The Five Squarable Lunes 

A planar region whose boundaries are arcs of two circles is called a lune (presumably because it resembles a crescent moon). Of course, it's impossible to "square” a complete circle by Euclidean methods – meaning that we can’t construct a square with the same area using just straightedge and compass – but, somewhat surprisingly, it is possible to “square” certain lunes. Hippocrates of Chios was the first to demonstrate such quadratures of lunes around 440 BC. It turns out that only five particular lunes can be squared by Euclidean methods. Three of these were described by Hippocrates himself, and two more were discovered in the mid 1700's. These last two are often credited to Euler in 1771, but according to Heath all five squarable lunes were given in a dissertation by Martin Johan Wallenius in 1766. The proof that these five are the only lunes squarable by Euclidean methods was given by Tchebatorew and Dorodnow. 

It's not too difficult to discover these five squarable lunes, especially with the help of modern trigonometric methods. In general, consider the lune described by arc segments of two offset circles, one of radius r and the other of radius R, as shown below. 


Under what conditions can we solve for the area of the lune (the shaded region) by Euclidean methods? This is equivalent to asked for the conditions in which we can solve for the shaded area by means of nothing more complex than quadratic equations and square roots. The area of the lune can be expressed as the difference between the two partcircular regions to the right of the vertical line, i.e., we have 



We can express the area of the region CFD as follows 



Now, since the area of a complete circle of radius r is πr^{2}, the area of the circular sector BCFD is simply 



Also, the area of the triangle BCD is 



where we have used the trigonometric identity 2sin(x)cos(x) = sin(2x). Therefore, the area of the region CFD is 



Similarly we have 


Hippocrates saw that we could eliminate the transcendental terms βr^{2} and αR^{2} from the expression for the area of the lune if we set those terms equal to each other. Thus, we restrict ourselves to only those cases where βr^{2} = αR^{2}. In these cases the area of the lune is simply 



Now, since we want to be able to construct this area from a unit length (such as taking r = 1) using only quadratic operations, the length of R must be constructible, so we will require it to be of the form R^{2} = ur^{2} so that R equals r times the square root of some rational number u. It follows from the relation βr^{2} = αR^{2} that β = uα, so we can make these substitutions into the expression for the lune area to give 



Of course, from the original diagram we can equate the vertical heights to give the relation r sin(β) = R sin(α), which implies 



Recalling the identities sin(2x) = 2sin(x)cos(x) and cos(x)^{2} = 1  sin(x)^{2} we can rewrite the equation for the area of the lune entirely in terms of sin(α), which we will abbreviate as s. 



Therefore, we can "square the lune" (with apologies to Debussey) by Euclidean methods if we can determine s = sin(α) by solving nothing more complicated than quadratics. Clearly equation (1) can be solved for sin(α) by means of quadratics and square roots only for certain values of the rational number u. To see this, recall the wellknown trigonometric identities for multiple angles 



and so on. If we set u = 2 in equation (1), and again denote sin(α) as simply s, we have 



and so s^{2} = 1/2 and . Substituting this into (2), and setting r = 1, we get the area A = 1. This was the first case found by Hippocrates, and is shown below. 


Notice that the area of the lune equals the area of the major triangle to the left of the vertical line. 

Now consider what happens if we set u = 3. In this case equation (1) becomes 



and so we have 



Substituting into equation (2) gives the area of the lune 



This lune is shown in the figure below. 


For the third squarable lune, consider what happens if we set u = 3/2. In this case we can define the halfangle ω = α/2 so we have α = 2ω and uα = 3ω. Equation (1) then becomes 



We can now use the trigonometric multiple angle formulas to expand this into an equation in terms of S = sin(ω) as follows 



Squaring both sides and simplifying gives a quadratic in S^{2} 



which gives 


(The other root leads to a complex result.) This is sin(ω)^{2}, which equals sin(2α)^{2}, so we need to convert this to an expression for sin(α) using the halfangle formula, which gives 



and so we have the angle α = 0.935929456... On this basis the area of the lune is 



This lune is illustrated below. 


Apparently the above three cases were the only ones known in antiquity. The next case for which equation (1) reduces to a quadratic is with u = 5, which leads to 



where s again denotes sin(α). Solving this for s^{2} gives 



The root with the "+" sign leads to a complex angle, but with the "–" sign we get α = 0.409090011..., which produces the squarable lune shown below. 


The only other value of u for which equation (1) reduces to a quadratic equation is u = 5/3. In this case we write the equation in terms of ω = α/3, so letting S denote sin(ω) we have 



This is a quadratic in S^{2}, so we can solve it to give 



Taking 3 times the inverse sine of the square root of these quantities gives the two possible values of the angle, of which only the first is relevant, so we have the angle 



This last squarable lune is illustrated below. 

