The Five Squarable Lunes
Considering that it's impossible to "square the circle" by Euclidean
methods, it's interesting that some regions whose boundaries are
circular arcs CAN be "squared" by Euclidean methods - meaning that we
can construct a square with the same area using just straightedge and
compass - whereas most such regions cannot. Hippocrates of Chios was
the first to demonstrate such "quadratures" (around 440 BC) for lunes.
It turns out that only five particular lunes can be "squared". 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. It's now
known (Tschebatorew and Dorodnow) that these five cases are the ONLY
lunes that are squarable by Euclidean methods.
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 part-circular regions to the right of the vertical
line, i.e., we have
Area of Lune = [Area of CFD] - [Area of CED]
We can express the area of the region CFD as follows
Area of CFD = [Area of sector BCFD] - [Area of triangle BCD]
Now, since the area of a cmplete circle of radius r is pi r^2, the
area of the circular sector BCFD is simply
2b
--- [pi r^2] = b r^2
2pi
Also, the area of the triangle BCD is
1
[r sin(b)][r cos(b)] = --- r^2 sin(2b)
2
where we have used the trigonometric identity 2sin(x)cos(x) = sin(2x).
Therefore, the area of the region CFD is
1
Area of CFD = b r^2 - --- r^2 sin(2b)
2
Similarly we have
1
Area of CED = a R^2 - --- R^2 sin(2a)
2
Now, the trick that Hippocrates saw was that we could eliminate the
transcendental terms b r^2 and a R^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
b r^2 = a R^2
In these cases the area of the lune is simply
1 1
Area of lune = --- R^2 sin(2a) - --- r^2 sin(2b)
2 2
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 = u r^2
so that R = sqrt(u) r for some rational number u. If then follows
from the relation br^2 = aR^2 that
b = ua
so we can make these substitutions into the expression for the lune
area to give
r^2 / \
Area of lune = --- ( u sin(2a) - sin(2ua) )
2 \ /
Of course, from the original diagram we can equate the vertical heights
to give the relation
r sin(b) = R sin(a)
which implies
__
sin(ua) = /u sin(a) (1)
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(a), which we will abbreviate as "s".
/ _______ _ ________ \
Area of lune = s r^2 ( u /1 - s^2 - /u /1 - us^2 ) (2)
\ /
Therefore, we can "square the lune" (apologies to Debussey) by
Euclidean methods if we can determine s = sin(a) by solving nothing
more complicated than quadratics. It's not hard to see that equation
(1) can be solved for sin(a) by means of quadratics and square roots
only for certain values of the rational number u. To see this, recall
the well-known trigonometric identities for multiple angles
sin(2x) = 2sin(x)cos(x)
sin(3x) = 3sin(x) - 4sin(x)^3
sin(4x) = cos(x)[4sin(x) - 8sin(x)^3]
sin(5x) = 5sin(x) - 20sin(x)^3 + 16sin(x)^5
sin(6x) = cos(x)[6sin(x) - 32sin(x)^3 + 32sin(x)^5]
sin(7x) = 7sin(x) - 56sin(x)^3 + 112sin(x)^5 - 64sin(x)^7
and so on. If we set u=2 in equation (1) (and again denote sin(a)
as simply "s") we have
_______ _
/1 - s^2 = /2
and so s^2 = 1/2 and s = 1/sqrt(2). Substituting this into (2), and
setting r=1, gives 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
_
3 - 4s^2 = /3
and so we have s^2 = (3-sqrt(3))/4 and s = sqrt[3-sqrt(3)]/2.
Substituting into equation (2) gives the area of the lune
_ ______ _____________ _
1 | / _ / _ |
A = --- | / 18 /3 - / 42 /3 - 72 |
4 |_ / / _|
= 1.179959679570986...
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 half-angle w = a/2 so we have a=2w and
ua = 3w. Equation (1) then becomes
___
sin(3w) = /3/2 sin(2w)
We can now use the trig multiple angle formulas to expand this into
an equation in terms of S = sin(w) as follows
____ _________
3 - 4S^2 = 2 / 3/2 / 1 - S^2
Squaring both sides and simplifying gives a quadratic in S^2
16 S^4 - 18 S^2 + 3 = 0
which gives S^2 = [9 - sqrt(33)]/16. (The other root leads to a
complex result.) This is sin(w)^2, which equals sin(2a)^2, so we
need to convert this to an expression for sin(a) using the half-angle
formula, which gives sin(a) = sqrt[30 + 2sqrt(33)] / 8, and so we
have the angle a = 0.935929456... On this basis the area of the
lune is
/ _________ ___________ \
1 / / __ _ / __ \
A = --- ( 3 / 111 + /33 - /3 / 93 - 13 /33 )
32 \ / --------- / ----------- /
\ / 2 / 2 /
= 0.552446605462519...
This lune is illustrated below.
Apparently the above 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
16s^4 - 20s^2 + [5 - sqrt(5)] = 0
where "s" again denotes sin(a). Solving this for s^2 gives
__________
/ _
5 +- / 5 + 4 /5
s^2 = ------------------
8
The root with the "+" sign leads to a complex angle, but with the "-"
sign we get a = 0.409090011..., which produces the squareable 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 w = a/3, so letting S denote sin(w) we have
5 - 30S^2 + 16S^4 = sqrt(5/3) [ 3 - 4S^2]
This is a quadratic in S^2, so we can solve it to give
_______________
__ / ____
15 - /15 +- / 60 + 6/ 15
S^2 = --------------------------------
24
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 relavant, so we have the angle
a = 0.87932759...
This last squarable lune is illustrated below.
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