A}^{2} - 1/2 R_{A}^{2} 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 R_{B}^{2} cos R_{B}^{2} sin ) = 1/2 R_{B}^{2} sin 2 = 1/2 R_{B}^{2} sin

The area of the sector is (for in radians): 1/2 R_{B}^{2}

Therefore, the area of the segment B: 1/2 R_{B}^{2} - 1/2 R_{B}^{2} sin

(diagram 5) The area of the lunule = segment A - segment B

= (1/2 R

_{A}^{2}- 1/2 R_{A}^{2}sin ) - (1/2 b R_{B}^{2}- 1/2 R_{B}^{2}sin )= 1/2 ( R

_{A}^{2}- R_{B}^{2}) + 1/2 (R_{B}^{2}sin - R_{A}^{2}sin )

(diagram 6) Note that R_{B}^{2} sin > R_{A}^{2} 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:

R

_{A}^{2}= R_{B}^{2}

or

R

_{A}^{2}: R_{B}^{2}= :

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 ( R

_{A}^{2}- R_{B}^{2}) = 0

The area of the lunule = the difference between the triangle of A and the triangle of B

= 1/2 (R_{B}^{2} sin - R_{A}^{2} 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 R_{A} sin = 2 R_{B} sin y

Let p : 1 = R_{A}^{2} : R_{B}^{2} = : = : , where p is a real number.

Then R_{A}^{2} = p R_{B}^{2} and = p . Hence,

2 R_{B}√p sin = 2 R_{B} sin p

Since p is a real number, for convenience, let's think of

= m

= n

R_{A} = √n

R_{B} = √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