Department of Mathematics

J. Wilson, EMAT 6690

**Five Quadratures of Lunes**

By Jan White

First we need to know what quadrature and lune actually means so here are the definitions.

** Quadrature: **We mean that we can construct a square of
some plane figure having the same area as the plane figure with only a
compass and straightedge.

** Lune:** A lune is the area between two curves. Think lunar
like the phases of the moon.

There are five squarable lunes, three of which Hippocrates of Chios found around the year 440 BC. The other two quadratures of lunes are attributed to Leonhard Euler (1707-1783) around 1771, but all five may have been given in a dissertation in 1766 by Martin Johan Wallenius.

Before discussing the three quadratures of lunes found by Hippocrates let's first look at how a quadrature of a rectangle and triangle can be shown.

Let ABCD be a rectangle. First we extend AB and equal distance to BD by using a compass and marking this segment BF. Next we find the midpoint of AF and mark off a semicircle as shown. We then construct a line perpendicular to AF and mark the intersection with the semicircle L. From this we construct the square BLMK. See figure 1. Does the area of ABCD = BLMK?

__Proof:__

By the Pythagorean theorem, GL^2=BL^2 +GB^2 or GL^2-GB^2=BL^2

AG=GL=GF are all radii of the semicircle

BF= GF-GB = GL-GB and AB=GA +GB = GL+GB

The area of the rectangle ABCD = AB x BD

=AB x BF (since BD=BF by construction)

=(GL + GB)(GL-GB)

=GL^2-GB^2=BL^2=Area of the square BLMK

Now for the quadrature of a triangle.

Let ABC be a triangle and construct a perpedicular segment from B to the segment AD.

The area of the triangle is 1/2 BD x AC. We the find the midpoint of
BD at E and construct a rectangle with sides GH=AC and IG=1/2 BD=ED. We
know the area of the rectangle is GH x IG= AC x ED= 1/2 BD x AC. Using
the previous example (figure 1) we can find the area of a square equal
to this rectangle and so the area of the square would be equal to the triangle.

The lune that Hippocrates was able to square is the green area below
AEBF. He knew three geaometric ideas that helped him to prove the
green area was quadrable.

1. The Pythagorean
theorem.

2. An angle inscribed
in a semicircle is a right angle.

3. The area of two
circles are in a ratio equal to the ratio of their diameters squared.

Angle ABC is a right angle
because it is inscribed in a semicircle. Triangles ADB and DBC are
congruent by SAS so

AB = BC. Using the Pythagorean theorem (AC)^{2} = (AB)^{2}
+ (BC)^{2} = 2(AB)^{2}.

__Area (AEB) = (AB) ^{2}__
=

Area (ABC) = (AC)

The area of the quadrant AFBD is half that of the semicircle ABC, so the area of semicircle AEB = quadrant AFBD.

Area (AEB)-Area (AFBG) = Area (AFBD) - Area (AFBG)

Area (Lune AEBF) = Area (Triangle ABD) and triangles are quadrable from figure #2.

In a similar fashion Hippocrates was able to prove lunes created by
inscribing a 30, 60 degree triangle in a semicircle.

There are two lunes left that are quadrable. These can be found using trigonometry. Let's give it try.

The next few pages uses a program called "mimio" where images are tranfered
from a white board to the computer and then saved as html files that can
be viewed of over the internet. Go There.

Hippocrates implied that the circle was also quadrable
but he never gave the proof. Maybe because the circle is not quadrabable,
here is a proof that the circle is not Quadrable.

References:

Dunham, William, *Journey Through Genius*, Penquin
Books, NY, 1991

http://www.mathpages.com/home/kmath171.htm